Bogolyubov transformation of the $O(\varepsilon )$ part

Here, we show how Bogolyubov transformation works on $H_{f,\varepsilon }^{(1)}$ . We consider the terms of Eq. (72) starting with

$$- \int\left[\check{a}(\textbf{x}\cdot\nabla_\textbf{x}\mu)\check{a}... ...nu)(\check{a}\check{a}_-+\check{a}^*\check{a}^*_-)\right]d\textbf{k}d\textbf{x}$$

$$= - \int(ub +vb_-^*)(\textbf{x}\cdot\nabla_\textbf{x}\mu)(ub^*+vb_-)d... ..._\textbf{x}\nu)(u b + v b_-^*)(u b_- +v b^*) d\textbf{k}d\textbf{x}+c.c.\right]$$

$$= - \int b \left[ (u^2+v^2)(\textbf{x}\cdot\nabla_\textbf{x}\mu_{ev})... ...d}) + 2uv(\textbf{x}\cdot\nabla_\textbf{x}\nu)\right]b^*d\textbf{k}d\textbf{x}-$$

$$~~~-\int(u v (\textbf{x}\cdot\nabla_{\textbf{x}} \mu)+\frac{1}{2}... ... (\textbf{x}\cdot \nabla_\textbf{x}\nu))(b b_- +b^*b_-^*)d\textbf{k}d\textbf{x}$$

$$= \int\left[-b(\textbf{x}\cdot\nabla_{\textbf{x}} \omega)b^*+\frac{... ...\mu_{ev}^2}}\right) \left(bb_- + b^* b_-^*\right)\right]d\textbf{k}d\textbf{x}.$$ (92)

Here we used the following equalities

$$u^2+v^2=\cosh(2\xi),\ \ 2uv=\sinh(2\xi),\ \ u^2-v^2=1,$$

$$\sinh(2\xi)=\partial_{\nu}\omega,\ \ \cosh(2\xi)=\partial_{\mu_{ev}}\omega,\ \ 1=\partial_{\mu_{od}}\omega,$$

$$\sinh(2\xi)\nabla_\textbf{x}\mu_{ev}+\cosh(2\xi)\nabla_\textbf{x}\nu=-\frac{\mu_{ev}^2}{\nu}\nabla_\textbf{x}\sqrt{1-\frac{\nu^2}{\mu_{ev}^2}}.$$ (93)

Next, we consider the terms of Eq. (72) with the Poisson bracket

$$\int i\check{a}\{\mu,\check{a}^*\}d\textbf{k}d\textbf{x}+\frac{1}... ...{\mu,\check{a}^*\}+\check{a}\{\nu,\check{a}_-\}\Big)d\textbf{k}d\textbf{x}+c.c.$$

$$=\frac{i}{2}\int(u b + v b_-^*)\Big(\{\mu,u b^* + v b_-\}+\{\nu,u b_- + v b^*\}\Big)d\textbf{k}d\textbf{x}+c.c.$$ (94)

Note that

• $bb^*$ terms give zero because their coefficients are purely imaginary and we add c.c. values in the end,
• $b^*\nabla b$ terms can be obtained as c.c. of $b\nabla b^*$ terms.

Here, $\nabla$ denotes a gradient either with respect to $\textbf{x}$ or with respect to $\textbf{k}$ . Now, let us consider (104) term by term.

1. $b\nabla b^*$ and $b^*\nabla b$ :
   
   

   $$\frac{i}{2}\int \left[u^2b\{\mu,b^*\}+\underbrace{v^2b_-^*\{\mu,b... ...f{k}\rightarrow-\textbf{k}$}} +uvb\{\nu,b^*\}\right]d\textbf{x}d\textbf{k}+c.c.$$

   $$=\frac{i}{2}\int \Big(b\{\omega ,b^*\}-b^*\{\omega ,b\}\Big)d\textbf{x}d\textbf{k}=i\int b\{\omega,b^*\}d\textbf{x}d\textbf{k}.$$ (95)

   
   Here, we used the fact that $\int b\{\omega ,b^*\}d\textbf{k}d\textbf{x}=-\int b^*\{\omega ,b\}d\textbf{k}d\textbf{x}$
2. $bb_-$ and $b^*b_-^*$ :
   
   

   $$\frac{i}{2}\int\left[ubb_-\{\mu,v\}+vb_-^*b^*\{\mu,u\}+\underbrac... ...v\}}_ {\mbox{give 0 because $\nu$\ is even}}\right] d\textbf{k}d\textbf{x}+c.c.$$

   $$=\frac{i}{2}\int\Big[(bb_--b^*b_-^*)(u\{\mu,v\}-v\{\mu,u\})\Big]d... ...\mu_{od},\xi\}}_{\mbox{only $\mu_{od}$\ survives}}\right]d\textbf{k}d\textbf{x}$$

   $$=\frac{i}{2}\int\Big[(bb_--b^*b_-^*)\{\mu_{od},\xi\}\Big]d\textbf{k}d\textbf{x}$$ (96)

   
   Here, we have used Eq. (101).
3. $b\nabla b_-$ and $b^*\nabla b_-^*$ :
   
   

   $$\frac{i}{2}\int\Big[uvb_-^*\{\mu,b^*\}+uvb\{\mu,b_-\}+u^2b\{\nu,b_-\}+v^2b_-^*\{\nu,b^*\}\Big] d\textbf{k}d\textbf{x}+c.c.$$

   $$=\frac{i}{2}\int\Big[b\sinh(2\xi)\{\mu_{ev},b_-\}+b\cosh(2\xi)\{\nu,b_-\}\Big]d\textbf{k}d\textbf{x}+c.c.$$

   $$=\frac{i}{2}\int b \left[\frac{\mu_{ev}^2}{\nu}\left\{\sqrt{1-\frac{\nu^2}{\mu_{ev}^2}},b_-\right\}\right]d\textbf{k}d\textbf{x}+c.c.$$ (97)

   
   We used Eq. (103) here.

Finally, the rest of $O(\varepsilon )$ terms are

$$\frac{i}{2}\int\tilde{\nu}aa_-d\textbf{k}d\textbf{x}+c.c.=\frac{i}{2}\int\tilde{\nu}bb_-d\textbf{k}d\textbf{x}+c.c.$$ (98)

Combining Eqs. (102), (105), (106), (107), and (108), we obtain the $O(\varepsilon )$ part of the Hamiltonian given by Eq. (83).