Only 4 possible single-qbit states?

[georgir]

Doesn't quantum teleportation show us that there are in fact at most 4 possible quantum states for a single qbit?
I mean, you are guaranteed to reproduce the completely arbitrary and unknown teleported state by doing one of 4 possible operations with your side of the entangled pair. You can even go ahead and do it at random before you receive the other party's measurement, and you have 25% chance to get it right.

 

[Nugatory]

Four isn't the number of possible states of the qubit, it's the number of ways you have of processing the qubit. Your entangled particle can be any of an infinite number of states before you apply one of the four operations, and no matter which of the four operations you choose, there are an infinite number of possible results.

It's as if I'm sending you a message encrypted with any of four possible keys. The restriction on the number of keys doesn't constrain the number of possible messages.

 

[georgir]

But with one of the 4 operations you are guaranteed to get the original completely arbitrary state, even before it has been teleported yet. You just don't know which of the 4 is needed until then...

 

[Nugatory]

But with one of the 4 operations you are guaranteed to get the original completely arbitrary state, even before it has been teleported yet. You just don't know which of the 4 is needed until then...

Right, and that's analogous to saying that when you receive the encrypted message, one of the four possible keys is guaranteed to correctly decrypt it, and you don't know which until you've received the two-bit classical channel message that tells you which one to use. That doesn't mean that there are only four possible messages that I could have sent you. There were an infinite number of messages that I could have encrypted in any of four different ways.

 

[Strilanc]

Doesn't quantum teleportation show us that there are in fact at most 4 possible quantum states for a single qbit?

Superdense coding, which transmit 2 bits (i.e. one of four states) via a single pre-entangled qubit, is a much better example of your point than quantum teleportation.

Technically you can encode any finite number of states into a single qubit, since it has a continuous state space instead of a discrete state space. The problem comes when you want to read the information back out, and you need to distinguish those states. That's where you're really limited to $2^n$ states for $n$ qubits.

Superdense coding and quantum teleportation can use and distinguish four states only because they involve two qubits. It's interesting and useful that you can write all four states using just one of the qubits involved in a bell pair, but reading what you wrote still requires combining both qubits and there's no way to make it work without that second qubit around.

Personally, I often find myself thinking of a qubit as being "worth" up to two bits, but it's important to understand that this is an intuition that breaks down in many cases (e.g. you need pre-existing entanglement).

*edit*: I think I inverted the meaning of your question, and answered "Doesn't QT mean a single qubit can hold 4 distinguishable states?". The answer to your actual question is that it's kind of complicated what you can and can't put in qubits, but we do know that 2 bits per qubit is an absolute maximum when it comes to transmitting classical information over a quantum channel (even with arbitrary pre-existing entanglement). This is different from "qubits can hold at most 4 states" though. The state space of a qubit is continuous, not discrete, and our predictions about how they behave would be wrong if we tried to reduce that down to just 4 points.

 

[georgir]

Right, and that's analogous to saying that when you receive the encrypted message, one of the four possible keys is guaranteed to correctly decrypt it, and you don't know which until you've received the two-bit classical channel message that tells you which one to use. That doesn't mean that there are only four possible messages that I could have sent you. There were an infinite number of messages that I could have encrypted in any of four different ways.

But its not "when you receive the encrypted message". Its well before the message even gets encrypted. You still have 25% chance to get it right back then. 25% chance to get your unknown qbit or "message" exactly right from basically nothing...

Edit: well, "basically nothing" = the promise that in the future you'll mix up your message qbit (i.e. "send" it) with the other end of the entangled pair I used, or some such.

Edit 2: In other words, it's a bit like 25% chance to perform teleportation into the past... I must be missing something.

 

[georgir]

And yes, I guess this can also be thought of as a variation on superdense coding, since its more or less the same thing in another direction.
Imagine that in superdense coding instead of using the received qbit to do your decode, you use some other qbit. Arbitrary or completely specifically prepared, doesn't matter. You'll get a "decode" that has 25% chance to match the original 2 bits, nothing surprising on first thought. What is surprising though is that in those 25% of cases, your replacement qbit that you used apparently happened to be exactly identical to the one you received and were actually supposed to use. So a random qbit that you chose has 25% chance to turn out to be identical to the one you got sent...

 

[georgir]

If I think of superdense coding as a quantum teleportation into the past, it creates a closed timelike curve... My head hurts now.