# 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? – ahelwer Apr 17 '19 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 – CFRedDemon Apr 17 '19 at 20:12

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? – CFRedDemon Apr 18 '19 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 $I$). – Danylo Y Apr 18 '19 at 22:10