A Fidelity for quantum state at t=0

[deepalakshmi]

TL;DR Summary

Fidelity for same pure quantum state

fidelity for pure state with respect to t=0 is 1. My teacher told me this.
But I am not getting this.
This is my detailed question
the initial state(t=0)$|\psi\rangle=|\alpha\rangle|0\rangle$
the final state (t) $|\chi\rangle= |i\alpha\sin(t)\rangle|cos(t)\alpha\rangle$
Fidelity between the states I got is $e^{-|\alpha|^2}e^{-|\alpha\sin(t)|^2}e^{-|\alpha\cos(t)|^2}$
when I tried to find fidelity for t=0, I got answer as 0.00004. But my teacher said that fidelity should be 1 here. I don't know how?

 

[DrClaude]

Fidelity is the "distance" between two quantum states, so the statement

fidelity for pure state at t=0 is 1

is meaningless by itself. My guess is that fidelity is being measured here with respect to the initial (t=0) state.

 

[deepalakshmi]

Fidelity is the "distance" between two quantum states, so the statement

is meaningless by itself. My guess is that fidelity is being measured here with respect to the initial (t=0) state.

I have edited the question

 

[Demystifier]

Fidelity between the states I got is $e^{-|\alpha|^2}e^{-|\alpha\sin(t)|^2}e^{-|\alpha\cos(t)|^2}$

If the third factor was $e^{+|\alpha\cos(t)|^2}$, then you would get fidelity 1 at $t=0$. So check your signs!

 

[deepalakshmi]

No, the last factor is -(minus)
Here I have attached my calculation for your reference
$F=|\rangle\psi|\chi\rangle|^{2}$
the initial state(t=0)$|\psi\rangle=|\alpha\rangle|0\rangle$

the final state (t) $|\chi\rangle= |i\alpha\sin(t)\rangle|cos(t)\alpha\rangle$
$F=|\langle|i\alpha\sin(t)\rangle\langle0|\alpha\cos(t)\rangle|^{2}$
On simplifying$\langle\alpha|i\alpha\sin(t)\rangle$, I will get $e^{{-|\alpha|^2}/2}e^{{-|\alpha\sin(t)|^2}/2}e^{-|\alpha|^2sin(t)}$
Similarly on simplifying $\langle0|\alpha\cos(t)\rangle$, I will get $e^{{-|\alpha\cos(t)|^2}/2}$
so finally my fidelity will be $e^{-|\alpha|^2}e^{-|\alpha\sin(t)|^2}e^{-|\alpha\cos(t)|^2}$

 

[Demystifier]

Here I have attached my calculation

That's not the calculation, that's just your results, so I cannot tell you where exactly is your error.

 

[deepalakshmi]

That's not the calculation, that's just your results, so I cannot tell you where exactly is your error.

Ok I will attach my detailed calculation on cos term alone

 

[Haborix]

I'm only doing math in my head, but are you sure you have the order of the kets correct in the final state? I would think you'd be able to just take $t=0$ in $\chi$ and recover the initial state.

 

[deepalakshmi]

even if I change the final state order, I am still getting the same fidelity

 

[Haborix]

In that case, I'm with @Demystifier. We will need to see your calculations to know what's going wrong. Because if you switch the order of the kets and take $t=0$, then you get exactly the initial state.

 

[deepalakshmi]

My detailed calculation
$F=|\langle\psi|\chi\rangle|^{2}$
the initial state(t=0)$|\psi\rangle=|\alpha\rangle|0\rangle$

the final state (t) $|\chi\rangle= |i\alpha\sin(t)\rangle|\cos(t)\alpha\rangle$
$F=|\langle|i\alpha\sin(t)\rangle\langle0|\alpha\cos(t)\rangle|^{2}$
On simplifying$\langle\alpha|i\alpha\sin(t)\rangle$
$\langle\alpha|i\alpha\sin(t)\rangle$

$$=e^{{-|\alpha|^2}/2}\sum_n{ \frac{(\alpha*)^n}{\sqrt{n!}} \langle n| e^{{-|\alpha \sin(t)|^2}/2}\sum_m{ \frac{(\alpha i \sin(t))^m}{\sqrt{m!}}| m\rangle$$

