2.1 Estimation of the non-bosonizable terms

Now we are ready to remove the non-bosonizable terms, namely, the terms involving operators $D_k$ in the decomposition (1.28) of the interaction operator:

(2.21) $$ \begin{align} k_{F}^{-1} H_{\text{int}}^{\prime} = \sum_{k\in\mathbb{Z}_{+}^{3}}\left(H_{\text{int}}^{k}-\frac{\hat{V}_{k}k_{F}^{-1}}{(2\pi)^{3}}\left|L_{k}\right|\right) +\frac{k_{F}^{-1}}{(2\pi)^{3}}\sum_{k\in\mathbb{Z}_{\ast}^{3}}\hat{V}_{k}\left(\tilde{B}_{k}^{\ast}D_{k}+D_{k}^{\ast}\tilde{B}_{k}+\frac{1}{2}D_{k}^{\ast}D_{k}\right), \end{align} $$

where $H_{\text {int}}^{k}$ is defined in (1.29). Moreover, for technical reasons, we will also impose a momentum cutoff in the bosonizable terms. Recall the set $S_C$ in (1.76). Define

(2.22) $$ \begin{align} \mathcal{E}_{\mathrm{NB}} = k_{F}^{-1} H_{\text{int}}^{\prime} - \sum_{k\in S_C} \left(H_{\mathrm{ int}}^{k}-\frac{\hat{V}_{k}k_{F}^{-1}}{(2\pi)^{3}}\left|L_{k}\right|\right). \end{align} $$

Proposition 2.4. Let $\sum _{k\in \mathbb {Z}^{3}}\hat {V}_{k}|k|<\infty $ . Then for all $\gamma \in (0,1/9)$ in $S_C$ , we have

$$ \begin{align*} \pm \mathcal{E}_{\mathrm{NB}} \le C k_F^{-\gamma/2} \Big(H_{\mathrm{kin}}' + k_F^{-1} \mathcal{N}_E H_{\mathrm{kin}}' + k_F\Big). \end{align*} $$

Here, the constant $C>0$ depends only on V (in particular, it is independent of $k,k_F$ and $\lambda $ ).

We write $\pm X\le Y$ for two operator inequalities $X\le Y$ and $-X\le Y$ .

Proof. For the bosonizable terms, by (2.6), Proposition 2.2 and Proposition 2.1, we can bound

(2.23) $$ \begin{align} \pm ( \{\tilde{B}_{k}^{\ast},\tilde{B}_{k}\} - |L_k| ) = \pm \left( 2 \tilde{B}_{k}^{\ast}\tilde{B}_{k} - \sum_{p\in L_{k}}c_{p}^{\ast}c_{p}-\sum_{p\in L_{k}}c_{p-k}c_{p-k}^{\ast} \right) \le 2 \tilde{B}_{k}^{\ast}\tilde{B}_{k} + \mathcal{N}_E\le C k_F H_{\mathrm{kin}}' \end{align} $$

for all $k\in \mathbb {Z}^3_{*}$ . Moreover, by the Cauchy–Schwarz inequality,

(2.24) $$ \begin{align} \pm (\tilde{B}_{k}^* \tilde{B}_{-k}^* + \tilde{B}_{-k}\tilde{B}_{k} ) \le |k|^{-1/2} \tilde{B}_{-k} \tilde{B}_{-k}^* + |k|^{1/2} \tilde{B}_{k}^* \tilde{B}_{k} \le C |k|^{1/2} k_F (H_{\mathrm{kin}}' + k_F) \end{align} $$

for all $k\in \mathbb {Z}^3_{*}$ . Combining (2.23) and (2.24), we find that

(2.25) $$ \begin{align} \pm \sum_{k \in \mathbb{Z}_{+}^{3} \backslash S_C} \left(H_{\mathrm{ int}}^{k}-\frac{\hat{V}_{k}k_{F}^{-1}}{(2\pi)^{3}}\left|L_{k}\right|\right) &\le C (H_{\mathrm{kin}}' + k_F) \sum_{k \in \mathbb{Z}_{+}^{3} \backslash S_C} \hat V_k |k|^{1/2}\nonumber\\ & \le C (H_{\mathrm{kin}}' + k_F) k_F^{-\gamma/2} \sum_{k \in \mathbb{Z}^{3}} \hat V_k |k|. \end{align} $$

For the non-bosonizable terms, by the Cauchy–Schwarz inequality and Proposition 2.2, we have

(2.26) $$ \begin{align} \pm \left( \tilde{B}_{k}^{\ast}D_{k}+D_{k}^{\ast}\tilde{B}_{k} \right) \le k_F^{-\gamma/2} |k| \tilde{B}_{k}^{\ast} \tilde{B}_{k} + k_F^{\gamma/2} |k|^{-1} D_{k}^{\ast} D_{k} \le C k_F^{1-\gamma/2} |k| H_{\mathrm{kin}}' + k_F^{\gamma/2} |k|^{-1} D_{k}^{\ast} D_{k}, \end{align} $$

and hence,

(2.27) $$ \begin{align} \pm \sum_{k\in\mathbb{Z}_{\ast}^{3}}\hat{V}_{k} \left( \tilde{B}_{k}^{\ast}D_{k}+D_{k}^{\ast}\tilde{B}_{k} + D_{k}^{\ast} D_{k} \right) \le C k_F^{1-\gamma/2} H_{\mathrm{kin}}' + \sum_{k\in\mathbb{Z}_{\ast}^{3}}\hat{V}_{k} ( k_F^{\gamma/2} |k|^{-1} +1 ) D_{k}^{\ast} D_{k}. \end{align} $$

Let us decompose the sum on the right-hand side of (2.27) into the high-momenta $|k|> k_F^{\gamma /2}$ and the low-momenta $|k|\le k_F^{\gamma /2}$ . For the high-momenta, from the simple bound (2.7), we get

(2.28) $$ \begin{align} \sum_{k\in\mathbb{Z}_{\ast}^{3}, |k|> k_F^{\gamma/2}}\hat{V}_{k} ( k_F^{\gamma/2} |k|^{-1} +1 ) D_{k}^{\ast} D_{k} \le C k_F^{-\gamma/2} \mathcal{N}_E^2 \sum_{k\in\mathbb{Z}^{3}}\hat{V}_{k} |k|. \end{align} $$

For the low-momenta, using Proposition 2.3 with $\lambda =k_F^{\gamma }/|k|^{2}$ , we have

(2.29) $$ \begin{align} D_{k}^{\ast}D_{k} \le C\left( k_F^{2\gamma} + k_F^\gamma |k|^{2/3} (\log k_F)^{\frac 2 3} k_F^{2/3} + |k|^{3+2/3} (\log k_F)^{\frac 2 3} k_F^{2/3}\right) |k|\mathcal{N}_E +Ck_F^{-\gamma/2} |k| \mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime}, \end{align} $$

and hence,

(2.30) $$ \begin{align} \sum_{k\in\mathbb{Z}_{\ast}^{3}, |k|\le k_F^{\gamma/2}}\hat{V}_{k} D_{k}^{\ast}D_{k} &\le C \left( \sum_{k\in\mathbb{Z}^{3}}\hat{V}_{k} |k| \right) \left( k_F^{(3+\frac 2 3) \frac \gamma 2+ \frac 2 3} (\log k_F)^{\frac 2 3} \mathcal{N}_E + k_F^{-\frac \gamma 2} \mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime}\right) \nonumber\\ &\le C k_F^{-\gamma/2} \left( k_F \mathcal{N}_E +\mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime}\right) \end{align} $$

for all $\gamma \in (0,1/7)$ . Moreover, using Proposition 2.3 with $\lambda =k_F^{2\gamma }/|k|^{4}$ , we have

(2.31) $$ \begin{align} k_F^{\gamma/2} |k|^{-1} D_{k}^{\ast}D_{k} &\leq C\left( k_F^{4\gamma} + k_F^{(2+\frac 1 2)\gamma} (\log k_F)^{\frac 2 3} k_F^{2/3} + k_F^{\frac \gamma 2}|k|^{2+2/3} (\log k_F)^{\frac 2 3} k_F^{2/3}\right) |k|\mathcal{N}_E \nonumber\\ &\qquad +Ck_F^{-\gamma/2} |k| \mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime}, \end{align} $$

and hence,

(2.32) $$ \begin{align} \sum_{k\in\mathbb{Z}_{\ast}^{3}, |k|\le k_F^{\gamma/2}}\hat{V}_{k} k_F^{\gamma/2} |k|^{-1} D_{k}^{\ast}D_{k} &\le C \left( \sum_{k\in\mathbb{Z}^{3}}\hat{V}_{k} |k| \right) \left( k_F^{(2+\frac 1 2) \gamma + \frac 2 3} (\log k_F)^{\frac 2 3} \mathcal{N}_E + k_F^{-\frac \gamma 2} \mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime}\right) \nonumber\\ &\le C k_F^{-\gamma/2} \left( k_F \mathcal{N}_E +\mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime}\right) \end{align} $$

for all $\gamma \in (0,1/9)$ . Inserting (2.28), (2.30) and (2.32) in (2.27) and using Proposition 2.1, we conclude that

(2.33) $$ \begin{align} \pm \frac{k_{F}^{-1}}{(2\pi)^{3}} \sum_{k\in\mathbb{Z}_{\ast}^{3}}\hat{V}_{k} \left( \tilde{B}_{k}^{\ast}D_{k}+D_{k}^{\ast}\tilde{B}_{k} + D_{k}^{\ast} D_{k} \right) \le C k_F^{-\gamma/2} \Big( H_{\mathrm{kin}}' + k_F^{-1}\mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime} \Big) \end{align} $$

for all $\gamma \in (0,1/9)$ . The conclusion follows from (2.25) and (2.33).

3.1 Quadratic Hamiltonians

Similarly to how we can to any $\varphi \in V$ associate the two operators $a(\varphi )$ and $a^{\ast }(\varphi )$ , we may also associate two types of symmetric operators on $\mathcal {F}^{+}\left(V\right)$ to any symmetric operator on V. For the definition, we let $(e_{i})_{i=1}^{n}$ denote an orthonormal basis of V. Given any symmetric operator $A:V\rightarrow V$ , we then define the operator $Q_{1}(A)$ on $\mathcal {F}^{+}\left(V\right)$ by

(3.3) $$ \begin{align} Q_{1}(A)=\sum_{i,j=1}^{n}\left\langle e_{i},Ae_{j}\right\rangle \left(a^{\ast}(e_{i})a(e_{j})+a(e_{j})a^{\ast}(e_{i})\right), \end{align} $$

and likewise, for any symmetric operator $B:V\rightarrow V$ , we define the operator $Q_{2}(B)$ by

(3.4) $$ \begin{align} Q_{2}(B)=\sum_{i,j=1}^{n}\left\langle e_{i},Be_{j}\right\rangle \left(a^{\ast}(e_{i})a^{\ast}(e_{j})+a(e_{j})a(e_{i})\right). \end{align} $$

These definitions are independent of the basis chosen, and we can write equivalently

(3.5) $$ \begin{align} Q_{1}(A) & =\sum_{i=1}^{n}\left(a^{\ast}(Ae_{i})a(e_{i})+a(e_{i})a^{\ast}(Ae_{i})\right)\\ Q_{2}(B) & =\sum_{i=1}^{n}\left(a^{\ast}\left(Be_{i}\right)a^{\ast}(e_{i})+a(e_{i})a\left(Be_{i}\right)\right).\nonumber \end{align} $$

Thus, for real, symmetric $A,B:V\rightarrow V$ , we can define a quadratic Hamiltonian on $\mathcal {F}^{+}\left(V\right)$ by

(3.6) $$ \begin{align} H=Q_{1}(A)+Q_{2}(B). \end{align} $$

Note that by the CCR, we may express $Q_{1}(A)$ as

(3.7) $$ \begin{align} Q_{1}(A) =2\sum_{i,j=1}^{n}\left\langle e_{i},Ae_{j}\right\rangle a^{\ast}(e_{i})a(e_{j})+\text{tr}(A)=2\,\text{d}\Gamma(A)+\text{tr}(A), \end{align} $$

where $\text {d}\Gamma (A)$ denotes the second quantization of $A:V\rightarrow V$ . Sometimes in the literature, in particular in infinite dimensions, quadratic Hamiltonians are defined by $\text {d}\Gamma (A)+Q_{2}(B)$ , which is the same to our definition up to the constant $\text {tr}(A)$ . Here, we prefer to use $Q_1(A)$ instead of $\text {d}\Gamma (\cdot )$ ; the reason for this is that the relations of Proposition 3.4 below are symmetric in the Q’s.

Note that the basis-independence is a nice property of the real space setting. In general, if V is a complex Hilbert space and B is symmetric, then the definition of $Q_{2}(B)$ in (3.4) may depend on the basis. In fact, we can obtain a basis-independent formulation in the complex case, but the mapping $B\mapsto Q_{2}(B)$ is not to be defined for symmetric linear operators B, but rather symmetric anti-linear operators B to make up for the fact that in the complex case the assignment $\varphi \mapsto a(\varphi )$ is also anti-linear. This is unimportant for our application, which is why we only consider real Hilbert spaces in this section, for the sake of simplicity.

3.2 Bogolubov transformations

In this subsection, we review an explicit construction of a Bogolubov transformation $\mathcal {U}:\mathcal {F}^{+}\left(V\right)\rightarrow \mathcal {F}^{+}\left(V\right)$ that diagonalizes the quadratic Hamiltonian $H= Q_1(A)+Q_2(B)$ , namely,

(3.8) $$ \begin{align} \mathcal{U}H\mathcal{U}^{\ast}=Q_{1}(E) \end{align} $$

for a real, symmetric operator $E:V\rightarrow V$ . Such a construction is well known; see, for example, [Reference Dereziński16] for a recent review. We consider a unitary transformation $\mathcal {U}=e^{\mathcal {K}}$ where $\mathcal {K}$ is an anti-symmetric operator on $\mathcal {F}^{+}\left(V\right)$ of the following form:

(3.9) $$ \begin{align} \mathcal{K}=\frac{1}{2}\sum_{i,j=1}^{n}\langle e_{i},Ke_{j}\rangle \left(a(e_{i})a(e_{j})-a^{\ast}(e_{j})a^{\ast}(e_{i})\right)=\frac{1}{2}\sum_{i=1}^{n}\left(a(Ke_{i})a(e_{i})-a^{\ast}(e_{i})a^{\ast}(Ke_{i})\right). \end{align} $$

Here, $K:V\rightarrow V$ is a symmetric operator (called the transformation kernel) and $(e_{i})_{i=1}^{n}$ denotes any orthonormal basis of V (as with $Q_{1}(\cdot )$ and $Q_{2}(\cdot )$ this definition is independent of the basis).

In this subsection, we discuss the following:

Theorem 3.1. Let $A,B:V\rightarrow V$ be real, symmetric operators such that $A\pm B>0$ (namely, $A+B>0$ and $A-B>0$ ). Consider the Bogolubov transformation $e^{\mathcal {K}}$ where $\mathcal {K}$ is given in (3.9) with

$$\begin{align*}K=-\frac{1}{2}\log\left((A-B)^{-\frac{1}{2}}\left((A-B)^{\frac{1}{2}}(A+B)(A-B)^{\frac{1}{2}}\right)^{\frac{1}{2}}(A-B)^{-\frac{1}{2}}\right). \end{align*}$$

Then

$$ \begin{align*}e^{\mathcal{K}} (Q_1(A)+Q_2(B))e^{-\mathcal{K}}=Q_{1}(E)=2\,\mathrm{d}\Gamma(E)+\mathrm{tr}(E), \end{align*} $$

where

$$ \begin{align*}E=e^{K}(A+B)e^{K}=e^{-K}(A-B)e^{-K}. \end{align*} $$

Moreover, the diagonalizing K is uniquely determined by this.

In the following, we will prove Theorem 3.1 by using a generalization and simplification of the argument used in [Reference Grech and Seiringer23, Reference Benedikter, Nam, Porta, Schlein and Seiringer5]. We will first discuss the action of the Bogolubov transformation with a general kernel K and then explain where the diagonalization condition comes from.

Let us start with some basic properties of $\mathcal {K}$ .

Proposition 3.2. For any symmetric operator $K:V\rightarrow V$ , the operator $\mathcal {K}$ defined by (3.9) is an anti-symmetric operator on $\mathcal {F}^{+}\left(V\right)$ and obeys the commutators

$$ \begin{align*} \left[\mathcal{K},a(\varphi )\right] =a^{\ast}\left(K\varphi\right), \quad \left[\mathcal{K},a^{\ast}(\varphi )\right] =a\left(K\varphi\right), \quad \forall \varphi\in V. \end{align*} $$

Thus, $\left[\mathcal {K},\cdot \right]$ acts on the creation and annihilation operators by ‘swapping’ each type into the other and applying the operator K to their arguments. From this, one can now deduce that the unitary transformation $e^{\mathcal {K}}$ acts on the creation and annihilation operators according to

(3.10) $$ \begin{align} e^{\mathcal{K}}a(\varphi )e^{-\mathcal{K}} & =a\left(\cosh\left(K\right)\varphi\right)+a^{\ast}\left(\sinh\left(K\right)\varphi\right)\\ e^{\mathcal{K}}a^{\ast}(\varphi )e^{-\mathcal{K}} & =a^{\ast}\left(\cosh\left(K\right)\varphi\right)+a\left(\sinh\left(K\right)\varphi\right),\nonumber \end{align} $$

since by the Baker-Campbell-Hausdorff formula,

(3.11) $$ \begin{align} e^{\mathcal{K}}a(\varphi )e^{-\mathcal{K}} & =a(\varphi )+\frac{1}{1!}\left[\mathcal{K},a(\varphi )\right]+\frac{1}{2!}\left[\mathcal{K},\left[\mathcal{K},a(\varphi )\right]\right]+\frac{1}{3!}\left[\mathcal{K},\left[\mathcal{K},\left[\mathcal{K},a(\varphi )\right]\right]\right]+\cdots\nonumber \\ & =a(\varphi )+\frac{1}{1!}a^{\ast}\left(K\varphi\right)+\frac{1}{2!}a\left(K^{2}\varphi\right)+\frac{1}{3!}a^{\ast}\left(K^{3}\varphi\right)+\cdots\\ & =a\left(\varphi+\frac{1}{2!}K^{2}\varphi+\cdots\right)+a^{\ast}\left(\frac{1}{1!}K\varphi+\frac{1}{3!}K^{3}\varphi+\cdots\right)\nonumber \\ & =a\left(\cosh\left(K\right)\varphi\right)+a^{\ast}\left(\sinh\left(K\right)\varphi\right),\nonumber \end{align} $$

and the identity for $e^{\mathcal {K}}a^{\ast }(\varphi )e^{-\mathcal {K}}$ then follows immediately by taking the adjoint.

Now let us consider $e^{\mathcal {K}}Q_{1}(\cdot )e^{-\mathcal {K}}$ and $e^{\mathcal {K}}Q_{2}(\cdot )e^{-\mathcal {K}}$ . For this, we will first make an observation on their structure which will greatly simplify computations: namely, we note that the operators $Q_{1}(A)$ and $Q_{2}(B)$ are both of a ‘trace-form’ in the sense that we can write, say, $Q_{1}(A)=\sum _{i=1}^{n}q\left(e_{i},Ae_{i}\right)$ , where

(3.12) $$ \begin{align} q\left(x,y\right)=a^{\ast}\left(y\right)a (x )+a (x )a^{\ast}\left(y\right) \end{align} $$

defines a bilinear mapping from $V\times V$ into the space of operators on $\mathcal {F}^{+}\left(V\right)$ , similar to how the trace of an operator T is $\text {tr}(T)=\sum _{i=1}^{n}q\left(e_{i},Te_{i}\right)$ for $q\left(x,y\right)=\left\langle x,y\right\rangle $ . This abstract viewpoint is worth noting because all such expressions are both basis-independent and obey an additional property, which for the trace is just the familiar cyclicity property. Since we will encounter such ‘trace-form’ expressions repeatedly during computations throughout this paper, we state this property in full generality. In the following, we take sesquilinear to mean anti-linear in the first argument and linear in the second (we note that in the present real case a sesquilinear mapping is of course just a bilinear mapping, but stating it in this generality will prove useful later).

Lemma 3.3. Let $\left(V,\left\langle \cdot ,\cdot \right\rangle \right)$ be an n-dimensional Hilbert space and let $q:V\times V\rightarrow W$ be a sesquilinear mapping into a vector space W. Let $(e_{i})_{i=1}^{n}$ be an orthonormal basis for V. Then for any linear operators $S,T:V\rightarrow V$ , it holds that

$$\begin{align*}\sum_{i=1}^{n}q\left(Se_{i},Te_{i}\right)=\sum_{i=1}^{n}q\left(ST^{\ast}e_{i},e_{i}\right). \end{align*}$$

As a consequence, the expression $\sum _{i=1}^{n}q\left(e_{i},e_{i}\right)$ is independent of the chosen basis.

