A Fidelity between initial and final states

[deepalakshmi]

TL;DR

Evolution of state |0>|alpha>

I have a state |0>|alpha>. Now I want to evolve this state at any time t and find the fidelity between the initial and final states. Any ideas how to do that? My main problem is that I don't know how to evolve this state.

 

[Haorong Wu]

Do you have the Hamiltonian of the system? If so, you can get the final state by solving Schrodinger's equation.

 

[deepalakshmi]

I have evolved the state and the final state is $|\alpha isin\theta>$ |\alpha cos\theta>$. Now how to find the fidelity between initial and final state?

 

[f95toli]

I am not sure what you mean by fidelity here.
Typically "fidelity" means approximately "how likely something is to work"
That is, the fidelity for an operation (aka gate) that takes you from |0> to |1> would be 1 if you always end up in the |1> state.
This means that calculating the fidelity only really makes sense if there is some "randomness" if in your calculation, either because you are averaging over many runs with slightly different parameters; or because you have included noise and/or decoherence in your calculations (in the simplest case using something like a Lindblad approach)

 

[deepalakshmi]

Can I omit sin and cos terms and consider only alpha terms? Can I do like that?

 

[deepalakshmi]

Another doubt is that is my initial and final state pure state?

 

[Haorong Wu]

Another doubt is that is my initial and final state pure state?

Since they are expressed in some certain states, they are both pure.

I think purity, fidelity as you need, coherence, and entanglement can be studied via density matrices.

Now that you mentioned pure states, I strongly suggest that you should learn density matrices or density operators.

 

[Haorong Wu]

Can I omit sin and cos terms and consider only alpha terms? Can I do like that?

Your states seem to be a bipartite system. Now you want to study the first particle. This can be done via, again, density matrices. Just construct them for your states, and then partial trace out the second particle, and you will get what you want. Generally, the matrices you got will represent mixed states.

 

[deepalakshmi]

Here I have two state in initial state. How to find density operator for biparitite system?

 

[deepalakshmi]

Your states seem to be a bipartite system. Now you want to study the first particle. This can be done via, again, density matrices. Just construct them for your states, and then partial trace out the second particle, and you will get what you want. Generally, the matrices you got will represent mixed states.

I can't understand. Can you explain it briefly?

 

[Haorong Wu]

No, you only have one state for the initial state, i.e., $\left | \psi \right >=\left | 0 \right >\left | \alpha \right >$ as a whole.

You can easily construct its corresponding density matrix by $\rho=\left | \psi \right > \left <\psi \right |$ since it is a pure state.

I can't understand. Can you explain it briefly?

I refer you to Nielsen M A, Chuang I. Quantum computation and quantum information[J]. 2002. section 2.4 and Sakurai J J, Commins E D. Modern quantum mechanics, revised edition[J]. 1995. section 3.4.

I think they are sufficient for you.

 

[deepalakshmi]

Then what about final state?

 

[Haorong Wu]

Then what about final state?

Still $\rho=\left | \psi \right >\left<\psi \right |$ because you gave a pure state.

 

[deepalakshmi]

Since here both the states are pure should I use F=|<\psi| \fi >|^2 ?

 

[Haorong Wu]

Since here both the states are pure should I use F=|<\psi| \fi >|^2 ?

I am not very familiar with this concept. But according to Wikipedia, it seems correct.

https://en.wikipedia.org/wiki/Fidelity_of_quantum_states

 

[deepalakshmi]

Now according to my question psi is the initial state and fi is the final state. So I have to find the inner product of $|\alpha>|o> with |\alpha>|\alpha>$. How to find that?

 

[Haorong Wu]

I am not sure what are |\alpha>|o>and |\alpha>|\alpha>.

But when calculating the inner product, always match states for the same particle together.

For example, the inner product for $\left | 0 \right >_1 \left | 1 \right >_2$ and $\left | 2 \right >_1 \left | 3 \right >_2$ is given by $$\left < 0 \right |_1 \left < 1 \right |_2 \left | 2 \right >_1 \left | 3 \right >_2= \left < 0 \right . \left | 2 \right >_1 \left < 1 \right . \left | 3 \right >_2 .$$

 

[deepalakshmi]

Like you said i tried to find the inner product of |α⟩|0⟩ and |α⟩|α⟩
This is my calculation
<α|<0| |α⟩|α⟩= | <α|α⟩ <0|α⟩ |^2 where <α|α⟩=1. But what is <0|α⟩? Is my calculation correct?

 

[PeterDonis]

I have a state |0>|alpha>.

What is $\ket{\alpha}$? Without that information we know nothing at all useful about whatever problem it is you are trying to solve.

 

[PeterDonis]

what is <0|α⟩?

How is anyone supposed to know that if we don't know what $\ket{\alpha}$ is?

 

[deepalakshmi]

How is anyone supposed to know that if we don't know what $\ket{\alpha}$ is?

$\ket{\alpha}$ is coherent state

 

[PeterDonis]

$\ket{\alpha}$ is coherent state

That narrows it down, but there is more than one possible "coherent state". Can you be more specific?

 

[deepalakshmi]

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

 

[PeterDonis]

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

You have just answered the question of what $\braket{0 \vert \alpha}$ is; just plug $n = 0$ into the above formula.

 

[deepalakshmi]

You have just answered the question of what $\braket{0 \vert \alpha}$ is; just plug $n = 0$ into the above formula.

can't understand. How can I just replace n with 0?

 

[PeterDonis]

I have evolved the state and the final state is $|\alpha isin\theta>$ |\alpha cos\theta>$.

As you write it, this state doesn't make sense; these don't look like meaningful kets. What Hamiltonian are you using?

 

[deepalakshmi]

$H=(\hat{a}^\dagger \hat{b}+\hat{b}^\dagger \hat{a})$

 

[PeterDonis]

can't understand. How can I just replace n with 0?

You have a formula for $\ket{\alpha}$ in terms of the number operator eigenstates $\ket{n}$. The number operator eigenstate for $n = 0$ is $\ket{0}$. Since number operator eigenstates are orthogonal, when you evaluate $\braket{0 \vert \alpha}$, the only term in $\alpha$ that matters is the $\ket{0}$ term, whose coefficient is given by your formula with $n = 0$.

 

[PeterDonis]

$H=(\hat{a}^\dagger \hat{b}+\hat{b}^\dagger \hat{a})$

The $\hat{a}$ operator and its counterpart make sense here (as the ladder operators for the number operator eigenstates $\ket{n}$, but what do the $\hat{b}$ operators act on?

 

[PeterDonis]

@deepalakshmi it might help if you would give more context about the problem you are trying to solve. Where did this initial state and the Hamiltonian come from? What physical situation are you modeling?

 

[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.