# Do Feynman path integrals satisfy Bell locality assumption?

There are generally two basic ways to solve physics models:

1. Directional, e.g. Euler-Lagrange equation in CM, Schrödinger equation in QM. We evolve some initial conditions in some direction, can imagine boundary conditions are e.g. in $$t_0 = -\infty$$.
2. Symmetric, e.g. the least action principle in CM, Feynman path integrals in QM, or of Feynman diagrams in QFT. Like choosing shape of a membrane, statistics inside Ising sequence - with symmetric boundary conditions, we can imagine they are e.g. in $$t_0 = -\infty$$ and $$t_1=+\infty$$ (measurement is between boundary conditions).

Having a solution found with 1. or 2., we can transform it into the second, but generally solutions originally found using 1. or 2. seem to have a bit different properties - for example regarding "hidden variables" in Bell theorem.

Transforming solution found in a symmetric way 2) to directional perspective 1), its state was originally chosen also accordingly to all future measurements - like in superdeterminism.

The directional ones (1.) like Schrödinger equation usually satisfy assumptions used to derive Bell inequality, which is violated by physics - what is seen as contradiction of local realistic "hidden variables" models. Does it also concern the symmetric ones (2.)?

We successfully use classical field theories like electromagnetism or general relativity, which assume existence of objective state of their field - how does this field differ from local realistic "hidden variables"?

Wanting to resolve this issue, there are e.g. trials to undermine the locality assumption by proposing faster-than-light communication, but these classical field theories don't allow for that.

So I would like to ask about another way to dissatisfy Bell's locality assumption: there is general belief that physics is CPT-symmetric, so maybe it solves its equations in symmetric ways (2.) like through Feynman path integrals?

Good intuitions for solving in symmetric way provides Ising model, where asking about probability distribution inside such Boltzmann sequence ensemble, we mathematically get $$Pr(u) = (\psi_u)^2$$ where one amplitude comes from left, second from right, such Born rule allows for Bell-violation construction. Instead of single "hidden variable", due to symmetry we have two: from both directions.

From perspective of e.g. general relativity, we usually solve it through Einstein's equation, which is symmetric - spacetime is kind of "4D jello" there, satisfying this this local condition for intrinsic curvature. It seems tough (?) to solve it in directional way like through Euler-Lagrange, what would require to "unroll" spacetime.

Assuming physics solves its equations in symmetric way, e.g. QM with Feynman path integrals instead of Schrödinger equation, do Bell's assumptions hold - are local realistic "hidden variables" still disproven?

Update: Born-like formulas from symmetry in Ising model (Boltzmann sequence ensemble): $$Pr(i)=(\psi_i)^2$$ where one amplitude ("hidden variable") comes from left, second from right:

• Does Bell's theorem rule out hidden-variable models? Experimental confirmation has ruled out local models, but AFAK there is nothing to prevent a nonlocal hidden-variable model. – Guy Inchbald Feb 15 at 11:40
• @GuyInchbald, it ruled out if using Schrodinger equation - but here I ask what if physics solves QM using Feynman path integrals instead? In contrast to Schrodinger, path integrals are time symmetric, what is completely different - "hidden variables" are not just in the past, but in both past and future. We could transform such solution to Schrodinger, but its state would be already optimized for all future measurements - as in superdeterminism. – Jarek Duda Feb 15 at 11:59
• Sorry Jarek but the question in bold makes no sense. Feynman path integral is simply a way to compute certain amplitudes, like $\langle x| e^{-i tH}| y \rangle$. It's not a different interpretation of quantum mechanics or anything like that. You can use whatever method you like to compute the ingredients that enter the Bell's theorem. Usually the latter is presented in CHSH form, i.e. for two qubits. Hence the computation boils down to computing expectation values of a $4\times 4$ matrix. Using the path integral (which for spin is not even rigorously defined) seems a bit a waste of resources. – lcv Feb 20 at 9:25
• @Icv, in path integrals you find solution based on very different boundary conditions - this x,y you have writen, which are time symmetric: x is in the past, y is in the future. Beside Bell's hidden variable in 'x', there is not included additional information hidden in y - literally after the measurement. It is more intuitive to think about it using spatial symmetry instead - using Ising model which has nearly the same math: Feynman -> Boltzmann path ensemble. Please think about probability distribution inside Ising sequence - getting Pr(i)=(psi_i)^2 from symmetry (e.g. diagram above). – Jarek Duda Feb 20 at 10:37
• @Jarek Duda Local realism and hidden-variable models are not necessarily the same thing. For example Bohm's hidden-variable "implicate order" model is compatible with the nonlocality demonstrated by tests of Bell's theorem. I think you may be able to clarify your question by removing all mention of hidden variables. – Guy Inchbald Feb 20 at 14:24

I’m answering the question you asked at the end:

Assuming physics solves its equations in symmetric way, e.g. QM with Feynman path integrals instead of Schrödinger equation, do Bell's assumptions hold - are local realistic "hidden variables" still disproven?