Proof. By orthonormal expansion, we find that

(3.13) $$ \begin{align} \sum_{i=1}^{n}q\left(Se_{i},Te_{i}\right) & =\sum_{i=1}^{n}q\left(Se_{i},\sum_{j=1}^{n}\left\langle e_{j},Te_{i}\right\rangle e_{j}\right) =\sum_{j=1}^{n}q\left(\sum_{i=1}^{n}\left\langle Te_{i},e_{j}\right\rangle Se_{i},e_{j}\right) \nonumber\\ & =\sum_{j=1}^{n}q\left(S\sum_{i=1}^{n}\left\langle e_{i},T^{\ast}e_{j}\right\rangle e_{i},e_{j}\right)=\sum_{i=1}^{n}q\left(ST^{\ast}e_{i},e_{i}\right). \end{align} $$

The basis independence follows from the fact that for all unitary transformation $U:V\rightarrow V$ ,

(3.14) $$ \begin{align} \sum_{i=1}^{n}q\left(Ue_{i},Ue_{i}\right)=\sum_{i=1}^{n}q\left(UU^{\ast}e_{i},e_{i}\right)=\sum_{i=1}^{n}q\left(e_{i},e_{i}\right). \end{align} $$

The lemma thus allows us to move a mapping from one argument to the other when under a sum, which will be immensely useful when simplifying expressions. As mentioned, this can indeed be seen as a generalization of the cyclicity property of the trace, since the lemma implies

(3.15) $$ \begin{align} \text{tr}\left(ST\right)=\sum_{i=1}^{n}\left\langle e_{i},STe_{i}\right\rangle =\sum_{i=1}^{n}\left\langle S^{\ast}e_{i},Te_{i}\right\rangle =\sum_{i=1}^{n}\left\langle S^{\ast}T^{\ast}e_{i},e_{i}\right\rangle =\sum_{i=1}^{n}\left\langle e_{i},TSe_{i}\right\rangle =\text{tr}\left(TS\right), \end{align} $$

but it is important to note that cyclicity is not a general property of trace-form sums; the assignments $A\mapsto Q_{1}(A)$ and $B\mapsto Q_{2}(B)$ do not obey such a property.

With the lemma, we can now easily derive the commutator of $\mathcal {K}$ with $Q_{1}(\cdot )$ and $Q_{2}(\cdot )$ :

Proposition 3.4. For any real, symmetric operators $A,B,K:V\rightarrow V$ , the operator $\mathcal {K}$ defined by equation (3.9) obeys the following commutators on $\mathcal {F}^+(V)$ :

$$ \begin{align*} \left[\mathcal{K},Q_{1}(A)\right] & =Q_{2}\left(\left\{ K,A\right\} \right)\\ \left[\mathcal{K},Q_{2}(B)\right] & =Q_{1}\left(\left\{ K,B\right\} \right). \end{align*} $$

Proof. We compute using the commutators of Proposition 3.2 that

(3.16) $$ \begin{align} \left[\mathcal{K},Q_{1}(A)\right] & =\sum_{i=1}^{n}\left(\left[\mathcal{K},a^{\ast}(Ae_{i})a(e_{i})\right]+\left[\mathcal{K},a(e_{i})a^{\ast}(Ae_{i})\right]\right)\nonumber\\ & =\sum_{i=1}^{n}(a^{\ast}(Ae_{i})\left[\mathcal{K},a(e_{i})\right] \nonumber \\ &\quad +\left[\mathcal{K},a^{\ast}(Ae_{i})\right]a(e_{i})+a(e_{i})\left[\mathcal{K},a^{\ast}(Ae_{i})\right]+\left[\mathcal{K},a(e_{i})\right]a^{\ast}(Ae_{i}))\nonumber \\ & =\sum_{i=1}^{n}\left(a^{\ast}(Ae_{i})a^{\ast}(Ke_{i})+a(KAe_{i})a(e_{i})+a(e_{i})a(KAe_{i})+a^{\ast}(Ke_{i})a^{\ast}(Ae_{i})\right). \end{align} $$

As the assignments $\varphi ,\psi \mapsto a(\varphi )a(\psi ),a^{\ast }(\varphi )a^{\ast }(\psi )$ are bilinear, we can apply Lemma 3.3 to see that

(3.17) $$ \begin{align} \left[\mathcal{K},Q_{1}(A)\right] & =\sum_{i=1}^{n}\left(a^{\ast}\left(AK^{\ast}e_{i}\right)a^{\ast}(e_{i})+a(e_{i})a\left(\left(KA\right)^{\ast}e_{i}\right)+a(e_{i})a(KAe_{i})+a^{\ast}\left(KA^{\ast}e_{i}\right)a^{\ast}(e_{i})\right)\nonumber \\ & =\sum_{i=1}^{n}\left(a^{\ast}\left(AKe_{i}\right)a^{\ast}(e_{i})+a(e_{i})a\left(AKe_{i}\right)+a(e_{i})a(KAe_{i})+a^{\ast}(KAe_{i})a^{\ast}(e_{i})\right)\nonumber \\ & =\sum_{i=1}^{n}\left(a^{\ast}\left((AK+KA)e_{i}\right)a^{\ast}(e_{i})+a(e_{i})a\left((AK+KA)e_{i}\right)\right)=Q_{2}\left(\left\{ K,A\right\} \right), \end{align} $$

where we also used that A and K are symmetric. The computation of $\left[\mathcal {K},Q_{2}(B)\right]$ is similar.

Note the similarity between this result and that of Proposition 3.2. Again we see that that $\left[\mathcal {K},\cdot \right]$ acts by ‘swapping the types and applying K to the argument’, although now the relevant types are $Q_{1}(\cdot )$ , and $Q_{2}(\cdot )$ and the application of K is taking the anticommutator.

We can now appeal to the Baker-Campbell-Hausdorff formula again to conclude that

(3.18) $$ \begin{align} &e^{\mathcal{K}}Q_{1}(A)e^{-\mathcal{K}} =Q_{1}(A)+\frac{1}{1!}\left[\mathcal{K},Q_{1}(A)\right]+\frac{1}{2!}\left[\mathcal{K},\left[\mathcal{K},Q_{1}(A)\right]\right]+\frac{1}{3!}\left[\mathcal{K},\left[\mathcal{K},\left[\mathcal{K},Q_{1}(A)\right]\right]\right]+\cdots \nonumber\\ & =Q_{1}(A)+\frac{1}{1!}Q_{2}\left(\left\{ K,A\right\} \right)+\frac{1}{2!}Q_{1}\left(\left\{ K,\left\{ K,A\right\} \right\} \right)+\frac{1}{3!}Q_{2}\left(\left\{ K,\left\{ K,\left\{ K,A\right\} \right\} \right\} \right)+\cdots \nonumber\\ & =Q_{1}\left(A+\frac{1}{2!}\left\{ K,\left\{ K,A\right\} \right\} +\cdots\right)+Q_{2}\left(\frac{1}{1!}\left\{ K,A\right\} +\frac{1}{3!}\left\{ K,\left\{ K,\left\{ K,A\right\} \right\} \right\} +\cdots\right), \end{align} $$

but to succeed, we must identify the sums of these iterated anticommutators. First, we note that we can rephrase this in a manner closer to that of equation (3.10) for $e^{\mathcal {K}}a(\varphi )e^{-\mathcal {K}}$ . One may view the anticommutator with K as a linear mapping $A\mapsto \left\{ K,A\right\} $ on the space of operators on V, $\mathcal {B}\left(V\right)$ – denote this mapping by $\mathcal {A}_{K}:\mathcal {B}\left(V\right)\rightarrow \mathcal {B}\left(V\right)$ (i.e., $\mathcal {A}_{K}(\cdot )=\left\{ K,\cdot \right\} $ ). Then we may phrase the above identity as

(3.19) $$ \begin{align} e^{\mathcal{K}}Q_{1}(A)e^{-\mathcal{K}}=Q_{1}\left(\cosh\left(\mathcal{A}_{K}\right)(A)\right)+Q_{2}\left(\sinh\left(\mathcal{A}_{K}\right)(A)\right) \end{align} $$

and likewise

(3.20) $$ \begin{align} e^{\mathcal{K}}Q_{2}(B)e^{-\mathcal{K}}=Q_{2}\left(\cosh\left(\mathcal{A}_{K}\right)(B)\right)+Q_{1}\left(\sinh\left(\mathcal{A}_{K}\right)(B)\right) \end{align} $$

so that the arguments again involve hyperbolic functions of linear operators, but now acting on $\mathcal {B}\left(V\right)$ rather than V itself. We then note the following ‘anticommutator Baker-Campbell-Hausdorff formula’:

Proposition 3.5. Let $\left(V,\left\langle \cdot ,\cdot \right\rangle \right)$ be an n-dimensional Hilbert space, let $K:V\rightarrow V$ be a self-adjoint operator and let $\mathcal {A}_{K}(\cdot )=\left\{ K,\cdot \right\} :\mathcal {B}\left(V\right)\rightarrow \mathcal {B}\left(V\right)$ denote the anticommutator with K. Then for any linear operator $T:V\rightarrow V$ ,

$$\begin{align*}e^{\mathcal{A}_{K}}(T)=\sum_{m=0}^{\infty}\frac{1}{m!}\mathcal{A}_{K}^{m}(T)=e^{K}Te^{K}. \end{align*}$$

Consequently,

$$ \begin{align*} \cosh\left(\mathcal{A}_{K}\right)(T) & =\frac{1}{2}(e^{K}Te^{K}+e^{-K}Te^{-K}),\\ \sinh(\mathcal{A}_{K})(T) & =\frac{1}{2}(e^{K}Te^{K}-e^{-K}Te^{-K}). \end{align*} $$

Proof. Let $(x_{i})_{i=1}^{n}$ be an eigenbasis for K with associated eigenvalues $\left(\lambda _{i}\right)_{i=1}^{n}$ . Denote $P_{i,j}=|x_j\rangle \langle x_i|$ , namely, $P_{i,j}x=\left\langle x_{i},x\right\rangle x_{j}$ for all $x\in V$ . It is well known that for any orthonormal basis $(x_{i})_{i=1}^{n}$ of V, the collection $\left(P_{i,j}\right)_{i,j=1}^{n}$ form an orthonormal basis for $\left(\mathcal {B}\left(V\right),\left\langle \cdot ,\cdot \right\rangle _{\text {HS}}\right)$ . Moreover, for any $x\in V$ and $1\leq i,j\leq n$ , by self-adjointness of K,

(3.21) $$ \begin{align} \mathcal{A}_{K}\left(P_{i,j}\right)x & =\left\{ K,P_{i,j}\right\} x=\left\langle x_{i},x\right\rangle Kx_{j}+\left\langle x_{i},Kx\right\rangle x_{j}=\left\langle x_{i},x\right\rangle \lambda_{j}x_{j}+\left\langle \lambda_{i}x_{i},x\right\rangle x_{j}\\ & =\left(\lambda_{i}+\lambda_{j}\right)\left\langle x_{i},x\right\rangle x_{j}=\left(\lambda_{i}+\lambda_{j}\right)P_{i,j}x.\nonumber \end{align} $$

Thus, $\{P_{i,j}\}_{i,j=1}^n$ an eigenbasis for $\mathcal {A}_{K}$ with associated eigenvalues $\left(\lambda _{i}+\lambda _{j}\right)_{i,j=1}^{n}$ .

Hence, it suffices to verify the identity $e^{\mathcal {A}_{K}}(T)=e^{K}Te^{K}$ with the eigenbasis $\left(P_{i,j}\right)_{i,j=1}^{n}$ :

(3.22) $$ \begin{align} e^{\mathcal{A}_{K}}\left(P_{i,j}\right) x & =e^{\lambda_{i}+\lambda_{j}}P_{i,j}=e^{\lambda_{i}+\lambda_{j}}\left\langle x_{i},x\right\rangle x_{j}=\left\langle e^{\lambda_{i}}x_{i},x\right\rangle e^{\lambda_{j}}x_{j}=\left\langle e^{K}x_{i},x\right\rangle e^{K}x_{j}\\ & =\left\langle x_{i},e^{K}x\right\rangle e^{K}x_{j}=e^{K}P_{i,j}e^{K}x.\nonumber \end{align} $$

The statements regarding $\cosh \left(\mathcal {A}_{K}\right)$ and $\sinh \left(\mathcal {A}_{K}\right)$ follow from the identities

(3.23) $$ \begin{align} \cosh (x )=\frac{1}{2}\left(e^{x}+e^{-x}\right), \quad \sinh (x )=\frac{1}{2}\left(e^{x}-e^{-x}\right), \quad \text{and }\left(-\mathcal{A}_{K}\right)=\mathcal{A}_{-K}. \end{align} $$

By these formulas, we thus deduce the quadratic operator analogue of equation (3.10):

(3.24) $$ \begin{align} e^{\mathcal{K}}Q_{1}(A)e^{-\mathcal{K}} & =\frac{1}{2}Q_{1}(e^{K}Ae^{K}+e^{-K}Ae^{-K})+\frac{1}{2}Q_{2}(e^{K}Ae^{K}-e^{-K}Ae^{-K}) \nonumber\\ e^{\mathcal{K}}Q_{2}(B)e^{-\mathcal{K}} & =\frac{1}{2}Q_{1}(e^{K}Be^{K}-e^{-K}Be^{-K})+\frac{1}{2}Q_{2}(e^{K}Be^{K}+e^{-K}Be^{-K}). \end{align} $$

Diagonalization condition

We can now finally describe how to diagonalize a quadratic Hamiltonian using a Bogolubov transformation of the form $e^{\mathcal {K}}$ . By the transformation identities above, we find that under $e^{\mathcal {K}}$ , the quadratic Hamiltonian $H=Q_{1}(A)+Q_{2}(B)$ transforms as

(3.25) $$ \begin{align} e^{\mathcal{K}}He^{-\mathcal{K}} &=\frac{1}{2}Q_{1}(e^{K}Ae^{K}+e^{-K}Ae^{-K})+\frac{1}{2}Q_{2}(e^{K}Ae^{K}-e^{-K}Ae^{-K})\nonumber \\ &\quad +\frac{1}{2}Q_{1}(e^{K}Be^{K}-e^{-K}Be^{-K})+\frac{1}{2}Q_{2}(e^{K}Be^{K}+e^{-K}Be^{-K})\nonumber\\ & =\frac{1}{2}Q_{1}(e^{K}(A+B)e^{K}+e^{-K}(A-B)e^{-K})+\frac{1}{2}Q_{2}(e^{K}(A+B)e^{K}-e^{-K}(A-B)e^{-K}). \end{align} $$

Therefore, the diagonalization condition on K is

(3.26) $$ \begin{align} e^{K}(A+B)e^{K}=e^{-K}(A-B)e^{-K}. \end{align} $$

If we can find such a K, then

(3.27) $$ \begin{align} e^{\mathcal{K}}He^{-\mathcal{K}}=Q_{1}(E)=2\,\text{d}\Gamma(E)+\text{tr}(E), \end{align} $$

where

(3.28) $$ \begin{align} E=e^{K}(A+B)e^{K}=e^{-K}(A-B)e^{-K}. \end{align} $$

There remains the question of existence and uniqueness of such a K:

Conclusion of the proof of Theorem 3.1 .

Write $A_{\pm }=A\pm B>0$ for brevity. Then we may write the diagonalization condition as

(3.29) $$ \begin{align} e^{-2K}A_{-}e^{-2K}=A_{+}. \end{align} $$

Multiplying by $A_{-}^{\frac {1}{2}}$ on both sides yields

(3.30) $$ \begin{align} \big(A_{-}^{\frac{1}{2}}e^{-2K}A_{-}^{\frac{1}{2}}\big)^{2}=A_{-}^{\frac{1}{2}}e^{-2K}A_{-}e^{-2K}A_{-}^{\frac{1}{2}}=A_{-}^{\frac{1}{2}}A_{+}A_{-}^{\frac{1}{2}}, \end{align} $$

which is equivalent to

(3.31) $$ \begin{align} A_{-}^{\frac{1}{2}}e^{-2K}A_{-}^{\frac{1}{2}}=\big(A_{-}^{\frac{1}{2}}A_{+}A_{-}^{\frac{1}{2}}\big)^{\frac{1}{2}}, \, \text{ namely } e^{-2K}=A_{-}^{-\frac{1}{2}}\big(A_{-}^{\frac{1}{2}}A_{+}A_{-}^{\frac{1}{2}}\big)^{\frac{1}{2}}A_{-}^{-\frac{1}{2}}. \end{align} $$

This implies the existence and uniqueness of the diagonalizing K as the operator exponential is a bijection between the real, symmetric operators and the real, symmetric, positive-definite operators.

4.1 Quadratic Hamiltonian

Let us define the pair excitation operators

(4.1) $$ \begin{align} b_{k,p}=c_{p-k}^{\ast}c_{p},\quad b_{k,p}^{\ast}=c_{p}^{\ast}c_{p-k},\quad k\in\mathbb{Z}_{\ast}^{3},\quad p\in L_{k}. \end{align} $$

We remark that in contrast to the bosonic case, the fermionic creation and annihilation operators are bounded (in fact, $\Vert c_{p,\sigma }\Vert _{\text {Op}}=\Vert c_{p,\sigma }^{\ast }\Vert _{\text {Op}}=1$ ), and therefore so are the operators $b_{k,p}^{\ast }$ , $b_{k,p}$ .

Then $H_{\text {int}}^{k}$ in (1.34) is exactly given by

(4.2) $$ \begin{align} H_{\text{int}}^{k} & =\sum_{p,q\in L_{k}}\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}\left(b_{k,p}^{\ast}b_{k,q}+b_{k,q}b_{k,p}^{\ast}\right)+\sum_{p,q\in L_{-k}}\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}\left(b_{-k,p}^{\ast}b_{-k,q}+b_{-k,q}b_{-k,p}^{\ast}\right)\nonumber \\ &\quad +\sum_{p\in L_{k}}\sum_{q\in L_{-k}}\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}\left(b_{k,p}^{\ast}b_{-k,q}^{\ast}+b_{-k,q}b_{k,p}\right) \nonumber \\ &\quad +\sum_{p\in L_{-k}}\sum_{q\in L_{k}}\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}\left(b_{-k,p}^{\ast}b_{k,q}^{\ast}+b_{k,q}b_{-k,p}\right). \end{align} $$

Thus, the natural one-body Hilbert space associated to $H_{\text {int}}^{k}$ is $\ell ^2 (L_k \cup L_{-k})$ . To free us from having to explicitly write sums over $L_{k}$ and $L_{-k}$ separately, we introduce some more notation. First, we will denote this union of lunes by

(4.3) $$ \begin{align} L_{k}^{\pm}=L_{k}\cup L_{-k}, \quad \ell^{2}\left(L_{k}^{\pm}\right)=\ell^{2}\left(L_{k}\cup L_{-k}\right)=\ell^{2}(L_{k})\oplus\ell^{2}(L_{-k}), \quad k\in\mathbb{Z}_{+}^{3}.\end{align} $$

Here, we used the fact that $L_{k}\cap L_{-k}=\emptyset $ for any $k\in \mathbb {Z}_{+}^{3}$ , since if $p\in L_{k}\cap L_{-k}$ , then

$$ \begin{align*}2|p|^2 \geq |p-k|^2 + |p+k|^2 = 2|p|^2 + 2|k|^2> 2|p|^2, \end{align*} $$

which is a contradiction. It is also convenient to introduce the ‘bar-notation’

(4.4) $$ \begin{align} \overline{k,p}=\begin{cases} k,p & p\in L_{k}\\ -k,p & p\in L_{-k} \end{cases},\quad\overline{p-k}=\begin{cases} p-k & p\in L_{k}\\ p+k & p\in L_{-k} \end{cases} \end{align} $$

to automatically encode the appropriate sign of k depending on $p\in L_{k}^{\pm }=L_{k}\cup L_{-k}$ (this will allow us to avoid expanding all our terms on a case-by-case basis when this is irrelevant).

In analogy with the definitions (3.3) and (3.4) we now define, for any $k\in \mathbb {Z}_{+}^{3}$ and symmetric operators $A,B:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ , the quadratic operators $Q_{1}^{k}(A),Q_{2}^{k}(B):\mathcal {H}_{N}\rightarrow \mathcal {H}_{N}$ by

(4.5) $$ \begin{align} Q_{1}^{k}(A) & =\sum_{p,q\in L_{k}^{\pm}}\left\langle e_{p},Ae_{q}\right\rangle \left(b_{\overline{k,p}}^{\ast}b_{\overline{k,q}}+b_{\overline{k,q}}b_{\overline{k,p}}^{\ast}\right),\nonumber\\ Q_{2}^{k}(B) & =\sum_{p,q\in L_{k}^{\pm}}\left\langle e_{p},Be_{q}\right\rangle \left(b_{\overline{k,p}}^{\ast}b_{\overline{k,q}}^{\ast}+b_{\overline{k,q}}b_{\overline{k,p}}\right). \end{align} $$

In order to cast $H_{\text {int}}^{k}$ as given by equation (4.2) into this form, we must identify the relevant operators A and B. Define the (un-normalized) rank-one projection $P_{v_k}: \ell ^2(L_k)\to \ell ^2(L_k)$ by

(4.6) $$ \begin{align} P_{v_k} =|v_k\rangle \langle v_k|, \quad v_{k}=\sqrt{\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}}\sum_{p\in L_{k}}e_{p} \in \ell^2(L_k), \end{align} $$

where $(e_{p})_{p\in L_{k}}$ denotes the standard orthonormal basis of $\ell ^{2}(L_{k})$ . Put differently, the matrix elements of $P_{v_k}$ are $\left\langle e_{p},P_{v_{k}}e_{q}\right\rangle =\frac {1}{2\left(2\pi \right)^{3}} \hat {V}_{k}k_{F}^{-1}$ for all $p,q\in L_{k}$ . Next, we define the operators

(4.7) $$ \begin{align} A_{k}^{\oplus},B_{k}^{\oplus}:\ell^{2}\left(L_{k}^{\pm}\right)\rightarrow\ell^{2}\left(L_{k}^{\pm}\right), \quad A_{k}^{\oplus}=\left(\begin{array}{cc} P_{v_{k}} & 0\\ 0 & P_{v_{k}} \end{array}\right),\quad B_{k}^{\oplus}=\left(\begin{array}{cc} 0 & P_{v_{k}}\\ P_{v_{k}} & 0 \end{array}\right) \end{align} $$

with respect to the decomposition $\ell ^{2}(L_{k}^{\pm })=\ell ^{2}(L_{k})\oplus \ell ^{2}(L_{-k})$ and the identification $\ell ^{2}(L_{k})\cong \ell ^{2}(L_{-k})$ (under $e_{p}\mapsto e_{-p}$ ).

