Can superdense coding be achieved without quantum memory?

[Armchair]

In superdense coding how can BSA be performed on the second of a pair of hyperentangled photons if the photon that was sent sometime earlier was measured on arrival? Would not the measurement have destroyed the entanglement?
I’m trying to understand superdense coding as set out in this paper.
It is suggested here that quantum memory is not required.

Hyperentanglement for advanced quantum communication
Julio T. Barreiro and Paul G. Kwiat
Department of Physics, University of Illinois at Urbana-Champaign, Urbana, IL 61801-3080,
USA

 

[Strilanc]

For some reason, this paper thinks that "superdense coding" means sending two classical bits of information with a single photon. It actually means sending two classical bits per qubit. Since the experimenters can put two qubits into their photons (one for polarization, one for spatial mode), they would actually have to send four bits of information per photon to be properly superdense coding.

The experiment they describe is:

1. A polarization-and-spatial-mode-entangled photon pair is generated with one part sent to Alice and the other part to Bob.
2. Alice uses her part to do a measurement with a desired preparation as one of the results.
3. If the measurement returns the wrong answer, give up and try again.
4. If the measurement gave the right answer, Bob's photon was prepared into a state where its polarization is entangled with its spatial mode.
5. Bob encodes two bits into the photon. Bob only operates on the polarization of the photon, which would normally limit him to encoding one bit, but the entanglement with the spatial mode allows for 2. This is the only sense in which superdense coding is used.
6. Bob sends the photon to Alice, who decodes it.

In other words: instead of using superdense coding to increase bandwidth, they're using it to decrease control requirements.

Also, I personally find it silly to claim anything related to increased communication capacity when performing steps that require post-selection (step 3). If they're preparing two qubit states, then the retry rate is probably 75% and any bandwidth-per-photon that they report probably needs to be divided by 4.