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Given Hamiltonian of Quantum Harmonic Oscillator,

$$H = \frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2-\gamma x$$

I have to find a constant $\beta$ such that the unitary operator $U=\exp(\beta a^{\dagger} - \beta^*a)$ diagonalize $H$ respect to the eigenstates of $N=a^{\dagger}a$, so that:

$$\bar{H} = U^{\dagger}HU$$

is diagonal respect to $|n \rangle$.

As you can understand, $a$ and $a^{\dagger}$ are annihilation and creator operators.

Second question. If $| 0 \rangle$ is the eigenket of $a^{\dagger}a$ with eigenvalue zero, suppose the particle is in $| 0 \rangle$ at $t=0$. Find the probability that the particle is in $| 0 \rangle$ at time $t>0$.

I'll show you my attempt.

Firstly, I wrote hamiltonian in terms of annihilation and creator operators and this is the result:

$$H = \hbar \omega(a^{\dagger}a+1/2)-\gamma\sqrt{\frac{m\omega}{2\hbar}}(a^{\dagger}+a)$$

and with the Baker-Hausdorff formula

$$e^ABe^{-A} = B+[A,B]+\frac{1}{2!}[A,[A,B]]+...$$

I write the result, only first three terms are not zero, and we have:

$$\bar{H} = \hbar\omega(a^{\dagger}a+1/2)-\gamma\sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a)+\gamma\sqrt{\frac{\hbar}{m\omega}}(\beta+\beta^*)-\hbar\omega(\beta a^{\dagger}+\beta^* a)+\hbar\omega|\beta|^2$$

Since that the first term and the constant are irrelevant, because $|n \rangle$ of course are eigenkets of, It must be:

$$-\gamma\sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a)=0$$ and $$-\hbar\omega(\beta a^{\dagger}+\beta^* a)=0$$

and i found $\beta=\frac{\gamma}{\hbar\omega}\sqrt{\frac{\hbar}{2m\omega}}$.

I'll show my attempt for the second question below on answer or comment.

My question is: obviously $U$ is a translation opertor, and my hamiltonian is shifted respect of classical quantum harmonic oscillator. Is true that also the eingenstates are shifted? And for the second question: is correct that the eigenkets of $H$ are $U| n \rangle$? This is my attempt to demonstrate it:

$$H=U\bar{H}U^{\dagger}$$

$$H(U|n \rangle) = U\bar{H}U^{\dagger}U|n \rangle =U\bar{H}|n \rangle= UE_n |n \rangle= E_n (U|n \rangle)$$

when $E_n$ are eigenvalue of $\bar{H}$. If I want to find the probability $|\langle 0(t)|0 \rangle|^2$ I have to write $|0 \rangle$ in a basis of $H$. Another attempt is to find $|0(t) \rangle$ using time-operator, but it seems more complicated mathematically. How I can solve the second question?

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  • $\begingroup$ Hello, I checked anymore my calculations. These are the results: $$e^AHe^{-A} = \hbar \omega(a^{\dagger}a+1/2)-\gamma\sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a)+\gamma\sqrt{\frac{\hbar}{2m\omega}}(\beta+\beta^*)-\hbar\omega(\beta a^{\dagger}-\beta^*a)+\hbar\omega |\beta|^2$$ Now, the constant are irrelevant, just because are diagonal and of course $|n \rangle$ are their eigenvalues and the same thing for $ \hbar \omega(a^{\dagger}a+1/2)$. It must be: $$\gamma\sqrt{\frac{\hbar}{2m\omega}}(a^{\dagger}+a)-\hbar\omega(\beta a^{\dagger}-\beta^*a)=0$$ $\endgroup$
    – Damark
    Commented Nov 1, 2023 at 6:42
  • $\begingroup$ Please edit the question with what you have done (the most important steps) and clarify where exactly you are stuck. And then try to reopen... Comments are usually not for discussions. $\endgroup$ Commented Nov 1, 2023 at 7:03
  • $\begingroup$ Your question(s) seem a bit confused. Consider to re-arrange and possibly shorten it a bit. For example, do you mean that you turn on the perturbation at $t=0$ and work within the sudden approximation? $\endgroup$ Commented Nov 1, 2023 at 12:03
  • $\begingroup$ It seems more understandable now. Please, is my answer (and interpretation) for first question correct? By the way, I'm blocked about the second question and I don't now how to solve it. I'm sorry for confusion. $\endgroup$
    – Damark
    Commented Nov 1, 2023 at 16:18
  • $\begingroup$ As I said, comments are really not intended for discussions. But here are some thoughts: Your expression for $\bar H$ is correct? If so, you just have to choose $\beta$ such that the second and fourth terms cancel, as I already said in a previous comment; I don't understand your two lines below, though, they make no sense to me. To the second question, yes the eigenstates of $H$ are $U|n\rangle$ in your notation (for the correct value of $\beta$). There are several ways to proceed now, but a comment is not suitable for explaining, sorry. Perhaps the question gets re-opened. $\endgroup$ Commented Nov 1, 2023 at 18:47

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