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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 answer is $U| n \rangle$? Or else, the shifted eigenstates or the "old" hamiltonian (without the $-\gamma x$ addendum) are eigenstates of the shifted hamiltonian? And how I can solve the second question? I'll show you my attempt, but of course are not correct.

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?

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.

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 answer is $U| n \rangle$? Or else, the shifted eigenstates or the "old" hamiltonian (without the $-\gamma x$ addendum) are eigenstates of the shifted hamiltonian? And how I can solve the second question? I'll show you my attempt, but of course are not correct.

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.

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.