Harmonic Crystal and Bogoliubov trasformation

[Satana]

Homework Statement

I just need a hint to solve the fifth question of this exercise. I have already solved the others

Relevant Equations

$$a_q'=\frac{1}{\sqrt {2}}\left(u_q\sqrt{\frac{M\omega_q'}{\hbar}}+ip_{-q}\frac{1}{\sqrt{M\omega_q'\hbar}}\right)
\,\,\,\,\,\,\,\,\,\,

a_q'^\dagger=\frac{1}{\sqrt {2}}\left(u_-q\sqrt{\frac{M\omega_q'}{\hbar}}-ip_q\frac{1}{\sqrt{M\omega_q'\hbar}}\right) $$

I thought about writing $$a_q'|0'> =0$$ then develop the U operator in series, after I don't know how to proceed

 

[member 553083]

.One approach is to use the definition of the derivative operator, $\hat{a}_q' = \frac{\mathrm{d}}{\mathrm{d}q}\hat{a}_q$. Then, by applying the chain rule, we can write $$\hat{a}_q'|0\rangle = \frac{\mathrm{d}}{\mathrm{d}q}\left(\hat{U}^\dagger(q)\hat{a}_0\hat{U}(q)|0\rangle\right) = \hat{U}^\dagger(q)\frac{\mathrm{d}}{\mathrm{d}q}\left(\hat{a}_0\hat{U}(q)|0\rangle\right).$$Using the fact that $\frac{\mathrm{d}}{\mathrm{d}q}\hat{U}(q)=\dot{\hat{U}}(q)\hat{U}^{-1}(q)$, and the fact that $\hat{U}|0\rangle=|0\rangle$ for all $q$, we can then rewrite the above equation as$$\hat{a}_q'|0\rangle = \hat{U}^\dagger(q)\dot{\hat{U}}(q)\hat{U}^{-1}(q)\hat{a}_0|0\rangle.$$Now, since $\hat{U}|0\rangle=|0\rangle$, we have $\hat{U}^{-1}|0\rangle=|0\rangle$ as well, which means that $\hat{a}_0|0\rangle=0$. Thus, we can conclude that$$\hat{a}_q'|0\rangle = 0.$$