Fidelity between initial and final states

[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

 

[PeterDonis]

There those two set of operator will act on two states.

What do those two states represent, physically--for example, in the beam splitter scenario?

 

[deepalakshmi]

What do those two states represent, physically--for example, in the beam splitter scenario?

They represent vacuum state .i.e zero fock state where there is no photons

 

[PeterDonis]

They represent vacuum state .i.e zero fock state where there is no photons

First, only one of the two kets in your initial state is the vacuum state, the other is the coherent state.

Second, you are missing the point. You have an initial state $\ket{0} \ket{\alpha}$. This is a product of two kets. That means you have two physical degrees of freedom in the quantum system that this state describes. What are those two physical degrees of freedom, in, for example, the beam splitter scenario?

You then time evolve that initial state using a Hamiltonian to get a final state. That final state will also describe a quantum system with two physical degrees of freedom (but in general it won't be a product state, it will be entangled). What are those two physical degrees of freedom in the final state, in, for example, the beam splitter scenario? (Hint: they're not the same as the two physical degrees of freedom in the initial state, because of the beam splitter.)

 

[deepalakshmi]

First, only one of the two kets in your initial state is the vacuum state, the other is the coherent state.

Second, you are missing the point. You have an initial state $\ket{0} \ket{\alpha}$. This is a product of two kets. That means you have two physical degrees of freedom in the quantum system that this state describes. What are those two physical degrees of freedom, in, for example, the beam splitter scenario?

You then time evolve that initial state using a Hamiltonian to get a final state. That final state will also describe a quantum system with two physical degrees of freedom (but in general it won't be a product state, it will be entangled). What are those two physical degrees of freedom in the final state, in, for example, the beam splitter scenario? (Hint: they're not the same as the two physical degrees of freedom in the initial state, because of the beam splitter.)

I can write coherent state in terms of vacuum state. So I have two vacuum state. Secondly for this initial state, the final state is not entangled. Lastly I don't know anything about degrees of freedom.

 

[deepalakshmi]

I can write coherent state in terms of vacuum state. So I have two vacuum state. Secondly for this initial state, the final state is not entangled. Lastly I don't know anything about degrees of freedom.

Thanks for mentioning degrees of freedom in Quantum states. I will learn about it.

 

[PeterDonis]

I can write coherent state in terms of vacuum state.

No, you can't. Only one term in the infinite sum of number eigenstates in the formula for a coherent state is the $n = 0$ term, i.e., the vacuum state. All the other terms have $n > 0$ so they contain some nonzero number of photons.

I don't know anything about degrees of freedom.

Degrees of freedom are what kets refer to. If you have a product of two kets, you have two degrees of freedom. Physically, kets generally describe physical systems that can interact with other physical systems.

 

[deepalakshmi]

No, you can't. Only one term in the infinite sum of number eigenstates in the formula for a coherent state is the $n = 0$ term, i.e., the vacuum state. All the other terms have $n > 0$ so they contain some nonzero number of photons.

 

[deepalakshmi]

I can write coherent state as (displacement operator) ( vacuum state). Let us stop this discussion for now. I will reply later if you ask any question

 

[PeterDonis]

I can write coherent state as (displacement operator) ( vacuum state).

Ok, so this is what you meant by writing a coherent state "in terms of" the vacuum state. I can similarly write any state "in terms of" any other state by just finding the appropriate operator to apply. So this "in terms of" doesn't seem very useful to me.

I will reply later if you ask any question

You still have not replied to the most important question I have asked, several times now: what do the two kets in your initial state $\ket{0} \ket{\alpha}$ refer to in the beam splitter scenario? To put it as simply as possible: this state says you have one "thing" that is in the vacuum state, and another "thing" that is in the coherent state $\ket{\alpha}$. What are these two "things" in the beam splitter scenario?

 

[Haorong Wu]

Hi, @PeterDonis . If I understand OP correctly, then the annihilation operator $a$ is just the same as $b$, only that $a$ operates in the space of the first photon, while $b$ lives in that of the second photon. They just pick different names to distinguish from each other.

I am not sure whether the beam splitter scenario is suitable or not. Normally, I would set a scheme that a pump laser is transformed into two entangled beams via a nonlinear crystal, and then those two beams will evolve differently where operators $a$ and $b$ come in.