I am given the following Hamiltonian, $$H=H_1=\frac{p^2}{2m}+\frac{1}{2}m\omega_1^2x^2$$ for $t<0$ and $$H=H_2=\frac{p^2}{2m}+\frac{1}{2}m\omega_2^2x^2$$ for $t\geq0$. For some time $t_1(<0)$, we know the eigenstates of the system as $\{|n_1\rangle\}$ which are eigenstates of $H_1$ and for some time $t_2(>0)$, we know the eigenstates of the system as $\{|n_2\rangle\}$ which are eigenstates of $H_2$.

Now, my question is, at time $t_1$, we have a whole list of operators which can act on the state like $a_1$ and $a_1^{\dagger}$ i.e the annihilation and creation operators and similarly at time $t_2$ as $a_2$ and $a_2^{\dagger}$. Now, if I act $a_1$ on $|0_2\rangle$, or perform similar operations, what do I get? Will the eigenvectors of $a_1$ also be the eigenvectors of $a_2$?


1 Answer 1


$a_1$ is not hermitian so be careful about its eigenvectors. Most importantly, the definition of the creation and destruction operators involves the frequency $\omega$ so the creation operator on both sides of $t=0$ will not be the same so the harmonic oscillator eigenstates, i.e. the number states, will not be the same. This is evident because the length scale for the number states depends also on $\omega$.

In particular, the change of scale is equivalent to a squeezing transformation so the vacuum state for $t_2$ will not be the vacuum state for $t_1$. You will need to expand $\vert 0_2\rangle$ in terms of $t_1$-states in order compute the action of $a_1$, i.e. given $\lambda_1=\sqrt{\frac{m\omega_1}{\hbar}}$ $$ \psi_n(x)=\frac{1}{\sqrt{2^n n!}} \left(\frac{\lambda^2_1}{\pi}\right)^{1/4}e^{-\lambda_1^2 x^2/2} H_n(\lambda_1 x) $$ and $$ \phi_n(x)=\frac{1}{\sqrt{2^n n!}} \left(\frac{\lambda^2_2}{\pi}\right)^{1/4}e^{-\lambda_2^2 x^2/2} H_n(\lambda_2 x) $$ the simplest way to proceed is to write $$ \vert n_2\rangle =\sum_{n_1} \vert n_1\rangle \langle n_1\vert n_2\rangle $$ with $$ \langle n_1\vert n_2\rangle =\int_{-\infty}^\infty \phi_{n_2}(\lambda_2 x) \phi_{n_1}(\lambda_1 x) $$ Thus, for instance, $$ \langle 0_1\vert 0_2\rangle= \sqrt{\frac{2\lambda_1\lambda_2}{\lambda_1^2+\lambda_2^2}}\, , \qquad \langle 2_1\vert 0_2\rangle = \frac{\sqrt{\lambda_1\lambda_2}(\lambda_1^2-\lambda_2^2)}{(\lambda_1^2+\lambda_2^2)^{3/2}} $$

  • $\begingroup$ Thanks for the fabulous ans. Please can u also tell me what is squeezing transformation? And mention some resources to read this from? $\endgroup$ Commented Jun 6, 2018 at 13:14
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    $\begingroup$ @NamanAgarwal see Eqs.(24a-b) of this paper arxiv.org/pdf/1401.4118.pdf. The scaling of $x\to x'=e^{-r}x$ and $p\to p'=e^{r}p$ will produce $H\to H'=e^{2r}(p^2/2m+e^{-4r}m\omega^2 x^2/2)$ which is same as a scaling of the original frequency $\omega\to \omega'=e^{-2r}\omega$ (or something like that). Lots of quantum optics books describe this as well, v.g. the book by Gerry & Knight or the book by Fox. $\endgroup$ Commented Jun 6, 2018 at 14:57
  • $\begingroup$ I am still stuck up with the math. Can you tell me the solution to how can we express the ground state of $H_2$ in terms of $\{|n_1\rangle\}$ because $|0_2\rangle=\sum_{n=0}^{\infty} \frac{|n_1\rangle}{\sqrt{n_1 !}}\langle0_1|a_1^n|0_2\rangle$, I can't evaluate $\langle0_1|a_1^n|0_2\rangle$. $\endgroup$ Commented Jun 8, 2018 at 6:11
  • $\begingroup$ @NamanAgarwal added some material $\endgroup$ Commented Jun 8, 2018 at 10:28

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