Thus, the operator $H_{\text {int}}^{k}$ is concisely expressed as

(4.8) $$ \begin{align} H_{\text{int}}^{k}=Q_{k}^{1}\left(A_{k}^{\oplus}\right)+Q_{k}^{2}\left(B_{k}^{\oplus}\right). \end{align} $$

It remains to consider the kinetic operator. The equality (1.74) bids us to think of $H_{\operatorname {\mathrm {kin}}}^{\prime }$ as it were

(4.9) $$ \begin{align} H_{\operatorname{\mathrm{kin}}}^{\prime} &\sim \sum_{k\in\mathbb{Z}_{\ast}^{3}}\sum_{p\in L_{k}}\left(|p|^{2}-|p-k|^{2}\right)b_{k,p}^{\ast}b_{k,p} \nonumber\\ &= \sum_{k\in\mathbb{Z}_{+}^{3}} \left(\sum_{p\in L_{k}}\left(|p|^{2}-|p-k|^{2}\right)b_{k,p}^{\ast}b_{k,p}+\sum_{p\in L_{-k}}\left(|p|^{2}-\left|p+k\right|^{2}\right)b_{-k,p}^{\ast}b_{-k,p}\right) \end{align} $$

in an appropriate sense. To put this in the same framework as $H_{\mathrm {int}}^k$ , let us introduce (for every $k\in \mathbb {Z}_{+}^{3}$ ) the operator $h_{k}:\ell ^{2}(L_{k})\rightarrow \ell ^{2}(L_{k})$ by

(4.10) $$ \begin{align} h_{k}e_{p} =\lambda_{k,p}e_{p},\quad\;\;\lambda_{k,p}=\frac{1}{2}(|p|^{2}-|p-k|^{2}). \end{align} $$

Using again the identification $\ell ^{2}(L_{k})\cong \ell ^{2}(L_{-k})$ (under $e_{p}\mapsto e_{-p}$ ), we define the operators $h_{k}^{\oplus }:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ by

(4.11) $$ \begin{align} h_{k}^{\oplus}=\left(\begin{array}{cc} h_{{k}} & 0\\ 0 & h_{{k}} \end{array}\right). \end{align} $$

Then we can rewrite (4.9) as

(4.12) $$ \begin{align} H_{\operatorname{\mathrm{kin}}}^{\prime} \sim \sum_{k\in\mathbb{Z}_{+}^{3}} \Big( Q_1^k (h_k^{\oplus}) - 2\,\mathrm{tr} (h_k)\Big). \end{align} $$

Recall that $S_{C}=\overline {B}\left(0,k_{F}^{\gamma }\right)\cap \mathbb {Z}_{+}^{3}$ for an exponent $1\ge \gamma>0$ which is to be optimized over at the end. As far as the lower bound is concerned, we may replace $\sum _{k\in \mathbb {Z}_+}$ by $\sum _{k\in S_C}$ (the upper bound is easier and will be explained separately). In summary, we arrive at the following quasi-bosonic expression for the bosonizable terms:

(4.13) $$ \begin{align} H_{\operatorname{\mathrm{kin}}}^{\prime} + \sum_{k\in S_C} H_{\mathrm{int}}^k \sim \sum_{k\in S_C} \Big( Q_1^k (h_k^{\oplus} + A_k^{\oplus}) + Q_2^k (B_k^{\oplus}) - 2\,\mathrm{tr}(h_k)\Big). \end{align} $$

Note that unlike the bosonic case, the operators on the right side of (4.13) are bounded.

4.2 Generalized pair operators

For every $k\in \mathbb {Z}_{+}^{3}$ and $\varphi \in \ell ^2(L_k^{\pm })$ , we define the operators

(4.14) $$ \begin{align} b_{k}(\varphi )=\sum_{p\in L_{k}^{\pm}}\left\langle \varphi,e_{p}\right\rangle b_{\overline{k,p}},\quad b_{k}^{\ast}(\varphi )=\sum_{p\in L_{k}^{\pm}}\left\langle e_{p},\varphi\right\rangle b_{\overline{k,p}}^{\ast}. \end{align} $$

They obey the quasi-bosonic commutation relations (for $k,l\in \mathbb {Z}_{+}^{3}$ and $\varphi \in \ell ^{2}(L_{k}^{\pm })$ , $\psi \in \ell ^{2}\left(L_{l}^{\pm }\right)$ )

(4.15) $$ \begin{align} \left[b_{k}(\varphi ),b_{l}\left(\psi\right)\right] & =\left[b_{k}^{\ast}(\varphi ),b_{l}^{\ast}\left(\psi\right)\right]=0, \nonumber\\ \left[b_{k}(\varphi ),b_{l}^{\ast}\left(\psi\right)\right] & =\delta_{k,l}\left\langle \varphi,\psi\right\rangle +\varepsilon_{k,l}\left(\varphi;\psi\right), \end{align} $$

where the correction term is

(4.16) $$ \begin{align} \varepsilon_{k,l}\left(\varphi;\psi\right) & =\sum_{p\in L_{k}^{\pm}}\sum_{q\in L_{l}^{\pm}}\left\langle \varphi,e_{p}\right\rangle \left\langle e_{q},\psi\right\rangle \varepsilon\left(\overline{k,p};\overline{l,q}\right), \nonumber\\ \varepsilon\left(\overline{k,p};\overline{l,q}\right) & =-\left(\delta_{p,q}c_{\overline{q-l}}c_{\overline{p-k}}^{\ast}+\delta_{\overline{p-k},\overline{q-l}}c_{q}^{\ast}c_{p}\right). \end{align} $$

We simply have $b_{k}(e_{p})=b_{\overline {k,p}}$ and the quadratic operators in (4.5) can be expressed as

(4.17) $$ \begin{align} Q_{1}^{k}(A) & =\sum_{p\in L_{k}^{\pm}}\left(b_{k}^{\ast}(Ae_{p})b_{k}(e_{p})+b_{k}(e_{p})b_{k}^{\ast}(Ae_{p})\right) \nonumber\\ Q_{2}^{k}(B) & =\sum_{p\in L_{k}^{\pm}}\left(b_{k}^{\ast}\left(Be_{p}\right)b_{k}^{\ast}(e_{p})+b_{k}(e_{p})b_{k}\left(Be_{p}\right)\right) \end{align} $$

in analogy with equation (3.5). In order to justify the quasi-bosonic interpretation, we need rigorous estimates for the correction term in (4.16). Let us start with the following:

Proposition 4.1. For all $k\in \mathbb {Z}_{+}^{3}$ and $\varphi \in \ell ^{2}(L_{k}^{\pm })$ , it holds that $\varepsilon _{k,k}\left(\varphi ,\varphi \right)\leq 0$ , namely,

$$ \begin{align*}b_{k}(\varphi )b_{k}^{\ast}(\varphi ) \le b_{k}^{\ast}(\varphi )b_{k}(\varphi )+\left\Vert \varphi\right\Vert ^{2}. \end{align*} $$

Note that the observation of the error term $\varepsilon _{k,k}$ being non-positive also appeared in [Reference Benedikter, Nam, Porta, Schlein and Seiringer5, Proof of Lemma 4.2] in the context of different bosonic operators.

Proof. We expand the term

(4.18) $$ \begin{align} \varepsilon_{k,k}\left(\varphi;\varphi\right) & =\sum_{p,q\in L_{k}^{\pm}}\left\langle \varphi,e_{p}\right\rangle \left\langle e_{q},\varphi\right\rangle \varepsilon\left(\overline{k,p};\overline{k,q}\right) \nonumber\\ &=-\sum_{p,q\in L_{k}^{\pm}}\left\langle \varphi,e_{p}\right\rangle \left\langle e_{q},\varphi\right\rangle \left(\delta_{p,q}c_{\overline{q-k}}c_{\overline{p-k}}^{\ast}+\delta_{\overline{p-k},\overline{q-k}}c_{q}^{\ast}c_{p}\right)\nonumber \\ & =-\sum_{p\in L_{k}^{\pm}}\left|\left\langle e_{p},\varphi\right\rangle \right|^{2}c_{\overline{p-k}}c_{\overline{p-k}}^{\ast}-\sum_{p,q\in L_{k}^{\pm}}\delta_{\overline{p-k},\overline{q-k}}\left\langle \varphi,e_{p}\right\rangle \left\langle e_{q},\varphi\right\rangle c_{q}^{\ast}c_{p} \\ &\le - \sum_{p,q\in L_{k}^{\pm}}\delta_{\overline{p-k},\overline{q-k}}\left\langle \varphi,e_{p}\right\rangle \left\langle e_{q},\varphi\right\rangle c_{q}^{\ast}c_{p}.\nonumber \end{align} $$

We treat the terms of the last sum on a case-by-case basis according to which of $L_{k}$ and $L_{-k}$ , p and q lie in: if p and q lie in the same lune, then $\delta _{\overline {p-k},\overline {q-k}}=\delta _{p\mp k,q\mp k}=\delta _{p,q}$ and so

(4.19) $$ \begin{align} A =\left(\sum_{p,q\in L_{k}}+\sum_{p,q\in L_{-k}}\right)\delta_{\overline{p-k},\overline{q-k}}\left\langle \varphi,e_{p}\right\rangle \left\langle e_{q},\varphi\right\rangle c_{q}^{\ast}c_{p}=\sum_{p\in L_{k}^{\pm}}\left|\left\langle e_{p},\varphi\right\rangle \right|^{2}c_{p}^{\ast}c_{p} \ge 0. \end{align} $$

However, by the Cauchy–Schwarz inequality,

(4.20) $$ \begin{align} &\pm \left( \sum_{p\in L_{k}}\sum_{q\in L_{-k}} + \sum_{p\in L_{-k}}\sum_{q\in L_{k}} \right) \delta_{\overline{p-k},\overline{q-k}}\left\langle \varphi,e_{p}\right\rangle \left\langle e_{q},\varphi\right\rangle c_{q}^{\ast}c_{p}\nonumber\\ &\le \left( \sum_{p\in L_{k}}\sum_{q\in L_{-k}} + \sum_{p\in L_{-k}}\sum_{q\in L_{k}} \right) \frac{1}{2}\Big( \delta_{\overline{p-k},\overline{q-k}} |\left\langle \varphi,e_{p}\right\rangle|^2 c_p^* c_p + |\left\langle e_{q},\varphi\right\rangle|^2 c_{q}^{\ast}c_{q} \Big) \\ &\le \sum_{p\in L_{k}^{\pm}}\left|\left\langle e_{p},\varphi\right\rangle \right|^{2}c_{p}^{\ast}c_{p} = A. \nonumber \end{align} $$

We thus conclude that $\varepsilon _{k,k}\left(\varphi ,\varphi \right)\leq 0$ as claimed.

Next, we have the following:

Proposition 4.2. For all $k\in \mathbb {Z}_{+}^{3}$ , $\varphi \in \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in \mathcal {H}_{N}$ , it holds that

$$\begin{align*}\left\Vert b_{k}(\varphi )\Psi\right\Vert \leq\left\Vert \varphi\right\Vert \sqrt{\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle },\quad\left\Vert b_{k}^{\ast}(\varphi )\Psi\right\Vert \leq\left\Vert \varphi\right\Vert \sqrt{\left\langle \Psi,\left(1+\mathcal{N}_{E}\right)\Psi\right\rangle }. \end{align*}$$

The bounds here are similar to [Reference Benedikter, Nam, Porta, Schlein and Seiringer5, Lemma 4.2]. Recall that in our quasi-bosonic setting the excitation number operator

(4.21) $$ \begin{align} \mathcal{N}_{E}=\sum_{p\in B_{F}^{c}}c_{p}^{\ast}c_{p}=\sum_{p\in B_{F}}c_{p}c_{p}^{\ast} \end{align} $$

plays the role that the usual number operator $\mathcal {N}$ does in the exact bosonic case. Thus, Proposition 4.2 is the analogue of the well-known bosonic estimate

(4.22) $$ \begin{align} \left\Vert a(\varphi )\Psi\right\Vert \leq\left\Vert \varphi\right\Vert \sqrt{\left\langle \Psi,\mathcal{N}\Psi\right\rangle },\quad\left\Vert a^{\ast}(\varphi )\Psi\right\Vert \leq\left\Vert \varphi\right\Vert \sqrt{\left\langle \Psi,\left(1+\mathcal{N}\right)\Psi\right\rangle }. \end{align} $$

Proof. By the Cauchy-Schwarz inequality,

(4.23) $$ \begin{align} \left\Vert b_{k}(\varphi )\Psi\right\Vert & = \left\Vert \sum_{p\in L_{k}^{\pm}}\left\langle \varphi,e_{p}\right\rangle b_{\overline{k,p}}\Psi\right\Vert \leq\sqrt{\sum_{p\in L_{k}^{\pm}}\left|\left\langle \varphi,e_{p}\right\rangle \right|^{2}}\sqrt{\sum_{p\in L_{k}^{\pm}}\left\Vert b_{\overline{k,p}}\Psi\right\Vert ^{2}}\nonumber \\ & \leq\left\Vert \varphi\right\Vert \sqrt{\sum_{p\in L_{k}^{\pm}}\left\Vert c_{p}\Psi\right\Vert ^{2}}\leq\left\Vert \varphi\right\Vert \sqrt{\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle }. \end{align} $$

The second bound follows from the first and Proposition 4.1.

We remark that the above estimate is also valid for $\Psi \in \mathcal {H}_{M}$ when $M\neq N$ , provided $\mathcal {N}_{E}$ is understood as $\sum _{p\in B_{F}^{c}}c_{p}^{\ast }c_{p}$ acting on $\mathcal {H}_{M}$ (in (4.23) we used $L_k^{\pm } \subset B_F^c$ ). One must be precise here as the identity $\mathcal {N}_{E}=\sum _{p\in B_{F}}c_{p}c_{p}^{\ast }$ does not hold on $\mathcal {H}_{M}$ . In fact, the estimate also holds if $\mathcal {N}_{E}$ is understood as $\sum _{p\in B_{F}}c_{p}c_{p}^{\ast }$ , up to an additional factor of $\sqrt {2}$ due to the necessary overcounting of the holes,Footnote ⁷ namely, from $\left\Vert b_{\overline {k,p}}\Psi \right\Vert =\left\Vert c_{\overline {p-k}}^{\ast }c_{p}\Psi \right\Vert \leq \left\Vert c_{\overline {p-k}}^{\ast }\Psi \right\Vert $ with $\overline {p-k} \in B_F$ , we get

(4.24) $$ \begin{align} \left\Vert b_{k}(\varphi )\Psi\right\Vert \leq\left\Vert \varphi\right\Vert \sqrt{\sum_{p\in L_{k}^{\pm}}\left\Vert c_{\overline{p-k}}^{\ast}\Psi\right\Vert ^{2}}\leq\sqrt{2}\left\Vert \varphi\right\Vert \sqrt{\left\langle \Psi,\left(\sum_{p\in B_{F}}c_{p}c_{p}^{\ast}\right)\Psi\right\rangle }. \end{align} $$

This is a point that we must consider, since below we will also encounter expressions such as $\left\Vert b_{k}(\varphi )c_{p}\Psi \right\Vert $ for $\Psi \in \mathcal {H}_{N}$ (so that $c_{p}\Psi \in \mathcal {H}_{N-1}$ ). For this, we denote by $\mathcal {N}_{E}^{\left(-1\right)}:\mathcal {H}_{N-1}\rightarrow \mathcal {H}_{N-1}$ and $\mathcal {N}_{E}^{\left(+1\right)}:\mathcal {H}_{N+1}\rightarrow \mathcal {H}_{N+1}$ the operators

(4.25) $$ \begin{align} \mathcal{N}_{E}^{\left(-1\right)}=\sum_{p\in B_{F}^{c}}c_{p}^{\ast}c_{p},\quad\mathcal{N}_{E}^{\left(+1\right)}=\sum_{p\in B_{F}}c_{p}c_{p}^{\ast}. \end{align} $$

This choice is motivated by the following identities:

Lemma 4.3. For all $p\in B_{F}^{c}$ and $q\in B_{F}$ , it holds that

$$ \begin{align*} \mathcal{N}_{E}c_{p}^{\ast} &=c_{p}^{\ast}\mathcal{N}_{E}^{\left(-1\right)}+c_{p}^{\ast}, \quad c_p\mathcal{N}_{E}^{(-1)}c_{p}^{\ast} \le \mathcal{N}_E,\\ \mathcal{N}_{E}c_{q} &=c_{q}\mathcal{N}_{E}^{\left(+1\right)}+c_{q}, \quad c_{q}^{\ast}\mathcal{N}_{E}^{\left(+1\right)} c_{q} \le \mathcal{N}_E. \end{align*} $$

Consequently,

$$\begin{align*}\sum_{p\in B_{F}^{c}}c_{p}^{\ast}\mathcal{N}_{E}^{\left(-1\right)}c_{p}=\mathcal{N}_{E}^{2}-\mathcal{N}_{E}=\sum_{p\in B_{F}}c_{p}\mathcal{N}_{E}^{\left(+1\right)}c_{p}^{\ast}. \end{align*}$$

Proof. This follows directly by the CAR, as for all $p\in B_{F}^{c}$ ,

(4.26) $$ \begin{align} \mathcal{N}_{E}c_{p}^{\ast}=\sum_{q\in B_{F}^{c}}c_{q}^{\ast}c_{q}c_{p}^{\ast}=\sum_{q\in B_{F}^{c}}c_{p}^{\ast}c_{q}^{\ast}c_{q}+\sum_{q\in B_{F}^{c}}\left(c_{q}^{\ast}\left\{ c_{q},c_{p}^{\ast}\right\} -\left\{ c_{q}^{\ast},c_{p}^{\ast}\right\} c_{q}\right)=c_{p}^{\ast}\mathcal{N}_{E}^{\left(-1\right)}+c_{p}^{\ast}. \end{align} $$

Consequently, using $\|c_p\|_{\mathrm {Op}}=1$ and $[\mathcal {N}_{E},c_{p}^{\ast }c_p]=0$ , we have

(4.27) $$ \begin{align} \mathcal{N}_{E} \ge \mathcal{N}_{E}c_{p}^{\ast}c_p = c_{p}^{\ast}\mathcal{N}_{E}^{\left(-1\right)}c_p + c_{p}^{\ast} c_p \ge c_{p}^{\ast}\mathcal{N}_{E}^{\left(-1\right)}c_p. \end{align} $$

Likewise, for all $q\in B_{F}$ ,

(4.28) $$ \begin{align} \mathcal{N}_{E}c_{q}=\sum_{p\in B_{F}}c_{p}c_{p}^{\ast}c_{q}=\sum_{p\in B_{F}}c_{q}c_{p}c_{p}^{\ast}+\sum_{p\in B_{F}}\left(c_{p}\left\{ c_{p}^{\ast},c_{q}\right\} -\left\{ c_{p},c_{q}\right\} c_{p}^{\ast}\right)=c_{q}\mathcal{N}_{E}^{\left(+1\right)}+c_{q}, \end{align} $$

and hence, $\mathcal {N}_{E} \ge c_{q}^{\ast }\mathcal {N}_{E}^{\left(+1\right)} c_{q}$ . Moreover,

(4.29) $$ \begin{align} \sum_{p\in B_{F}^{c}}c_{p}^{\ast}\mathcal{N}_{E}^{\left(-1\right)}c_{p} &=\sum_{p\in B_{F}^{c}}\left(\mathcal{N}_{E}c_{p}^{\ast}-c_{p}^{\ast}\right)c_{p}=\mathcal{N}_{E}^{2}-\mathcal{N}_{E}=\sum_{p\in B_{F}}\left(\mathcal{N}_{E}c_{p}-c_{p}\right)c_{p}^{\ast}\nonumber \\ &=\sum_{p\in B_{F}}c_{p}\mathcal{N}_{E}^{\left(+1\right)}c_{p}^{\ast}. \end{align} $$

In some cases, it is important to refine error estimates by using the kinetic operator $H_{\operatorname {\mathrm {kin}}}^{\prime }$ rather than $\mathcal {N}_{E}$ . We can implement the kinetic estimate of Proposition 2.2 in the generalized setting:

Proposition 4.4. For all $k\in \mathbb {Z}_{+}^{3}$ , $\varphi \in \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in D\left(H_{\operatorname {\mathrm {kin}}}^{\prime }\right)$ , it holds that

$$\begin{align*}\left\Vert b_{k}(\varphi )\Psi\right\Vert \leq\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\varphi\right\Vert \sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle },\quad\left\Vert b_{k}^{\ast}(\varphi )\Psi\right\Vert \leq\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\varphi\right\Vert \sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle }+\left\Vert \varphi\right\Vert \left\Vert \Psi\right\Vert. \end{align*}$$

