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Green's function differential equation

  1. Jun 27, 2016 #1
    Hello I'm doing some problems in QM scattering regarding the Green's function.
    1. The problem statement, all variables and given/known data
    Determine the differential equation of [itex] G(\vec{r},\vec{r}',\omega) [/itex]

    2. Relevant equations
    I've been given the Fourier transform for the case where the Hamiltonian is time independent:
    [itex] G(\vec{r},\vec{r}',t-t')=\int \frac{d\omega}{2\pi} G(\vec{r},\vec{r}',\omega)e^{-i\omega(t-t')} [/itex]
    and the DE for [itex] G(\vec{r},\vec{r}',t,t')[/itex]:
    [itex] (i\hbar\partial_t -\hat{H})G(\vec{r},\vec{r}',t,t')=\delta(\vec{r}-\vec{r}')\delta(t-t')[/itex]

    3. The attempt at a solution
    I thought I would just plug the Fourier Transform into the given DE and get:
    [itex] (i\hbar\partial_t -\hat{H})\int \frac{d\omega}{2\pi} G(\vec{r},\vec{r}',\omega)e^{-i\omega(t-t')}=\delta(\vec{r}-\vec{r}')\delta(t-t')[/itex]
    I couldn't come up with anything else to do.
    Any hints are very appreciated :)
     
  2. jcsd
  3. Jun 27, 2016 #2

    Charles Link

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    I think I have something that may help. The partial w.r.t. to t can be put inside the integral and you get a factor ## - i \omega ##, so that you get a ## \hbar \omega ## factor. The ## - \hat{H} ## on another term can also be moved inside the integral. In order to get this delta function, we already have the makings of a delta function inside the integral : ## \int (1/(2 \pi)) exp(-i \omega (t-t') d \omega=\delta (t-t') ##. This means that the other factor inside the integral must be equal to ## \delta (\vec{r}-\vec{r}') ## i.e. ## (\hbar \omega -\hat{H})G(\vec{r},\vec{r}',\omega)=\delta (\vec{r}-\vec{r}') ## . This is your differential equation. (One additional detail: I think you have the sign reversed in this equation, i.e. it should read ## (\hat{H} -\hbar \omega ) G(\vec{r}, \vec{r}', \omega )=\delta(\vec{r}-\vec{r}') ## .) Note: I did a little editing. The first version wasn't completely correct...
     
    Last edited: Jun 27, 2016
  4. Jun 27, 2016 #3

    Charles Link

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    Please read post #2 again=I edited it because my first version contained a couple errors.
     
  5. Jun 27, 2016 #4
    Thanks alot for the help.
    I could follow you this far: [itex] (\hbar \omega - \hat{H})\int \frac{d\omega}{2\pi}G(\vec{r},\vec{r}',\omega)e^{-i\omega(t-t')}=\delta(\vec{r}-\vec{r}')\delta(t-t') [/itex]

    but why can I write it like this: [itex] (\hbar \omega - \hat{H}) G(\vec{r},\vec{r}',\omega) \int \frac{d\omega}{2\pi}e^{-i\omega(t-t')}=\delta(\vec{r}-\vec{r}')\delta(t-t') [/itex] so that [itex] (\hbar \omega - \hat{H}) G(\vec{r},\vec{r}',\omega) =\delta(\vec{r}-\vec{r}') [/itex]
    What I mean is, could you elaborate your last step please? :)

    (I haven't reverted the sign, I've written it as it's given in the problem statement. Maybe they have :) )
     
  6. Jun 27, 2016 #5

    Charles Link

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    The whole thing goes inside the integral. The integral is over ## \omega ## so that you can differentiate the integrand over t first, leaving any results, such as the ##-i \omega ## inside the integral. The ## \hat{H} ## can also be done inside the integrand and processed before the integration. Now here's the tricky part but its also kind of simple. You already have a factor inside the integrand that gives you ## \delta (t-t') ##. This means whatever else is there,(multiplying) inside the integrand must equal ## \delta (\vec{r}-\vec{r}') ## which is the other factor on the right side of the equation. This allows us to write down the differential equation without any additional computations. One puzzle is that this factor ## (\hbar \omega-\hat{H}) G(\vec{r}, \vec{r}',\omega)##, (which must equal ## \delta (\vec{r}-\vec{r}') ## ) appears to be independent of ## \omega ##, but that does not seem to be problematic. In fact, because you already have the makings of ## \delta(t-t') ## inside the integral, this factor must be independent of ## \omega ##.
     
