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Dirac's K

  1. Jul 15, 2006 #1
    In http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html" [Broken], he mentions this:

    I'm trying to get the same result, but I'm stuck. Has anyone done this before?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 16, 2006 #2


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    Write [tex]L= \frac{1}{2} (\frac{x-x'}{\lambda})^{2} - V(x)[/tex] , change integration variable to [tex]x-x'=y[/tex] , then expand [tex]V(x),\phi(x)[/tex] about [tex]y=0[/tex] and keep first order terms in [tex]\lambda[/tex] ;
    [tex]\phi(x',t+\lambda)= A \int dy exp(\frac{iy^{2}}{2\lambda}) exp[-i\lambda (V(x') + yV'(x') + ..)][\phi(x',t) + y \frac{\partial {\phi(x')}}{\partial{x'}} + \frac{1}{2} y^{2}\frac{\partial^{2} \phi(x',t)}{\partial {x'}^{2}} +...][/tex]
    [tex]\phi(x',t+\lambda) = A[1 - i \lambda V(x')] \int dy exp[\frac{iy^{2}}{2\lambda}][\phi(x',t) + (1/2) y^{2} \frac{\partial^{2}\phi}{\partial {x'}^{2}}][/tex]
    we dropped the factor
    because when the integration over y is done this will be of order
    we also used
    [tex] \int dy y exp(-a y^{2}) = 0[/tex]

    Now use the integrals:
    [tex] \int dy exp(-a y^{2}) = (\pi/a)^{1/2}[/tex]
    [tex] \int dy y^{2} exp(-a y^{2}) = (\pi/2)^{1/2}(1/2a)[/tex]

    Schrodinger eq follows when you choose A so that the coefficient of
    [itex]\phi(x',t)[/itex] is one, and take the limit [itex]\lambda\rightarrow 0[/itex];

    [tex]\frac{\phi(x',t+\lambda)-\phi(x',t)}{\lambda} = \frac{\partial}{\partial{t}} \phi(x',t)[/tex]


    Last edited by a moderator: May 2, 2017
  4. Jul 17, 2006 #3
    Thanks for the reply!
    I didn't know the bounds are infinities... Also, I couldn't get [tex]\lambda[/tex] to the 3/2th power, for [tex]-iV'(x') y \lambda[/tex], but as you did, I used only the first term for potential.

    Anyway, I couldn't see how Schrödinger equation comes out from what you've written, so
    I've tried to work out the details for myself, but I ended up a weird equation that is far from Schrödinger's. Here it goes:

    (You seem to've cranked up [tex]\hbar[/tex] and m to 1, but for keeping a better track of things, I'm leaving them there).

    [tex]L = \frac{m}{2} {(\frac{x-x'}{\epsilon})}^2 - V(x)[/tex]
    [tex]y = x-x'[/tex], keeping [tex]x'[/tex] fixed, [tex]dy = dx[/tex] (I guess this was what you meant).


    [tex]\varphi(x',t+\epsilon) = A \int_{-\infty}^{\infty} e^{\frac{im}{\hbar \epsilon}y^2} e^{-\frac{i}{\hbar} \epsilon V(x')} (\varphi(x',t) + y\frac{\partial \varphi(x',t)}{\partial x} + \frac{y^2}{2}\frac{\partial^2 \varphi(x',t)}{\partial x^2}) dy[/tex]

    [tex]\varphi(x',t+\epsilon) = A e^{-\frac{i}{\hbar} \epsilon V(x')} (\varphi(x',t)\sqrt{\frac{\pi}{-im/\hbar \epsilon}} + \frac{1}{2}\frac{\partial^2 \varphi(x',t)}{\partial x^2} \sqrt{\frac{\pi}{2}} \frac{1}{2(-im/\hbar \epsilon)})[/tex]

    To make coefficient of [tex]\varphi(x',t)[/tex] 1, A should obey:

    [tex]A e^{-\frac{i}{\hbar} \epsilon V(x')} \sqrt{\frac{\pi}{-im/\hbar \epsilon}} = 1[/tex]

    By making A so, I got:

    [tex]i\hbar \frac{\partial \varphi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \varphi}{\partial x^2} \frac{i}{2\sqrt2} \sqrt{\frac{\hbar \epsilon}{im}}[/tex]

    which has some weird coefficients for spatial derivative, and missing the potential term. Am I (or maybe you) doing something wrong??
    Last edited: Jul 17, 2006
  5. Jul 17, 2006 #4


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  6. Jul 17, 2006 #5
    Ups. I missed that bit. So my A became

    [tex]A \sqrt{\frac{\pi}{-im/\hbar \epsilon}} = 1[/tex]. [tex]V(x') \frac{\partial^2 \varphi}{\partial x^2}...[/tex] term is dropped since it has an second order [tex]\epsilon[/tex]. OK, by doing so, I had the potential term nicely, but I still have a weird extra coefficent in the same place. This time, it's this:

    [tex]\sqrt{\frac{im}{8 \hbar \epsilon}}[/tex].

    Can you spot the error this time?..
  7. Jul 17, 2006 #6
    Umm.it seems too easy :rolleyes: :rolleyes: if all physics were so easy we shouldn't have problem finding Quantum Gravity:

    As far as i know the Integral equation satisfied by the wave is:

    [tex] \psi(x',t')=\int_{-\infty}^{\infty}dxdtG(x,x',t,t')\psi(x,t) [/tex]

    with [tex] G(x,x',t,t')=A\int D[x]e^{iS(x)]/\hbar [/tex]

    [tex] S(x)=\int_{a}^{b}dt [(1/2)m(\dot x)^{2}-V(x)] [/tex]

    However the Functional integral above can't be done except for "Semiclassical Approach" or when V(x)=0 or V(x)=x^{2}
  8. Jul 17, 2006 #7


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    the error was in
    [tex] \int dy e^{-a y^{2}} = (\pi/2)^{1/2}(1/2a)[/tex]
    it should be
    [tex] \int dy e^{-a y^{2}} = (\pi/a)^{1/2}(1/2a)[/tex]
    I think my miss-keying is to blame :frown:


  9. Jul 18, 2006 #8
    Horray! :approve: Huge thanks, finally Schrödinger equation did come out!
    One last question, how could you (or Feynman) see which term you had to expand? I rather tried expanding [tex]e^{iLdt/\hbar}[/tex] directly, without expanding the potential term around a fixed point, and without leaving a [tex]e^{imv^2/2\hbar}[/tex]. It was all a mess. I really wonder how you did see?

    eljose, is that equation for QG?
  10. Jul 18, 2006 #9


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