How does the teleportation step work in entanglement dilution?

Question

Consider the entanglement dilution protocol for pure states, as described in Preskill's notes (Link to pdf, see around page 32). The context is that Alice and Bob share $k=n(S(\rho_A)+\delta)$ Bell pairs, with $S(\rho_A)$ the von Neumann entropy of $\rho_A\equiv \operatorname{Tr}_2(|\psi\rangle\!\langle\psi|)$, while Alice holds $n$ copies of $|\psi\rangle$. Their goal is to end up, performing only LOCC operations, sharing $n$ copies of $|\psi\rangle$. They thus want to somehow "convert" the entangled pairs they share into another form of entanglement.

As described in the notes, the idea of the protocol is roughly as follows:

1. Notice that asymptotically in $n$, Alice's states have the form $$|\psi\rangle_{AC}^{\otimes n} \sim \sum_{\vec x}\sqrt{p(\vec x)}|\vec x\rangle_{A^n}\otimes |\vec x\rangle_{C^n},$$ where the sum is over typical sequences $\vec x$, that is, roughly speaking, sequences containing a number of 0s and 1s close to their expected value, so that also $p(\vec x)\sim 2^{-n S(\rho_A)}$. Here, $A$ and $C$ are two degrees of freedom held by Alice.

2. Alice teleports the $C^n$ half of (the projection onto the typical subspace of) $|\psi\rangle_{AC}^{\otimes n}$ to Bob, using only the $\sim n H(\rho_A)$ Bell pairs.

3. Bob performs a decoding operation on his register, and the result is them sharing $|\psi\rangle^{\otimes n}$.

I'm not quite sure how the teleportation step works here. In particular, why can Alice manage to teleport the $2^n$ qubits in the $C^n$ register, using only $\sim n H(\rho_A)$ Bell pairs?

Answer 1

Preskill's notes explain this: prior to teleportation, Alice compresses $C^n$ into fewer qubits using Schumacher compression (p. 27, especially eq. 10.144 and surrounding text).
For a finite-copy illustration, consider $\vert\psi\rangle = \sqrt{\frac{3}{4}}\vert00\rangle + \sqrt{\frac{1}{4}}\vert11\rangle$. Then a typical sequence for $\vert\psi\rangle^{\otimes 4}$ has one 1 and three 0s and so to share $\vert\psi\rangle^{\otimes 4}$ Alice only teleports half of the subspace projection $\vert\phi\rangle = \frac{1}{2}(\vert0001\rangle\vert0001\rangle+\vert0010\rangle\vert0010\rangle + \vert0100\rangle\vert0100\rangle + \vert1000\rangle\vert1000\rangle)_{12345678}$.

This can first be compressed via Alice applying a unitary mapping $U_{5678}$ to the last 4 qubits, e.g., where $U_{5678}\vert\phi\rangle = \frac{1}{2}(\vert0001\rangle\vert0000\rangle+\vert0010\rangle\vert0100\rangle + \vert0010\rangle\vert1000\rangle + \vert1000\rangle\vert1100\rangle)_{12345678}$, discarding qubits 7 and 8, and teleporting qubits 5 and 6, using only 2 Bell pairs. Bob can then restore his half of $\vert\phi\rangle$ by introducing 2 ancillas in state $\vert0\rangle$ to the teleported state and then applying $U_{5678}^{-1}$.

(Of course, in this case $\vert\phi\rangle$ is a poor approximation to $\vert\psi\rangle^{\otimes 4}$ but that's due to the low numbers of copies involved.)