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 global quantum state is not a function of the reduced density matrices.1Tim Maudlin in fact argued that because of this quantum mechanics is nonlocal, and therefore one shouldn’t be bothered by the nonlocality of Bohmian mechanics.

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 a function 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 the Choi-Jamiołkowski representation of $U$2If you want to be pedantic, it is not the Choi-Jamiołkowski representation of $U$, but of the linear map $\mathcal U(A) = U A U^\dagger$.
\[ \lambda(U) := \sum_{ij} \ket{i}\bra{j} \otimes U \ket{i}\bra{j} U^\dagger.\]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:
\[ \lambda^{AB}(U) := \sum_{ijkl} \ket{i}\bra{j} \otimes \ket{k}\bra{l} \otimes U \big( \ket{i}\bra{j} \otimes \ket{k}\bra{l} \big)U^\dagger,\] and we name the tensor factors as $A_I,B_I,A_O,B_O$, in that order. The ontic state of Alice and Bob’s parts are then defined as
\[ \lambda^{A}(U) := \tr_{B_O} \lambda^{AB}(U)\quad\text{and}\quad\lambda^{B}(U) := \tr_{A_O} \lambda^{AB}(U).\]If you use the identity $(\id \otimes U)\ket{\phi^+} = (U^T \otimes \id) \ket{\phi^+}$ they get the convenient representation
\[ \lambda^{A}(U) = \sum_{ij} {U}^T \big( \ket{i}\bra{j} \otimes \id \big){U}^* \otimes \ket{i}\bra{j},\]with the analogous formula holding for Bob.

Now we deliver the goods. 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 $(\id \otimes V) U$. Alice’s ontic state is
\begin{align*}
\lambda^{A}\big((\id \otimes V)U\big) &= \sum_{ij} {U}^T(\id \otimes V^T) \big( \ket{i}\bra{j} \otimes \id \big)(\id\otimes V^*){U}^* \otimes \ket{i}\bra{j} \\
&= \sum_{ij} {U}^T \big( \ket{i}\bra{j} \otimes \id \big){U}^* \otimes \ket{i}\bra{j},
\end{align*} so Bob cannot change it, whatever he does.

Next we show that the ontic state of the global system is a function of the ontic states of the parts. In fact it’s just the product of them, with some strategically placed identities:
\[\lambda^{AB}(U) = \big[\lambda^{A}(U)\otimes \id_{B_O}\big]\big[\lambda^{B}(U)\otimes \id_{A_O}\big].\] 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
\[ \rho^{AB} := \bra{0}_{A_IB_I} \lambda^{AB}(U) \ket{0}_{A_IB_I} = U\ket{0}\bra{0}U^\dagger, \] and the reduced density matrix are given by the analogous formula:
\[ \rho^{A} := \bra{0}_{A_IB_I} \lambda^{A}(U) \ket{0}_{A_IB_I} = \tr_{B_O}\big(U\ket{0}\bra{0}U^\dagger\big).\]

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 global 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 $\sim d^n/n$ 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.