Proof. We start by applying the Cauchy-Schwarz inequality

(4.30) $$ \begin{align} \left\Vert b_{k}(\varphi )\Psi\right\Vert =\left\Vert \sum_{p\in L_{k}^{\pm}}\left\langle \varphi,e_{p}\right\rangle b_{\overline{k,p}}\Psi\right\Vert \leq\sqrt{\sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}^{-1}\left|\left\langle \varphi,e_{p}\right\rangle \right|^{2}}\sqrt{\sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}\left\Vert b_{\overline{k,p}}\Psi\right\Vert ^{2}}. \end{align} $$

As the vectors $(e_{p})_{p\in L_{k}^{\pm }}$ obey $h_{k}^{\oplus }e_{p}=\lambda _{\overline {k,p}}e_{p}$ , we recognize the first sum on the right-hand side as

(4.31) $$ \begin{align} \sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}^{-1}\left|\left\langle \varphi,e_{p}\right\rangle \right|^{2}=\left\langle \varphi,\left(h_{k}^{\oplus}\right)^{-1}\varphi\right\rangle =\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\varphi\right\Vert ^{2}. \end{align} $$

For the second sum, we have by equation (2.4) that

(4.32) $$ \begin{align} \sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}\left\Vert b_{\overline{k,p}}\Psi\right\Vert ^{2}=\sum_{p\in L_{k}}\lambda_{k,p}\left\Vert c_{p-k}^{\ast}c_{p}\Psi\right\Vert ^{2}+\sum_{p\in L_{-k}}\lambda_{-k,p}\left\Vert c_{p+k}^{\ast}c_{p}\Psi\right\Vert ^{2}\leq\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle, \end{align} $$

which implies the first claim. The second bound follows from the first and Proposition 4.1:

(4.33) $$ \begin{align} \left\Vert b_{k}^{\ast}(\varphi )\Psi\right\Vert \leq\sqrt{\left\langle \Psi,\left(\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\varphi\right\Vert ^{2}H_{\operatorname{\mathrm{kin}}}^{\prime}+\left\Vert \varphi\right\Vert ^{2}\right)\Psi\right\rangle }\leq\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\varphi\right\Vert \sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle }+\left\Vert \varphi\right\Vert \left\Vert \Psi\right\Vert. \end{align} $$

4.3 Preliminary estimates for quadratic operators

In this subsection, we provide some basic bounds on the quadratic operators $Q_{1}^{k}(A)$ and $Q_{1}^{k}(B)$ defined in (4.5) for any $k\in \mathbb {Z}_{+}^{3}$ . First, for $Q_{1}^{k}(A)$ , we can normal order as follows:

(4.34) $$ \begin{align} Q_{1}^{k}(A) & =\sum_{p\in L_{k}^{\pm}}\left(2\,b_{k}^{\ast}(Ae_{p})b_{k}(e_{p})+\left[b_{k}(e_{p}),b_{k}^{\ast}(Ae_{p})\right]\right)\nonumber \\ & =2\sum_{p\in L_{k}^{\pm}}b_{k}^{\ast}(Ae_{p})b_{k}(e_{p})+\sum_{p\in L_{k}^{\pm}}\left\langle e_{p},Ae_{p}\right\rangle +\sum_{p\in L_{k}^{\pm}}\varepsilon_{k,k}\left(e_{p};Ae_{p}\right)\\ & =2\,\tilde{Q}_{1}^{k}(A)+\text{tr}(A)+\varepsilon_{k}(A)\nonumber, \end{align} $$

where for brevity, we have defined the notation

(4.35) $$ \begin{align} \tilde{Q}_{1}^{k}(A)=\sum_{p\in L_{k}^{\pm}}b_{k}^{\ast}(Ae_{p})b_{k}(e_{p}),\quad\varepsilon_{k}(A)=\sum_{p\in L_{k}^{\pm}}\varepsilon_{k,k}\left(e_{p};Ae_{p}\right). \end{align} $$

The term $\tilde {Q}_{1}^{k}(A)$ plays the same role of $\text {d}\Gamma (A)$ in the exact bosonic case, whereas $\varepsilon _{k}(A)$ is a correction term in the quasi-bosonic case.

Proposition 4.5. For all $k\in \mathbb {Z}_{+}^{3}$ , symmetric $A:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in \mathcal {H}_{N}$ , it holds that

$$ \begin{align*} \left|\left\langle \Psi,\tilde{Q}_{1}^{k}(A)\Psi\right\rangle \right| &\leq \left\Vert A\right\Vert _{\operatorname{\mathrm{Op}}}\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle ,\\ \left|\left\langle \Psi,\varepsilon_{k}(A)\Psi\right\rangle \right| &\leq 3\left\Vert A\right\Vert _{\operatorname{\mathrm{Op}}}\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle. \end{align*} $$

If furthermore, $A\geq 0$ , then also $\tilde {Q}_{1}^{k}(A)\geq 0$ .

Proof. Let $(x_{i})_i$ be an eigenbasis for A with eigenvalues $\left(\lambda _{i}\right)_i$ . Noting that the mapping $x,y\mapsto b_{k}^{\ast }\left(Ax\right)b_{k}\left(y\right)$ is bilinear, we may invoke Lemma 3.3 (the part of basis independence) to write

(4.36) $$ \begin{align} \tilde{Q}_{1}^{k}(A)=\sum_{i}b_{k}^{\ast}\left(Ax_{i}\right)b_{k}(x_{i})=\sum_{i}\lambda_{i}b_{k}^{\ast}(x_{i})b_{k}(x_{i}). \end{align} $$

Clearly, if $A\ge 0$ , then all $\lambda _i\ge 0$ , and hence, $\tilde {Q}_{1}^{k}(A)\ge 0$ . In general, we always have $|\lambda _i | \le \left\Vert A\right\Vert _{\operatorname {\mathrm {Op}}}$ for all i. Hence, using Lemma 3.3 again and $b_{\overline {k,p}}^* b_{\overline {k,p}} \le c_p^* c_p$ , we have

(4.37) $$ \begin{align} \pm \tilde{Q}_{1}^{k}(A) \le \left\Vert A\right\Vert _{\text{Op}} \sum_{i}b_{k}^{\ast}(x_{i})b_{k}(x_{i})= \left\Vert A\right\Vert _{\text{Op}} \sum_{p\in L_{k}^{\pm}}b_{\overline{k,p}}^{\ast}b_{\overline{k,p}} \le \left\Vert A\right\Vert _{\text{Op}} \sum_{p\in L_{k}^{\pm}} c_p^* c_p \le \left\Vert A\right\Vert _{\text{Op}} \mathcal{N}_{E}. \end{align} $$

Similarly,

(4.38) $$ \begin{align} \pm \varepsilon_{k}(A)&= \pm \sum_{i}\varepsilon_{k,k}\left(x_{i};Ax_{i}\right)= \pm \sum_{i}\lambda_{i}\varepsilon_{k,k}\left(x_{i};x_{i}\right) \nonumber\\ &\le - \|A\|_{\mathrm{Op}}\sum_{i} \varepsilon_{k,k}\left(x_{i};x_{i}\right) = - \|A\|_{\mathrm{Op}} \sum_{p\in L_{k}^{\pm}}\varepsilon_{k,k}\left(e_{p};e_{p}\right), \end{align} $$

where in the first inequality we used the fact that $\varepsilon _{k,k}\left(x_{i};x_{i}\right)\le 0$ as shown in the proof of Proposition 4.1. Using $\varepsilon _{k,k}(e_{p};e_{p})=\varepsilon (\overline {k,p};\overline {l,q})$ and the definition (4.16), we get

(4.39) $$ \begin{align} - \sum_{p\in L_{k}^{\pm}}\varepsilon\left(\overline{k,p};\overline{k,p}\right) &=\sum_{p\in L_{k}^{\pm}}\left(c_{\overline{p-k}}c_{\overline{p-k}}^{\ast}+c_{p}^{\ast}c_{p}\right) \leq 2 \sum_{p\in B_{F}} c_{p}c_{p}^{\ast} +\sum_{p\in B_{F}^{c}} c_{p}^{\ast}c_{p} =3 \mathcal{N}_{E}, \end{align} $$

which implies the desired claim.

From these results and equation (4.34), we immediately obtain the following:

Proposition 4.6. For all $k\in \mathbb {Z}_{+}^{3}$ , symmetric $A:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in \mathcal {H}_{N}$ , it holds that

$$\begin{align*}\left|\left\langle \Psi,\left(Q_{1}^{k}(A)-\operatorname{\mathrm{tr}}(A)\right)\Psi\right\rangle \right|\leq5\left\Vert A\right\Vert _{\operatorname{\mathrm{Op}}}\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle. \end{align*}$$

Next, we turn to $Q_{2}^{k}(B)$ .

Proposition 4.7. For all $k\in \mathbb {Z}_{+}^{3}$ , symmetric $B:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in \mathcal {H}_{N}$ , it holds that

$$\begin{align*}\left|\left\langle \Psi,Q_{2}^{k}(B)\Psi\right\rangle \right|\leq2\left\Vert B\right\Vert _{\operatorname{\mathrm{HS}}}\sqrt{\left\langle \Psi,\left(1+\mathcal{N}_{E}\right)\Psi\right\rangle \left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle }\leq2\left\Vert B\right\Vert _{\operatorname{\mathrm{HS}}}\left\langle \Psi,\left(1+\mathcal{N}_{E}\right)\Psi\right\rangle. \end{align*}$$

Proof. We have (using that the $b_{k}$ operators commute)

(4.40) $$ \begin{align} \left\langle \Psi,Q_{2}^{k}(B)\Psi\right\rangle &=\sum_{p\in L_{k}^{\pm}}\left\langle \Psi,\left(b_{k}^{\ast}\left(Be_{p}\right)b_{k}^{\ast}(e_{p})+b_{k}(e_{p})b_{k}\left(Be_{p}\right)\right)\Psi\right\rangle \nonumber\\ &=2\sum_{p\in L_{k}^{\pm}}\mathrm{Re} \left\langle b_{k}^{\ast}\left(Be_{p}\right)\Psi,b_{k}(e_{p})\Psi\right\rangle, \end{align} $$

so using the estimates of Proposition 4.2 and the Cauchy-Schwarz inequality, we conclude that

(4.41) $$ \begin{align} \left|\left\langle \Psi,Q_{2}^{k}(B)\Psi\right\rangle \right| &\leq 2\sum_{p\in L_{k}^{\pm}}\left\Vert b_{k}^{\ast}\left(Be_{p}\right)\Psi\right\Vert \left\Vert b_{k}(e_{p})\Psi\right\Vert \leq2\sqrt{\left\langle \Psi,\left(1+\mathcal{N}_{E}\right)\Psi\right\rangle }\sum_{p\in L_{k}^{\pm}}\left\Vert Be_{p}\right\Vert \left\Vert b_{\overline{k,p}}\Psi\right\Vert \nonumber\\ & \leq2\sqrt{\left\langle \Psi,\left(1+\mathcal{N}_{E}\right)\Psi\right\rangle }\sqrt{\sum_{p\in L_{k}^{\pm}}\left\Vert Be_{p}\right\Vert ^{2}}\sqrt{\sum_{p\in L_{k}^{\pm}}\left\Vert b_{\overline{k,p}}\Psi\right\Vert ^{2}} \nonumber \\ &\leq2\left\Vert B\right\Vert _{\text{HS}}\left\langle \Psi,\left(1+\mathcal{N}_{E}\right)\Psi\right\rangle, \end{align} $$

where we again used that $\left\Vert b_{\overline {k,p}}\Psi \right\Vert \leq \left\Vert c_{p}\Psi \right\Vert $ .

Kinetic estimates for quadratic operators

Finally, let us improve the estimates in this subsection by using the kinetic operator $H_{\operatorname {\mathrm {kin}}}^{\prime }$ instead of the number operator $\mathcal {N}_E$ .

Proposition 4.8. For all $k\in \mathbb {Z}_{+}^{3}$ , symmetric $A:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in D\left(H_{\operatorname {\mathrm {kin}}}^{\prime }\right)$ , it holds that

$$\begin{align*}\left|\left\langle \Psi,\tilde{Q}_{1}^{k}(A)\Psi\right\rangle \right|\leq\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}A\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\right\Vert _{\operatorname{\mathrm{Op}}}\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle. \end{align*}$$

Proof. Let $(x_{i})_{i}$ be an eigenbasis for $(h_{k}^{\oplus })^{-\frac {1}{2}}A(h_{k}^{\oplus })^{-\frac {1}{2}}$ with eigenvalues $\left(\mu _{i}\right)_i$ . By Lemma 3.3, we then see that we may write $\tilde {Q}_{1}^{k}(A)$ as

(4.42) $$ \begin{align} \tilde{Q}_{1}^{k}(A) & =\sum_{i}b_{k}^{\ast}\left(Ax_{i}\right)b_{k}(x_{i})=\sum_{i}b_{k}^{\ast}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}A\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)b_{k}(x_{i})\nonumber \\ & =\sum_{i}b_{k}^{\ast}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}A\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}x_{i}\right)b_{k}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)\\ & =\sum_{i}\mu_{i}b_{k}^{\ast}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)b_{k}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right),\nonumber \end{align} $$

and so we can estimate

(4.43) $$ \begin{align} \left|\left\langle \Psi,\tilde{Q}_{1}^{k}(A)\Psi\right\rangle \right| &\leq\left(\max_{1\leq i\leq\left|L_{k}^{\pm}\right|}\left|\mu_{i}\right|\right)\sum_{i}\left\langle \Psi,b_{k}^{\ast}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)b_{k}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)\Psi\right\rangle \\ & =\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}A\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\right\Vert _{\text{Op}}\left\langle \Psi,\sum_{i}b_{k}^{\ast}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)b_{k}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)\Psi\right\rangle .\nonumber \end{align} $$

Applying Lemma 3.3 again, we also see that

(4.44) $$ \begin{align} \sum_{i}b_{k}^{\ast}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right)b_{k}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}x_{i}\right) &=\sum_{p\in L_{k}^{\pm}}b_{k}^{\ast}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}e_{p}\right)b_{k}\left(\left(h_{k}^{\oplus}\right)^{\frac{1}{2}}e_{p}\right)\nonumber \\ &=\sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}b_{\overline{k,p}}^{\ast}b_{\overline{k,p}}, \end{align} $$

so by equation (4.32), we obtain the desired bound of

(4.45) $$ \begin{align} \left|\left\langle \Psi,\tilde{Q}_{1}^{k}(A)\Psi\right\rangle \right|\leq\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}A\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\right\Vert _{\text{Op}}\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle. \end{align} $$

Next are the $\varepsilon _{k}(A)$ terms. These we cannot estimate in terms of $H_{\operatorname {\mathrm {kin}}}^{\prime }$ , but for A of diagonal form, we can still control them strongly:

Proposition 4.9. For all $k\in \mathbb {Z}_{+}^{3}$ , symmetric $A^{\oplus }=\left(\begin {array}{cc} A & 0\\ 0 & A \end {array}\right):\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in \mathcal {H}_{N}$ , it holds that

$$\begin{align*}\left|\left\langle \Psi,\varepsilon_{k}\left(A^{\oplus}\right)\Psi\right\rangle \right|\leq3\left(\max_{p\in L_{k}}\left|\left\langle e_{p},Ae_{p}\right\rangle \right|\right)\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle. \end{align*}$$

Proof. By the assumed form of $A^{\oplus }$ , we may write $\varepsilon _{k}\left(A^{\oplus }\right)$ as

(4.46) $$ \begin{align} \varepsilon_{k}\left(A^{\oplus}\right) & =\sum_{p\in L_{k}^{\pm}}\varepsilon_{k,k}\left(e_{p};A^{\oplus}e_{p}\right)=\sum_{p,q\in L_{k}^{\pm}}\left\langle e_{q},A^{\oplus}e_{p}\right\rangle \varepsilon_{k,k}\left(\overline{k,p};\overline{k,q}\right)\nonumber \\ & =-\sum_{p,q\in L_{k}^{\pm}}\left\langle e_{q},A^{\oplus}e_{p}\right\rangle \left(\delta_{p,q}c_{\overline{q-k}}c_{\overline{p-k}}^{\ast}+\delta_{\overline{p-k},\overline{q-k}}c_{q}^{\ast}c_{p}\right)\nonumber \\ & =-\sum_{p,q\in L_{k}}\left\langle e_{q},Ae_{p}\right\rangle \left(\delta_{p,q}c_{q-k}c_{p-k}^{\ast}+\delta_{p-k,q-k}c_{q}^{\ast}c_{p}\right)\\ &\quad -\sum_{p,q\in L_{-k}}\left\langle e_{-q},Ae_{-p}\right\rangle \left(\delta_{p,q}c_{q+k}c_{p+k}^{\ast}+\delta_{p+k,q+k}c_{q}^{\ast}c_{p}\right)\nonumber \\ & =-\sum_{p\in L_{k}}\left\langle e_{p},Ae_{p}\right\rangle \left(c_{p-k}c_{p-k}^{\ast}+c_{p}^{\ast}c_{p}\right)-\sum_{p\in L_{-k}}\left\langle e_{-p},Ae_{-p}\right\rangle \left(c_{p+k}c_{p+k}^{\ast}+c_{p}^{\ast}c_{p}\right)\nonumber \end{align} $$

since the terms with $p\in L_{k},q\in L_{-k}$ or $p\in L_{-k},q\in L_{k}$ vanish (because $L_k\cap L_{-k}=\emptyset $ and there are $\delta _{p,q}$ , $\delta _{p-k,q-k}$ in the summand). We can thus estimate

(4.47) $$ \begin{align} \left|\left\langle \Psi,\varepsilon_{k}\left(A^{\oplus}\right)\Psi\right\rangle \right| & \leq\sum_{p\in L_{k}}\left|\left\langle e_{p},Ae_{p}\right\rangle \right|\left\langle \Psi,\left(c_{p-k}c_{p-k}^{\ast}+c_{p}^{\ast}c_{p}\right)\Psi\right\rangle \nonumber \\ &\quad +\sum_{p\in L_{-k}}\left|\left\langle e_{-p},Ae_{-p}\right\rangle \right|\left\langle \Psi,\left(c_{p+k}c_{p+k}^{\ast}+c_{p}^{\ast}c_{p}\right)\Psi\right\rangle \\ & \leq\left(\max_{p\in L_{k}}\left|\left\langle e_{p},Ae_{p}\right\rangle \right|\right)\left\langle \Psi,\left(\sum_{p\in L_{k}^{\pm}}c_{p}^{\ast}c_{p}+\sum_{p\in L_{k}-k}c_{p}c_{p}^{\ast}+\sum_{p\in L_{-k}+k}c_{p}c_{p}^{\ast}\right)\Psi\right\rangle \nonumber \\ & \leq3\left(\max_{p\in L_{k}}\left|\left\langle e_{p},Ae_{p}\right\rangle \right|\right)\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle .\nonumber \end{align} $$

Lastly, we consider the $Q_{2}^{k}(B)$ terms:

Proposition 4.10. For all $k\in \mathbb {Z}_{+}^{3}$ , symmetric $B:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in D\left(H_{\operatorname {\mathrm {kin}}}^{\prime }\right)$ , it holds that

$$\begin{align*}\left|\left\langle \Psi,Q_{2}^{k}(B)\Psi\right\rangle \right|\leq2\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}B\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\right\Vert _{\operatorname{\mathrm{HS}}}\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle +2\left\Vert B\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}\right\Vert _{\operatorname{\mathrm{HS}}}\sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle }\left\Vert \Psi\right\Vert. \end{align*}$$

Proof. By the Cauchy-Schwarz inequality and Proposition 4.4, we have

(4.48) $$ \begin{align} & \left|\left\langle \Psi,Q_{2}^{k}(B)\Psi\right\rangle \right| =\left|2\sum_{p\in L_{k}^{\pm}}\mathrm{ Re}\left(\left\langle \Psi,b_{k}\left(Be_{p}\right)b_{k}(e_{p})\Psi\right\rangle \right)\right|\leq2\sum_{p\in L_{k}^{\pm}}\left\Vert b_{k}^{\ast}\left(Be_{p}\right)\Psi\right\Vert \left\Vert b_{k}(e_{p})\Psi\right\Vert \nonumber \\ & \leq2\sum_{p\in L_{k}^{\pm}}\left(\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Be_{p}\right\Vert \sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle }+\left\Vert Be_{p}\right\Vert \left\Vert \Psi\right\Vert \right)\left\Vert b_{k}(e_{p})\Psi\right\Vert \\ & \leq2\sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle }\sum_{p\in L_{k}^{\pm}}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Be_{p}\right\Vert \left\Vert b_{k}(e_{p})\Psi\right\Vert +2\left\Vert \Psi\right\Vert \sum_{p\in L_{k}^{\pm}}\left\Vert Be_{p}\right\Vert \left\Vert b_{k}(e_{p})\Psi\right\Vert .\nonumber \end{align} $$

For the first sum, we can again apply the Cauchy-Schwarz inequality and (4.32):

(4.49) $$ \begin{align} \sum_{p\in L_{k}^{\pm}}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Be_{p}\right\Vert \left\Vert b_{k}(e_{p})\Psi\right\Vert & \leq\sqrt{\sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}^{-1}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Be_{p}\right\Vert ^{2}}\sqrt{\sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}\left\Vert b_{\overline{k,p}}\Psi\right\Vert ^{2}}\\ & \leq\sqrt{\sum_{p\in L_{k}^{\pm}}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}B\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}e_{p}\right\Vert ^{2}}\sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle },\nonumber \end{align} $$

and we likewise estimate the second sum as

(4.50) $$ \begin{align} \sum_{p\in L_{k}^{\pm}}\left\Vert Be_{p}\right\Vert \left\Vert b_{k}(e_{p})\Psi\right\Vert & \leq\sqrt{\sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}^{-1}\left\Vert Be_{p}\right\Vert ^{2}}\sqrt{\sum_{p\in L_{k}^{\pm}}\lambda_{\overline{k,p}}\left\Vert b_{\overline{k,p}}\Psi\right\Vert ^{2}}\\ & \leq\sqrt{\sum_{p\in L_{k}^{\pm}}\left\Vert B\left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}e_{p}\right\Vert ^{2}}\sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle }.\nonumber \end{align} $$

The claim now follows by recognizing the Hilbert-Schmidt norms.

