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Path integral Quantum Mechanics

  1. May 3, 2014 #1

    Matterwave

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    Hi guys, I have a few questions regarding Feynman's formulation of quantum mechanics.

    Given the propagator K(x',t';x,t), when is this propagator equal to:

    $$K(x',t';x,t)=Ae^{\frac{i}{\hbar}S_{cl}(x',t';x,t)}$$

    Where S_cl is the classical action evaluated along the classical path of motion (i.e. where δS=0), and A is some normalization factor (which we we will get to in a moment)? I have seen arguments from method of stationary phase of the path integral:

    $$K(x',t';x,t)=\mathcal{N}\int_x^{x'} e^{\int_t^{t'}\mathcal{L}dt}\mathcal{D}x$$

    Where ##\mathcal{N}## is another normalization constant (let's not worry too much about convergence of this integral at this point, since I have only basic questions). To show that the most contributions will come about for paths along stationary action if the action is large compared to ##\hbar##. But this is an approximation.

    We know; however, from direct integration of the Schrodinger equation that the first equation is identically true (not just as an approximation) for e.g. the free particle propagator. So, this prompts me to ask, when is the first equation true in general and not just as an approximation?

    I have seen an argument that the first equation will be identically true, and not just approximately true, given that the potential V(x) is at most quadratic in x (so it would work for e.g. the Harmonic oscillator as well). Is this so?

    My next question is in regards to the normalization constant A. I have seen it given, without proof, as:

    $$A=\sqrt{\frac{i}{2\pi\hbar}\frac{\partial^2 S_{cl}(x',t';x,t)}{\partial x'\partial x}}$$

    How general is this formula? Is this also for cases where the potential is at most quadratic in x?
     
  2. jcsd
  3. May 3, 2014 #2
    Yes, it is true if the Lagrangian is at most quadratic in ##x## and ##\dot x##. In this case the path integral is an infinite-dimensional version of a Gaussian integral, and evaluates to the expression you gave.
     
  4. May 3, 2014 #3

    Matterwave

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    And in cases where the Lagrangian is more than quadratic in ##x##, the equation is only approximately true for large actions?

    This point has always confused me in many treatments of the path integral formulation, for example, the formulation found in Gordon Baym's lecture notes. He goes and first gets the first equation for a free-particle, and then he goes on to break the interval up into many little pieces to derive the path integral, and then arrives at the exact same answer he started out with. I always wondered, why in the world, if you have the solution already for the entire interval, would you go back and break up this interval up into little pieces and make things 100 times more complicated for yourself? -.-
     
  5. May 3, 2014 #4

    bhobba

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    Its the standard calculus thing - by splitting the interval and going to zero you get an integral in the action.

    You start out with <x'|x> then you insert a ton of ∫|xi><xi|dxi = 1 in the middle to get ∫...∫<x|x1><x1|......|xn><xn|x> dx1.....dxn. Now <xi|xi+1> = ci e^iSi so rearranging you get
    ∫.....∫c1....cn e^ i∑Si.

    Focus in on ∑Si. Define Li = Si/Δti, Δti is the time between the xi along the jagged path they trace out. ∑ Si = ∑Li Δti. As Δti goes to zero the reasonable physical assumption is made that Li is well behaved and goes over to a continuum so you get ∫L dt.

    Now Si depends on xi and Δxi. But for a path Δxi depends on the velocity vi = Δxi/Δti so its very reasonable to assume when it goes to the continuum L is a function of x and the velocity v.

    Its a bit of fun working through the math with Taylor approximations seeing its quite a reasonable process.

    In this way you see the origin of the Lagrangian. And by considering close paths we see most cancel and you are only left with the paths of stationary action.

    Go and get copy of Landau - Mechanics where all of Classical mechanics is derived from this alone - including the existence of mass and that its positive. Strange but true. Actually some other assumptions are also made, but its an interesting exercise first seeing what they are, and secondly their physical significance. Then from that going through Chapter 3 of Ballentine: QM - A Modern Development.

    You will emerge shaking your head at the awesome power of symmetry in physics. Don't take my word for it - experience the revelation for youself - its life changing.

    Thanks
    Bill
     
    Last edited: May 4, 2014
  6. May 3, 2014 #5

    Matterwave

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    Ok, thanks for the recommendation.
     
  7. May 3, 2014 #6

    bhobba

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    The convergence of the path integral is a very very difficult issue, even its existence is a problem.

    Some recent work on so called Hida distributions has shed light on it.

    The issue though is Hida distributions were not developed in physics, but in the applied math area of stochastic modelling so it may take mathematical physicists a little while to incorporate its insights. I first came across this years ago when I was into Stochastic modelling, so it all may be sorted out now.

    Thanks
    Bill
     
  8. May 3, 2014 #7

    Matterwave

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    I'm aware of the convergence issues for the path integrals. I am, in fact, reading a monograph at this moment about these such issues. I am not able to follow all the arguments in my first read-through, so I wanted to clarify some pre-existing issues I had with the formulation first, which is why I posted here.

    But regardless of the convergence of the path integral, we should be able to obtain the propagator itself by solving the Schrodinger equation (it is, after all, a solution for the Schrodinger equation with ##\psi(x,t_0)=\delta(x-x_0)##) right? So the propagator itself is always well defined (as long as the S.E. has a solution), but the path integral may not be.

    It would just be formally enlightening if it is indeed always possible for us to express the propagator in such a path-integral expression, so work should/is being done on trying to define this path integral for a wider class of applications?
     
