Explicitly local quantum mechanics

I’ve just attended a nice online talk by Charles Bédard about his somewhat recent paper, where he studies the explicitly local model of quantum mechanics developed by Deutsch and Hayden, with a twist by Raymond-Robichaud. I think it’s a neat idea encumbered by horrible notation, so I’ll try to expose it in a more intelligible way here.

First of all, what is the problem? Quantum mechanics is local. Very local in fact, if you look at QFT. It’s Lorentz covariant, by construction. Non-relativistic quantum mechanics is not very local, as it is still non-relativistic, but still much more local than people usually think: the only outrageously nonlocal thing about it is the collapse of the wavefunction, but everybody knows that it is not real. Even Bell’s theorem doesn’t imply nonlocality, as long as you don’t pretend there is a wavefunction collapse happening.

Nevertheless, quantum states seem rather nonlocal. If Alice and Bob share an entangled state, and Alice acts locally on it, the quantum state changes in a seemingly global way, oblivious to the locality of her action. Furthermore, you can’t separate it in Alice’s part of the state and Bob’s part of the state, it only exists as an indivisible whole. Sure, you can define reduced density matrices for Alice and for Bob, but in this way you lose a lot of information, the quantum state is much more than the union of these parts.1

What Deutsch and Hayden showed is that it doesn’t need to be like this. It is possible to define a kind of quantum state that you can separate in Alice’s part and Bob’s part, in such a way that the global quantum state is just the union of both parts, and they both evolve in a perfectly local way: whatever Alice does cannot change Bob’s part and vice-versa. I’ll call them ontic states, because perfectly local states seem perfectly real.

The key idea is to define an ontic state based on the dynamics of quantum mechanics, since they are local: instead of focusing on a quantum state $\ket{\psi}$ we focus on the unitary $U$ such that
\[\ket{\psi} = U\ket{0},\]where $\ket{0}$ is a standard fixed state, like how on quantum circuits we usually assume that all qubits are initialized on the $\ket{0}$ state. The ontic state of the global system is then just a Choi-Jamiołkowski representation of $U$2
\[ \lambda(U) := \sum_{ij} \ket{i}\bra{j} \otimes U^\dagger \ket{j}\bra{i} U.\]So far so boring. Things get interesting when we represent the ontic states of the parts of the quantum system. First we rewrite the global ontic state explicitly as a composite system:
\[ \lambda^{AB}(U) := \sum_{ijkl} \ket{i}\bra{j} \otimes \ket{k}\bra{l} \otimes U^\dagger \big( \ket{j}\bra{i} \otimes \ket{l}\bra{k} \big)U,\] and we name the tensor factors as $A_1,B_1,A_2,B_2$, in that order. The ontic state of Alice’s part is then
\[ \lambda^{A}(U) := \tr_{B_1} \lambda^{AB}(U) = \sum_{ij} \ket{i}\bra{j} \otimes U^\dagger \big(\ket{j}\bra{i} \otimes \id\big) U,\] and the ontic state of Bob’s part is
\[ \lambda^{B}(U) := \tr_{A_1} \lambda^{AB}(U) = \sum_{kl} \ket{k}\bra{l} \otimes U^\dagger \big( \id \otimes \ket{l}\bra{k}\big) U.\]First we show that the ontic states of the parts do evolve locally: if Bob applies an unitary $V$ to his part of the system, the global unitary is now $U’ = (\id \otimes V) U$. Alice’s ontic state is
\begin{align*}
\lambda^{A}(U’) &= \sum_{ij} \ket{i}\bra{j} \otimes U^\dagger (\id \otimes V^\dagger)\big(\ket{j}\bra{i} \otimes \id\big) (\id \otimes V)U \\
&= \sum_{ij} \ket{i}\bra{j} \otimes U^\dagger \big(\ket{j}\bra{i} \otimes V^\dagger V \big) U \\
&= \sum_{ij} \ket{i}\bra{j} \otimes U^\dagger \big(\ket{j}\bra{i} \otimes \id \big) U \\
&=\lambda^{A}(U),
\end{align*} so Bob cannot change it, whatever he does.

Now we show that the ontic state of the global system is just the union of the ontic states of the parts. In fact it’s just the product of them, with some strategically placed indentities:
\[\lambda^{AB}(U) = \big[\lambda^{A}(U)\otimes \id_{B_1}\big]\big[\lambda^{B}(U)\otimes \id_{A_1}\big].\] That’s it. A demonstration in a couple of lines of something that was widely regarded as impossible. Neat, no?

We should also remark that these ontic states do give you back the usual quantum states. The global quantum state is
\[ \rho^{AB} := \bra{0}_{A_2B_2} \lambda^{AB}(U) \ket{0}_{A_2B_2} = U\ket{0}\bra{0}U^\dagger, \] and the reduced density matrices are given by the analogous formulas:
\[ \rho^{A} := \bra{0}_{A_2B_2} \lambda^{A}(U) \ket{0}_{A_2B_2} = \tr_B\big(U\ket{0}\bra{0}U^\dagger\big), \] and
\[ \rho^{B} := \bra{0}_{A_2B_2} \lambda^{B}(U) \ket{0}_{A_2B_2} = \tr_A\big(U\ket{0}\bra{0}U^\dagger\big).\]

One might wonder if it is possible to do a similar trick using just the quantum state $\ket{\psi} = U\ket{0}$ instead of the unitary $U$ that produces it. As Deutsch and Hayden argue (and Raymond-Robichaud gives a formal proof), the answer is no. The crucial problem is that $(X \otimes \id)\ket{\phi^+} = (\id \otimes X)\ket{\phi^+}$. That is, Alice can act on a quantum state in a completely local way, but you can’t tell from looking at resulting state whether it was changed by Alice or by Bob. This never happens if we look at $U$ instead: if $(S\otimes \id)U = (\id \otimes T)U$ then $S$ and $T$ must be just identity with a phase.