5.1 Transformation of quadratic operators

By exploiting the similarity of our quasi-bosonic definitions with the exact bosonic case, we can now easily deduce the analogues of Propositions 3.2 and 3.4:

Proposition 5.1. For all $k\in S_{C}$ , $\varphi \in \ell ^{2}(L_{k}^{\pm })$ and symmetric operators $(K_{l}^{\oplus })_{l\in S_{C}}$ , it holds that

$$ \begin{align*} \left[\mathcal{K},b_{k}(\varphi )\right] & =b_{k}^{\ast}\left(K_{k}^{\oplus}\varphi\right)+\mathcal{E}_{k}(\varphi ),\\ \left[\mathcal{K},b_{k}^{\ast}(\varphi )\right] & =b_{k}\left(K_{k}^{\oplus}\varphi\right)+\mathcal{E}_{k}(\varphi )^{\ast}, \end{align*} $$

where

$$\begin{align*}\mathcal{E}_{k}(\varphi )=\frac{1}{2}\sum_{l\in S_{C}}\sum_{q\in L_{l}^{\pm}}\left\{ b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{k,l}\left(\varphi;e_{q}\right)\right\}. \end{align*}$$

Proof. We calculate using the commutation relations of (4.15) that

(5.3) $$ \begin{align} \left[\mathcal{K},b_{k}(\varphi )\right] & =\frac{1}{2}\sum_{l\in S_{C}}\sum_{q\in L_{l}^{\pm}}\left(\left[b_{l}\left(K_{l}^{\oplus}e_{q}\right)b_{l}(e_{p}),b_{k}(\varphi )\right]-\left[b_{l}^{\ast}\left(e_{q}\right)b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),b_{k}(\varphi )\right]\right)\nonumber \\ & =\frac{1}{2}\sum_{l\in S_{C}}\sum_{q\in L_{l}^{\pm}}\left(b_{l}^{\ast}\left(e_{q}\right)\left[b_{k}(\varphi ),b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\right]+\left[b_{k}(\varphi ),b_{l}^{\ast}\left(e_{q}\right)\right]b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\right)\nonumber \\ & =\frac{1}{2}\sum_{l\in S_{C}}\sum_{q\in L_{l}^{\pm}}b_{l}^{\ast}\left(e_{q}\right)\left(\delta_{k,l}\left\langle \varphi,K_{l}^{\oplus}e_{q}\right\rangle +\varepsilon_{k,l}\left(\varphi;K_{l}^{\oplus}e_{q}\right)\right)\nonumber\\ &\quad +\frac{1}{2}\sum_{l\in S_{C}}\sum_{q\in L_{l}^{\pm}}\left(\delta_{k,l}\left\langle \varphi,e_{q}\right\rangle +\varepsilon_{k,l}\left(\varphi;e_{q}\right)\right)b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\nonumber \\ & =\frac{1}{2}b_{k}^{\ast}\left(\sum_{q\in L_{k}^{\pm}}\left\langle \varphi,K_{k}^{\oplus}e_{q}\right\rangle e_{q}\right)+\frac{1}{2}b_{k}^{\ast}\left(K_{k}^{\oplus}\sum_{q\in L_{k}^{\pm}}\left\langle \varphi,e_{q}\right\rangle e_{q}\right)+\mathcal{E}_{k}(\varphi )\nonumber \\ & =b_{k}^{\ast}\left(K_{k}^{\oplus}\varphi\right)+\mathcal{E}_{k}(\varphi ) \end{align} $$

for $\mathcal {E}_{k}(\varphi )$ given by

(5.4) $$ \begin{align} \mathcal{E}_{k}(\varphi ) & =\frac{1}{2}\sum_{l\in S_{C}}\sum_{q\in L_{l}^{\pm}}\left(b_{l}^{\ast}\left(e_{q}\right)\varepsilon_{k,l}\left(\varphi;K_{l}^{\oplus}e_{q}\right)+\varepsilon_{k,l}\left(\varphi;e_{q}\right)b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\right)\nonumber\\ & =\frac{1}{2}\sum_{l\in S_{C}}\sum_{q\in L_{l}^{\pm}}\left\{ b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{k,l}\left(\varphi;e_{q}\right)\right\} , \end{align} $$

where we used Lemma 3.3 to simplify the expression (as $x,y\mapsto b_{l}^{\ast } (x )\varepsilon _{k,l}\left(\varphi ;y\right)$ is bilinear for fixed $\varphi $ and $K_{k}^{\oplus }$ is symmetric). The commutator $\left[\mathcal {K},b_{k}^{\ast }(\varphi )\right]$ follows by taking the adjoint.

From this, we easily deduce the commutator of $\mathcal {K}$ with quadratic operators:

Proposition 5.2. For all $k\in S_{C}$ and symmetric operators $A,B:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ , it holds that

$$ \begin{align*} \left[\mathcal{K},Q_{1}^{k}(A)\right] & =Q_{2}^{k}\left(\left\{ K_{k}^{\oplus},A\right\} \right)+\mathcal{E}_{1}^{k}(A)\\ \left[\mathcal{K},Q_{2}^{k}(B)\right] & =Q_{1}^{k}\left(\left\{ K_{k}^{\oplus},B\right\} \right)+\mathcal{E}_{2}^{k}(B), \end{align*} $$

where

$$ \begin{align*} \mathcal{E}_{1}^{k}(A) & =\frac{1}{2}\sum_{l\in S_{C}}\sum_{p\in L_{k}^{\pm}}\sum_{q\in L_{l}^{\pm}}\left(\left\{ b_{k}^{\ast}(Ae_{p}),\left\{ b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{k,l}(e_{p};e_{q})\right\} \right\} \right.\\ & \quad\left.+\left\{ \left\{ \varepsilon_{l,k}\left(e_{q};e_{p}\right),b_{l}\left(K_{l}^{\oplus}e_{q}\right)\right\} ,b_{k}(Ae_{p})\right\} \right)\\ \mathcal{E}_{2}^{k}(B) & =\frac{1}{2}\sum_{l\in S_{C}}\sum_{p\in L_{k}^{\pm}}\sum_{q\in L_{l}^{\pm}}\left(\left\{ b_{k}^{\ast}\left(Be_{p}\right),\left\{ b_{l}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{l,k}\left(e_{q};e_{p}\right)\right\} \right\} \right.\\ & \quad \left.+\left\{ \left\{ \varepsilon_{k,l}(e_{p};e_{q}),b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\right\} ,b_{k}\left(Be_{p}\right)\right\} \right). \end{align*} $$

Proof. We compute using the commutators of the previous proposition (and Lemma 3.3, to simplify the resulting expressions) that

(5.5) $$ \begin{align} \left[\mathcal{K},Q_{1}^{k}(A)\right] & =\sum_{p\in L_{k}^{\pm}}\left(\left[\mathcal{K},b_{k}^{\ast}(Ae_{p})b_{k}(e_{p})\right]+\left[\mathcal{K},b_{k}(e_{p})b_{k}^{\ast}(Ae_{p})\right]\right)\nonumber \\ & =\sum_{p\in L_{k}^{\pm}}\left(b_{k}^{\ast}(Ae_{p})\left[\mathcal{K},b_{k}(e_{p})\right]+\left[\mathcal{K},b_{k}^{\ast}(Ae_{p})\right]b_{k}(e_{p})\right)\nonumber \\ &\quad +\sum_{p\in L_{k}^{\pm}}\left(b_{k}(e_{p})\left[\mathcal{K},b_{k}^{\ast}(Ae_{p})\right]+\left[\mathcal{K},b_{k}(e_{p})\right]b_{k}^{\ast}(Ae_{p})\right)\nonumber\\ & =\sum_{p\in L_{k}^{\pm}}\left(b_{k}^{\ast}(Ae_{p})\left(b_{k}^{\ast}\left(K_{k}^{\oplus}e_{p}\right)+\mathcal{E}_{k}(e_{p})\right)+\left(b_{k}\left(K_{k}^{\oplus}Ae_{p}\right)+\mathcal{E}_{k}(Ae_{p})^{\ast}\right)b_{k}(e_{p})\right)\nonumber \\ &\quad +\sum_{p\in L_{k}^{\pm}}\left(b_{k}(e_{p})\left(b_{k}\left(K_{k}^{\oplus}Ae_{p}\right)+\mathcal{E}_{k}(Ae_{p})^{\ast}\right)+\left(b_{k}^{\ast}\left(K_{k}^{\oplus}e_{p}\right)+\mathcal{E}_{k}(e_{p})\right)b_{k}^{\ast}(Ae_{p})\right)\nonumber \\ & =\sum_{p\in L_{k}^{\pm}}\left(b_{k}^{\ast}\left(\left(AK_{k}^{\oplus}+K_{k}^{\oplus}A\right)e_{p}\right)b_{k}^{\ast}(e_{p})+b_{k}(e_{p})b_{k}\left(\left(K_{k}^{\oplus}A+K_{k}^{\oplus}A\right)e_{p}\right)\right)\nonumber \\ &\quad +\sum_{p\in L_{k}^{\pm}}\left(b_{k}^{\ast}(Ae_{p})\mathcal{E}_{k}(e_{p})+\mathcal{E}_{k}(e_{p})^{\ast}b_{k}(Ae_{p})+b_{k}(Ae_{p})\mathcal{E}_{k}(e_{p})^{\ast}+\mathcal{E}_{k}(e_{p})b_{k}^{\ast}(Ae_{p})\right)\nonumber \\ & =Q_{2}^{k}\left(\left\{ K_{k}^{\oplus},A\right\} \right)+\sum_{p\in L_{k}^{\pm}}\left(\left\{ b_{k}^{\ast}(Ae_{p}),\mathcal{E}_{k}(e_{p})\right\} +\left\{ \mathcal{E}_{k}(e_{p})^{\ast},b_{k}(Ae_{p})\right\} \right) \end{align} $$

and

(5.6) $$ \begin{align} & \sum_{p\in L_{k}^{\pm}}\left(\left\{ b_{k}^{\ast}(Ae_{p}),\mathcal{E}_{k}(e_{p})\right\} +\left\{ \mathcal{E}_{k}(e_{p})^{\ast},b_{k}(Ae_{p})\right\} \right)\nonumber \\ &\quad =\frac{1}{2}\sum_{l\in S_{C}}\sum_{p\in L_{k}^{\pm}}\sum_{q\in L_{l}^{\pm}}\left(\left\{ b_{k}^{\ast}(Ae_{p}),\left\{ b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{k,l}(e_{p};e_{q})\right\} \right\} \right.\\ & \qquad\left.+\left\{ \left\{ \varepsilon_{l,k}\left(e_{q};e_{p}\right),b_{l}\left(K_{l}^{\oplus}e_{q}\right)\right\} ,b_{k}(Ae_{p})\right\} \right)=\mathcal{E}_{1}^{k}(A) \nonumber \end{align} $$

as $\varepsilon _{k,l}(e_{p};e_{q})^{\ast }=\varepsilon _{l,k}\left(e_{q};e_{p}\right)$ . The computation of $Q_{2}^{k}(B)$ is similar.

Action of $e^{\mathcal {K}}$ on quadratic operators

With the commutators calculated, we are now ready to determine the full action of $e^{\mathcal {K}}$ on the quadratic operators $Q_{1}^{k}(\cdot )$ and $Q_{2}^{k}(\cdot )$ . Rather than appeal to the Baker-Campbell-Hausdorff formula, which would also require describing the commutators $\left[\mathcal {K},\mathcal {E}_{1}^{k}(A)\right]$ , etc., we will employ a ‘Duhamel-type’ argument which allows us to more selectively expand the operator $e^{\mathcal {K}}$ .

As in Section 3, we use the notation $\mathcal {A}_{K_{k}^{\oplus }}=\left\{ K_{k}^{\oplus },\cdot \right\}$ for anticommutators with $K_{k}^{\oplus }$ .

Before stating the proposition, we must make a remark. To use these identities, we will need to take limits, and to justify those limits, we need some general estimates on operators of the form $Q_{1}^{k}(\cdot ),Q_{2}^{k}(\cdot ),\mathcal {E}_{1}^{k}(\cdot ),\mathcal {E}_{2}^{k}(\cdot )$ . The Propositions 4.6, 4.7 establish these for $Q_{1}^{k}(\cdot )$ and $Q_{2}^{k}(\cdot )$ , while Proposition 6.4 will establish these for $\mathcal {E}_{1}^{k}(\cdot )$ and $\mathcal {E}_{2}^{k}(\cdot )$ .

The statement follows:

Proposition 5.3. For all $k\in S_{C}$ and symmetric $A,B:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ , it holds that

$$ \begin{align*} e^{\mathcal{K}}Q_{1}^{k}(A)e^{-\mathcal{K}} & =\frac{1}{2}Q_{1}^{k}\left(e^{K_{k}^{\oplus}}Ae^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}Ae^{-K_{k}^{\oplus}}\right)+\frac{1}{2}Q_{2}^{k}\left(e^{K_{k}^{\oplus}}Ae^{K_{k}^{\oplus}}-e^{-K_{k}^{\oplus}}Ae^{-K_{k}^{\oplus}}\right)\\ &\quad +\int_{0}^{1}e^{t\mathcal{K}}\left(\mathcal{E}_{1}^{k}\left(\cosh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)(A)\right)+\mathcal{E}_{2}^{k}\left(\sinh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)(A)\right)\right)e^{-t\mathcal{K}}\,dt\\ e^{\mathcal{K}}Q_{2}^{k}(B)e^{-\mathcal{K}} & =\frac{1}{2}Q_{1}^{k}\left(e^{K_{k}^{\oplus}}Be^{K_{k}^{\oplus}}-e^{-K_{k}^{\oplus}}Be^{-K_{k}^{\oplus}}\right)+\frac{1}{2}Q_{2}^{k}\left(e^{K_{k}^{\oplus}}Be^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}Be^{-K_{k}^{\oplus}}\right)\\ &\quad +\int_{0}^{1}e^{t\mathcal{K}}\left(\mathcal{E}_{1}^{k}\left(\sinh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)(B)\right)+\mathcal{E}_{2}^{k}\left(\cosh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)(B)\right)\right)e^{-t\mathcal{K}}\,dt, \end{align*} $$

the integrals being Riemann integrals of bounded operators.

Proof. We consider $e^{\mathcal {K}}Q_{1}^{k}(A)e^{-\mathcal {K}}$ , with the argument for $e^{\mathcal {K}}Q_{2}^{k}(B)e^{-\mathcal {K}}$ being similar. We first claim that for any $n\in \mathbb {N}$ ,

(5.7) $$ \begin{align} e^{\mathcal{K}}Q_{1}^{k}(A)e^{-\mathcal{K}} &=Q_{1}^{k}\left(\sum_{m=0}^{n_{1}}\frac{1}{\left(2m\right)!}\mathcal{A}_{K_{k}^{\oplus}}^{2m}(A)\right)+Q_{2}^{k}\left(\sum_{m=0}^{n_{2}}\frac{1}{\left(2m+1\right)!}\mathcal{A}_{K_{k}^{\oplus}}^{2m+1}(A)\right)\\ &\quad +\int_{0}^{1}e^{t\mathcal{K}}\left(\mathcal{E}_{1}^{k}\left(\sum_{m=0}^{n_{1}}\frac{1}{\left(2m\right)!}\mathcal{A}_{(1-t)K_{k}^{\oplus}}^{2m}(A)\right)+\mathcal{E}_{2}^{k}\left(\sum_{m=0}^{n_{2}}\frac{1}{\left(2m+1\right)!}\mathcal{A}_{(1-t)K_{k}^{\oplus}}^{2m+1}(A)\right)\right)e^{-t\mathcal{K}}\,dt\nonumber \\ &\quad +\frac{1}{(n-1)!}\int_{0}^{1}e^{t\mathcal{K}}Q_{\overline{n-1}}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right)e^{-t\mathcal{K}}(1-t)^{n-1}\,dt,\nonumber \end{align} $$

where, for brevity, $\overline {n-1}=n-1\ \mod 2$ and $n_{1},n_{2}$ are the largest integers such that $2n_{1}<n$ and $2n_{2}+1<n$ , respectively.

We proceed by induction. For $n=1$ , we find by the fundamental theorem of calculus that

(5.8) $$ \begin{align} e^{\mathcal{K}}Q_{1}^{k}(A)e^{-\mathcal{K}} & =Q_{1}^{k}(A)+\int_{0}^{1}\frac{d}{dt}\left(e^{t\mathcal{K}}Q_{1}^{k}(A)e^{-t\mathcal{K}}\right)\,dt=Q_{1}^{k}(A)+\int_{0}^{1}e^{t\mathcal{K}}\left[\mathcal{K},Q_{1}^{k}(A)\right]e^{-t\mathcal{K}}\,dt\nonumber \\ & =Q_{1}^{k}(A)+\int_{0}^{1}e^{t\mathcal{K}}\left(Q_{2}^{k}\left(\left\{ K_{k}^{\oplus},A\right\} \right)+\mathcal{E}_{1}^{k}(A)\right)e^{-t\mathcal{K}}\,dt\\ & =Q_{1}^{k}(A)+\int_{0}^{1}e^{t\mathcal{K}}\mathcal{E}_{1}^{k}(A)e^{-t\mathcal{K}}\,dt+\int_{0}^{1}e^{t\mathcal{K}}Q_{2}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}(A)\right)e^{-t\mathcal{K}}\,dt\nonumber \end{align} $$

by the commutator of Proposition 5.2, which is the statement for $n=1$ (in this case, $n_1=0$ and $n_2=-1$ , so $\sum _{m=0}^{n_1}$ contains one term and $\sum _{m=0}^{n_2}$ is empty).

For the inductive step, we now assume that case n holds. Integrating the last term of equation (5.7) by parts, we find that

(5.9) $$ \begin{align} & \frac{1}{(n-1)!}\int_{0}^{1}e^{t\mathcal{K}}Q_{\overline{n-1}}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right)e^{-t\mathcal{K}}(1-t)^{n-1}dt\nonumber \\ &\quad =\frac{1}{(n-1)!}\left[e^{t\mathcal{K}}Q_{\overline{n-1}}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right)e^{-t\mathcal{K}}\left(-\frac{(1-t)^{n}}{n}\right)\right]_{0}^{1}\nonumber \\ &\qquad -\frac{1}{(n-1)!}\int_{0}^{1}e^{t\mathcal{K}}\left[\mathcal{K},Q_{\overline{n-1}}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right)\right]e^{-t\mathcal{K}}\left(-\frac{(1-t)^{n}}{n}\right)dt\\ &\quad =\frac{1}{n!}Q_{\overline{n-1}}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right)+\frac{1}{n!}\int_{0}^{1}e^{t\mathcal{K}}\left(Q_{\overline{n}}^{k}\left(\left\{ K_{k}^{\oplus},\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right\} \right)+\mathcal{E}_{\overline{n-1}}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right)\right)e^{-t\mathcal{K}}(1-t)^{n}dt\nonumber \\ &\quad =Q_{\overline{n-1}}^{k}\left(\frac{1}{n!}\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\right)+\int_{0}^{1}e^{t\mathcal{K}}\mathcal{E}_{\overline{n-1}}^{k}\left(\frac{1}{n!}\mathcal{A}_{(1-t)K_{k}^{\oplus}}^{n}(A)\right)e^{-t\mathcal{K}}\,dt\nonumber \\ &\qquad +\frac{1}{n!}\int_{0}^{1}e^{t\mathcal{K}}Q_{\overline{n}}^{k}\left(\mathcal{A}_{K_{k}^{\oplus}}^{n+1}(A)\right)e^{-t\mathcal{K}}(1-t)^{n}dt,\nonumber \end{align} $$

where we also used that

(5.10) $$ \begin{align} (1-t)^{n}\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)=\big((1-t)\mathcal{A}_{K_{k}^{\oplus}}(A)\big)^{n}=\mathcal{A}_{(1-t)K_{k}^{\oplus}}^{n}(A). \end{align} $$

Inserting this into (5.7) and collecting like terms yields the statement for case $n+1$ .

