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15316005 No.15316005 [Reply] [Original]

How to define the quantum mechanical path integral rigorously? I heard physicists use hand-waving when it comes to path integration but somehow it still works. How can a calculation without fundamental rigorous explaination just work?

>> No.15316031

Under some assumption regarding the potential you can express the path integral via Feynman-Kac formula as an integral with respect to Wiener measure.

>> No.15316083

>>15316031
>Wiener measure
will never not be amusing

>> No.15316105

There are some open questions in functional analysis which make it impossible to define it rigorously with current tools, and there are simpler problems like the i^{N/2} phase which is taken as real for any N in the Feynman formulation. In fact, i^{N/2} is only real for some N.

>> No.15316110
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15316110

>>15316105
>In fact, i^{N/2} is only real for some N.
#60

>> No.15316152

>>15316005
>How to define the quantum mechanical path integral rigorously?
Look up the derivation of the path integral for ordinary quantum mechanics (where operators only depend on time not space). Everything is completely rigorous. The path integral is defined in terms of discretized time steps of length \epsilon and any terms thrown away vanish in the limit of vanishing \epsilon. All fussy details about path integral measure are related to how you interpret terms in the path integral like p q, where p and q are operators that don't commute. If you look at the discretized form of the path integral there is no ambiguity because p and q are evaluated at different ends of the timestep.

Path integrals in quantum field theory are a little harder to be rigorous because of extra singularities when two operators are evaluated at the same point in space, but there is really nothing to worry about.

>> No.15316166

>>15316105
>There are some open questions in functional analysis which make it impossible to define it rigorously with current tools
Like what?

If I can calculate expectation values of any product of operators in the path integral approach and take the limit as the time step goes to zero, then it seems to me I have a well defined mathematical object. You can do this in ordinary quantum mechanics, and quantum field theory regulated on a lattice (which is basically ordinary quantum mechanics). Any difficulties of the continuum limit of quantum field theory are just regular renormalization issues which we know how to deal with perturbatively at least. Are mathematicians worried about something more interesting than this?

>> No.15316180

>>15316166
Nta, but the usual argument in this approach is to use a sequence of equipartitions of the time interval so you can use Trotter product formula. Can you get the same limit with any arbitrary sequence of partitions? (I'm genuinely too lazy right now to investigate this.) Also in what sense can the resulting measure on a function space be interpreteted as a uniform distribution as intended by Feynman when there is no equivalent to Lebesgue measure in infinite dimensions?

>> No.15316201

>>15316180
>Can you get the same limit with any arbitrary sequence of partitions?
Why does it need to be an arbitrary partition if equipartitions work to define it?

>Also in what sense can the resulting measure on a function space
I don't really care about thinking of the measure on function space directly. I'm saying the path integral gives a well-defined recipe for calculating expectation values of time ordered products, and the measure in so far as it needs to exist is defined implicitly by that. If it violates some axiom of a measure (or is difficult to prove) just invent a new concept of measure to correspond to the well-defined expectation values.

>> No.15316233

>>15316166
>Like what?
For example, off the top of my head, I do not believe it has been shown rigorously that the Dyson series converges to the path integral of a time dependent Hamiltonian that doesn't commute with itself at different times. Even if has, there are other things I can't remember right now. Actually, another one I can remember is that it has never been rigorously demonstrated that a quantum Yang-Mills theory can exist in 4D at all, or N=4, or whatever the Millennium Prize asks about.

Your comment "we do this in QM" suggests you have a some kind of rigor in mind other than mathematical rigor. I agree that path integrals work perfectly well for physics. Rigor isn't a question about physics in the usual semantics.

>> No.15316241
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15316241

>>15316233
>>15316233
https://www.claymath.org/sites/default/files/yangmills.pdf

>> No.15316243

>>15316201
>Why does it need to be an arbitrary partition if equipartitions work to define it?
Because all analytic constructions of integrals share this degree of freedom that they're invariant under choice of partitions. If Feynman's integral doesn't, then you'd need an argument why only equipartitions are physically meaningful.


>If it violates some axiom of a measure (or is difficult to prove) just invent a new concept of measure to correspond to the well-defined expectation values.
I guess that's what qualifies as hand-wavy.

>> No.15316258

>>15316233
>Your comment "we do this in QM" suggests you have a some kind of rigor in mind other than mathematical rigor.
Maybe, but I am under the impression that in ordinary quantum mechanics (this means 1 dimensional path integrals) everything is completely rigorous even to the most boring mathematician. Correct me if I'm wrong.

In quantum field theory (greater than 1 dimensional path integrals) there may be problems, but my claim is that you can turn this into a 1 dimensional path integral by discretizing space and putting the system in a finite volume. The connection between this regularization and the genuine quantum field theory is just the renormalization issue which physicists understand, even if it is not rigorous. But I was wondering if mathematicians that care about this stuff have anything more say.

> I do not believe it has been shown rigorously that the Dyson series converges to the path integral of a time dependent Hamiltonian that doesn't commute with itself at different times.
Physicists know that quite generically the Dyson series does not converge and is only an asymptotic series. Under certain situations you can use Borel summation to define it, but often not.

>> No.15316268

>>15316243
>I guess that's what qualifies as hand-wavy
If the expectation values are well-defined so is the roundabout definition of the measure. Mathematicians in other fields flip the definition like this too, see e.g. non-commutative probability.

>> No.15316300

>>15316258
>Correct me if I'm wrong.
Well, to show that the momentum operator is the derivative times a constant, you have to throw out some O(dx^2) terms in a Taylor series and that always bugged me since those terms are not "rigorously" equal to zero. Taking it as an axiom of QM that that operator is the derivative works, but that kind of sidesteps rigor rather than delivering it. I haven't thought about this much lately but there's probably other stuff too. For instance, doesn't the 1D path integral diverge when it's not restricted to arbitrary finite bounds? That's not what I would call "rigorous."

>> No.15316309

>>15316258
>path integral by discretizing space and putting the system in a finite volume.
Yes, I agree with your claim and I claim that same thing myself. That has literally nothing to do with rigor, however, IMO. Putting the equals sign between the integral and the discretized integral is probably a mathematical error and you **definitely** can't prove that it isn't an error, which would be required for what I am calling "mathematical rigor."

>> No.15316312

>>15316258
>if it is not rigorous.
This is what OP was asking about though:
>>15316005
>How to define the quantum mechanical path integral rigorously?

>> No.15316315

>>15316258
>Physicists know that quite generically the Dyson series does not converge and is only an asymptotic series. Under certain situations you can use Borel summation to define it, but often not.
That's a big problem for physics with magnetism then.

>> No.15316334

Basically, I think OP wants to know if a quantum Yang-Mills theory can exist in 4D. No one knows if it can, unless I missed something new.