Let the fidelity between two quantum states be defined as

$$F(\rho, \sigma) = \|\sqrt{\rho}\sqrt{\sigma}\|_1.$$

If $\rho = \vert\psi\rangle\langle\psi\vert$, then $F(\rho, \sigma) = \sqrt{\langle\psi\vert\sigma\vert\psi\rangle}$.

Let us look at a teleportation protocol of a Bell state, where the two classical bits are sent with an error at most $\varepsilon$. That is, the set of messages from $\{00,01,10,11\}$ is received correctly with probability at least $1-\varepsilon$ and incorrectly with probability at most $\varepsilon$.

If the state to be sent was the Bell state $\vert\psi\rangle\langle\psi\vert$, this implies that the teleported state is simply

$$\omega = (1-\varepsilon)\vert\psi\rangle\langle\psi\vert + \varepsilon\vert\psi^\perp\rangle\langle\psi^\perp\vert$$

The fidelity of the received state with respect to the intended state is

$$F(\omega, \vert\psi\rangle\langle\psi\vert) \geq \sqrt{1-\varepsilon}$$

Now, let's look at the reversed situation. Suppose one has a state transmission protocol that works with error at most $\sqrt{1-\varepsilon}$ in fidelity. That is, given a Bell state $\vert\psi\rangle\langle\psi\vert$, the state transmission protocol outputs a state $\omega$ such that

$$F(\omega, \vert\psi\rangle\langle\psi\vert) \geq \sqrt{1-\varepsilon}.$$

If one uses this protocol for superdense coding, what is the worst case error in decoding a message in the set $\{00,01,10,11\}$? I am not sure how to determine the worst case error post-measurement from the fidelity. Using Fuchs van de Graaf, the trace distance error is $\sqrt{\varepsilon}$ and therefore my post-measurement outcome has error $\sqrt{\varepsilon}$. But if teleportation and superdense coding were "reversible", it should be $\varepsilon$ here, shouldn't it?

So what is the error of correctly decoding messages in a superdense coding protocol if the Bell state is transmitted with fidelity of $\sqrt{1-\varepsilon}$? Did I "lose too much" in my analysis above to obtain the looser bound of $\sqrt{\varepsilon}$?


1 Answer 1


Imagine we have some two-qubit state $\omega$ which Bob holds after the transmission of a qubit from Alice. For superdense coding, he's going to measure it in the Bell basis. Let's assume the answer is supposed to be $|\psi\rangle$. Hence, the probability of correct communication is $$ \langle\psi|\omega|\psi\rangle. $$ Of course, this is just $F(\omega,|\psi\rangle\langle\psi|)^2$, so it all matches up. The error probability is $\epsilon$.

That said, I would recommend being a bit more careful with notation. Your $\omega$ and $|\psi\rangle$ in the teleportation and superdense coding parts are very different things - the first is one qubit, the second is two qubit - so you want to make sure that you are expecting the right thing for the right reason (and not through a cross-over of notation)


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