We now deduce the statement from (5.7) by taking $n\to \infty $ . Recall the identities

(5.11) $$ \begin{align} \cosh\big(\mathcal{A}_{K_{k}^{\oplus}}\big)(T) & =\frac{1}{2}\big(e^{K_{k}^{\oplus}}Te^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}Te^{-K_{k}^{\oplus}}\big) \\ \sinh\big(\mathcal{A}_{K_{k}^{\oplus}}\big)(T) & =\frac{1}{2}\big(e^{K_{k}^{\oplus}}Te^{K_{k}^{\oplus}}-e^{-K_{k}^{\oplus}}Te^{-K_{k}^{\oplus}}\big)\nonumber \end{align} $$

from Proposition 3.5 and note that $\left((n-1)!\right)^{-1}\mathcal {A}_{K_{k}^{\oplus }}^{n}(A)\to 0$ as $n\to \infty $ . By Proposition 4.6,

$$ \begin{align*}Q_{1}^{k}\left(\sum_{m=0}^{n_{1}}\frac{1}{\left(2m\right)!}\mathcal{A}_{K_{k}^{\oplus}}^{2m}(A)\right)\to \frac{1}{2}Q_{1}^{k}\left(e^{K_{k}^{\oplus}}Ae^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}Ae^{-K_{k}^{\oplus}}\right) \end{align*} $$

and

$$ \begin{align*}\frac{1}{(n-1)!}\int_{0}^{1}e^{t\mathcal{K}}Q_{1}^{k}\big(\mathcal{A}_{K_{k}^{\oplus}}^{n}(A)\big)e^{-t\mathcal{K}}(1-t)^{n-1}\,dt \to 0. \end{align*} $$

Similar convergence for $Q_2$ is justified by Proposition 4.7. The convergence for $\mathcal {E}_{1}^{k}$ and $\mathcal {E}_{2}^{k}$ follows from Proposition 6.4.

Remark on the transformation of excitation operators

Let us make a quick remark on why we choose to approach the Bogolubov transformation from the point of view of quadratic operators rather than the usual creation and annihilation operator approach. Recall that in the exact bosonic case the creation and annihilation operators transformed under a Bogolubov transformation as

(5.12) $$ \begin{align} e^{\mathcal{K}}a(\varphi )e^{-\mathcal{K}} & =a\left(\cosh\left(K\right)\varphi\right)+a^{\ast}\left(\sinh\left(K\right)\varphi\right)\\ e^{\mathcal{K}}a^{\ast}(\varphi )e^{-\mathcal{K}} & =a^{\ast}\left(\cosh\left(K\right)\varphi\right)+a\left(\sinh\left(K\right)\varphi\right).\nonumber \end{align} $$

In the quasi-bosonic setting, we can use the commutators of Proposition 5.1 and a similar Duhamel-type argument to what we just applied to conclude that

(5.13) $$ \begin{align} e^{\mathcal{K}}b_{k}(\varphi )e^{-\mathcal{K}} & =b_{k}\left(\cosh\left(K_{k}^{\oplus}\right)\varphi\right)+b_{k}^{\ast}\left(\sinh\left(K_{k}^{\oplus}\right)\varphi\right)\\ &\quad +\int_{0}^{1}e^{t\mathcal{K}}\left(\mathcal{E}_{k}\left(\cosh\left((1-t)K_{k}^{\oplus}\right)\varphi\right)+\mathcal{E}_{k}\left(\sinh\left((1-t)K_{k}^{\oplus}\right)\varphi\right)^{\ast}\right)e^{-t\mathcal{K}}\,dt\nonumber \end{align} $$

with a similar expression for $e^{\mathcal {K}}b_{k}^{\ast }(\varphi )e^{-\mathcal {K}}$ . This is a more cumbersome expression to work with, and if we were to describe $e^{\mathcal {K}}Q_{1}^{k}(A)e^{-\mathcal {K}}$ by transforming the individual terms of $Q_{1}^{k}(A)$ like this rather than transforming $Q_{1}^{k}(A)$ as a whole, the error terms would not only go from being under a single integral to involving the product of two integrals, it would also involve cross terms between the bosonic terms and the error terms of equation (5.13). These cross terms, in particular, would severely reduce the quality of the final error estimate. Hence, we prefer the quadratic operator approach in the quasi-bosonic setting.

5.2 Transformation of the kinetic operator

There remains the task of describing the action of $e^{\mathcal {K}}$ on the localized kinetic operator $H_{\operatorname {\mathrm {kin}}}^{\prime }$ . For this, we must first formulate $H_{\operatorname {\mathrm {kin}}}^{\prime }$ – or rather the commutator $[H_{\operatorname {\mathrm {kin}}}^{\prime },b_{k,p}^{\ast }]$ calculated in (1.74) – within the general framework that we have introduced in this section. Recalling the operators $h_{k}^{\oplus }:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ in (4.11), then by (1.74) and linearity it follows that

(5.14) $$ \begin{align} \left[H_{\operatorname{\mathrm{kin}}}^{\prime},b_{k}(\varphi )\right]=-2\,b_{k}\left(h_{k}^{\oplus}\varphi\right),\quad\left[H_{\operatorname{\mathrm{kin}}}^{\prime},b_{k}^{\ast}(\varphi )\right]=2\,b_{k}^{\ast}\left(h_{k}^{\oplus}\varphi\right) \end{align} $$

for all $\varphi \in \ell ^{2}(L_{k}^{\pm })$ . (The factor of $2$ is introduced here because in the analogy of equation (4.9), $H_{\operatorname {\mathrm {kin}}}^{\prime }$ appears like a $\text {d}\Gamma (\cdot )=\frac {1}{2}Q_{1}(\cdot )-\frac {1}{2}\text {tr}(\cdot )$ term rather than a pure $Q_{1}(\cdot )$ term.)

We now calculate $\left[\mathcal {K},H_{\operatorname {\mathrm {kin}}}^{\prime }\right]$ as follows:

Proposition 5.4. $H_{\operatorname {\mathrm {kin}}}^{\prime }$ obeys

$$\begin{align*}\left[\mathcal{K},H_{\operatorname{\mathrm{kin}}}^{\prime}\right]=\sum_{k\in S_{C}}Q_{2}^{k}\left(\left\{ K_{k}^{\oplus},h_{k}^{\oplus}\right\} \right). \end{align*}$$

Proof. We compute, using the commutators of equation (5.14) and Lemma 3.3, that

(5.15) $$ \begin{align} \left[\mathcal{K},H_{\operatorname{\mathrm{kin}}}^{\prime}\right] & =\frac{1}{2}\sum_{k\in S_{C}}\sum_{p\in L_{k}^{\pm}}\left(\left[b_{k}\left(K_{k}^{\oplus}e_{p}\right)b_{k}(e_{p}),H_{\operatorname{\mathrm{kin}}}^{\prime}\right]-\left[b_{k}^{\ast}(e_{p})b_{k}^{\ast}\left(K_{k}^{\oplus}e_{p}\right),H_{\operatorname{\mathrm{kin}}}^{\prime}\right]\right)\nonumber \\ & =\sum_{k\in S_{C}}\sum_{p\in L_{k}^{\pm}}\left(b_{k}^{\ast}(e_{p})b_{k}^{\ast}\left(h_{k}^{\oplus}K_{k}^{\oplus}e_{p}\right)+b_{k}^{\ast}\left(h_{k}^{\oplus}e_{p}\right)b_{k}^{\ast}\left(K_{k}^{\oplus}e_{p}\right)\right)\\ &\quad +\sum_{k\in S_{C}}\sum_{p\in L_{k}^{\pm}}\left(b_{k}\left(K_{k}^{\oplus}e_{p}\right)b_{k}\left(h_{k}^{\oplus}e_{p}\right)+b_{k}\left(h_{k}^{\oplus}K_{k}^{\oplus}e_{p}\right)b_{k}(e_{p})\right)\nonumber \\ & =\sum_{k\in S_{C}}Q_{2}^{k}\left(\left\{ K_{k}^{\oplus},h_{k}^{\oplus}\right\} \right).\nonumber \end{align} $$

Note that because the commutator $\left[H_{\operatorname {\mathrm {kin}}}^{\prime },b_{k}^{\ast }(\varphi )\right]=2\,b_{k}^{\ast }\left(h_{k}^{\oplus }\varphi \right)$ exactly mirrors the bosonic case (in that there is no additional error term), the commutator $\left[\mathcal {K},H_{\operatorname {\mathrm {kin}}}^{\prime }\right]$ is likewise ‘purely bosonic’, being simply a sum of $Q_{2}^{k}(\cdot )$ terms without error terms such as those appearing in the statement of Proposition 5.2. With the groundwork laid, we can now easily deduce the following:

Proposition 5.5. $H_{\operatorname {\mathrm {kin}}}^{\prime }$ obeys

$$ \begin{align*} & e^{\mathcal{K}}H_{\operatorname{\mathrm{kin}}}^{\prime}e^{-\mathcal{K}}=H_{\operatorname{\mathrm{kin}}}^{\prime}\\ & +\sum_{k\in S_{C}}\left(\frac{1}{2}Q_{1}^{k}\left(e^{K_{k}^{\oplus}}h_{k}^{\oplus}e^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}h_{k}^{\oplus}e^{-K_{k}^{\oplus}}-2h_{k}^{\oplus}\right)+\frac{1}{2}Q_{2}^{k}\left(e^{K_{k}^{\oplus}}h_{k}^{\oplus}e^{K_{k}^{\oplus}}-e^{-K_{k}^{\oplus}}h_{k}^{\oplus}e^{-K_{k}^{\oplus}}\right)\right)\\ & +\sum_{k\in S_{C}}\int_{0}^{1}e^{t\mathcal{K}}\left(\mathcal{E}_{1}^{k}\left(\cosh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)\left(h_{k}^{\oplus}\right)-h_{k}^{\oplus}\right)+\mathcal{E}_{2}^{k}\left(\sinh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)\left(h_{k}^{\oplus}\right)\right)\right)e^{-t\mathcal{K}}\,dt. \end{align*} $$

Proof. By adding and subtracting, we have

(5.16) $$ \begin{align} e^{\mathcal{K}}H_{\operatorname{\mathrm{kin}}}^{\prime}e^{-\mathcal{K}}=\sum_{k\in S_{C}}e^{\mathcal{K}}Q_{1}^{k}\left(h_{k}^{\oplus}\right)e^{-\mathcal{K}}+e^{\mathcal{K}}\left(H_{\operatorname{\mathrm{kin}}}^{\prime}-\sum_{k\in S_{C}}Q_{1}^{k}\left(h_{k}^{\oplus}\right)\right)e^{-\mathcal{K}}, \end{align} $$

and the first term on the right-hand side is by Proposition 5.3,

(5.17) $$ \begin{align} &\sum_{k\in S_{C}}e^{\mathcal{K}}Q_{1}^{k}\left(h_{k}^{\oplus}\right)e^{-\mathcal{K}} \\ &\quad =\sum_{k\in S_{C}}\left(\frac{1}{2}Q_{1}^{k}\left(e^{K_{k}^{\oplus}}h_{k}^{\oplus}e^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}h_{k}^{\oplus}e^{-K_{k}^{\oplus}}\right)+\frac{1}{2}Q_{2}^{k}\left(e^{K_{k}^{\oplus}}h_{k}^{\oplus}e^{K_{k}^{\oplus}}-e^{-K_{k}^{\oplus}}h_{k}^{\oplus}e^{-K_{k}^{\oplus}}\right)\right) \nonumber\\ &\qquad +\sum_{k\in S_{C}}\int_{0}^{1}e^{t\mathcal{K}}\left(\mathcal{E}_{1}^{k}\left(\cosh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)\left(h_{k}^{\oplus}\right)\right)+\mathcal{E}_{2}^{k}\left(\sinh\left(\mathcal{A}_{(1-t)K_{k}^{\oplus}}\right)\left(h_{k}^{\oplus}\right)\right)\right)e^{-t\mathcal{K}}\,dt,\nonumber \end{align} $$

while the second is calculated using the commutators of the Propositions 5.2 and 5.4 to be

(5.18) $$ \begin{align} &e^{\mathcal{K}}\left(H_{\operatorname{\mathrm{kin}}}^{\prime}-\sum_{k\in S_{C}}Q_{1}^{k}\left(h_{k}^{\oplus}\right)\right)e^{-\mathcal{K}} - \left( H_{\operatorname{\mathrm{kin}}}^{\prime}-\sum_{k\in S_{C}}Q_{1}^{k}\left(h_{k}^{\oplus}\right) \right) \nonumber\\ &\quad =\int_{0}^{1}e^{t\mathcal{K}}\left[\mathcal{K},H_{\operatorname{\mathrm{kin}}}^{\prime}-\sum_{k\in S_{C}}Q_{1}^{k}\left(h_{k}^{\oplus}\right)\right]e^{-t\mathcal{K}}\,dt = -\sum_{k\in S_{C}}\int_{0}^{1}e^{t\mathcal{K}}\mathcal{E}_{1}^{k}\left(h_{k}^{\oplus}\right)e^{-t\mathcal{K}}\,dt , \end{align} $$

which yields the claim.

5.3 Fixing the transformation kernels

With all the transformation identities determined, we now choose the transformation kernels $(K_{k}^{\oplus })_{k\in S_{C}}$ such that $H_{\operatorname {\mathrm {kin}}}^{\prime }+\sum _{k\in S_{C}}H_{\text {int}}^{k}$ is diagonalized. For any choice of $(K_{k}^{\oplus })_{k\in S_{C}}$ , the Propositions 5.3 and 5.5 imply that

(5.19) $$ \begin{align} & e^{\mathcal{K}}\left(H_{\operatorname{\mathrm{kin}}}^{\prime}+\sum_{k\in S_{C}}H_{\text{int}}^{k}\right)e^{-\mathcal{K}} \nonumber\\ &\quad =\frac{1}{2}\sum_{k\in S_{C}}Q_{1}^{k}\left(e^{K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-K_{k}^{\oplus}}-2h_{k}^{\oplus}\right)\nonumber \\ &\qquad +\frac{1}{2}\sum_{k\in S_{C}}Q_{2}^{k}\left(e^{K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{K_{k}^{\oplus}}-e^{-K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-K_{k}^{\oplus}}\right)+H_{\operatorname{\mathrm{kin}}}^{\prime}+\text{error terms}. \end{align} $$

In analogy with the bosonic case, we consider this expression to be diagonalized provided the $Q_{2}^{k}(\cdot )$ terms vanish, whence the diagonalization condition is that

(5.20) $$ \begin{align} e^{K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{K_{k}^{\oplus}}=e^{-K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-K_{k}^{\oplus}}, \end{align} $$

which we note is the same as the diagonalization condition (equation (3.26)) of the exact bosonic quadratic Hamiltonian

(5.21) $$ \begin{align} H=Q_{1}\left(h_{k}^{\oplus}+A_{k}^{\oplus}\right)+Q_{2}\left(B_{k}^{\oplus}\right)\quad\text{on }\mathcal{F}^{+}\left(\ell^{2}\left(L_{k}^{\pm}\right)\right). \end{align} $$

Recalling the definitions of $h_k^{\oplus }$ , $A_k^{\oplus }$ and $B_k^{\oplus }$ from (4.11) and (4.7), we have

$$ \begin{align*} h_{k}^{\oplus}+A_{k}^{\oplus}\pm B_{k}^{\oplus} =\left(\begin{array}{cc} h_k+P_{v_k} & \pm P_{v_{k}}\\ \pm P_k & h_k + P_{v_{k}}\\ \end{array}\right)>0. \end{align*} $$

So by Theorem 3.1, the choice

$$ \begin{align*} K_{k}^{\oplus} & =-\frac{1}{2}\log\left(\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)^{-\frac{1}{2}}\left(\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)^{\frac{1}{2}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}\right.\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\;\left.\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)^{-\frac{1}{2}}\right) \end{align*} $$

is the unique diagonalizing kernel for the Hamiltonian. In this form, it is, however, not easy to see how $K_{k}^{\oplus }$ acts, so we will proceed slightly differently: we define $K_{k}^{\oplus }:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ by

(5.22) $$ \begin{align} K_{k}^{\oplus}=\left(\begin{array}{cc} 0 & K_{k}\\ K_{k} & 0 \end{array}\right), \end{align} $$

where the operator $K_{k}:\ell ^{2}(L_{k})\rightarrow \ell ^{2}(L_{k})$ is given by

(5.23) $$ \begin{align} K_{k}=-\frac{1}{2}\log\left(h_{k}^{-\frac{1}{2}}\Big(h_{k}^{\frac{1}{2}}\left(h_{k}+2P_{v_{k}}\right)h_{k}^{\frac{1}{2}}\Big)^{\frac{1}{2}}h_{k}^{-\frac{1}{2}}\right)=-\frac{1}{2}\log\left(h_{k}^{-\frac{1}{2}}\Big(h_{k}^{2}+2P_{h_{k}^{\frac{1}{2}}v_{k}}\Big)^{\frac{1}{2}}h_{k}^{-\frac{1}{2}}\right). \end{align} $$

A kernel similar to $K_k$ also appeared in [Reference Benedikter, Nam, Porta, Schlein and Seiringer5, Reference Benedikter, Nam, Porta, Schlein and Seiringer6]. Note that $K_{k}$ is precisely the diagonalizer of Theorem 3.1 for the exact bosonic quadratic Hamiltonian

(5.24) $$ \begin{align} H=Q_{1}\left(h_{k}^{\oplus}+P_{v_{k}}\right)+Q_{2}(P_{v_{k}})\quad\text{on }\mathcal{F}^{+}(\ell^{2}(L_{k})), \end{align} $$

rather than that of equation (5.21). Now we can verify that this $K_{k}^{\oplus }$ is, in fact, equal to the diagonalizing kernel:

Proposition 5.6. The operator $K_{k}^{\oplus }$ defined by the equations (5.22) and (5.23) satisfies

$$\begin{align*}e^{K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{K_{k}^{\oplus}}=e^{-K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-K_{k}^{\oplus}}=\left(\begin{array}{cc} E_{k} & 0\\ 0 & E_{k} \end{array}\right) \end{align*}$$

for $E_{k}=e^{-K_{k}}h_{k}e^{-K_{k}}$ .

Proof. It is easily verified that $e^{\pm K_{k}^{\oplus }}$ is given by

(5.25) $$ \begin{align} e^{\pm K_{k}^{\oplus}}=\left(\begin{array}{cc} \cosh\left(K_{k}\right) & \pm\sinh\left(K_{k}\right)\\ \pm\sinh\left(K_{k}\right) & \cosh\left(K_{k}\right) \end{array}\right), \end{align} $$

and so

(5.26) $$ \begin{align} & e^{\pm K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}\pm B_{k}^{\oplus}\right)e^{\pm K_{k}^{\oplus}}\nonumber\\ &\quad =\left(\begin{array}{cc} \cosh\left(K_{k}\right) &\quad \pm\sinh\left(K_{k}\right)\\ \pm\sinh\left(K_{k}\right) & \cosh\left(K_{k}\right) \end{array}\right)\left(\begin{array}{cc} h_{k}+P_{v_{k}} & \pm P_{v_{k}}\\ \pm P_{v_{k}} & h_{k}+P_{v_{k}} \end{array}\right)\left(\begin{array}{cc} \cosh\left(K_{k}\right) & \pm\sinh\left(K_{k}\right)\\ \pm\sinh\left(K_{k}\right) & \cosh\left(K_{k}\right) \end{array}\right)\nonumber \\ &\quad =\frac{1}{2}\left(\begin{array}{cc} e^{K_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{K_{k}}+e^{-K_{k}}h_{k}e^{-K_{k}} & \pm\left(e^{K_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{K_{k}}-e^{-K_{k}}h_{k}e^{-K_{k}}\right)\\ \pm\left(e^{K_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{K_{k}}-e^{-K_{k}}h_{k}e^{-K_{k}}\right) & e^{K_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{K_{k}}+e^{-K_{k}}h_{k}e^{-K_{k}} \end{array}\right). \end{align} $$

The condition

(5.27) $$ \begin{align} e^{K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{K_{k}^{\oplus}}=e^{-K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-K_{k}^{\oplus}} \end{align} $$

thus holds if and only if

(5.28) $$ \begin{align} e^{K_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{K_{k}}=e^{-K_{k}}h_{k}e^{-K_{k}}, \end{align} $$

which is the diagonalization condition for the bosonic Hamiltonian of equation (5.24). Theorem 3.1 asserts that this condition is satisfied for our choice of $K_{k}$ , and the claim follows.