The short answer is “Yes.” Solving QM with path integrals is equivalent to using Schrödinger’s equation, so the details of Bell’s inequalities are irrelevant. Whether you use path integrals or Schrödinger’s equation you will get the exact same results.

There are two sides to Bell's inequalities. On one side there is a quantum mechanical calculation and on the other side there is a calculation using an unspecified local hidden variable theory.

If you do look at the details of the quantum side of Bell’s inequalities, you will notice that its derivation only depends on the initial entangled state and calculations of correlations. Neither path integrals nor Schrödinger’s equation are being used.

• While we can transform between solutions found with Schrödinger and Feynman path integrals, how would the "hidden variables" look like in the latter? It uses path ensembles for symmetric boundary conditions: in the past and future - wouldn't it need "hidden variables" in both past and future? – Jarek Duda Feb 17 at 5:47
• Bell's inequality has nothing to do with entangled states or QM for that matter. It is precisely the opposite – lcv Feb 17 at 7:05
• @icv, these are inequalities derived from locality and realism assumptions, which are violated by physics, hence at least one of these assumptions is nonphysical. Usually the locality is suspected, but faster-than-light communication might not be the only way to dissatisfy, accepting time/CPT symmetry for solving such models seems another way (?) Indeed this is more general issue, e.g. also electromagnetism and general relativity seem local realistic "hidden variable" models - do Bell theorem also disprove them? No if solving them using the least action principle instead of Euler-Lagrange. – Jarek Duda Feb 17 at 8:38
• @icv, I added clarifications on the two sides of Bell's inequality. You are right that in the hidden variable side there is no QM. But there is also a QM calculation and it does involve an entangled state. – udifuchs Feb 18 at 6:44
• Richard Gill explains that symmetric way of solving models does not satisfy "no-conspiracy" hidden assumption of Bell theorem. It agrees with en.wikipedia.org/wiki/Counterfactual_definiteness quote: "no conspiracy (called also "asymmetry of time")" – Jarek Duda Feb 18 at 9:24

Assuming physics solves its equations in symmetric way, e.g. QM with Feynman path integrals instead of Schrödinger equation, do Bell's assumptions hold - are local realistic "hidden variables" still disproven?

Yes. To check this for spins, you can take a spin to a point $$\hat{n}$$ in the group manifold $$SU(2)$$. Then take wave function to be defined by the path integral over paths $$\mathcal{C}_{\hat n(t)}\equiv \mathcal{C}(\hat{n}_0(t_0)|\hat{n}(t))$$ where the system is a point $$\hat{n}_0$$ at time $$t_0$$ and at point $$\hat{n}$$ at time $$t$$.

$$\psi(\hat{n},t)\equiv\int\!d\mathcal{C}_{\hat n(t)} \exp\left(- \int_{\mathcal{C}_{\hat n(t)}}\hat{n}\cdot d\hat{n}\right).$$

The statistics of the the wavefunction $$\psi$$ are exactly those of quantum mechanics and will satisfy Bell's theorem at time $$t$$. This path integral is also exactly what you integrate over. Thus the path integral, on any time slice, contains Bell's theorem. In other words, Bell's theorem is a consquence of superpositions, which is exactly what you are constructing in the path integral.

• But what if measurement is between this time t and t0? Do Bell's "hidden variables" contain both information before (n0) and after measurement (n)? Both are required to solve what you have written, but I believe "hidden variables" use only information before (?) Think about Ising model (e.g. diagram before) - the math is nearly the same (Feynman -> Boltzmann path ensemble), but past/future become more intuitive left/right, still getting e.g. Pr(i)=(psi_i)^2 probability distribution inside. – Jarek Duda Feb 20 at 11:03
• All the measurements are at the time $t$. Just the order in which they are applied is important. – Eric David Kramer Feb 20 at 12:37
• Ok, in this case you are right. But I have meant solving (CPT symmetric) physics in time-symmetric way: imagine t0 = -infinity, t = +infinity, measurement is between them. Is Bell locality assumption still satisfied in this case? While it is highly nonintuitive, Ising model is great to gain intuitions here - as it uses spatial instead of temporal direction in analogous way. – Jarek Duda Feb 20 at 12:50
• The path integral has a slicing formula: $A(t_2,t_0)=\int\!dt_1\,A(t_2,t_1)A(t_1,t_0)$ so you can always insert your operators at $t_1$ and have a wave function there. – Eric David Kramer Feb 20 at 12:56
• Indeed, but if measurement is in your t1 in the center, we have symmetrically kind of two "hidden variables": one from past, one from future, as in TSVF: en.wikipedia.org/wiki/Two-state_vector_formalism or Ising model: one amplitude from left, one from right - we get Born rule from symmetry ... instead of standard Kolmogorov probability theory in which we can derive Bell-like inequalities. – Jarek Duda Feb 20 at 13:08