Fidelity between initial and final states

[deepalakshmi]

This is my calculation (final state using hamiltonian) for your reference:
The initial state is
$$|\psi(0)\rangle = |\alpha\rangle|0\rangle$$

The time evolved state is
$$|\psi(t)\rangle = e^{-i H t} |\alpha\rangle |0\rangle$$

$$ = e^{-i H t} e^{-\frac{|\alpha|^2}{2}} \sum_n{ \frac{(\alpha)^n}{\sqrt{n!}} \ket{n} } |0\rangle$$

$$ = e^{-i H t} e^{-\frac{|\alpha|^2}{2}} \sum_n{ \frac{(\alpha)^n}{\sqrt{n!}} \frac{(a^\dagger)^n}{\sqrt{n!}} } |0\rangle |0\rangle$$

Inserting $U U^\dagger$

$$= e^{-\frac{|\alpha|^2}{2}} \sum_n{ \frac{(\alpha)^n}{\sqrt{n!}} e^{-i H t} \frac{(a^\dagger)^n}{\sqrt{n!}} e^{i H t}e^{-i H t}} |0\rangle |0\rangle$$

We know that $U a^\dagger U^\dagger=(a^\dagger cos t+i b^\dagger sin t )^n$

$$|\psi(t)\rangle =e^{-\frac{|\alpha|^2}{2}} \sum_n \frac{(\alpha)^n}{\sqrt{n!}} \frac{(a^\dagger cos t+i b^\dagger sin t )^n}{\sqrt{n!}} |0\rangle |0\rangle$$

Using binomial expression and using the property ${(a^\dagger)^{n-p}} |0\rangle = \sqrt{(n-p)!}|(n-p)\rangle$

$$|\psi(t)\rangle = e^{-\frac{|\alpha|^2}{2}} \sum_n \sum_p \frac{\sqrt{n!}}{\sqrt{p!}\sqrt{n-p}!} ((isin(t))^p)((cost(t))^{n-p}\frac{(\alpha)^n}{\sqrt{n!}}|p\rangle|n-p\rangle$$

Substitute $\alpha^n = {\alpha^p}{\alpha}^{n-p}$

$$|\psi(t)\rangle = e^{-\frac{|\alpha|^2}{2}} \sum_p\frac{((\alpha)^p)(isin(t))^p}{\sqrt{p!}}|p\rangle\sum_{n-p}\frac{((\alpha)^{n-p})(cos(t))^{n-p}}{\sqrt{n-p!}}|n-p\rangle$$

$$|\psi(t)\rangle=|\alpha isin(t)\rangle |\alpha cos(t)\rangle$$

 

[deepalakshmi]

This is my calculation (final state using hamiltonian) for your reference:
The initial state is$|\psi(0)\rangle = |\alpha\rangle|0\rangle$
The time evolved state is
$|\psi(t)\rangle = e^{-i H t} |\alpha\rangle|0\rangle$
$= e^{-i H t}e^{-\frac{|\alpha|^2}{2}}\sum_n\frac{(\alpha)^n}{\sqrt{n!}}|\alpha\rangle$

$= e^{-i H t}e^{-\frac{|\alpha|^2}{2}}\sum_n\frac{(\alpha)^n}{\sqrt{n!}}\frac{(a^\dagger)^n}{{\sqrt{n!}}|0\rangle|0\rangle$

Inserting #U U\dagger#
$= e^{-\frac{|\alpha|^2}{2}}\sum_n\frac{(\alpha)^n}{\sqrt{n!}}e^{-i H t}\frac{(a^\dagger)^n}{{\sqrt{n!}}e^{i H t}e^{-i H t}|0\rangle|0\rangle$

We know that #U a\dagger U\dagger=(a\dagger cos t+i b\dagger sin t )^n
$|\psi(t)\rangle =e^{-\frac{|\alpha|^2}{2}}\sum_n\frac{(\alpha)^n}{\sqrt{n!}}\frac{(a^\dagger cos t+i b\dagger sin t )^n}{{\sqrt{n!}}0\rangle 0\rangle$