5.4 Full transformation of the bosonizable terms

With the above choice of transformation kernels, we thus conclude that

(5.29) $$ \begin{align} & e^{\mathcal{K}}\left(H_{\operatorname{\mathrm{kin}}}^{\prime}+\sum_{k\in S_{C}}H_{\text{int}}^{k}\right)e^{-\mathcal{K}}=H_{\operatorname{\mathrm{kin}}}^{\prime}+\text{error terms}\nonumber \\ &\qquad +\frac{1}{2}\sum_{k\in S_{C}}Q_{1}^{k}\left(e^{K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{K_{k}^{\oplus}}+e^{-K_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-K_{k}^{\oplus}}-2h_{k}^{\oplus}\right)\nonumber\\ &\quad =H_{\operatorname{\mathrm{kin}}}^{\prime}+\sum_{k\in S_{C}}Q_{1}^{k}\left(\begin{array}{cc} E_{k}-h_{k} & 0\\ 0 & E_{k}-h_{k} \end{array}\right)+\text{error terms}, \end{align} $$

and so we have succeeded in diagonalizing $H_{\operatorname {\mathrm {kin}}}^{\prime }+\sum _{k\in S_{C}}H_{\text {int}}^{k}$ while simultanously decoupling the spaces $\ell ^{2}\left(L_{\pm k}\right)\subset \ell ^{2}(L_{k}^{\pm })$ in a symmetric fashion. We still need to determine the exact form of the error terms, which we record in the following proposition:

Proposition 5.7. Let $S_{C}=\overline {B}\left(0,k_{F}^{\gamma }\right)\cap \mathbb {Z}_{+}^{3}$ with $\gamma \in (0,1]$ . Then the unitary transformation $e^{\mathcal {K}}:\mathcal {H}_{N}\rightarrow \mathcal {H}_{N}$ with $\mathcal {K}$ defined by (5.2), (5.22), (5.23) satisfies

$$ \begin{align*} & e^{\mathcal{K}}\left(H_{\operatorname{\mathrm{kin}}}^{\prime}+\sum_{k\in S_{C}}H_{\operatorname{\mathrm{int}}}^{k}\right)e^{-\mathcal{K}}=H_{\operatorname{\mathrm{kin}}}^{\prime}+\sum_{k\in S_{C}}Q_{1}^{k} (E_k^{\oplus} - h_k^{\oplus})\\ &\quad +\sum_{k\in S_{C}}\int_{0}^{1}e^{(1-t)\mathcal{K}}\left(\mathcal{E}_{1}^{k}(A_k^\oplus(t))+\mathcal{E}_{2}^{k} (B_k^\oplus(t))\right)e^{-(1-t)\mathcal{K}}\,dt, \end{align*} $$

where $\mathcal {E}_1(\cdot )$ , $\mathcal {E}_2(\cdot )$ are defined in Proposition 5.2 and

$$ \begin{align*}E_k^{\oplus} - h_k^{\oplus}= \left(\begin{array}{cc} E_{k}-h_{k} & 0\\ 0 & E_{k}-h_{k} \end{array}\right),\,\, A_{k}^\oplus (t) = \left(\begin{array}{cc} A_{k} (t ) & 0\\ 0 & A_{k} (t ) \end{array}\right),\,\, B_{k}^\oplus (t) = \left(\begin{array}{cc} 0 & B_{k} (t )\\ B_{k} (t ) & 0 \end{array}\right) \end{align*} $$

with $E_k= e^{-K_k} h_k e^{-K_k}$ and the operators $A_{k} (t ),B_{k} (t ):\ell ^{2}(L_{k})\rightarrow \ell ^{2}(L_{k})$ defined by

$$ \begin{align*} A_{k} (t ) & =\frac{1}{2}\left(e^{tK_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{tK_{k}}+e^{-tK_{k}}h_{k}e^{-tK_{k}}\right)-h_{k}\\ B_{k} (t ) & =\frac{1}{2}\left(e^{tK_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{tK_{k}}-e^{-tK_{k}}h_{k}e^{-tK_{k}}\right). \end{align*} $$

Proof. By the Propositions 5.3 and 5.5, the error terms are

$$ \begin{align*} & \sum_{k\in S_{C}}\int_{0}^{1}e^{(1-t)\mathcal{K}}\left(\mathcal{E}_{1}^{k}\left(\cosh\left(\mathcal{A}_{tK_{k}^{\oplus}}\right)\left(h_{k}^{\oplus}+A_{k}^{\oplus}\right)+\sinh\left(\mathcal{A}_{tK_{k}^{\oplus}}\right)\left(B_{k}^{\oplus}\right)-h_{k}^{\oplus}\right)\right)e^{-(1-t)\mathcal{K}}\,dt\\ &\quad +\sum_{k\in S_{C}}\int_{0}^{1}e^{(1-t)\mathcal{K}}\left(\mathcal{E}_{2}^{k}\left(\sinh\left(\mathcal{A}_{tK_{k}^{\oplus}}\right)\left(h_{k}^{\oplus}+A_{k}^{\oplus}\right)+\cosh\left(\mathcal{A}_{tK_{k}^{\oplus}}\right)\left(B_{k}^{\oplus}\right)\right)\right)e^{-(1-t)\mathcal{K}}\,dt, \end{align*} $$

where we have reparametrized the integral by $t\mapsto 1-t$ to simplify the arguments of the $\mathcal {E}_{1}^{k}(\cdot )$ and $\mathcal {E}_{2}^{k}(\cdot )$ operators. By (5.11), the arguments of $\mathcal {E}_{1}^{k}$ and $\mathcal {E}_{2}^{k}$ in each term above equal

(5.30) $$ \begin{align} \frac{1}{2}\left(e^{tK_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{tK_{k}^{\oplus}}+e^{-tK_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-tK_{k}^{\oplus}}\right)-h_{k}^{\oplus} \end{align} $$

and

(5.31) $$ \begin{align} \frac{1}{2}\left(e^{tK_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}+B_{k}^{\oplus}\right)e^{tK_{k}^{\oplus}}-e^{-tK_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}-B_{k}^{\oplus}\right)e^{-tK_{k}^{\oplus}}\right), \end{align} $$

respectively. By the same identities that we used in the preceding proposition, it holds that

(5.32) $$ \begin{align} & e^{\pm tK_{k}^{\oplus}}\left(h_{k}^{\oplus}+A_{k}^{\oplus}\pm B_{k}^{\oplus}\right)e^{\pm tK_{k}^{\oplus}}\\ &\quad =\frac{1}{2}\left(\begin{array}{cc} e^{tK_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{tK_{k}}+e^{-tK_{k}}h_{k}e^{-tK_{k}} & \pm\left(e^{tK_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{tK_{k}}-e^{-tK_{k}}h_{k}e^{-tK_{k}}\right)\\ \pm\left(e^{tK_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{tK_{k}}-e^{-tK_{k}}h_{k}e^{-tK_{k}}\right) & e^{tK_{k}}\left(h_{k}+2P_{v_{k}}\right)e^{tK_{k}}+e^{-tK_{k}}h_{k}e^{-tK_{k}} \end{array}\right)\nonumber, \end{align} $$

and the claim follows.

6.1 Reduction to simpler expressions

Recall that for $k\in S_C$ and symmetric operators $A,B:\ell ^{2}(L_{k}^{\pm })\rightarrow \ell ^{2}(L_{k}^{\pm })$ , we already defined $\mathcal {E}_{1}^{k}(A)$ and $\mathcal {E}_{2}^{k}(B)$ in Proposition 5.2. Since these expressions are complicated, it is helpful to discuss the general structure of $\mathcal {E}_{1}^{k}(A)$ and $\mathcal {E}_{2}^{k}(B)$ . Consider the first term of $\mathcal {E}_{1}^{k}(A)$ , which upon expansion is

(6.1) $$ \begin{align} & \left\{ b_{k}^{\ast}(Ae_{p}),\left\{ b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{k,l}\left(e_{p},e_{q}\right)\right\} \right\} \nonumber \\ &\quad =b_{k}^{\ast}(Ae_{p})\left\{ b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{k,l}\left(e_{p},e_{q}\right)\right\} +\left\{ b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right),\varepsilon_{k,l}\left(e_{p},e_{q}\right)\right\} b_{k}^{\ast}(Ae_{p})\\ &\quad =b_{k}^{\ast}(Ae_{p})b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\varepsilon_{k,l}\left(e_{p},e_{q}\right)+b_{k}^{\ast}(Ae_{p})\varepsilon_{k,l}\left(e_{p},e_{q}\right)b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\nonumber \\ &\qquad +b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)\varepsilon_{k,l}\left(e_{p},e_{q}\right)b_{k}^{\ast}(Ae_{p})+\varepsilon_{k,l}\left(e_{p},e_{q}\right)b_{l}^{\ast}\left(K_{l}^{\oplus}e_{q}\right)b_{k}^{\ast}(Ae_{p}),\nonumber \end{align} $$

which we may expand further using

(6.2) $$ \begin{align} \varepsilon_{k,l}\left(e_{p},e_{q}\right)=\varepsilon\left(\overline{k,p};\overline{l,q}\right)=-\left(\delta_{p,q}c_{\overline{q-l}}c_{\overline{p-k}}^{\ast}+\delta_{\overline{p-k},\overline{q-l}}c_{q}^{\ast}c_{p}\right) \end{align} $$

and then removing the delta on a case-by-case basis. This causes the sums over $p\in L_{k}^{\pm }$ and $q\in L_{l}^{\pm }$ of any of these terms to reduce to one of the schematic forms

(6.3) $$ \begin{align} &\sum_{p\in S}b_{k}^{\natural}\left(Te_{p_{1}}\right)b_{l}^{\natural}\left(K_{l}^{\oplus}e_{p_{2}}\right)\tilde{c}_{p_{3}}^{\ast}\tilde{c}_{p_{4}},\quad\sum_{p\in S}b_{k}^{\natural}\left(Te_{p_{1}}\right)\tilde{c}_{p_{3}}^{\ast}\tilde{c}_{p_{4}}b_{l}^{\natural}\left(K_{l}^{\oplus}e_{p_{2}}\right),\quad \nonumber\\ & \quad \sum_{p\in S}\tilde{c}_{p_{3}}^{\ast}\tilde{c}_{p_{4}}b_{k}^{\natural}\left(Te_{p_{1}}\right)b_{l}^{\natural}\left(K_{l}^{\oplus}e_{p_{2}}\right) \end{align} $$

subject to the following: S is a subset of $L_{k}^{\pm }\cap L_{l}^{\pm }$ , $b_{k}^{\natural }$ can denote either $b_{k}$ or $b_{k}^{\ast }$ , $\varepsilon _{k,l}(e_{p},e_{q})$ may instead be $\varepsilon _{l,k}(e_{q},e_{p})=\varepsilon _{k,l}(e_{p},e_{q})^{\ast }$ , T denotes either A or B, the terms $b_{k}^{\natural }(Te_{p})$ and $b_{l}^{\natural }(K_{l}^{\oplus }e_{q})$ may be interchanged, the notation

(6.4) $$ \begin{align} \tilde{c}_{p}=\begin{cases} c_{p} & p\in B_{F}^{c}\\ c_{p}^{\ast} & p\in B_{F} \end{cases} \end{align} $$

encodes the correct type of creation/annihilation operator depending on whether p corresponds to a hole state or an excited state, and $p_{1},p_{2},p_{3},p_{4}$ denote indices which depend on p.

The same decomposition holds for every term appearing in either $\mathcal {E}_{1}^{k}(A)$ or $\mathcal {E}_{2}^{k}(B)$ , so we must consider the forms of (6.3).

The only important feature of the dependency that the $p_{i}$ have with respect to p is that regardless of the term, when summing over $p\in S$ , $p_{i}$ ranges either exclusively over excited states (i.e., $p_{i}\in L_{k}^{\pm }$ ) or exclusively over hole states (i.e., $p_{i}\in \left(L_{k}-k\right)\cup \left(L_{-k}+k\right)$ or the analogous set for $L_{l}^{\pm }$ ), and that the assignments $p\mapsto p_{i}$ (for a given term) are injective. (Additionally, $p_{1}$ and $p_{2}$ will always be excited states.)

Therefore, when estimating, we can always expand the sum to either all of $B_{F}$ or all of $B_{F}^{c}$ , which is why the exact identities of S and the $p_{i}$ are of no importance to the estimation. For example,

(6.5) $$ \begin{align} \left|\sum_{p\in S}\left\langle \Psi,\tilde{c}_{p_{3}}^{\ast}\tilde{c}_{p_{4}}\Psi\right\rangle \right| \leq\sum_{p\in S}\left\Vert \tilde{c}_{p_{3}}\Psi\right\Vert \left\Vert \tilde{c}_{p_{4}}\Psi\right\Vert \leq\sqrt{\sum_{p\in S}\left\Vert \tilde{c}_{p_{3}}\Psi\right\Vert ^{2}}\sqrt{\sum_{p\in S}\left\Vert \tilde{c}_{p_{4}}\Psi\right\Vert ^{2}} \le \left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle \end{align} $$

independently of S, $p_{3}$ and $p_{4}$ . Here, the two situations when both $p_{3}$ and $p_{4}$ range over excited states, and when both $p_{3}$ and $p_{4}$ range over hole states, can be treated similarly thanks to the particle-hole symmetry (1.13).

Discussion of estimation strategy

We conclude that both $\mathcal {E}_{1}^{k}(A)$ and $\mathcal {E}_{2}^{k}(B)$ reduce to sums over $l\in S_{C}$ of finitely many terms of the schematic forms of equation (6.3), so it suffices to estimate these. To this end, we must first perform some additional algebraic manipulation.

To motivate our goal, let us first derive a simple but insufficent estimate for one of these terms:

(6.6) $$ \begin{align} \sum_{p\in S}b_{k}^{\ast}\left(Te_{p_{1}}\right)\tilde{c}_{p_{3}}^{\ast}\tilde{c}_{p_{4}}b_{l}\left(K_{l}^{\oplus}e_{p_{2}}\right). \end{align} $$

Using $\left\Vert c_{p}\right\Vert _{\text {Op}}=1$ , Proposition 4.4 and the Cauchy–Schwarz inequality, we find that

(6.7) $$ \begin{align} & \sum_{p\in S}\left|\left\langle \Psi,b_{k}^{\ast}\left(Te_{p_{1}}\right)\tilde{c}_{p_{3}}^{\ast}\tilde{c}_{p_{4}}b_{l}\left(K_{l}^{\oplus}e_{p_{2}}\right)\Psi\right\rangle \right| \leq\sum_{p\in S}\left\Vert b_{k}\left(Te_{p_{1}}\right)\Psi\right\Vert \left\Vert b_{l}\left(K_{l}^{\oplus}e_{p_{2}}\right)\Psi\right\Vert \\ &\quad \leq\sum_{p\in S}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Te_{p_{1}}\right\Vert \left\Vert \left(h_{l}^{\oplus}\right)^{-\frac{1}{2}}K_{l}^{\oplus}e_{p_{2}}\right\Vert \left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle \leq\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}T\right\Vert _{\operatorname{\mathrm{HS}}}\left\Vert \left(h_{l}^{\oplus}\right)^{-\frac{1}{2}}K_{l}^{\oplus}\right\Vert _{\operatorname{\mathrm{HS}}}\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle \nonumber \end{align} $$

for any $\Psi \in \mathcal {H}_{N}$ . To get a feeling for the quality of this estimate, we must know what to expect of the quantities on the right-hand side. We will see in the next sections that $\big\Vert (h_{l}^{\oplus })^{-\frac {1}{2}}K_{l}^{\oplus }\big\Vert _{\operatorname {\mathrm {HS}}} \leq O(k_{F}^{-\frac {1}{3} + \epsilon })$ . In general, what will take the place of T will be the $A_{k} (t )$ and $B_{k} (t )$ operators we defined in the last section, but as a simple example we consider

(6.8) $$ \begin{align} T= \left(\begin{array}{cc} P_{v_{k}} & 0\\ 0 & P_{v_{k}} \end{array}\right),\quad P_{v_{k}} = |v_k\rangle \langle v_k|, \quad v_{k}=\sqrt{\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}}\sum_{p\in L_{k}}e_{p} \in \ell^{2}(L_{k}) \end{align} $$

for which

(6.9) $$ \begin{align} \left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}T\right\Vert _{\operatorname{\mathrm{HS}}} & =2\left\Vert h_{k}^{-\frac{1}{2}}P_{v_{k}}\right\Vert _{\operatorname{\mathrm{HS}}}=2\sqrt{\operatorname{\mathrm{tr}}\left(P_{v_{k}}h_{k}^{-1}P_{v_{k}}\right)}\\ & =2\left\Vert v_{k}\right\Vert \sqrt{\left\langle v_{k},h_{k}^{-1}v_{k}\right\rangle }=\frac{\hat{V}_{k}k_{F}^{-1}}{(2\pi)^{3}}\left|L_{k}\right|^{\frac 1 2} \sqrt{\sum_{p\in L_{k}}\lambda_{k,p}^{-1}}\leq O\big(k_{F}^{\frac{1}{2}} \big)\nonumber \end{align} $$

when $|k|\sim 1$ . Here, we used $|L_k|\le C|k| k_F^2$ and the bound $\sum _{p\in L_{k}}\lambda _{k,p}^{-1}\leq Ck_{F}$ from Proposition A.2. Thus, for any state satisfying $\left\langle \Psi ,H_{\operatorname {\mathrm {kin}}}^{\prime }\Psi \right\rangle \le O(k_F)$ (c.f. Theorem 1.2), the overall estimate for the right side of (6.7) is $O(k_{F}^{\frac {7}{6} + \epsilon })$ which is insufficient as the correlation energy is of order $k_F$ .

The technical issue with the estimation in (6.7) lies in only using that $\left\Vert c_{p}\right\Vert _{\text {Op}}=1$ , for we may get better bounds by using $\left\langle \Psi ,\mathcal {N}_{E} H^{\prime }_{\mathrm {kin}}\Psi \right\rangle $ instead of $\left\langle \Psi ,H^{\prime }_{\mathrm {kin}}\Psi \right\rangle $ . For example,

(6.10) $$ \begin{align} &\quad\,\sum_{p\in S}\left|\left\langle \Psi,b_{k}^{\ast}\left(Te_{p_{1}}\right)\tilde{c}_{p_{3}}^{\ast}\tilde{c}_{p_{4}}b_{l} \left(K_{l}^{\oplus}e_{p_{2}}\right)\Psi\right\rangle \right| \end{align} $$

(6.11) $$ \begin{align} &=\sum_{p\in S}\left|\left\langle \Psi,\tilde{c}_{p_{3}}^{\ast}b_{k}^{\ast}\left(Te_{p_{1}}\right)b_{l}\left(K_{l}^{\oplus}e_{p_{2}}\right)\tilde{c}_{p_{4}}\Psi\right\rangle \right|\leq\sum_{p\in S}\left\Vert b_{k}\left(Te_{p_{1}}\right)\tilde{c}_{p_{3}}\Psi\right\Vert \left\Vert b_{l}\left(K_{l}^{\oplus}e_{p_{2}}\right)\tilde{c}_{p_{4}}\Psi\right\Vert \nonumber \\ & \leq\sum_{p\in S}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Te_{p_{1}}\right\Vert \left\Vert \left(h_{l}^{\oplus}\right)^{-\frac{1}{2}}K_{l}^{\oplus}e_{p_{2}}\right\Vert \sqrt{\left\langle \tilde{c}_{p_{3}}\Psi,H_{\operatorname{\mathrm{kin}}}^{\prime\left(\pm1\right)}\tilde{c}_{p_{3}}\Psi\right\rangle \left\langle \tilde{c}_{p_{4}}\Psi,H_{\operatorname{\mathrm{kin}}}^{\prime\left(\pm1\right)}\tilde{c}_{p_{4}}\Psi\right\rangle } \end{align} $$

$$\begin{align*} & \leq\left(\max_{p\in L_{k}^{\pm}}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Te_{p}\right\Vert \right)\sqrt{\sum_{p\in S}\left\Vert \left(h_{l}^{\oplus}\right)^{-\frac{1}{2}}K_{l}^{\oplus}e_{q}\right\Vert ^{2}}\sqrt{\sum_{p\in S}\left\langle \Psi,\tilde{c}_{p_{3}}^{\ast}H_{\operatorname{\mathrm{kin}}}^{\prime\left(\pm1\right)}\tilde{c}_{p_{3}}\Psi\right\rangle }\sqrt{\left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle }\nonumber \\ & \leq\left(\max_{p\in L_{k}^{\pm}}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Te_{p}\right\Vert \right)\left\Vert \left(h_{l}^{\oplus}\right)^{-\frac{1}{2}}K_{l}^{\oplus}\right\Vert _{\operatorname{\mathrm{HS}}}\sqrt{\left\langle \Psi,\mathcal{N}_{E}H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle \left\langle \Psi,H_{\operatorname{\mathrm{kin}}}^{\prime}\Psi\right\rangle },\end{align*}$$

where we used that $\left[\tilde {c}_{p},b_{k}(\cdot )\right]=0$ (as we will see in Proposition 6.1 below) and momentarily looked ahead to the definition (6.30) for $H_{\operatorname {\mathrm {kin}}}^{\prime \left(\pm 1\right)}$ and Lemma 6.6 (we take supremum over $p_4$ and sum over $p_3$ to get the second inequality). Considering again the example in (6.8), we find

(6.12) $$ \begin{align} \max_{p\in L_{k}^{\pm}}\left\Vert \left(h_{k}^{\oplus}\right)^{-\frac{1}{2}}Te_{p}\right\Vert =\sqrt{\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}}\left\Vert h_{k}^{-\frac{1}{2}}v_{k}\right\Vert =\frac{\hat{V}_{k}k_{F}^{-1}}{2(2\pi)^{3}}\sqrt{\sum_{k\in L_{k}}\lambda_{k,p}^{-1}}\leq O(k_{F}^{-\frac{1}{2}}). \end{align} $$

Thus, for any state satisfying $\left\langle \Psi ,H_{\operatorname {\mathrm {kin}}}^{\prime }\Psi \right\rangle \le O(k_F)$ and $\left\langle \Psi ,\mathcal {N}_{E}H_{\operatorname {\mathrm {kin}}}^{\prime }\Psi \right\rangle \le O(k_F^2)$ (c.f. Theorem 1.2), the right side of (6.10) is thus bounded by $O(k_{F}^{ \frac {2}{3} + \epsilon })$ , which is much smaller than the correlation energy.

Our goal is, therefore, to reduce the schematic forms of equation (6.3) to those of the form $\sum _{p\in S}\tilde {c}_{p_{3}}^{\ast }b_{k}^{\natural }(Te_{p_{1}})b_{l}^{\natural }(K_{l}^{\oplus }e_{p_{2}})\tilde {c}_{p_{4}}$ , which we may then estimate as above. While $\left[\tilde {c}_{p},b_{k}(\cdot )\right]=0$ , it is generally the case that $\left[\tilde {c}_{p},b_{k}^{\ast }(\cdot )\right]\neq 0$ , so this will also introduce additional commutator terms which we must then estimate separately.

