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?


1 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.)


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