A Fidelity between initial and final states

[deepalakshmi]

I will attach my work here

 

[PeterDonis]

I will attach my work here

This is not allowed. You need to post your work directly, including equations (use the PF LaTeX feature for that--a LaTeX Guide link is at the bottom left of the post window).

 

[deepalakshmi]

My calculation is big. I will summarize it down. I evolved the state using binomial distribution. After evolving I was asked to find the fidelity between initial and final state

 

[deepalakshmi]

Here initial state is given as$|\alpha> |0>$. I evolved it with time evolution operator.

 

[deepalakshmi]

$|\psi(t)>= time evolution operator(|\alpha>|0>)$
$=time evolution operator e^{-\frac{|\alpha|^2}{2}}\frac{(\alpha)^n}{\sqrt{n!}}|n>|0>$
since time evolution operator is a unitary operator inserting uudagger
B adagger B = adagger cos theta+ib dagger sin theta
from this equation I got sin and cos terms

 

[PeterDonis]

I evolved it with time evolution operator.

That still doesn't answer the question I asked in post #29 about the Hamiltonian.

 

[deepalakshmi]

That still doesn't answer the question I asked in post #29 about the Hamiltonian.

a, b are annihilation operator. b is similar to a. just change in variable

 

[PeterDonis]

b is similar to a. just change in variable

I don't understand. You only have one set of number operator states $\ket{n}$. That means you have only one annihilation operator and one creation operator. "Change in variable" makes no sense here.

 

[deepalakshmi]

Is there any other way to share my work. Calculation is 3 pages

 

[PeterDonis]

Is there any other way to share my work.

Not here.

Calculation is 3 pages

Then you will need to decide how much of it to post here. First, however, you might try answering post #38, and also post #30.

 

[deepalakshmi]

I don't understand. You only have one set of number operator states $\ket{n}$. That means you have only one annihilation operator and one creation operator. "Change in variable" makes no sense here.

I wrote the number terms in terms of creation operator. Then I inserted UUdagger which is basically time evolution operator. Later I applied BadaggerBdagger formula( baker campbell formula). I got a dagger cos theta+ ib dagger sin theta. I substituted this in place of creation operator. Later I used binomial expression and simplified

 

[deepalakshmi]

Not here.Then you will need to decide how much of it to post here. First, however, you might try answering post #38, and also post #30.

As I said earlier, the initial state and hamiltonian is given to me by my guide.

 

[deepalakshmi]

https://www.researchgate.net/publication/231004410_Generation_of_two-mode_entangled_coherent_states_via_a_cavity_QED_system .You can see this paper equation 8. It is similar

 

[Haorong Wu]

$e^{-\frac{|\alpha|^2}{2}}\frac{(\alpha)^n}{\sqrt{n!}}|n>$

It seems you are studying something close to quantum optics or QED.

Anyway, you did not write it correctly. The coherent state should be $\sum_{n=0} ^{\infty}e^{-\frac{|\alpha|^2}{2}}\frac{(\alpha)^n}{\sqrt{n!}}|n>$. Also, in quantum optics, $\left < n \right . \left | m \right >=\delta_{nm}$ where $\left | n \right >$ are eigenmodes of quantum electric fields.

Therefore $\left <0 \right . \left | \alpha \right >=e^{-\frac{|\alpha|^2}{2}}\frac{(\alpha)^0}{\sqrt{0!}}=e^{-\frac{|\alpha|^2}{2}}$.

 

[deepalakshmi]

Yeah I am studying quantum optics.
This is my final calculation:
$F=(\langle\psi|\chi\rangle|^2 )$
$=\langle\psi|\chi\rangle|=|(\langle\alpha|\langle 0|)(|\alpha\rangle |\alpha\rangle)|^2$
$=|(\langle\alpha|\alpha\rangle)(\langle 0|\alpha\rangle)|^2$
$=|e^{-\frac{|\alpha|^2}{2}}|^2$
$=|e^{-|\alpha|^2}|$
Is this correct?

 

[Haorong Wu]

Looks fine to me.

 

[Demystifier]

$e^{-\frac{|\alpha|^2}{2}}\frac{(\alpha)^n}{\sqrt{n!}}|n>$

You forgot the sum over $n$.

 

[deepalakshmi]

You forgot the sum over $n$.

Yeah I forgot

 

[deepalakshmi]

actually the fidelity is always between 0 to 1 right? But I got exponential terms with alpha

 

[Haorong Wu]

actually the fidelity is always between 0 to 1 right? But I got exponential terms with alpha

Think harder. It poses no contradiction.

 

[deepalakshmi]

Think harder. It poses no contradiction.

Ok. I understood. Since there is a exponential power minus, value always lies between 0 to 1. I also took an example and the value always lies between 0 to 1. Thankyou

 

[PeterDonis]

I wrote the number terms in terms of creation operator.

You are still not answering my question. Your formula for the Hamiltonian has two sets of ladder operations, the a's and the b's. But you only have one set of number eigenstates $\ket{n}$, so you should only have one set of ladder operators. Why are there two?

the initial state and hamiltonian is given to me by my guide.

Who is your "guide"?

 

[deepalakshmi]

You are still not answering my question. Your formula for the Hamiltonian has two sets of ladder operations, the a's and the b's. But you only have one set of number eigenstates $\ket{n}$, so you should only have one set of ladder operators. Why are there two?

Who is your "guide"?

Guide - teacher

 

[deepalakshmi]

Are you saying my question is wrong?

 

[deepalakshmi]

For example take beam splitter with input state as coherent and vacuum state. There also you have two sets of operator in Hamiltonian

 

[deepalakshmi]

Another thing is that ket n is called fock state in coherent state

 

[PeterDonis]

Are you saying my question is wrong?

I'm saying I don't understand why there are two sets of ladder operators in your Hamiltonian since you only have one set of number eigenstates. If you don't understand that either, perhaps you should ask your teacher.

take beam splitter with input state as coherent and vacuum state. There also you have two sets of operator in Hamiltonian

What do the two sets of operators operate on?

 

[PeterDonis]

you only have one set of number eigenstates

Note: I've said this several times now, but you should think carefully about whether you agree with it.

 

[deepalakshmi]

Note: I've said this several times now, but you should think carefully about whether you agree with it.

Ok. I will ask my teacher about it.

 

[deepalakshmi]

I'm saying I don't understand why there are two sets of ladder operators in your Hamiltonian since you only have one set of number eigenstates. If you don't understand that either, perhaps you should ask your teacher.What do the two sets of operators operate on?

There those two set of operator will act on two states. One is coherent state which I will write in terms of vacuum state and other is vacuum state. So basically the two operator will act on two vacuum state