Near-identity transformation

Lets us first apply the near-identity transformation to the $O(1)$ part of the Hamiltonian that is given by Eq. (81)

$$\int\omega b^*b~d\textbf{k}d\textbf{x}=\int\omega c^*c+\big(\omeg... ...^*cc_-+\omega \beta ^*c\{\gamma ^*,c_-\}+c.c.\big)d\textbf{k}d\textbf{x}+h.o.t.$$

To apply this transformation to the $O(\varepsilon )$ part we just need to substitute $b$ with $c$ in Eq. (82). The non-diagonal terms cancel in the Hamiltonian if

$$\int\big(\sigma+\omega \alpha ^*\big)cc_-+\left(\omega \beta ^*c\... ...,c_-\}+\frac{i\mu_{ev}^2}{2\nu}c\{\varphi,c_-\}\right)d\textbf{k}d\textbf{x}=0.$$ (106)

Let us choose

$$\gamma =\gamma ^*=\varphi=\frac{\omega _{ev}}{\mu_{ev}}.$$ (107)

Then we can rewrite Eq. (116) as

$$\int(\sigma+\omega\alpha ^*)cc_-+\left((\omega_{ev}+\omega_{od})\beta^*+\frac{i\mu_{ev}^2}{2\nu}\right)c\{\varphi,c_-\}d\textbf{k}d\textbf{x}=0.$$

Integrating by parts one can show that

$$\int\omega_{od}\beta^*c\{\varphi,c_-\}d\textbf{k}d\textbf{x}=\frac{1}{2}\int\{\omega_{od}\beta^*,\varphi\}cc_-d\textbf{k}d\textbf{x}.$$

Therefore, the diagonalizing condition becomes

$$\int\left(\sigma+\omega\alpha ^*+\frac{1}{2}\{\omega_{od}\beta^*,... ...\beta^*+\frac{i\mu_{ev}^2}{2\nu}\right)c\{\varphi,c_-\}d\textbf{k}d\textbf{x}=0$$

From Eq. (118), the condition on $\beta$ immediately follows

$$\beta =\frac{i\mu_{ev}^2}{2\nu\omega_{ev}}.$$ (108)

In order to obtain the condition on $\alpha$ , we expand the second term in the integral

$$\omega \alpha ^*=\omega_{od}\alpha_{od}^*+\omega_{ev}\alpha_{ev}^*+\omega_{od}\alpha_{ev}^*+\omega_{ev}\alpha_{od}^*.$$ (109)

Integral over the last two terms vanishes because these functions are odd. Therefore, we consider only the other two terms. Next, we insert this expansion into Eq. (118)

$$\int\left(\sigma+\omega_{od}\alpha_{od}^*+\omega_{ev}\alpha_{ev}^... ...+\left(\omega_{ev}\beta^*+\frac{i\mu_{ev}^2}{2\nu}\right)\{\varphi,c_-\}cdkdx=0$$

Then, we obtain the following diagonalizing conditions on $\alpha$

$$\alpha_{ev} = -\frac{\sigma^*}{\omega_{ev}}-\frac{\beta }{2\omega_{ev}}\left\{\omega_{od},\frac{\omega _{ev}}{\mu_{ev}}\right\},$$ (110)

$$\alpha_{od} = -\frac{1}{2}\left\{\beta ,\frac{\omega_{ev}}{\mu_{ev}}\right\}.$$ (111)

Let us prove that $\alpha _{od}=0$ . Substituting Eq. (118) into Eq. (121), we obtain

$$\alpha _{od}=-\frac{i}{4}\left\{\frac{\mu_{ev}^2}{\nu\omega_{ev}},\frac{\omega_{ev}}{\mu_{ev}}\right\}.$$

Expanding the Poisson bracket we find the following identity

$$\left\{\frac{\mu_{ev}^2}{\nu\omega_{ev}},\frac{\omega_{ev}}{\mu_{... ..._{ev}\}+\mu\{\omega _{ev},\nu\}+\omega _{ev}\{\nu,\mu_{ev}\}}{\nu^2\omega_{ev}}$$ (112)

According to the definition, $\omega _{ev}^2=\mu_{ev}^2-\nu^2$ . Differentiation of both sides of the last equality with respect to $\textbf{x}$ or $\textbf{k}$ yields

$$\nabla\omega _{ev}=\frac{\mu_{ev}}{\omega _{ev}}\nabla\mu_{ev}-\frac{\nu}{\omega _{ev}}\nabla\nu.$$ (113)

We use Eq. (123) to show that

$$\{\mu_{ev},\omega _{ev}\} = -\frac{\nu}{\omega _{ev}}\{\mu_{ev},\nu\},$$

$$\{\omega _{ev},\nu\} = \frac{\mu_{ev}}{\omega _{ev}}\{\mu_{ev},\nu\}.$$

Plugging Eq. (124) into Eq. (122) proves that $\alpha _{od}=0$ .