  9. May 4, 2014 #8

    bhobba

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    Yep - as far as I know anyway.

    Hida distributions do solve the issue.

    But beware - it advanced (you need to be familiar with and understand the language of Rigged Hilbert Spaces for example) and the tomes on it are from stochastic modelling - not physics - at least the ones I own anyway - I was into it at one time.

    Thanks
    Bill
     
  10. May 4, 2014 #9

    Matterwave

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    By "Hida distributions do solve the issue", do you mean they solve the convergence issues given any arbitrary Hamiltonian? I have seen Wiener measure regularizations in the phase space path integral which give a real Wiener measure for the path integrals to give the propagator in a coherent state representation, but these regularizations are limited at least to polynomial (in P and Q) Hamiltonians.

    Also as a follow up, I'm wondering if these issues of convergence of path integrals have anything to do with renormalization? I've only seen renormalization in a very basic introduction (integrals of momenta over loops in the Feynman diagrams).
     
  11. May 4, 2014 #10

    ftr

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    Just in case that you have not seen this book by Zee. It goes through path integral very thoroughly.


    https://www.google.com.kw/url?sa=t&...=_HuxvbAtzp1iaxj6uTiTgg&bvm=bv.65788261,d.ZGU


    It will take some time to download.
     
  12. May 4, 2014 #11

    bhobba

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    If you want to delve into Hida distributions be my guest eg:
    https://www.amazon.com/Introduction-Hida-Distributions-Si/dp/9812836888

    How it applies to Feynman Integrals:
    http://arxiv.org/abs/0805.3253

    Its been years since I was into this sort of stuff so I wont be able to assist - good luck.

    This has nothing to do with renormalisation which is entirely different:
    http://arxiv.org/pdf/hep-th/0212049.pdf

    The fundamental idea behind renormalisation is that some quantities in your equations blow up to infinity and yet you use them to perturb around - which obviously wont work. Quantities you perturb around should be small, and infinity is the worse choice of all - they are called the bare quantities. So what you do is instead perturb around a quantity that is in a tricky way small (this is guaranteed by the fact its fixed by measurement) - they are called renormalised.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  13. May 4, 2014 #12

    Matterwave

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    Ok Thanks for the sources. :)
     
  14. May 4, 2014 #13

    king vitamin

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    Yes, you can see this just by considering the propagator for energy eigenstates, which is trivial:

    [tex]K(n',t';n,t) = \delta_{n,n'}e^{-iE_n(t'-t)}[/tex]

    then you can get your desired propagator from a simple change of basis:

    [tex]K(x',t';x,t) = \sum_{n}\langle x'| n \rangle \langle n | x \rangle e^{-iE_n(t'-t)} .[/tex]

    So you have this representation of the propagator with a pretty small number of assumptions.
     
  15. Jul 3, 2014 #14

    RUTA

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    In Zee Chapter III.5 Field Theory without Relativity, first two pages, he obtains the free-particle Lagrangian density in ψ for Schrodinger's eqn from the non-relativistic limit of the free scalar field of the Klein-Gordon equation. What I'd like to do is relate the corresponding transition amplitude Z, computed using ψ, to the free-particle propagator K given in this thread. In QM, one computes the propagator K for a free particle to go from (x,t) to (x',t'). In QFT, one computes the transition amplitude Z for the creation of a particle at (x,t) and its later annihilation at (x',t'). I've looked at dozens of lecture notes and text excerpts online, but I haven't found anyone who relates K and Z for the simple free particle case. If QM can be obtained from QFT as Zee shows in this case, then I should be able to compute the free-particle ψ(x',t') directly from the free-particle propagator K (e.g., with ψ(x,t) = δ(x), as is commonly done) or using the non-relativistic free scalar field transition amplitude Z. Has anyone seen this done?
     
  16. Jul 3, 2014 #15

    atyy

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    I'm not sure for that particular formula, but the idea behind The Duck's comments is that in general the classical part is only the stationary phase or saddle point approximation that you mentioned.

    In some cases like non-Abelian gauge theories, the path integral is easier to calculate in. In general, one cannot just write down a path integral and hope that it will correspond to a quantum theory with operators and a Hilbert space. To solve the problem of mathematically defining the path integral, one rotates to imaginary time, so that the path integral becomes a classical stochastic process which is mathematically well-defined (eg. the old theory of Wiener measure), and then rotates the answer back to real time to get quantum mechanics instead of a stochastic process. The partition function of quantum mechanics is formally analogous to what probabilists call a generating functional, eg. http://arxiv.org/abs/1009.5966. The path integral should be seen as just an amazing calculational trick. One cannot just write down any path integral describing a stochastic process and hope that the corresponding quantum theory exists. In relativistic QFT, one set of conditions for the path integral to be a meaningful quantum theory are the Osterwalder-Schrader axioms http://www.einstein-online.info/spotlights/path_integrals. Strocchi's book has some information about these issues in QM https://www.amazon.com/Introduction-Mathematical-Structure-Quantum-Mechanics/dp/9812835229.
     
    Last edited by a moderator: May 6, 2017
  17. Jul 5, 2014 #16

    RUTA

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    Sorry, I was being stupid. ψ(x',t') is the solution to the homogeneous SE. To make use of Z(J) from the non-relativistic KG equation, specifically the Green's function aka the QFT free field propagator, one needs both ψ(x,t) = δ(x) as a Source J and V(x')ψ(x') as a sink J. See section 17.1 through equation 17.19 from http://www.phy.ohiou.edu/~elster/lectures/relqm_17.pdf
     
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