Using binomial expression and using the property {(a\dagger)^n-p}0\rangle={\sqrt{(n-p)!}(|n-p)!\rangle##

$|\psi(t)\rangle =e^{-\frac{|\alpha|^2}{2}}\sum_n\sum_p\frac{(\sqrt{n!})}{\sqrt{(p!)|sqrt(n-p)!}} ((isin(t))^p)((cost(t))^n-p)\frac{(\alpha)^n}{\sqrt{n!}}|p\rangle|n-p\rangle$
Substitute #\alpha^n={ alpha^p}{alpha}^n-p#
$|\psi(t)\rangle =e^{-\frac{|\alpha|^2}{2}}\sum_p\frac{((\alpha)^p)(isin(t))^p}{\sqrt{p!}}|p\rangle\sum_n-p\frac{((\alpha)^n-p)(cos(t))^n-p}{\sqrt{n-p!}}|n-p\rangle$
$|\psi(t)\rangle=|\alphaisin(t)\rangle|\alphacos(t)\rangle$

can someone edit this?

 

[PeterDonis]

the annihilation operator $a$ is just the same as $b$

This contradicts this:

only that $a$ operates in the space of the first photon, while $b$ lives in that of the second photon.

This means they are not the same operator, since they act on different degrees of freedom (each photon is a different degree of freedom). They are the same "kind" of operator (ladder operators on a one-photon Hilbert space), but they're not the same operator.

I am not sure whether the beam splitter scenario is suitable or not.

It can be; each of the two input arms of the beam splitter can be described by a one-photon Hilbert space, so the input state $\ket{0} \ket{\alpha}$ would just mean that input arm #1 is in the vacuum state (nothing is being pumped into it), while input arm #2 has coherent state $\alpha$ being pumped into it.

 

[PeterDonis]

can someone edit this?

I've edited the post as best I can, but some of the expressions don't make sense to me.

 

[PeterDonis]

I've edited the post as best I can

Btw, @deepalakshmi, I strongly suggest using the \ket operator in LaTeX instead of trying to manually put in vertical lines and right angle brackets. PF recently added support for the set of standard QM LaTeX codes, of which \ket is one, and it makes QM equations a lot easier to post.

 

[deepalakshmi]

This is my calculation (final state using hamiltonian) for your reference:
The initial state is
$$|\psi(0)\rangle = |\alpha\rangle|0\rangle$$

The time evolved state is
$$|\psi(t)\rangle = e^{-i H t} |\alpha\rangle |0\rangle$$

$$ = e^{-i H t} e^{-\frac{|\alpha|^2}{2}} \sum_n{ \frac{(\alpha)^n}{\sqrt{n!}} \ket{n} } |0\rangle$$

$$ = e^{-i H t} e^{-\frac{|\alpha|^2}{2}} \sum_n{ \frac{(\alpha)^n}{\sqrt{n!}} \frac{(a^\dagger)^n}{\sqrt{n!}} } |0\rangle |0\rangle$$

Inserting $U U^\dagger$

$$= e^{-\frac{|\alpha|^2}{2}} \sum_n{ \frac{(\alpha)^n}{\sqrt{n!}} e^{-i H t} \frac{(a^\dagger)^n}{\sqrt{n!}} e^{i H t}e^{-i H t}} |0\rangle |0\rangle$$

We know that $U a^\dagger U^\dagger=(a^\dagger cos t+i b^\dagger sin t )$

$$|\psi(t)\rangle =e^{-\frac{|\alpha|^2}{2}} \sum_n \frac{(\alpha)^n}{\sqrt{n!}} \frac{(a^\dagger cos t+i b^\dagger sin t )^n}{\sqrt{n!}} |0\rangle |0\rangle$$

