Fidelity between a pure state and an arbitrary state

[Emil_M]

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Homework Statement

Show that the Fidelity between one pure state $|\Psi\rangle$ and one arbitrary state $\rho$ is given by $F(|\Psi\rangle , \rho)=\sqrt{\langle\Psi|\rho|\Psi\rangle}$ .

Homework Equations

The quantum mechanical fidelity is defined as $\begin{equation*} F(\rho,\sigma):= Tr{ \sqrt{\rho^{\frac{1}{2}} \sigma \rho^{\frac{1}{2}}}} \end{equation*}$

The Attempt at a Solution

The operator corresponding to $|\Psi\rangle$ is $\sigma=|\Psi\rangle\langle\Psi|$.
We can make use of the spectral decomposition of $\sigma$ in order to write
$\begin{equation*} \sigma=\sum_{j=1}^N s_j |j\rangle\langle j| \end{equation*}$
It follows, that the Fidelity $F(\sigma, \rho)$ is given by
\begin{align*}
F(\sigma, \rho) & =Tr{\sqrt{\left(|\Psi\rangle\langle\Psi|\right)^{\frac{1}{2}}\rho\left(|\Psi\rangle\langle\Psi|\right)^{\frac{1}{2}}}} \\
&= Tr{\sqrt{|\Psi\rangle\langle\Psi|\rho|\Psi\rangle\langle\Psi|}}\\
&= Tr{\sqrt{\langle\Psi|\rho|\Psi\rangle\;|\Psi\rangle\langle\Psi|}}\\
&= Tr{\sqrt{\langle\Psi|\rho|\Psi\rangle\sum_{j=1}^N s_j |j\rangle\langle j|}}\\
&=\sqrt{\langle\Psi|\rho|\Psi\rangle} \; Tr \sum_{j=1}^N \sqrt{s_j} |j\rangle\langle j|\\
&=\sqrt{\langle\Psi|\rho|\Psi\rangle} \; \sum_{k,j=1}^N \sqrt{s_j} \langle k|j\rangle\langle j|k\rangle\\
&=\sqrt{\langle\Psi|\rho|\Psi\rangle} \; \sum_{j=1}^N \sqrt{s_j}
\end{align*}
This is not the expected result, so I have made a mistake somewhere along the way. I would be really thankful if somebody were to spot it!

Thanks in advance

 

[Fightfish]

Well, the easiest way to go about doing it is to realize that
$$\sqrt{\langle \psi | \rho | \psi\rangle |\psi\rangle \langle \psi| } = \sqrt{\langle \psi | \rho | \psi\rangle} \sqrt{ |\psi\rangle \langle \psi| } = \sqrt{\langle \psi | \rho | \psi\rangle} |\psi\rangle \langle \psi|$$
and so the trace becomes trivial.

Your method is confusing, but there isn't anything wrong with it. When you perform a spectral decomposition of a projector into its diagonal basis, well its just itself isn't it? So, all the coefficients are zero except for one which is unity. So, the summation in your final expression is indeed 1.

 

[Emil_M]

Thanks!

 

[Emil_M]

Well, the easiest way to go about doing it is to realize that
$$\sqrt{\langle \psi | \rho | \psi\rangle |\psi\rangle \langle \psi| } = \sqrt{\langle \psi | \rho | \psi\rangle} \sqrt{ |\psi\rangle \langle \psi| } = \sqrt{\langle \psi | \rho | \psi\rangle} |\psi\rangle \langle \psi|$$
and so the trace becomes trivial.

Your method is confusing, but there isn't anything wrong with it. When you perform a spectral decomposition of a projector into its diagonal basis, well its just itself isn't it? So, all the coefficients are zero except for one which is unity. So, the summation in your final expression is indeed 1.

I still don't get it. Why is it a given that $|\Psi\rangle \langle\Psi|$ is a projector?

Thanks!

 

[Fightfish]

Well, any operator of the form $\hat{P}_{\alpha} = |\alpha \rangle \langle \alpha |$, when acted upon any state ket $|\beta \rangle$, yields the component of $|\beta \rangle$ along the "direction" of $|\alpha \rangle$. So, the operator $\hat{P}_{\alpha}$ is said to project $|\beta \rangle$ into the "direction" of $|\alpha\rangle$, and that's why it is known as a projector.

 

[Emil_M]

Yes, but a projector has to respect the condition $P^2=P$. And the density matrix corresponding to a general qubit, for example, $|\Psi\rangle = a|0\rangle+b|1\rangle$ is thus not a projector, is it?

 

[Fightfish]

Yes, but a projector has to respect the condition $P^2=P$. And the density matrix corresponding to a general qubit, for example, $|\Psi\rangle = a|0\rangle+b|1\rangle$ is thus not a projector, is it?

$|\Psi\rangle \langle \Psi | \Psi \rangle \langle \Psi | = ?$

 

[Emil_M]

right :)
Thanks for taking the time, don't know why I was so confused.