PROOF: Quantum Fidelity with pure states

[Alex Dingo]

Hi, I'm currently working on showing the relation of quantum fidelity:

The quantum “fidelity” between two pure states ρ1 = |ψ1⟩⟨ψ1| and ρ2 = |ψ2⟩⟨ψ2| is given by |⟨ψ1|ψ2⟩|^2.
Show that this quantity may be written as Tr(ρ1ρ2).

I've been following the wikipedia page on fidelity but can't understand it.

If I assume the statement is true and start from the Tr(ρ1ρ2) = Tr(|ψ1⟩⟨ψ1|ψ2⟩⟨ψ2|) I'm trying to find how I can reduce this, perhaps with summations to arrive at |⟨ψ1|ψ2⟩|^2 - can anyone suggest which direction to go in?

 

[bhobba]

Try expressing both states in the same orthonormal basis.

Thanks
Bill

 

[Alex Dingo]

Try expressing both states in the same orthonormal basis.

Thanks
Bill

Hi Bill,

If I express Ψ1=∑√(p(x))|x> and Ψ2=∑√(q(x))|x> such that they are both expressed in the basis of x and compute |⟨ψ1|ψ2⟩|^2 I find Is equal to ∑[√p(x)√q(x)]^2 <x|x> such that I am left with Σp(x)q(x) which is a summation of their probabilities over x. Can I then relate the relation that the sum of p(x) and p(x) is 1 to make the link between the final result of Tr(p1p2) or have I made a mistake?
Thanks,
Alex.

 

[DrClaude]

If I express Ψ1=∑√(p(x))|x> and Ψ2=∑√(q(x))|x> such that they are both expressed in the basis of x and compute |⟨ψ1|ψ2⟩|^2 I find Is equal to ∑[√p(x)√q(x)]^2 <x|x> such that I am left with Σp(x)q(x) which is a summation of their probabilities over x.

First, get rid of those square roots, which don't make any sense, since the expansion coefficients are complex numbers:
$$
\psi_1 = \sum_n p_n | n \rangle
$$
with $p_n \in \mathbb{C}$.

After you have done that and calculated $| \langle \psi_1 | \psi_2 \rangle |^2$, do it also for the trace of $\rho_1 \rho_2$ over the same basis.

 

[Alex Dingo]

First, get rid of those square roots, which don't make any sense, since the expansion coefficients are complex numbers:
$$
\psi_1 = \sum_n p_n | n \rangle
$$
with $p_n \in \mathbb{C}$.

After you have done that and calculated $| \langle \psi_1 | \psi_2 \rangle |^2$, do it also for the trace of $\rho_1 \rho_2$ over the same basis.

Hi DrCloud,

So the result for calculating $| \langle \psi_1 | \psi_2 \rangle |^2$ I found it equat to $\Sigma p^2_n q^2_n$. As for the calculation with the trace. I find
$Tr(\rho_1 \rho_2) = Tr(\langle n | \Sigma p^2_n |n\rangle \langle n | \Sigma q^2_n |n\rangle$

From this point I can take the summations outside the trace and which leave $\Sigma p^2_n q^2_n Tr( \langle n|n \rangle \langle n|n \rangle)$
Although this final result with a number inside the trace I'm not sure I've done correctly.

Thanks,
Alex.

 

[DrClaude]

From this point I can take the summations outside the trace and which leave $\Sigma p^2_n q^2_n Tr( \langle n|n \rangle \langle n|n \rangle)$
Although this final result with a number inside the trace I'm not sure I've done correctly.

You seem to be holding on too long to the trace operation. Consider
$$
\mathrm{Tr}(\hat{A}) = \sum_n \langle n | \hat{A} | n \rangle
$$

 

[Alex Dingo]

You seem to be holding on too long to the trace operation. Consider
$$
\mathrm{Tr}(\hat{A}) = \sum_n \langle n | \hat{A} | n \rangle
$$

Hi,
Okay this makes more sense, I forgot about this relationship.

I have computed this with $\hat{A} = p1p2$
I find that $\mathrm{Tr}(p1p2)=\Sigma\langle n | p_n|n\rangle \langle n | p_n q_n |n \rangle \langle n| q_n | n \rangle$
which reduces to $\Sigma p_n^2 q_n^2$ as desired ?

 

[DrClaude]

I find that $\mathrm{Tr}(p1p2)=\Sigma\langle n | p_n|n\rangle \langle n | p_n q_n |n \rangle \langle n| q_n | n \rangle$

I'm not sure how you got there (that is, I can't vouch that you did everything right), but the final result is correct.