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Path integral formula so hard

  1. Apr 15, 2015 #1
    Here's the source:http://web.mit.edu/dvp/www/Work/8.06/dvp-8.06-paper.pdf

    Regarding page 5 of 14, I don't understand the multiple integrals thing. What is that supposed to mean? Ain't we supposed to sum up all the paths but why do they do the multiple integral thing?

    Also regarding page 4 of 14 section 2.1 for the first few sentence, why does the difference in action differs ∆S≈πh where h here is the Dirac constant?
     
  2. jcsd
  3. Apr 15, 2015 #2
    The path between point A and B has been made into a collection of paths t_i. In formula (5) the action is found for a free particle to propagate from one line segment to another, thus, since we will take a continuum limit, the total amplitude is a multiple integral over the collection of paths.
    Think interference of waves for understanding your second point, as the action acts as a phase factor in the path integral. Have a nice day.
     
  4. Apr 15, 2015 #3

    vanhees71

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    I refuse to read the paper, because already Eq. (1) is plain wrong or better said it doesn't make sense already from its syntax!
     
  5. Apr 15, 2015 #4
    Note that this treatment is one-dimensional, the small intervals are slices in time evolution. The particle can go forwards and then backwards infinite times in infinite places, while time goes forwards. This means the integral takes into consideration really all the possible paths. This approach can be then generalized into 3 dimensions.

    The sentence says "the corresponding contributions will cancel": if you plug in that value for the action delta into equation (4) you get a Pi phase shift, which means that when two of these phase vectors (complex numbers of norm 1) are summed the resulting amplitude is zero.

    It's called Planck's constant, not Dirac constant.

    I strongly suggest you follow the reference book given, Shankar's Principles of Quantum Mechanics, everything is explained in more detail there.
     
  6. Apr 16, 2015 #5

    vanhees71

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    And I hope in a correct way! I'm pretty sure, because Shanakr's text is recommended frequently in these forums. Of course, the standard source is

    Feynman, Hibbs, Quantum mechanics and path integrals

    There's a rather new edition with a lot of typos corrected. It's a great book.

    The most comprehensive book I know is

    Hagen Kleinert
    Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets 5th edition, World Scientific 2009
     
  7. Apr 16, 2015 #6
    The paper linked basically follows Shankar. Also I think Shankar really succeeds in making the topic reasonable for beginners without being simplistic.
     
  8. Apr 16, 2015 #7
    I was reading the book you recommend but then I was stuck at one part. What do they mean by the phase?
    E.g. For a classical particle of mass, say 1g, the action changes by roughly (1.6*10^26)(h), and the phase by roughly 1.6*10^26 rad as we move from the classical path x=t to the nonclassical path x=t^2.

    Where h is the planck constant
     
  9. Apr 16, 2015 #8
    A complex number ##a+ib## can also be expressed as ##\rho e^{i \phi}## where ##\rho## is called amplitude and ##\phi## the phase in the language of signals and waves.
     
  10. Apr 16, 2015 #9
    So for this, the phase is relates to the wave property displayed by the classical particle? You mean the trajectory of the classical particle can be thought of as the wave of that classical particle?
     
  11. Apr 16, 2015 #10

    DrClaude

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    Could you please explain, because I don't see the problem and would like to know what I'm missing.
     
  12. Apr 16, 2015 #11
    No, in this case the phase is simply that of the wavefunction representing the quantum particle.
     
  13. Apr 16, 2015 #12

    stevendaryl

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    Yeah, I didn't see what was wrong with it, either, except for a tiny flaw: The formula uses [itex]t_0[/itex] as an arbitrary starting time, but the expression [itex]e^{-i E t/\hbar}[/itex] only works if [itex]t_0 = 0[/itex]. The expression should really be: [itex]e^{-i E (t-t_0)/\hbar}[/itex]
     
  14. Apr 16, 2015 #13

    vanhees71

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    What's wrong is obviously to write an expression like ##|\psi(x,t) \rangle##. A state ket doesn't know a priory about any observables. It should read ##|\psi(t) \rangle## and to be complete you have to say in which picture of time evolution you work. Most QM texts work in the Schrödinger picture and only switch to the interaction picture when needed for perturbative scattering theory. Then the full time dependence is on the states, and operators representing observables are time in-dependent (except you have some explicit time dependence, but this we neglect here, and you rarely need it in practice anyway). Then also the eigenvectors of operators representing observables are, of course, independent of time.

    The wave function is a picture-independent quantity and always a (formal) scalar product between an eigenvector of an operator representing an observable (e.g., position) and a state ket:
    $$\psi(t,x)=\langle x|\psi(t) \rangle.$$
    This object "knows" something about the position of the particle, namely, according to Born's rule,
    $$P_{\psi}(t,x)=|\psi(t,x)|^2$$
    is the probability distribution for the position of the particle, given it is in the state represented by ##|\psi(t) \rangle## (or more precisely the state is reprented by the corresponding ray or the Statistical Operator ##|\psi(t) \rangle \langle \psi(t)|##.

    A notation like ##|\psi(x,t) \rangle## is confusing at best and leads to mistakes at worst!
     
  15. Apr 16, 2015 #14

    DrClaude

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    I completely agree. I even teach my students explicitly never to write that. But as you wrote "already Eq. (1) is plain wrong," I stopped reading at eq. (1) and couldn't figure out what was so wrong, but that ket only appears after eq. (1).
     
  16. Apr 16, 2015 #15

    vanhees71

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    Yep, it's equation (2). Sorry for the confusion!
     
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