Using binomial expression and using the property ${(a^\dagger)^{n-p}} |0\rangle = \sqrt{(n-p)!}|(n-p)\rangle$

$$|\psi(t)\rangle = e^{-\frac{|\alpha|^2}{2}} \sum_n \sum_p \frac{\sqrt{n!}}{\sqrt{p!}\sqrt{n-p}!} ((isin(t))^p)((cost(t))^{n-p}\frac{(\alpha)^n}{\sqrt{n!}}|p\rangle|n-p\rangle$$

Substitute $\alpha^n = {\alpha^p}{\alpha}^{n-p}$

$$|\psi(t)\rangle = e^{-\frac{|\alpha|^2}{2}} \sum_p\frac{((\alpha)^p)(isin(t))^p}{\sqrt{p!}}|p\rangle\sum_{n-p}\frac{((\alpha)^{n-p})(cos(t))^{n-p}}{\sqrt{n-p!}}|n-p\rangle$$

$$|\psi(t)\rangle=|\alpha isin(t)\rangle |\alpha cos(t)\rangle$$

This is how I got my final state

 

[Haorong Wu]

@PeterDonis Sorry, I stated it incorrectly.

@deepalakshmi . After utilizing $U a^\dagger U^\dagger$, where is the $e^{-i H t}$ before $|0\rangle |0\rangle$? Also, $U a^\dagger U^\dagger=(a^\dagger cos t+i b^\dagger sin t )^n$ is not consistent with its following equation. Maybe you mean $U a^\dagger U^\dagger=a^\dagger cos t+i b^\dagger sin t$?

 

[deepalakshmi]

e^{-iHt} on |0\rangle |0\rangle becomes| 0\rangle|0\rangle. Yes $U a^\dagger U^\dagger=(a^\dagger cos t+i b^\dagger sin t )$

 

[deepalakshmi]

Is my evolution of initial state correct?. My final state is looking odd because of sin and cos terms. But this is the answer I am getting.

 

[PeterDonis]

Is my evolution of initial state correct?

What would you expect the final state to look like, on physical grounds? As John Wheeler once said, never do a calculation unless you already know the answer.

 

[deepalakshmi]

What would you expect the final state to look like, on physical grounds? As John Wheeler once said, never do a calculation unless you already know the answer.

I don't know the answer. I just tried to solve it. If my calculation is right then my answer will be right.

 

[PeterDonis]

I don't know the answer.

Have you tried to think about what the answer might be, physically? That's what Wheeler was talking about: think about the physics first, before trying to do the math.

 

[deepalakshmi]

Have you tried to think about what the answer might be, physically? That's what Wheeler was talking about: think about the physics first, before trying to do the math.

I saw a paper which is similar to my question. There, the initial state was two coherent state which has my hamiltonian. Its final state was given as two coherent state in which each coherent state is mentioned as̃ α1 = α1 cos λt − iα2 sin λt, α 2 = α2 cos λt − iα1 sin λt. There was no calculation for this. Just answer was given in that paper. I had already given link to you for that paper.

 

[PeterDonis]

I saw a paper which is similar to my question. There, the initial state was two coherent state which has my hamiltonian. Its final state was given as two coherent state in which each coherent state is mentioned as̃ α1 = α1 cos λt − iα2 sin λt, α 2 = α2 cos λt − iα1 sin λt. There was no calculation for this. Just answer was given in that paper. I had already given link to you for that paper.

What does any of this have to do with the question I asked you?

 

[deepalakshmi]

What does any of this have to do with the question I asked you?

You asked me whether have I tried to think my answer? I am telling you that I took that paper as reference and got this answer.

 

[PeterDonis]

You asked me whether have I tried to think my answer?

Whether you have tried to think about what you would expect the answer to be, physically.

I am telling you that I took that paper as reference and got this answer.

So in other words, no, you have not tried to think about what you would expect the answer to be, physically. You are just accepting the authority of that paper without trying to think about what it means and whether it makes sense.

Based on that, I don't see any point in continuing this thread further. Thread closed.