# Eavesdropping in superdense coding: why can't the third party infer anything about the message?

There's an exercise in Nielsen & Chuang asking us to show that if some third party intercepts Alice's qubit she's sending to Bob while attempting to transmit (2 bits of) classical information, then this third party can't actually infer anything about the information she's trying to communicate. Effectively, this comes down to showing that $$\langle\psi|E_A \otimes I_B|\psi\rangle$$ is the same whenever $$\psi$$ is any of the Bell states and $$E$$ is a positive operator.

This is simple to show, but I wonder if this is an incomplete concept, since it supposedly requires $$E$$ to be a positive operator. If we just check one of them:

$$\langle\Phi^+|E \otimes I|\Phi^+\rangle = \langle 00 + 11|E \otimes I|00 + 11 \rangle = \langle 0|E|0 \rangle\langle 0|0 \rangle + \langle 0|E|1 \rangle\langle 0|1 \rangle + \langle 1|E|0 \rangle\langle 1|0 \rangle + \langle 1|E|1 \rangle\langle 1|1 \rangle = \langle 0|E|0 \rangle + \langle 1|E|1 \rangle.$$

This is the same for each Bell state. Couldn't $$E$$ just be Hermitian and we would obtain the same result?

Mathematically, this has nothing to do with the positivity of $$E$$. It doesn't really have anything to do with $$E$$ at all - it's a property of the Bell states themselves (you've probably not got there yet, but they have the same reduced density matrices). I presume the reason for specifying the positivity of $$E$$ in the question is to help you make the connection with the allowed operators that an eavesdropper could perform, such as measurements.

Incidentally, don't forget about normalisation factors. There's probably a factor of 1/2 missing from your final answer.

This happens for any maximally entangled state $$|\Psi\rangle$$ and operator $$E$$.

Indeed, a maximally entangled state is, by definition, one whose partial trace is the maximally mixed one. Writing $$|\Psi\rangle\equiv\sum_{ij}\psi_{ij}|ij\rangle$$, this means that $$\sum_i \psi_{ij}\bar\psi_{kj}=\delta_{ik}/D$$, with $$D$$ the dimension of each space ($$D=2$$ in your case). Thus you have (assume repeated indices are summed): $$\langle\Psi|E\otimes I|\Psi\rangle=\bar \psi_{ij} E_{ik} \psi_{kj}=\frac{1}{D}\delta_{ik}E_{ik}=\frac{1}{D}\mathrm{Tr}(E).$$