Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Osterwalder-Schrader theorem and non-relativistic QM

  1. Dec 3, 2017 #1

    king vitamin

    User Avatar
    Gold Member

    I have recently learned a bit about the Osterwalder-Schrader theorem. From my understanding, this tells you when a Euclidean path integral can be analytically continued to a valid relativistic Hermitian quantum field theory (one needs reflection positivity etc.).

    I am curious about corresponding results for non-relativistic theories. Let's say I have a Euclidean field theory where there isn't full rotational symmetry; if a definite example is needed, let's say I want it to Wick-rotate to a Galilean-invariant theory. Is there a corresponding theorem for what properties are needed in the Euclidean theory for the Wick rotation (which is used all the time) to guarantees a valid Hermitian quantum field theory?

    I don't have an extremely strong background in the mathematics behind AQFT. I did find this statement of the OS theorem by Urs Schreiber (who I know posts here), which gives rotational invariance as an assumption. So my question boils down to whether this assumption can be relaxed.
     
  2. jcsd
  3. Dec 4, 2017 #2

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    By a Wick rotation one obtains from the symmetry group ##ISO(4)## of the Euclidean path integral one with symmetry group ##ISO(1,3)##, which is the Poincare group and not the Galilei group. To get the latter you'd need to perform a contraction instead of a Wick rotation.
     
  4. Dec 4, 2017 #3

    rubi

    User Avatar
    Science Advisor

    In order to reconstruct a quantum theory from a path integral, you just need reflection positivity (in order to get the Hilbert space) and (imaginary) time-translation invariance, in order to reconstruct the (imaginary) time-evolution operators. If time is continuous and the time evolution is strongly continuous, you can reconstruct the Hamiltonian, which generates ordinary unitary time-evolution. All the other baggage in the OS axioms just ensures that the corresponding quantum theory is a Wightman quantum field theory, but if you don't care about that, then the subset of the axioms that I stated, is enough.
     
  5. Dec 4, 2017 #4

    king vitamin

    User Avatar
    Gold Member

    I don't follow you here. I am asking about Euclidean-signature theories which do not have ##ISO(4)## symmetry and their relation to quantum theories which do not have ##ISO(1,3)## symmetry. There is certainly a Wick rotation involved, as seen in any textbook. I assume by "contraction" you're referring to the Inönü-Wigner method for getting the Galilei group from the Poincaré? Since I'm not interested in Poincaré invariance I do not understand this comment either.

    This is really helpful, thanks. To repeat what I think I have understood of your post: if I want to apply the path integral to a quantum system with a finite Hilbert space (e.g. it is not a continuum field theory), I can formulate it in terms of a Euclidean path integral where the Euclidean theory requires reflection positivity and imaginary time-translation invariance. The other assumptions of the statement of OS are only needed if I want to guarantee that the corresponding theory is also a "Wightman QFT."

    Unfortunately I'm not really familiar with what a Wightman QFT is; I'm looking at the Scholarpedia article now. It seems that it is specifically a Poincaré invariant QFT. I am largely interested in field theories within condensed matter/stat mech where we do not worry much in the passage from a finite lattice to the continuum limit, so I'm not too worried these extra issues provided the finite-dimensional case holds. My question mostly stemmed from thinking of mappings between theories of classical and quantum critical phenomena.
     
  6. Dec 5, 2017 #5

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    One can get the Galilei group both from ISO(4) and from ISO(1,3) by Inönü-Wigner contraction.

    If the field theory you start with has no big symmetry group such as ISO(4) then Wick rotation also produces a theory with little symmetry only - not with the Galilei group.

    In this case you cannot get the Galilei-invariant theory you had asked for. It needs fields defined on 3 space variables.
     
  7. Dec 5, 2017 #6

    king vitamin

    User Avatar
    Gold Member

    But there are certainly Euclidean field theories which Wick rotate to Galilean invariant quantum field theories. I still don't see the point of starting with ##ISO(4)## symmetry. In any case, as I said in my first post, the Galilean symmetry was only mentioned if a definite example is needed.

    Let's forget the Galilei group. Assume I start with a continuum Euclidean field theory with full translation invariance along one of the dimensions, and no assumptions about the symmetry along/between the others. I now choose the translation-invariant direction as my "imaginary time" direction which will be Wick rotated. I'm curious about what assumptions are needed for the Wick rotated theory to be a valid QFT.

    My understanding from rubi's post is that there is some extra baggage in the OS theorem needed for us to get a "Wightman QFT", and while I don't think I really care about this for the applications I'm currently considering, I am curious how important the assumption of spacetime symmetry is.

    Yes, that much is obvious (perhaps you missed the context of the sentence you've quoted?).
     
  8. Dec 6, 2017 #7

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    You get a valid quantum theory but it has no symmetries apart from time translation symmetry. The relevent piece of theory is the Feynman-Kac formula. Whether the result is a valid QFT depends on your definition of the latter. (There is no standard definition in the nonrelativistic case.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Osterwalder-Schrader theorem and non-relativistic QM
Loading...