Quantum teleportation and the reality of quantum states

This question is perhaps philosophical but it's been confusing me. Suppose Alice is teleporting some qubit state $$|\phi\rangle$$ to Bob via the quantum teleportation protocol. After Alice applies the operations necessary to her qubit and her half of the Bell pair, but before she sends that classical information over to Bob, by the no-communication theorem the state of Bob's pair should be the mixed state $$1/2 (|0\rangle\langle0| + |1\rangle\langle1|)$$. After she transmits her measurement outcome to Bob, however, it seems like Bob's state changes to some pure state, e.g. $$|\phi\rangle$$ if Alice happened to measure $$00$$. It's not important what exactly the state is, just that it seems to have changed from a mixed state to a pure state.

This seems to favor the thought that quantum states are not "really real" but measures of subjective (dis)information about a system. However that troubles me, because I've thought of quantum states as being literally real aspects of the universe. Perhaps the trouble is that we should not think of Bob's state by itself, since it is entangled -- but then that seems to lead us to a non-manifestly local description of quantum theory, which I also find bothersome. I guess I'd just like to know what do people make of this.

The state of Bob's pair is not $$1/2 (|0\rangle\langle0| + |1\rangle\langle1|)$$. This is only his reduced density matrix. By definition, it is a representation of the locally accessible information Bob has. It's perfectly natural that it will change when Bob gets some nonlocal information from Alice.

The question remains, however, what is Bob's state. The standard answer is that there is no such thing; an entangled state can only be described as a whole. Since what Bob has is a part of an entangled state, we cannot assign a state to him alone. As you notice, this is rather unsatisfactory, as it makes quantum mechanics seem rather nonlocal, even though it is perfectly local in the operational sense.

There is, however, a solution: the Deutsch-Hayden model was designed to represent quantum mechanics (and quantum teleportation specifically) in a completely local way. It can assign "really real" states even to parts of entangled systems, and these states are local in the sense that they don't change with respect to what other parties do. The core idea behind it is that Bob's state doesn't encode only the locally accessible information, like the reduced density matrix, but also its correlations with the rest of the entangled system.

In quantum teleportation specifically, Bob's state doesn't change to $$|\phi\rangle$$ when he learns that Alice's measurement result is 00. Instead, $$|\phi\rangle$$ becomes locally accessible information for Bob when he can make measurements on his part of the entangled system together with Alice's measurement result (which is just another physical system).

The Deutsch-Hayden paper uses rather archaic notation, I find this review more intelligible. Appendix A specifically.

• Thanks very much for this reference! I did stumble upon the Deutsch-Hayden paper, and looked at it for a while but was confused about what exactly was it saying. This review should help me. – Pedro Oct 30 '20 at 14:40

Remember that mixed states can be a subjective description of a quantum state. In a teleportation operation, where Alice has made the measurement, but Bob has not yet received the measurement result, then Alice and Bob have different information, and therefore they have different descriptions. Alice knows exactly the state that Bob holds. Bob has no idea, and describes the state by the maximally mixed state.

When Bob learns the same thing as Alice, the new information he has updates his best description of the state so that it is the same as Alice's description.

This is no different to the classical world. Imagine a game where there are two doors. Behind one door, Alice places an amazing prize. The game contestant, Bob, gets to pick a door. Since Bob does not know which door the prize is behind, he assigns his subjective probability distribution: it's 50:50. But there's still an objective reality in the background of exactly which door the prize is behind in this particular run of the game.

Now, to complicate matters, density matrices can be doing other things as well. In particular if the global state is entangled, then the reduced density matrix is the best possible description anyone can give. It's an objective description (but it is still missing something that you can only capture by describing the entangled state).

• The situation in the classical world is completely different, precisely because then there is an objective reality about which door the prize is behind, and the probability distribution is subjective. This is not the case in quantum mechanics, there is no objective quantum state on Bob's side alone, unless you believe in objective, faster-than-light, collapse. Remember that when they're space-like separated there is a reference frame in which Alice hasn't done her measurement yet. – Mateus Araújo Oct 30 '20 at 10:44

Yes, you're right. That is why there is no commonly accepted interpretation of the quantum theory.

A simpler paradoxes, such as with Wigner's friend, also show that quantum state is not quite objective thing.

• I think Wigner's friend paradox can be resolved just fine via decoherence & MWI though. – Pedro Oct 29 '20 at 16:36

The ontology of pure states is tricky, but if you believe in pure states then mixed states are fairly straightforward, I think.

Teleportation of a third qubit seems unnecessary in this thought experiment. The same issue arises if Alice and Bob hold halves of a Bell pair and Alice sends an email to Bob stating the outcome of a measurement she performed on her half. The rest of this answer is about that simpler experiment.

The no-communication theorem means that if Bob is sure that he'll never have access to Alice's qubit again, then he can assume without loss of predictive power that she has already measured and discarded it, and the wavefunction has collapsed to some pure $$|\phi\rangle$$, but he doesn't know which.

His knowledge can then be represented by a classical Bayesian probability distribution over possible values of $$\phi$$. This needn't be a uniform distribution—maybe he knows that Alice prefers to measure in the Hadamard basis—but at the very least, unless he thinks that Alice has the power of postselection or might have rigged the qubit creation process, he should believe that $$|0\rangle$$ and $$|1\rangle$$ are equally likely, $$|-\rangle$$ and $$|+\rangle$$ are equally likely, etc. In that case Bob's beliefs about likely measurement axes have no effect on his prediction of the outcome of any experiment on his qubit. You can conclude this from the no-communication theorem again or by directly calculating the outcomes of measurements on all axes.

A mixed state is just a compact way of representing this "classical knowledge about a pure state modulo distinguishability via experiments on the state". It's updated in light of new information in the same way as an ordinary probability distribution because it's just a representation of an aspect of that distribution.

A mixed state can't replace the full probability distribution for all purposes, only for the purpose of answering questions about experiments on the state. For example, if someone offers to bet Bob that Alice will measure in the Hadamard basis, whether he should take that bet depends on his beliefs about Alice's basis preferences, and the density matrix doesn't contain that information. If Alice sends an email saying she measured her qubit and the result was $$1$$, but she doesn't say what gates she applied to it first, then Bob's beliefs about the results of experiments on his qubit after getting the email might be represented by a state like $$0.9|+\rangle\langle+|\,+\,0.1|-\rangle\langle-|$$. This can't be calculated from the mixed state derived from his beliefs prior to getting the email; you have to update his full set of beliefs and calculate a new mixed state from that.