Quantum teleportation with "noisy" entangled state

Question

This is actually an exercise from Preskill (chapter 4, new version 4.4). So they are asking about the fidelity of teleporting a random pure quantum state from Bob to Alice, who both have one qubit of the following system ("noisy" entangled state):

$$\rho = (1 − \lambda)|\psi ^-\rangle \langle \psi ^- | + \frac{1}{4} \lambda I $$

with $|\psi ^-\rangle$ one of the Bell states. In the notes of Preskill they show you the example of quantum teleportation with a Bell state $|ψ ^-\rangle_{AB}$, by uniting the random qubit with the system (Bell state) and then make Alice do some measurements on the system, from which Bob gets a "copy" (or to be correct: the qubit was teleported to Bob) of the random qubit.

Now in this example, we are presented with the density matrix $\rho$, from which we cannot just get one "state". As I can not follow the example of the book (where they manipulate the state $|\psi ^-\rangle$), but now have to deal with the density matrix, I have no idea where to begin. How can we proceed to calculate the fidelity with which Bob will have the correct teleported state if using the given noisy entangled system $\rho$ to teleport the random qubit?

They also state that a random 'guess' has a 1/2 chance (which refers to the identity part $I$ in the $\rho$ system). I also know the fidelity of a pure Bell state will be 1. But I suppose I can't just say that for the system $\rho$ the fidelity is the sum of the parts?:

$$F = (1 − λ) + \frac{1}{4} \lambda \times \frac{1}{2}$$

And if I can, why is this? How can I explicitly calculate this?

P.S: By the way, if anyone would have solutions to the exercises in the notes of Preskill, a link would be much appreciated.

Answer 1

I'm not sure what was the expected solution, but this also works.

First of all, note that $$ I = |\phi^+\rangle\langle\phi^+| + |\phi^-\rangle\langle\phi^-| + |\psi^+\rangle\langle\psi^+| + |\psi^-\rangle\langle\psi^-|,$$ where $|\phi^+\rangle, |\phi^-\rangle, |\psi^+\rangle, |\psi^-\rangle$ are Bell states and $I$ is 4-dimensional identity operator.

The second observation is that if you apply teleportation scheme (that supposed to correctly work with $|\psi^-\rangle$ entangled state) on the wrong entangled Bell state (for example, Alice and Bob could share $|\phi^+\rangle$ instead of $|\psi^-\rangle$), then you will end up with one of this $$ |r_0\rangle = a|0\rangle + b|1\rangle \\ |r_1\rangle = a|0\rangle - b|1\rangle \\ |r_2\rangle = a|1\rangle + b|0\rangle \\ |r_3\rangle = a|1\rangle - b|0\rangle \\ $$ where $|r_0\rangle$ is the teleported qubit. I'm sure it can be checked that for 4 possible Bell states a fixed teleportation scheme will give exactly 4 different results $|r_0\rangle, |r_1\rangle, |r_2\rangle, |r_3\rangle$.

So, if you will use teleportation scheme (specified to $|\psi^-\rangle$) with the entangled state $$ \rho = (1 − \lambda)|\psi ^-\rangle \langle \psi ^- | + \frac{1}{4} \lambda I = \\ = (1 − \lambda)|\psi ^-\rangle \langle \psi ^- | + \frac{1}{4} \lambda (|\phi^+\rangle\langle\phi^+| + |\phi^-\rangle\langle\phi^-| + |\psi^+\rangle\langle\psi^+| + |\psi^-\rangle\langle\psi^-|) $$ then the result will be $$ (1-\lambda)|r_0\rangle\langle r_0| + \frac{1}{4} \lambda (|r_0\rangle\langle r_0| + |r_1\rangle\langle r_1| + |r_2\rangle\langle r_2| + |r_3\rangle\langle r_3|) $$

But you can check that $$ |r_0\rangle\langle r_0| + |r_1\rangle\langle r_1| + |r_2\rangle\langle r_2| + |r_3\rangle\langle r_3| = 2I $$ for any random qubit $|r_0\rangle$.

So, the final teleported state will be $$ f = (1-\lambda)|r_0\rangle\langle r_0| + \frac{1}{4} \lambda \cdot 2I = (1-\lambda)|r_0\rangle\langle r_0| + \frac{1}{2} \lambda I $$

The fidelity between $r = |r_0\rangle\langle r_0|$ and $f$ is $$ \left(\text{Tr}\sqrt{\sqrt{r}f\sqrt{r}}\right)^2 = \left(\text{Tr}\sqrt{rfr}\right)^2 = \left(\text{Tr}\sqrt{r((1-\lambda)r + \frac{1}{2} \lambda I)} r \right)^2 = \\ = \left(\text{Tr}\sqrt{(1-\lambda)r^3 + \frac{1}{2} \lambda r^2)} \right)^2 = \left(\text{Tr}\sqrt{(1-\lambda)r + \frac{1}{2} \lambda r)} \right)^2 = \\ = \left(\text{Tr}\sqrt{\frac{1}{2}(2-\lambda)r} \right)^2 = \frac{1}{2}(2-\lambda) $$ Note, that the link you shared contain the answer on the page 45 (but there $\lambda$ is actually $1-\lambda$).