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Why this equation is an integral over all paths?

  1. Jan 19, 2012 #1
    Dear all,

    I have decided to begin working my way through Quantum Field Theory in a Nutshell (Zee), and hopefully thoroughly. Here is one simple question from page 11, relating to the path integral formulation of QM.

    Equation (4) is

    [tex]<q_F|e^{-iHT}|q_i>=\int{}Dq(t)e^{i\int_0^T dt \frac{1}{2}m{\dot{q}}^2} [/tex]

    where

    [tex]\int{}Dq(t)=\lim_{N\to\infty}\left(\frac{-im}{2\pi \delta t}\right)^{\frac{N}{2}}\left(\prod_{k=1}^{N-1}\int dq_k\right)[/tex]

    and then Zee states that

    I was just wondering how one can tell that this is an integral over all possible paths? As far as I can tell, it could just be a multi-dimensional volume integral (i.e. the [itex]\prod_{k=1}^{N-1}\int dq_k[/itex] term).

    Could someone please dot my "i"s?

    Ianhoolihan
     
  2. jcsd
  3. Jan 20, 2012 #2

    Matterwave

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    I think the previous few pages of Zee explain quite nicely why this is a sum over all paths. Did you read the previous few pages? He spends them all developing the idea that this is a sum over paths.
     
  4. Jan 20, 2012 #3
    Thanks Matterwave,

    I understand the first few pages, and this heuristic argument. However I was wondering how one tells from the mathematical form of the equation that it is a sum over paths?

    Ianhoolihan
     
  5. Jan 20, 2012 #4

    tom.stoer

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    Only the derivation and the physical context tell you that it is a sum over paths. A formula alone is not sufficient.

    Why is U(t) = exp[-iHt] a time-evolution operator? This can be understood only if you know what H is, how it acts on states and other observables (via Heisenberg equation of motion).
     
  6. Jan 20, 2012 #5
    Although an integral with a capital D is usually reserved for path integrals, D in this case is a functional and is sometimes notated as

    [tex]\int D[q] [/tex]
     
  7. Jan 20, 2012 #6
    Thanks maverick_starstrider,

    How do you know that

    [tex]\int{}Dq(t)=\lim_{N\to\infty}\left(\frac{-im}{2\pi \delta t}\right)^{\frac{N}{2}}\left(\prod_{k=1}^{N-1}\int dq_k\right)[/tex]

    is a functional? I know that

    [tex]S[q(t)]=\int_{t_1}^{t_2}L[q(t),\dot{q}(t),t]dt[/tex]

    is a functional, as for a given function [itex]q(t)[/itex] one can operate on it with [itex]S[/itex] to return a scalar [itex]S[q(t)][/itex]. I do not see how this follows, as I don't see how [itex]D[q(t)][/itex] is a functional?

    Is the solution to my problem that here the [itex]dq_k[/itex] are not variables of integration, such as [itex]dx,dy,dz[/itex], but actually are specific to the given curve in the sense of at each position [itex]k[/itex], you have a triangle at each point on the curve, of width [itex]\delta t[/itex] and height [itex]dq_k[/itex]? Then one takes the limit for [itex]N\to\infty[/itex] of the sume of the area of these triangles? So it is not a usual integration (where the "height" is [itex]h+dq_k[/itex] as such), but still takes a function to give a scalar?

    Ianhoolihan
     
  8. Jan 20, 2012 #7
    Because you're integrating over all paths through q_k for all q_k's
     
  9. Jan 20, 2012 #8
    I still don't see why it is over all paths instead of a single given path. As far as I can tell, in the equation

    [tex]<q_F|e^{-iHT}|q_i>=\int{}Dq(t)e^{i\int_0^T dt \frac{1}{2}m{\dot{q}}^2}[/tex]

    you take a given path [itex]q(t)[/itex] from [itex]q_i[/itex] to [itex]q_F[/itex] and evaluate the function [itex]D[q(t)][/itex] and then the remaining path integral [itex]\int_0^T dt \frac{1}{2}m{\dot{q}}^2[/itex] along [itex]q(t)[/itex]. So where does the "integrating over all possible paths" come from, given you are doing it for a given [itex]q(t)[/itex]?

    Ianhoolihan
     
  10. Jan 20, 2012 #9

    tom.stoer

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    No - but in order to understand that you have to understand Feynman's derivation. What he basically did is to insert

    [tex]\int dx\,|x,t\rangle\langle x,t|[/tex]

    for each timeslice t.

    So you do not start from a single path
     
  11. Jan 21, 2012 #10
    Is not my procedure in the previous post correct for evaluating [itex]<q_F|e^{-iHT}|q_i>[/itex] , in a purely mathematical sense? Where does it mathematically indicate to sum over all possible paths? If it does not, then how would one go about actually evaluating [itex]<q_F|e^{-iHT}|q_i>[/itex] over all possible paths?

    I understand how [itex]\int dx\,|x,t\rangle\langle x,t|[/itex] was added in between each time slice. However, I don't see how that is relevant (possibly as I don't fully understand its role). For example, in a normal functional, I could add [itex]x_i+1-x_i[/itex] as a factor between each [itex]dx[/itex] slice in the normal riemannian sum, and then take the limit, and obtain exactly the same standard integral [itex]\int dx[/itex]. It doesn't mean I have to sum over all possible paths.

