# Quantum teleportation with "noisy" entangled state

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.

• Side question for my own curiosity: what book by Preskill is this from exactly? Apr 17, 2019 at 16:44
• Preskill: Quantum Information, the lecture notes you can find online from him. If you google it you easily find his own website from Caltech where you can find these notes. link Apr 17, 2019 at 20:12
• – glS
Feb 27 at 8:01

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 $$1-\lambda$$).

• Thanks for the elaborate answer! This helps a lot. Yours is indeed a correct solution. I cannot thank you enough. My solution was, I just assumed that we had two subsystems, $\rho_{Bell}$ and the noise part $\rho_{noise}$. We have $\lambda$ chance to "use" the Bell state to teleport, which has fidelity 1 (perfect teleportation) and (1-$\lambda$) chance to use the noise, which is a random guess, which results into 1/2 fidelity. So we have F = $\lambda * 1 + (1-\lambda) * 1/2 = 1 - 1/2 \lambda$ which is the correct result. Is this a correct reasoning? Apr 18, 2019 at 21:51
• Indeed, you can calculate the final teleportation result this way, so $f = (1-\lambda)|r\rangle\langle r| + \frac{1}{2} \lambda I$. But I'm not sure about fidelity summation. I guess it works here because both outcomes commute ($|r\rangle\langle r|$ and $I$). Apr 18, 2019 at 22:10