With the introduction done, we can speak about Bédard’s talk itself. He was interested in the number of parameters that are necessary to describe the ontic states. He showed that global ontic state of $n$ quantum systems of dimension $d$ requires $d^{2n}-1$ parameters. This might be obvious from the Choi-Jamiołkowski representation, as it is just the number of parameters of the global unitary $U$ (the global phase doesn’t count), but it was not obvious from the Deutsch-Hayden notation he was working with. More interestingly, he showed that the ontic state of a single quantum system needs $(1-d^{-2})d^{2n}$ parameters. Proving this is difficult in either notation, but intuitively we can expect this number to be true, since the ontic state of a single system does not depend on unitaries that act on the other systems. Since these have $d^{2n-2}-1$ parameters, we subtract this from the number of parameters of the global unitary and voilà!

Now, the dimension of the ontic state of a single system grows exponentially with the number of systems. This might seem strange, but Bédard had a good argument about why this is actually necessary: the quantum state of $n$ systems of dimension $d$ already needs $2d^n-2$ parameters. Since we want the local ontic states to recover at least the global quantum state, if follows that each must have at least $\approx d^n/n$ parameters, if we assume they are uniformly spread.

There was a lively discussion after the talk about Wallace and Timpson’s criticism of the Deutsch-Hayden model. They notice that for any unitary W that acts on the whole Hilbert space except $\ket{0}$, the unitary $U(1\oplus W)$ will give rise to the same quantum states, and therefore to the same observations, but in general different ontic states. They argue that this $W$ is akin to a gauge freedom, and we should identify all these different ontic states, lest we suffer from indeterminism and (ontic) state underdetermination. If we do that, however, we end up back with just regular quantum states and their nonlocality, making the Deutsch-Hayden model completely pointless.

Wallace and Timpson missed a very important point, however. Doing quantum state tomography is not the only way to determine $U$; this would be a hopelessly positivist point of view. No, determining $U$ is determining what are the dynamics of our theory. Doing this is what science is all about, and we use all tools we can to do that, more importantly assuming that the fundamental dynamics do not change with time, and that we can reproduce experiments. Once we know that fundamental dynamics of a (closed) system, we know what $U$ is, there is no underdetermination; the fact that $U(1\oplus W)$ would also give rise to the same observations is not relevant, it’s just a fun fact. The transformation from $U$ to $U(1\oplus W)$ is not a gauge transformation, it is a transformation from the real dynamics to some incorrect dynamics.

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6 Responses to Explicitly local quantum mechanics

  1. Danylo says:

    Cool. I also want to note that a product of 2 local ontic states is commutative, so this is indeed can be considered as a union.

    Also, some small disentangled subsystem doesn’t need a huge number of parameters when considered as a part of some large system. If it’s disentangled then the corresponding global unitary can be written as a tensor product of unitaries, so the unitaries on the other parts are cancelled in the local ontic state of a disentangled part.

    But I still don’t understand what is “objective” in the quantum world :) Even if we agree on what are the local subsystems, how can we agree what are their ontic states? Any experiment or communication has some built in chance, which means the world splits every time when we get the result.

  2. Mateus Araújo says:

    Well, even if it weren’t commutative that wouldn’t be a problem for me, the only thing that would make the idea fail for me is if the global ontic state were not a function of the local ontic states, as the global quantum state is not a function of the reduced density matrices.

    It is not true that if the global quantum state is not entangled then the global unitary can be written as a tensor product of local unitaries. Take for example $U$ being the CNOT gate (or the SWAP gate, if you prefer). CNOT applied to $\ket{00}$ is still $\ket{00}$, so the global quantum state is not entangled, but CNOT cannot be written as a tensor product of unitaries.

    The world splitting when we make a measurement doesn’t pose any further problem to objectivity than the result of the measurement being fundamentally random (in a single world theory). Just as we can do quantum state tomography in a single world theory, and recover the real state with high probability, we can do it in Many-Worlds, and recover the real state in most worlds. We can always be wrong, in a single world theory, with some small probability, just as we will always be wrong, in Many-Worlds, in a small proportion of the worlds. Epistemology is not ontology: the theory is still perfectly objective, it will say that we will be wrong about the state, not that the state will be whatever we measure.

  3. Danylo says:

    I’ve meant we can find a global decomposable $U$ for a disentangled state. But if we want to keep track of a single global unitary then this will be the problem, yes. Small disentangled states would still require a huge number of parameters.

  4. Mateus Araújo says:

    Yep, that’s true. But I find it essential to keep track of the real unitary that actually generated the quantum state, otherwise we are vulnerable to Wallace and Timpson’s criticism.

  5. Doug McDonald says:

    I do not fully understand your notation, but what if U is the usual time evolution operator? What happens if the overall U requires the transfer of some of the n’s
    from A’s domain to B’s at some point in time? This is of course very common, at least it seems so. It there some theorem that says this is impossible, or can in all cases be removed by global redefinition without ruining the arguments you present? Doing that does not bother Bell’s Theorem. Does it?

  6. Mateus Araújo says:

    $U$ is the usual time evolution operator. I don’t know what you mean by “the n’s”, but it’s perfectly alright for $U$ to make A and B interact, either by swapping their states or entangling them. It’s also very common. The point of the Deutsch-Hayden model is that if the time evolution is not an interaction, e.g. acting only on A’s side, then it only changes A’s ontic state, while leaving B’s invariant.

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