# Quantum teleportation and the density matrix

1. May 18, 2013

### VantagePoint72

I'm re-reading some course notes on quantum teleportation, and something isn't making sense. In the description my instructor gave, we used the Bell state $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ for the entangled pair. So, suppose the state we want to teleport is $|\phi\rangle = \alpha|0\rangle + \beta|1\rangle$ and so the tripartite state is:
$|\psi\rangle = |\phi\rangle |\Phi^+\rangle = \frac{1}{\sqrt{2}}(\alpha|0\rangle + \beta|1\rangle)(|00\rangle + |11\rangle)$
Alice holds the first two qubits and Bob holds the third. This overall state can be rewritten as:
$|\psi\rangle = \frac{1}{2}\left[ |\Phi^+\rangle(\alpha|0\rangle + \beta|1\rangle) + |\Phi^-\rangle(\alpha|0\rangle - \beta|1\rangle) + |\Psi^+\rangle(\alpha|1\rangle + \beta|0\rangle) + |\Psi^-\rangle(\alpha|1\rangle - \beta|0\rangle) \right]$
(with the qubits in the same order). Hence, Alice determines the phase and parity bit of her Bell pair and communicates this to Bob, who uses the info to do a suitable unitary transformation of his qubit so that he winds up with $|\phi\rangle$.

What confuses me is the next line in my notes, which say that the density matrix for Alice's pair of qubits is:
$\rho_A = \frac{1}{4}\left( |\Phi^+\rangle \langle \Phi^+ | + |\Phi^-\rangle \langle \Phi^- | + |\Psi^+\rangle \langle \Psi^+ | + |\Psi^-\rangle \langle \Psi^- | \right)$
While this makes intuitive sense—afterall, Alice obtains each of the Bell states with probability $1/4$—it doesn't seem to work out computationally. In computing the overall density matrix for $|\psi\rangle$, there are a number of cross terms between the Bell states. Not all of these terms vanish when Bob's qubit is traced out in the computational Basis. For example, we'll be left with a term $(|\alpha|^2 - |\beta|^2)|\Phi^+\rangle \langle \Phi^- |$, which doesn't vanish for arbitrary $\alpha$ and $\beta$. The four pure states for Bob's qubit that appear in the second expression for $|\psi\rangle$ (i.e. $(\alpha|0\rangle + \beta|1\rangle$, $\alpha|0\rangle - \beta|1\rangle$ and so on) are not linearly independent (obviously—there's four of them), so they are not suitable as the basis for Bob's qubit over which the partial trace is taken.

So are my notes wrong in their expression for Alice's density matrix, or is a mistake elsewhere?

2. May 18, 2013

### VantagePoint72

Never mind. Compared with a classmate's notes; turns out this was Alice's reduced density matrix after measuring her two qubits (but before she's looked at the results of the measurement). That is, it's a genuine ignorance-based proper mixture.