# Computer analogy to non-locality in quantum mechanics

It is not uncommon to say that the non-locality of quantum mechanics is equivalent to the following computer analogy: if you are trying to model an entangled two spin system, then even if the spins are simulated to move apart from each other, you need to do the calculations in the same computer, that is, you cannot take one computer for each spin and separate them to do the calculations, the two computers have to remain together. I heard Leonard Susskind saying this in one of his classes, and there is also the quantum randi challenge https://arxiv.org/abs/1207.5294, which is used to debunk local theories of quantum mechanics.

My question is: can this analogy be formally proved from quantum mechanics, is it so obvious that it does not need any proof, or is it not even correct?

• What do you mean by you need to do the calculations in the same computer? Which calculations? Commented Aug 18, 2021 at 13:22
• Your algorithm for deciding if the measurement is a sping aligned or antialigned with your detector (the detector's direction is free to be chosen after separation)
– user65081
Commented Aug 18, 2021 at 13:24

In quantum mechanics, possible states of a system are described by a Hilbert space. If you have two systems, with state spaces $$\mathcal{H}_1, \mathcal{H}_2$$ then the combined system is described by $$\mathcal{H}_1 \otimes \mathcal{H}_2$$, which consists of linear combinations of product states like $$|\phi \rangle \otimes |\psi \rangle$$. So e.g. for a pair of qubits you can have states like

$$\frac{1}{\sqrt{2}} \left( |0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle \right)$$

which cannot be written as $$|\phi\rangle \otimes |\psi\rangle$$ for any choice of $$|\phi\rangle, |\psi\rangle$$. These states are called entangled.

The point is you cannot describe this state as putting together a state for the first qubit and a state for the second qubit. It only makes sense as a state for both at the same time. This is what Susskind means: you can't "store the state of the first qubit" in one computer and "store the state of the second qubit" in another computer, because they don't even have individual states: they are entangled.

This poses a challenge for thinking about locality in quantum mechanics, for what does locality even mean when your very description of nature is non-local? Nevertheless, 'operational' locality can be rescued, in the sense of the No Communication theorem for instance -- it is not possible to use entanglement to send messages faster than the speed of light.

• Thanks for your answer. I do not think it is that simple. The fact that an entangled state cannot be written as a product does not imply you cannot separate the qbits, you could still have local hidden variables If the experimenter were simulating measurements of spin in a single direction, you can used separated computers and get the right statistics. It is only when you measure the spin in 3 different directions that you need a single computer.
– user65081
Commented Aug 21, 2021 at 18:44
• In support of the above, remember bell's theorem. This theorem is important because it shows you cannot have local realism, and the theorem was needed, it was not enough to claim that an entangled state could not be local by looking at the equation describing a two spin fully entangled state.
– user65081
Commented Aug 21, 2021 at 18:45
• Wolphram jonny, Bell's theorem requires the hidden variables and detector's settings to be independent. Such an assumption is not justified in the context of a theory with long-range interactions, like classical electromagnetism. As explained in my answer below, interacting systems cannot be described in terms of independent subsystems. So, local-realism can still be compatible with QM's predictions. Also, it's good to remember that non-realism directly implies non-locality, as EPR argument proves. So, if local-realism does not work physics has to be non-local. Commented Aug 25, 2021 at 6:15

I think that it can be proved, and the proof is Bell's theorem (or a similar result). I suspect that Susskind had Bell's theorem in mind.

In Bell's original paper, the outcomes of the two measurements were determined by arbitrary functions $$A(\vec a, λ)$$ and $$B(\vec b, λ)$$, where $$λ$$ was arbitrary hidden state information and $$\vec a$$ and $$\vec b$$ were the measurement axes, and noncommunication was captured in the fact that each function had only one axis as an argument. In an attempt to make this more friendly, you could say that $$A$$ and $$B$$ are calculated by computers, $$λ$$ is perhaps the software installed on them, and the code that is told $$\vec a$$ can't communicate with the code that is told $$\vec b$$, which is most easily phrased as there being two (non-networked) computers doing the calculations. Those assumptions lead to a result that contradicts quantum mechanics. Therefore, to correctly simulate quantum mechanics, you need networked computers, or more simply a single computer. I imagine that Susskind's example came from a line of thought similar to that.

This argument breaks down if the computers are quantum and their initial memory states are entangled. You can use quantum computers if the initial states aren't entangled, but it's probably easier to require classical computers. Classical computers can simulate quantum computers anyway (as long as they aren't entangled with other quantum computers).

• What you say makes sense, thanks!
– user65081
Commented Aug 25, 2021 at 19:53
• Without computer, take $$A(a,\lambda)=-B(a,\lambda)=sign(\cos a\cos\lambda)$$. Then the correlation is $$C(a,b)= sign(\cos a\cos b)$$. The point is that a and b angles mean nothing separately but one need a physical reference direction, let say a. You get the relative angle correlation $$C(a-b)=sign(\cos(a-b))$$, a rectangle covariance. Then $$3C(\pi/4)-C(3\pi/4)=4$$ Commented Sep 12, 2021 at 19:14

Wolphram jonny,

The first thing you need to understand is that quantum mechanics deals with long-range interactions, mostly electromagnetism. This means that each part of the experiment is in continuous interaction with the other parts.