    When you answer, please quote the parts you addressing, so I don't get confused.

    Ianhoolihan
     
  12. Jan 21, 2012 #11

    tom.stoer

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    It seems that you don't understand the equation

    [tex]<q_F|e^{-iHT}|q_i>=\int{}Dq(t)e^{i\int_0^T dt \frac{1}{2}m{\dot{q}}^2}[/tex]

    and its derivation.

    There is no a priory (pure mathematical) reason for this equation to hold, it requires a proof. But exactly during this proof the insertion of [itex]\int dx\,|x,t\rangle\langle x,t|[/itex] IS relevant, and it's excatly this procedure which gives us the interpretation as a path integral. w/o this insertion you can neither claim the equatility to hold, nor can you present a reasonable interpretation.

    The l.h.s. of your equation need not be evaluated using path integrals; you can apply several different ideas. But there is one formal manipulation (Feynman's) which tells you that it can be interpreted as the r.h.s. Note that numerous physicists calculated expressions like the l.h.s. decades before Feynman invented the r.h.s.

    Are you aware of the fact that Feynman's derivation first gives you a Hamiltonian path integral which only via a Gaussian integration d[p] transforms into the Lagrangian one? It is by no means clear that this Gaussian integration always works b/c it depends crucially on the presence of a p² term. In more complicated cases for the kinetic energy e.g. for particles moving on curved manifolds, this d[p] integration is no longer trivial.
     
    Last edited: Jan 21, 2012
  13. Jan 21, 2012 #12
    I clearly don't understand the equation =). But what was wrong with my mathematical interpretation? If I had a function [itex]q(t)[/itex], could I not evaluate it as I have described previously? How would I evaluate it with two, or infinitely many, functions/paths?

    I agree --- and I followed the proof. As you suggest, I must be missing the point regarding

    [tex]\int dx\,|x,t\rangle\langle x,t|[/tex]

    Can you please explain why this implies a sum over all paths? All I know is that [itex]\int dx\,|x,t\rangle\langle x,t|=1[/itex].

    Ianhoolihan
     
  14. Jan 21, 2012 #13

    tom.stoer

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    We start with

    [tex]\langle x_f|e^{-iHT}|x_i\rangle = \langle x_f|e^{-iHT/2}e^{-iHT/2}|x_i\rangle = \int dy\,\langle x_f|e^{-iHT/2}|y\rangle\langle y|e^{-iHT/2}|x_i\rangle [/tex]

    Instead of having a particle travelling from xi at t=0 to xf at time t=T we now have a particle travelling
    i) from xi at t=0 to all points x at time t=T/2 and then
    ii) from each x at t=T/2 to xf at t=T
    This is now an integral over all contributions from all straight lines (the paths) from xi to and all straight lines from x to xf. Note that x can take any value.

    What we want to do is to get rid of the exponential exp(-iHt) and replace it with 1-iH*dt, therefore we do this slicing infinitly many times, i.e. we introduce x, x', x'', for t=0, t=t', t=t'' etc. where t''-t' = dt.

    That means we get straight lines from intermediate x'(t') to x''(t'') with t''-t' = dt, but x'' and x' (and all other x''', ...) are integrated over, and t', t'', ... act as indices labelling x', x'', ...

    This justifies the interpretation as a path integral.

    Note that I have omitted the dp integrations which are required for the derivation, but which drop out when doing the Gaussian integral d[p].
     
    Last edited: Jan 21, 2012
  15. Jan 21, 2012 #14
    [Solved] Why this equation is an integral over all paths?

    Aha! That was what I was looking for tom.stoer!

    Very good explanation as well,

    Ianhholihan
     
  16. Jan 22, 2012 #15

    tom.stoer

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    You're welcome.

    You are saying that you are reading 'Quantum Field Theory in a Nutshell' from Zee - and that my explanation was what you was looking for. So it's not explained that way in that book? Strange b/c that's the basic explanation in QM when introducing PIs.
     
  17. Jan 22, 2012 #16

    vanhees71

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    This book is more confusing than illuminating anyway. I've never understood the hype made about it, given the fact that there are many good qft books around. My favorites are:

    To begin with: Ryder, Bailin-Love
    To learn all the details: Weinberg, Quantum Theory of Fields
     
  18. Jan 22, 2012 #17

    tom.stoer

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    I don't know Bailin-Love, but I studied Ryder intensively and I know Weinberg, so I can fully agree .
     
  19. Jan 22, 2012 #18
    I've read the first few chapters of Weinberg, and my lecturer considers it one of the only good textbooks on QFT. He thinks Ryder has errors, and to be honest, after trying to make sense of his introduction to spinors, I passed it over.

    I passed on Weinberg for something a little more approachable and heuristic. Zee seems good so far. He did describe Feynman's path integral formalism well, but missed the key point regarding why adding [itex]\int dx\,|x,t\rangle\langle x,t|[/itex] results in an integral over all paths.

    I haven't heard of Bailin-Love --- I'll look into it.

    Thanks again,

    Ianhoolihan
     
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