Taking into account whether $b_{k}^{\natural }=b_{k}$ or $b_{k}^{\natural }=b_{k}^{\ast }$ , the schematic forms of equation (6.3) are either of the form (supressing the summation, the arguments and the subscripts for brevity)

(6.13) $$ \begin{align} b^{\ast}b^{\ast}\tilde{c}^{\ast}\tilde{c},\quad b^{\ast}b\tilde{c}^{\ast}\tilde{c},\quad bb^{\ast}\tilde{c}^{\ast}\tilde{c},\quad bb\tilde{c}^{\ast}\tilde{c},\quad b^{\ast}\tilde{c}^{\ast}\tilde{c}b,\quad b\tilde{c}^{\ast}\tilde{c}b,\quad b\tilde{c}^{\ast}\tilde{c}b^{\ast}, \end{align} $$

or reduce to one of these by taking the adjoint, which as we will estimate, $\mathcal {E}_{1}^{k}(A)$ and $\mathcal {E}_{2}^{k}(B)$ as bilinear forms does not matter. Using that commutators of the form $\left[b,\tilde {c}\right]$ , $\left[b^{\ast },\tilde {c}^{\ast }\right]$ and $\left[b,\left[b,\tilde {c}^{\ast }\right]\right]$ vanish (verified below), these schematic forms reduce to

(6.14) $$ \begin{align} b^{\ast}b^{\ast}\tilde{c}^{\ast}\tilde{c} & =\tilde{c}^{\ast}b^{\ast}b^{\ast}\tilde{c} ,\nonumber \\ b^{\ast}b\tilde{c}^{\ast}\tilde{c} & =\tilde{c}^{\ast}b^{\ast}b\tilde{c}+\left[b,\tilde{c}^{\ast}\right]b^{\ast}\tilde{c}+\left[b^{\ast},\left[b,\tilde{c}^{\ast}\right]\right]\tilde{c},\nonumber \\ bb^{\ast}\tilde{c}^{\ast}\tilde{c} & =\tilde{c}^{\ast}bb^{\ast}\tilde{c}+\left[b,\tilde{c}^{\ast}\right]b^{\ast}\tilde{c},\nonumber \\ bb\tilde{c}^{\ast}\tilde{c} & =\tilde{c}^{\ast}bb\tilde{c}+\left[b,\tilde{c}^{\ast}\right]b\tilde{c}+\left[b,\tilde{c}^{\ast}\right]b\tilde{c},\\ b^{\ast}\tilde{c}^{\ast}\tilde{c}b & =\tilde{c}^{\ast}b^{\ast}b\tilde{c},\nonumber \\ b\tilde{c}^{\ast}\tilde{c}b & =\tilde{c}^{\ast}bb\tilde{c}+\left[b,\tilde{c}^{\ast}\right]b\tilde{c},\nonumber \\ b\tilde{c}^{\ast}\tilde{c}b^{\ast} & =\tilde{c}^{\ast}bb^{\ast}\tilde{c}+\left[b,\tilde{c}^{\ast}\right]b^{\ast}\tilde{c}+\tilde{c}^{\ast}b\left[\tilde{c},b^{\ast}\right]+\left[b,\tilde{c}^{\ast}\right]\left[\tilde{c},b^{\ast}\right].\nonumber \end{align} $$

Reintroducing the $b^{\natural }$ notation outside the commutators and using once more our freedom to take adjoints, we find that every term on the right-hand sides of the two equations above takes one of the four schematic forms

(6.15) $$ \begin{align} \tilde{c}^{\ast}b^{\natural}b^{\natural}\tilde{c},\quad\left[\tilde{c},b^{\ast}\right]^{\ast}b^{\natural}\tilde{c},\quad\left[\left[\tilde{c},b^{\ast}\right],b\right]^{\ast}\tilde{c},\quad\left[\tilde{c},b^{\ast}\right]^{\ast}\left[\tilde{c},b^{\ast}\right]. \end{align} $$

These are the final forms which we will explicitly estimate.

6.2 Preliminary commutator estimates

In addition to the general estimates which we derived at the start of this section, we will also need estimates on the commutator terms which appear in the schematic forms of equation (6.23), which we now derive. First, we must, however, verify that the commutators $\left[b,\tilde {c}\right]$ , $\left[b^{\ast },\tilde {c}^{\ast }\right]$ and $\left[b,\left[b,\tilde {c}^{\ast }\right]\right]$ vanish, which we relied upon in our reduction procedure:

Proposition 6.1. For all $k,l\in \mathbb {Z}_{+}^{3}$ , $\varphi \in \ell ^{2}(L_{k}^{\pm })$ , $\psi \in \ell ^{2}\left(L_{l}^{\pm }\right)$ and $p\in \mathbb {Z}^{3}$ it holds that

$$ \begin{align*} \left[b_{k}(\varphi ),\tilde{c}_{p}\right]=\left[b_{k}^{\ast}(\varphi ),\tilde{c}_{p}^{\ast}\right] =0,\quad \left[b_{l}\left(\psi\right),\left[b_{k}(\varphi ),\tilde{c}_{p}^{\ast}\right]\right] =0. \end{align*} $$

Proof. We compute from the definitions that for any $q\in L_{k}^{\pm }$ ,

(6.16) $$ \begin{align} \left[b_{\overline{k,q}},\tilde{c}_{p}\right] & =\left[c_{\overline{q-k}}^{\ast}c_{q},\tilde{c}_{p}\right]=c_{\overline{q-k}}^{\ast}\left\{ c_{q},\tilde{c}_{p}\right\} -\left\{ c_{\overline{q-k}}^{\ast},\tilde{c}_{p}\right\} c_{q}\nonumber \\ & =\begin{cases} c_{\overline{q-k}}^{\ast}\left\{ c_{q},c_{p}\right\} -\left\{ c_{\overline{q-k}}^{\ast},c_{p}\right\} c_{q} ,& p\in B_{F}^{c}\nonumber\\ c_{\overline{q-k}}^{\ast}\left\{ c_{q},c_{p}^{\ast}\right\} -\left\{ c_{\overline{q-k}}^{\ast},c_{p}^{\ast}\right\} c_{q} ,& p\in B_{F} \end{cases}\\ & =0 \end{align} $$

as all anticommutators on the second line vanish either directly by the CAR or by disjointness of $B_{F}$ and $B_{F}^{c}$ . By linearity, $\left[b_{k}(\varphi ),\tilde {c}_{p}\right]=0$ , and $\left[b_{k}^{\ast }(\varphi ),\tilde {c}_{p}^{\ast }\right]=-\left[b_{k}(\varphi ),\tilde {c}_{p}\right]^{\ast }=0$ .

For the double commutator, we first compute $[b_{\overline {k,q}},\tilde {c}_{p}^{\ast }]$ . As above, we find

$$\begin{align*}\left[b_{\overline{k,q}},\tilde{c}_{p}^{\ast}\right] =\begin{cases} c_{\overline{q-k}}^{\ast}\left\{ c_{q},c_{p}^{\ast}\right\} -\left\{ c_{\overline{q-k}}^{\ast},c_{p}^{\ast}\right\} c_{q} ,& p\in B_{F}^{c}\\ c_{\overline{q-k}}^{\ast}\left\{ c_{q},c_{p}\right\} -\left\{ c_{\overline{q-k}}^{\ast},c_{p}\right\} c_{q}, & p\in B_{F} \end{cases} =\begin{cases} \delta_{q,p}\tilde{c}_{\overline{q-k}} ,& p\in B_{F}^{c}\\ -\delta_{\overline{q-k},p}\tilde{c}_{q}, & p\in B_{F} \end{cases}\end{align*}$$

so

(6.17) $$ \begin{align} \left[b_{k}(\varphi ),\tilde{c}_{p}^{\ast}\right] & =\sum_{q\in L_{k}^{\pm}}\left\langle \varphi,e_{q}\right\rangle \left[b_{\overline{k,q}},\tilde{c}_{p}^{\ast}\right]=\sum_{q\in L_{k}^{\pm}}\left\langle \varphi,e_{q}\right\rangle \begin{cases} \delta_{q,p}\tilde{c}_{\overline{q-k}} & p\in B_{F}^{c} \nonumber\\ -\delta_{\overline{q-k},p}\tilde{c}_{q} & p\in B_{F} \end{cases}\\ & =1_{L_{k}^{\pm}}(p)\left\langle \varphi,e_{p}\right\rangle \tilde{c}_{\overline{p-k}}-1_{L_{k}-k}(p)\left\langle \varphi,e_{p+k}\right\rangle \tilde{c}_{p+k}-1_{L_{k}+k}(p)\left\langle \varphi,e_{p-k}\right\rangle \tilde{c}_{p-k}, \end{align} $$

where $1_{S}(\cdot )$ denotes the indicator function of a set S. Observing that $\left[b_{k}(\varphi ),\tilde {c}_{p}^{\ast }\right]$ is a linear combination of $\tilde {c}_{p}$ terms, we conclude that $\left[b_{l}(\psi ),\left[b_{k}(\varphi ),\tilde {c}_{p}^{\ast }\right]\right]=0$ by the first part.

We now move into the estimation of the nonvanishing commutators. We begin with the single commutator; we state the estimate and make a remark:

Proposition 6.2. For all $k\in \mathbb {Z}_{+}^{3}$ , sequences $\left(\varphi _{p}\right)_{p\in \mathbb {Z}^{3}}\in \ell ^{2}(L_{k}^{\pm })$ and $\Psi \in \mathcal {H}_{N}$ , it holds that

$$ \begin{align*} \sum_{p\in\mathbb{Z}^{3}}\left\Vert \left[\tilde{c}_{p},b_{k}^{\ast}\left(\varphi_{p}\right)\right]\Psi\right\Vert ^{2} & \leq3\left(\sum_{p\in L_{k}^{\pm}}\max_{q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left\Vert \Psi\right\Vert ^{2}\\ \sum_{p\in\mathbb{Z}^{3}}\left\Vert \left[\tilde{c}_{p}^{\ast},b_{k}\left(\varphi_{p}\right)\right]\Psi\right\Vert ^{2} & \leq4\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle. \end{align*} $$

Proof. Taking the adjoint of equation (6.17) yields

(6.18) $$ \begin{align} \left[\tilde{c}_{p},b_{k}^{\ast}(\varphi )\right]=1_{L_{k}^{\pm}}(p)\left\langle e_{p},\varphi\right\rangle \tilde{c}_{\overline{p-k}}^{\ast}-1_{L_{k}-k}(p)\left\langle e_{p+k},\varphi\right\rangle \tilde{c}_{p+k}^{\ast}-1_{L_{k}+k}(p)\left\langle e_{p-k},\varphi\right\rangle \tilde{c}_{p-k}^{\ast}, \end{align} $$

and so we can for any $\Psi \in \mathcal {H}_{N}$ estimate by the (squared) triangle inequality, using also that $L_{k}^{\pm }$ and $\left(L_{k}-k\right)\cap \left(L_{-k}+k\right)$ are disjoint and $\left\Vert \tilde {c}_{p}^{\ast }\right\Vert _{\text {Op}}=1$ , that

(6.19) $$ \begin{align} &\sum_{p\in\mathbb{Z}^{3}}\left\Vert \left[\tilde{c}_{p},b_{k}^{\ast}\left(\varphi_{p}\right)\right]\Psi\right\Vert ^{2} \nonumber \\ &\leq\sum_{p\in L_{k}^{\pm}}\left|\left\langle e_{p},\varphi_{p}\right\rangle \right|^{2}\left\Vert \tilde{c}_{\overline{p-k}}^{\ast}\Psi\right\Vert ^{2} +2\sum_{p\in L_{k}-k}\left|\left\langle e_{p+k},\varphi_{p}\right\rangle \right|^{2}\left\Vert \tilde{c}_{p+k}^{\ast}\Psi\right\Vert ^{2} +2\sum_{p\in L_{-k}+k}\left|\left\langle e_{p-k},\varphi_{p}\right\rangle \right|^{2}\left\Vert \tilde{c}_{p-k}^{\ast}\Psi\right\Vert ^{2}\nonumber \\ & \leq\left(\sum_{p\in L_{k}^{\pm}}\left|\left\langle e_{p},\varphi_{p}\right\rangle \right|^{2}+2\sum_{p\in L_{k}-k}\left|\left\langle e_{p+k},\varphi_{p}\right\rangle \right|^{2}+2\sum_{p\in L_{-k}+k}\left|\left\langle e_{p-k},\varphi_{p}\right\rangle \right|^{2}\right)\left\Vert \Psi\right\Vert ^{2} \nonumber\\ & \leq\left(\sum_{p\in L_{k}^{\pm}}\max_{q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}+2\sum_{p\in L_{k}}\max_{q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}+2\sum_{p\in L_{-k}}\max_{q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left\Vert \Psi\right\Vert ^{2}\nonumber \\ & =3\left(\sum_{p\in L_{k}^{\pm}}\max_{q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left\Vert \Psi\right\Vert ^{2}, \end{align} $$

which implies the first estimate. For the second estimate, we find in a similar manner (now directly from equation (6.17)) that

(6.20) $$ \begin{align} &\sum_{p\in\mathbb{Z}^{3}}\left\Vert \left[\tilde{c}_{p}^{\ast},b_{k}\left(\varphi_{p}\right)\right]\Psi\right\Vert ^{2} \leq\sum_{p\in L_{k}^{\pm}}\left|\left\langle e_{p},\varphi_{p}\right\rangle \right|^{2}\left\Vert \tilde{c}_{\overline{p-k}}\Psi\right\Vert ^{2} +2\sum_{p\in L_{k}-k}\left|\left\langle e_{p+k},\varphi_{p}\right\rangle \right|^{2}\left\Vert \tilde{c}_{p+k}\Psi\right\Vert ^{2} \nonumber \\ &\qquad +2\sum_{p\in L_{-k}+k}\left|\left\langle e_{p-k},\varphi_{p}\right\rangle \right|^{2}\left\Vert \tilde{c}_{p-k}\Psi\right\Vert ^{2} \nonumber\\ &\quad \leq\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left(\sum_{p\in L_{k}^{\pm}}\left\Vert \tilde{c}_{\overline{p-k}}\Psi\right\Vert ^{2}+2\sum_{p\in L_{k}-k}\left\Vert \tilde{c}_{p+k}\Psi\right\Vert ^{2}+2\sum_{p\in L_{-k}+k}\left\Vert \tilde{c}_{p-k}\Psi\right\Vert ^{2}\right)\nonumber \\ &\quad =\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left(\sum_{p\in L_{k}^{\pm}}\left\Vert c_{\overline{p-k}}^{\ast}\Psi\right\Vert ^{2}+2\sum_{p\in L_{k}}\left\Vert c_{p}\Psi\right\Vert ^{2}+2\sum_{p\in L_{-k}}\left\Vert c_{p}\Psi\right\Vert ^{2}\right)\nonumber \\ &\quad \leq4\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle. \end{align} $$

Lastly, we estimate the double commutator:

Proposition 6.3. For all $k,l\in \mathbb {Z}_{+}^{3}$ , sequences $\left(\varphi _{p}\right)_{p\in \mathbb {Z}^{3}}\subset \ell ^{2}(L_{k}^{\pm })$ and $\left(\psi _{p}\right)_{p\in \mathbb {Z}^{3}}\subset \ell ^{2}\left(L_{l}^{\pm }\right)$ , and $\Psi \in \mathcal {H}_{N}$ , it holds that

$$\begin{align*}\sum_{p\in\mathbb{Z}^{3}}\left\Vert \left[\left[\tilde{c}_{p},b_{k}^{\ast}\left(\varphi_{p}\right)\right],b_{l}\left(\psi_{p}\right)\right]\Psi\right\Vert ^{2}\leq12\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\psi_{q}\right\rangle \right|^{2}\right)\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle. \end{align*}$$

Proof. From (6.18), we have that

(6.21) $$ \begin{align} \left[\left[\tilde{c}_{p},b_{k}^{\ast}\left(\varphi_{p}\right)\right],b_{l}\left(\psi_{p}\right)\right] & =1_{L_{k}^{\pm}}(p)\left\langle e_{p},\varphi\right\rangle \left[\tilde{c}_{\overline{p-k}}^{\ast},b_{l}\left(\psi_{p}\right)\right]-1_{L_{k}-k}(p)\left\langle e_{p+k},\varphi\right\rangle \left[\tilde{c}_{p+k}^{\ast},b_{l}\left(\psi_{p}\right)\right]\\ &\quad -1_{L_{k}+k}(p)\left\langle e_{p-k},\varphi\right\rangle \left[\tilde{c}_{p-k}^{\ast},b_{l}\left(\psi_{p}\right)\right]\nonumber, \end{align} $$

and so, by the triangle inequality and the second estimate of Proposition 6.2,

(6.22) $$ \begin{align} & \sum_{p\in\mathbb{Z}^{n}}\left\Vert \left[\left[\tilde{c}_{p},b_{k}^{\ast}\left(\varphi_{p}\right)\right],b_{l}\left(\psi_{p}\right)\right]\Psi\right\Vert ^{2}\leq\sum_{p\in L_{k}^{\pm}}\left|\left\langle e_{p},\varphi_{p}\right\rangle \right|^{2}\left\Vert \left[\tilde{c}_{\overline{p-k}}^{\ast},b_{l}\left(\psi_{p}\right)\right]\Psi\right\Vert ^{2}\nonumber \\ &\qquad +2\sum_{p\in L_{k}-k}\left|\left\langle e_{p+k},\varphi_{p}\right\rangle \right|^{2}\left\Vert \left[\tilde{c}_{p+k}^{\ast},b_{l}\left(\psi_{p}\right)\right]\Psi\right\Vert ^{2}+2\sum_{p\in L_{-k}+k}\left|\left\langle e_{p-k},\varphi_{p}\right\rangle \right|^{2}\left\Vert \left[\tilde{c}_{p-k}^{\ast},b_{l}\left(\psi_{p}\right)\right]\Psi\right\Vert ^{2}\nonumber \\ &\quad \leq\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left(\sum_{p\in L_{k}^{\pm}}\left\Vert \left[\tilde{c}_{\overline{p-k}}^{\ast},b_{l}\left(\psi_{p}\right)\right]\Psi\right\Vert ^{2}+2\sum_{p\in L_{k}}\left\Vert \left[\tilde{c}_{p}^{\ast},b_{l}\left(\psi_{p-k}\right)\right]\Psi\right\Vert ^{2}\right.\nonumber \\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.+2\sum_{p\in L_{-k}}\left\Vert \left[\tilde{c}_{p}^{\ast},b_{l}\left(\psi_{p+k}\right)\right]\Psi\right\Vert ^{2}\right) \nonumber\\ &\quad \leq12\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\varphi_{q}\right\rangle \right|^{2}\right)\left(\max_{p\in L_{k}^{\pm},q\in\mathbb{Z}^{3}}\left|\left\langle e_{p},\psi_{q}\right\rangle \right|^{2}\right)\left\langle \Psi,\mathcal{N}_{E}\Psi\right\rangle. \end{align} $$

6.3 Final estimation of the exchange terms

Now we are ready to derive bounds for the exchange terms $\mathcal {E}_{1}^{k}(A)$ and $\mathcal {E}_{2}^{k}(B)$ defined in Proposition 5.2. Recall that we have reduced the estimation of these complicated operators to the task of obtaining a uniform estimate for the four explicit forms

(6.23) $$ \begin{align} & \sum_{l\in S_{C}}\sum_{p\in S}\tilde{c}_{p_{3}}^{\ast}b_{k}^{\natural}\left(Te_{p_{1}}\right)b_{l}^{\natural}\left(K_{l}^{\oplus}e_{p_{2}}\right)\tilde{c}_{p_{4}},\qquad\quad\;\;\sum_{l\in S_{C}}\sum_{p\in S}\left[\tilde{c}_{p_{3}},b_{k}^{\ast}\left(Te_{p_{1}}\right)\right]^{\ast}b_{l}^{\natural}\left(K_{l}^{\oplus}e_{p_{2}}\right)\tilde{c}_{p_{4}} \nonumber\\ & \sum_{l\in S_{C}}\sum_{p\in S}\left[\left[\tilde{c}_{p_{3}},b_{k}^{\ast}\left(Te_{p_{1}}\right)\right],b_{l}\left(K_{l}^{\oplus}e_{p_{2}}\right)\right]^{\ast}\tilde{c}_{p_{4}},\quad\sum_{l\in S_{C}}\sum_{p\in S}\left[\tilde{c}_{p_{3}},b_{k}^{\ast}\left(Te_{p_{1}}\right)\right]^{\ast}\left[\tilde{c}_{p_{4}},b_{l}^{\ast}\left(K_{l}^{\oplus}e_{p_{2}}\right)\right], \end{align} $$

subject to the following rules: $b_{k}^{\natural }$ denotes either $b_{k}$ or $b_{k}^{\ast }$ , T denotes either A or B, and $b_{k}^{\natural }\left(Te_{p_{1}}\right)$ and $b_{l}^{\natural }(K_{l}^{\oplus }e_{p_{2}})$ may be interchanged. Furthermore, the notation $\tilde {c}_{p}$ denotes either $c_{p}$ or $c_{p}^{\ast }$ as appropriate for p, and the set S is such that the assignments $p\mapsto p_{1},p_{2},p_{3},p_{4}$ are injective and map exclusively into $B_{F}$ or $B_{F}^{c}$ .