When a diatomic molecule splits, creating an entangled pair of atoms, that split takes place in the presence of the electric and magnetic fields associated with the rest of the experimental setup, which includes the detectors and everything else. One would therefore expect those spins to be correlated with the experimental environment. This has nothing to do with non-locality, it's a consequence of the fact that long-range interactions are present.

It is not surprising that electromagnetic phenomena cannot be described in terms of non-interacting parts. You cannot explain induction with billiard balls that only interact by contact forces.

The same happens in gravity. You cannot describe a planetary system if you assume that no interaction takes place between the star and the planets. Each object, if isolated, should just travel in a straight line. So, if you use separate computers, you will fail to simulate planetary orbits. Separate computers will also fail to describe an electromagnetic system.

If you have N non-interacting objects their states will be solutions to N 1-body problems. You can describe the situation using N isolated computers.

If you have N interacting objects, their state would be described by a solution of the N-body problem. Solving the N-body problem requires the knowledge of all N objects so it cannot be done using N isolated computers.

Entanglement is an electromagnetic phenomenon. All particles interact so you cannot use isolated computers to correctly model the physics of the experiment. There is no reason to believe that faster than light effects are involved. The so-called non-locality is a red-herring.

• Thanks for your answer. There are no explicit interactions between the two particles in the two spin system, your claim that entanglement is an electromagnetic phenomenon seems non-mainstream physics. I am looking for an answer based on standard quantum mechanics.
– user65081
Commented Aug 24, 2021 at 18:14
• QM deals with atoms, electrons, molecules, EM radiation and the like. Those are electromagnetic systems, this is as "mainstream" as you can get. Standard quantum mechanics does not make any statement about locality, so you need to make supplementary assumptions, not part of QM, in order to decide the local/non-local character of physics. The purpose of my answer was to show you that the inability to describe a system as a sum of independent/isolated parts is a direct consequence of long-range interactions and it's not a proof of non-locality. To continue... Commented Aug 25, 2021 at 5:58
• Cont: There is indeed no EM interaction between two photons, but there is such interaction between their source and the polarizers. In order to model the experiment correctly you need, as explained in my answer, consider the whole experimental setup (source+detectors). If you do that, no proof of non-locality can be sustained. Commented Aug 25, 2021 at 6:04

I think non-locality comes from the mathematical approach used to solve your problem.

Suppose the heat equation similar to Schrödinger's :

$$\partial^2_{xx}f(x,t)=\partial_t f(x,t)$$

With an initial condition, and boundary condition for f to be zero. Moreover we admit an intermediate condition that the function shall vanish on a central interval, the initial condition being non 0 on one side of the intermediate condition.

Analytically there is only 1 solution : an analytical function being 0 on an interval is 0 everywhere, since all derivatives vanish on it and that an entire function is the same as its Taylor expansion. This comes as a corollary of locality and analyticity.

So there is no solution if the initial condition is not 0, analytically, since the question is bad posed and self contradicting.

If you now apply a finite element method in space and finite difference in time you shall write your function at discretized points as : $$f(x,t)=\sum_i a_i(t)\phi_i(x)$$, with special triangle wave basis functions, which by the way are not analytic, but that underspin bothdomains and vanish in the intermediate one (this looks quite artificial and thus the following is to be taken as artificial too)

The weak formulation of FEM implies a matrix equation with two matrices A and B (left to the reader) :

$$A\vec{a}(t)=B\partial_t\vec{a}(t)$$

Now a finite difference scheme for the time dependence implies :

$$\vec{a}(t+h)=(\mathbb{1}+hB^{-1}A)\vec{a}(t)$$

Hence

1. there exist a non vanishing solution
2. the inverse of matrix B can be a full one, so 2a) the solution at every point at the next time step depends on the values at all points at the previous time step, hence there is non locality, even if it is not a quantum equation. 2b) the function on the other side of the intermediate condition changes even if it was zero on that domain and even if the solution does not pass through the intermediate interval, which could be called a true tunneling, explained as a quantum space jump.

So holism, or non-locality is not a quantum physical only phenomenon here, but rather a property on how the problem is solved mathematically or numerically.

If this corresponds to a physical transfer is an ongoing problem since EPR in quantum.

• Thank you for your answer, I confess I could not see how this relates to my question. Also, your statement "non-locality is not purely a quantum physical phenomenon here, but rather a property on how the problem is solved mathematically or numerically" seems to be non mainstream physics
– user65081
Commented Sep 12, 2021 at 15:49
• What I don't understand in Bell's theorem is what is a in $A(a,\lambda)$. I think if it is an angle then it is between 2 directions, so nothing prevents to choose one of them so that it is 0, so that it depends only on the relative angle. Or: we choose a reference frame. Else we have 4 directions which means we choose one reference frame for every side ? Commented Feb 26, 2022 at 7:51