Can quantum error correcting codes compensate for post-selected quantum teleportation?

The process for quantum teleportation ends with two fixup operations, applied by Bob, conditional on the respective results of two measurements done by Alice.

Sometimes, when you read about experimental demonstrations of quantum teleportation, you find that the experimenters just skipped the fixup operations. Instead, they throw out the 3/4 of ruined runs where fixup operations were needed and focus on the remaining 1/4. Call this 'post-selected quantum teleportation'. I usually find post-selected quantum teleportation protocols a bit silly, since they're basically useless for most communication tasks. You only see effects after the fact, when cherry-picking. In the moment all Bobs get is useless noise.

But that's a bit unfair. At least, unlike with noise, Alice can tell if Bob ends up with the right state or not. And you can do useful things with just "it worked" information, like gradually building up entanglement. (Another example: it's how one of the 'loophole-free' Bell test experiments was done.)

Which raises the question: suppose Alice and Bob are doing quantum teleportation, but are only told which runs were ruined and which worked. Alice doesn't get to see both fixup-determining measurement results $x$ and $y$, she's only given $z = x \lor y$. Can they still communicate?

Using heralded 25%-success-rate teleportation, can Alice send quantum information to Bob?

• Can one rephrase your question as follows: Given a channel which with 25% probability works perfectly and otherwise erases the input and outputs an "erase" flag (i.e., and erasure channel) and no classical side channel, can this be used to transport quantum information (i.e., an unknown quantum state) from Alice to Bob? – Norbert Schuch Sep 30 '16 at 21:15
• @NorbertSchuch There's one minor difference w.r.t. that situation: the "erasing" is biased away from the correct value. When $|\psi\rangle$ is 'erased' by the channel, the received qubit could be $X |\psi\rangle$ or $Y |\psi\rangle$ or $Z |\psi\rangle$, but definitely not $I |\psi\rangle$. – Craig Gidney Sep 30 '16 at 21:27
• Got it. So it is in fact slightly "nicer" (=less erasing) than an erasure channel. – Norbert Schuch Oct 1 '16 at 8:09
• Question: Are Alice and Bob allowed classical communication in both directions (except for Alice telling Bob the results of her Bell measurement -- let's say that the display of her measurement apparatus is somewhat broken). In that case, they can use the channel to send quantum information, see arxiv.org/abs/quant-ph/9701015 – Norbert Schuch Oct 1 '16 at 8:13
• ... I also would suspect that in the other case, they cannot use the channel to send information since probably one can build their channel from an erasure channel with no capacity (using the same paper). Didn't go through the math, though (but it is straightforward). – Norbert Schuch Oct 1 '16 at 8:51