  7. Jun 27, 2016 #6

    Charles Link

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    Besides my post #5, it might be worthwhile to review the Green's function methodology. The homogeneous differential equation is ## (i \hbar \partial_t-\hat{H})\Psi(\vec{r},t)=0 ##. The Green's function is used for the inhomogeneous equation where the right side has ## V(\vec{r},t)\Psi(\vec{r},t) ##. The Green's function solution writes a convolution integral of the Green's function with the inhomogeneous term (which is the often the driving term such as a force f(t) in other simpler differential equations such as a mass on a spring etc.) (This explains the minus sign=if the ## V ## here is positive on the right side, the sign gets reversed.) Meanwhile the Green's function is the solution/system response, when the driving force is a delta function. For a function f(t) as the driving function (instead of a delta function), this f(t) will wind up in a convolution integral with the Green's function to get the system response. Just as the case for simpler differential equations, the inhomogeneous solution for the wave function needs to be added to the homogeneous solution to get the complete solution for the wave function.
     
    Last edited: Jun 27, 2016
  8. Jun 27, 2016 #7

    Charles Link

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    One additional comment to posts #5 and #6: Normally when you encounter an integral that is taking the inverse Fourier transform of an expression, it is usually necessary to take the Fourier transform of that integral to recover the expression. In this case though, when the inverse Fourier transform of the expression was simply a delta function times another delta function that is essentially a constant, it was clear that the expression needed to be equal to that constant,( ## \delta(\vec{r}-\vec{r}') ##),without needing to do a Fourier transform on the integral. (No doubt you could also do it the long way (by taking the F.T.), but this one could be evaluated by inspection)...editing.. yes, that way works too but its a little longer: (take the F.T. of both sides of the equation). Beginning with ## G(\vec{r},\vec{r}', t'') ## where ## t''=t-t' ##, take ## \int G(\vec{r},\vec{r}',t")exp^{i \omega_o t"} dt" ## and you recover the expression with ## \omega=\omega_o ##.
     
    Last edited: Jun 27, 2016
  9. Jun 28, 2016 #8
    Okay, I think I understand it now. I do the derivative: [itex]
    \int \frac{d\omega}{2\pi} (\hbar \omega - \hat{H})G(\vec{r},\vec{r}',\omega)e^{-i\omega(t-t')}=\delta(\vec{r}-\vec{r}')\delta(t-t')
    [/itex]
    Then I do the inverse FT on both sides:
    [itex]
    \frac{1}{2\pi} (\hbar \omega - \hat{H})G(\vec{r},\vec{r}',\omega)= \delta(\vec{r}-\vec{r}') \int d\tau \delta(\tau) e^{i \omega \tau}
    [/itex]

    [itex]
    \frac{1}{2\pi} (\hbar \omega - \hat{H})G(\vec{r},\vec{r}',\omega)= \delta(\vec{r}-\vec{r}')
    [/itex]

    I think this is correct. Thanks alot for the help. I'll probably be back with more problems regarding this. (This was only a part of the whole sheet. Your post #6 clarifies some later problem thank you :) )
     
  10. Jun 28, 2016 #9
    I have another question: How do I show that the poles of [itex]
    G(\vec{r},\vec{r}',\omega)
    [/itex] are at the eigenvalues of the Hamiltonian (The energies).
    I tried to expand
    [itex]
    G(\vec{r},\vec{r}',\omega)
    [/itex] into the eigenfunctions of H and using the equation from above, I got this:
    [itex]
    \frac{1}{2\pi} \Sigma_n (\hbar \omega - E_n)G_n(\vec{r}',\omega)\phi_n(\vec{r})= \delta(\vec{r}-\vec{r}')
    [/itex]
    The problem statement says I should also use the completeness relation: [itex] \Sigma_n\phi_n(\vec{r})\phi^*_n(\vec{r}')=\delta(\vec{r}-\vec{r}') [/itex]
    but now I'm stuck again. I can't multiply by [itex]\phi^*_n[/itex] to get to use the relation.
     
  11. Jun 28, 2016 #10

    Charles Link

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    When taking the FT with ## exp^{i \omega_o \tau} d \tau ## (you called it the inverse, but i think it's simply the FT), it's best to do it with ## \omega_o ## because you already have an integral with ## d \omega ##. You don't want to mix the two. (Let ## \tau=t-t' ##). Switch the order of integration and do ## d \tau ## first. (Save ## d \omega ## for last.) The ## \int (1/(2 \pi))exp^{i (\omega_o-\omega) \tau} d \tau ## gives you ## \delta(\omega-\omega_o) ## so when you integrate over ## d \omega ## you get ## (\hbar \omega_o-\hat{H})G(\vec{r},\vec{r}',\omega_o) ## (without the ## 1/(2 \pi) ## ). Once you have done all the algebra, you can then switch ## \omega_o ## to ## \omega ##. Meanwhile, I'm going to need to work on your question in post #9. It's probably not real difficult, but I don't have all the details at my fingertips.
     
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