$= (e^{{-|\alpha|^2}/2}e^{{-|\alpha \sin(t)|^2}/2}\sum_n\sum_m{ \frac{(\alpha*)^n}{(\alpha i \sin(t))^m}{\sqrt{n!m!}} \langle n|m\rangle$

$$=(e^{{-|\alpha|^2}/2}e^{{-|\alpha \sin(t)|^2}/2}\sum_n{ \frac{(\alpha*\alpha i \sin(t))^n}{\sqrt{n!}}$$

$e^{{-|\alpha|^2}/2}e^{{-|\alpha \sin(t)|^2}/2}e^{{i|\alpha|^2}sin(t)}$

on simplifying $\langle0|\alpha\cos(t)\rangle$

$\langle0|\alpha\cos(t)\rangle$
=$\langle0|(e^{{-|\alpha|^2}/2}\sum_n{ \frac{(\alpha\cos(t))^n}{\sqrt{n!}} \langle n|$

=(e^{{-|\alpha|^2}/2}\sum_n{ \frac{(\alpha\cos(t))^n}{\sqrt{n!}}\langle 0|n\rangle

= e^{{-|\alpha\cos(t)|^2}/2}

putting both the simplified equation on F

$$F=|(e^{{-|\alpha|^2}/2})e^{{-|\alpha \sin(t)|^2}/2}e^{{i|\alpha|^2}sin(t)}e^{{-|\alpha\cos(t)|^2}/2}|^{2}$$

$$=e^{-|\alpha|^2}e^{-|\alpha \sin(t)|^2}e^{{i|\alpha|^2}sin(t)}e^{-{i|\alpha|^2}sin(t)}e^{-|\alpha\cos(t)|^2}$$

So finally my fidelity will be $$e^{-|\alpha|^2}e^{-|\alpha\sin(t)|^2}e^{-|\alpha\cos(t)|^2}$$

 

[deepalakshmi]

My detailed calculation
$F=|\rangle\psi|\chi\rangle|^{2}$
the initial state(t=0)$|\psi\rangle=|\alpha\rangle|0\rangle$

the final state (t) $|\chi\rangle= |i\alpha\sin(t)\rangle|\cos(t)\alpha\rangle$
F=$|\langle|i\alpha\sin(t)\rangle\langle0|\alpha\cos(t)\rangle|^{2}$
On simplifying$\langle\alpha|i\alpha\sin(t)\rangle$
$\langle\alpha|i\alpha\sin(t)\rangle$=$(e^{{-|\alpha|^2}/2}\sum_n{ \frac{(\alpha*)^n}{\sqrt{n!}} \langle n|)(e^{{-|\alpha \sin(t)|^2}/2}\sum_m{ \frac{(\alpha i \sin(t))^m}{\sqrt{m!}}| m\rangle$
$=(e^{{-|\alpha|^2}/2}e^{{-|\alpha \sin(t)|^2}/2}\sum_n\sum_m{ \frac{(\alpha*)^n}{(\alpha i \sin(t))^m}{\sqrt{n!m!}} \langle n|m\rangle$
= $(e^{{-|\alpha|^2}/2}e^{{-|\alpha \sin(t)|^2}/2}\sum_n{ \frac{(\alpha*\alpha i \sin(t))^n}{\sqrt{n!}}$
=$(e^{{-|\alpha|^2}/2}e^{{-|\alpha \sin(t)|^2}/2}e^{{i|\alpha|^2}sin(t)}$
on simplifying $\langle0|\alpha\cos(t)\rangle$
$\langle0|\alpha\cos(t)\rangle$=$\langle0|(e^{{-|\alpha|^2}/2}\sum_n{ \frac{(\alpha\cos(t))^n}{\sqrt{n!}} \langle n|$
=$(e^{{-|\alpha|^2}/2}\sum_n{ \frac{(\alpha\cos(t))^n}{\sqrt{n!}}\langle 0|n\rangle$
= $e^{{-|\alpha\cos(t)|^2}/2}$
putting both the simplified equation on F
$F=|(e^{{-|\alpha|^2}/2})e^{{-|\alpha \sin(t)|^2}/2}e^{{i|\alpha|^2}sin(t)}e^{{-|\alpha\cos(t)|^2}/2}|^{2}$
$=e^{-|\alpha|^2}e^{-|\alpha \sin(t)|^2}e^{{i|\alpha|^2}sin(t)}e^{-{i|\alpha|^2}sin(t)}e^{-|\alpha\cos(t)|^2}$
$=e^{-|\alpha|^2}e^{-|\alpha\sin(t)|^2}e^{-|\alpha\cos(t)|^2}$

Can someone edit this? I need help. I am trying to edit, but i can't

 

[deepalakshmi]

Pls help with editing. It is taking more than 1 hour for me. Still I can't properly edit it

 

[deepalakshmi]

@Demystifier @Haborix I have sent my full calculation. I am sure there is no mistake in it. But why am I not getting 1 when I substitute with t=0?

