# Strange inequivalence between superdense coding and teleportation

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 $$\sqrt{\varepsilon}$$?

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)