 

[deepalakshmi]

I'm only doing math in my head, but are you sure you have the order of the kets correct in the final state? I would think you'd be able to just take $t=0$ in $\chi$ and recover the initial state.

As you have said, I changed the order of ket, I got the following fidelity
$$=e^{-|\alpha|^2}e^{-|\alpha\cos(t)|^2}|e^{|\alpha|^2cos(t)}|^2 e^{-|\alpha\sin(t)|^2}$$
$$=e^{-|\alpha|^2} e^{-|\alpha\cos(t)|^2}e^{|\alpha|^4cos^2(t)}e^{-|\alpha\sin(t)|^2}$$
$$=e^{-|\alpha|^2} e^{-(|\alpha|^2cos^2(t))+|\alpha|^4cos^2(t)} e^{-|\alpha\sin(t)|^2}$$
$$=e^{-|\alpha|^2} e^{|\alpha|^2cos^2(t)}e^{-|\alpha\sin(t)|^2}$$
Now if I substitute t=0, I am getting F=1.
Is my substitution correct?(mainly 4th equation)

 

[Demystifier]

I am sure there is no mistake in it.

If you make errors in editing that you don't see, how can you be sure that you don't make errors in computation that you don't see?

 

[deepalakshmi]

If you make errors in editing that you don't see, how can you be sure that you don't make errors in computation that you don't see?

Yes. I am wrong

 

[Demystifier]

Can someone edit this? I need help. I am trying to edit, but i can't

I could do that but I will not, for didactic reasons. You have to train your mind to spot all the tiny errors that you make, both in editing and in calculations.

 

[deepalakshmi]

I could do that but I will not, for didactic reasons. You have to train your mind to spot all the tiny errors that you make, both in editing and in calculations.

Ok. Thanks.

 

[Haborix]

As you have said, I changed the order of ket, I got the following fidelity
$$=e^{-|\alpha|^2}e^{-|\alpha\cos(t)|^2}|e^{|\alpha|^2cos(t)}|^2 e^{-|\alpha\sin(t)|^2}$$
$$=e^{-|\alpha|^2} e^{-|\alpha\cos(t)|^2}e^{|\alpha|^4cos^2(t)}e^{-|\alpha\sin(t)|^2}$$
$$=e^{-|\alpha|^2} e^{-(|\alpha|^2cos^2(t))+|\alpha|^4cos^2(t)} e^{-|\alpha\sin(t)|^2}$$
$$=e^{-|\alpha|^2} e^{|\alpha|^2cos^2(t)}e^{-|\alpha\sin(t)|^2}$$
Now if I substitute t=0, I am getting F=1.
Is my substitution correct?(mainly 4th equation)

Your substitution of $t=0$ looks fine, but you make an algebra error moving from the first to second line. Maybe you corrected the mistake later.

 

[deepalakshmi]

Your substitution of $t=0$ looks fine, but you make an algebra error moving from the first to second line. Maybe you corrected the mistake later.

Can you point that algebra error? And what is $|e^{|\alpha|^2cos(t)}|^2$?

 

[deepalakshmi]

Your substitution of $t=0$ looks fine, but you make an algebra error moving from the first to second line. Maybe you corrected the mistake later.

$$F=e^{-|\alpha|^2}e^{-|\alpha\cos(t)|^2}|e^{|\alpha|^2cos(t)}|^2 e^{-|\alpha\sin(t)|^2}$$
$$=e^{-|\alpha|^2} e^{-|\alpha\cos(t)|^2}e^{|\alpha|^2cos(t)}e^{|\alpha|^2cos(t)}e^{-|\alpha\sin(t)|^2}$$
$$ =e^{-|\alpha|^2} e^{-|\alpha|^2cos^2(t)}e^{2|\alpha|^2cos(t)} e^{-|\alpha\sin(t)|^2}$$Is it correct now?