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Homework Help: Mathematical methods of physics problem

  1. Jul 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Here's the problem. Verify the operator identity
    x - d/dx = -exp (-x^2 / 2) d /dx exp (-x^2 / 2)
    Can someone please help me? Thanks :)
    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 19, 2011 #2

    hunt_mat

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    I would apply both sides of the equation to a function [itex]f(x)[/itex] and compare the results.

    I sense the integrating factor method...
     
  4. Jul 19, 2011 #3
    ok.thanks for help :)
     
  5. Jul 19, 2011 #4

    Ray Vickson

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    This seems wrong. You have the same factor exp(-x^2/2) both outside and inside the d/dx. The result (when applied to f(x)) will have a factor exp(-x^2) on the right, but not on the left. I think one of the factors on the right should be exp(+x^2/2), so that the two "exp's" finally cancel.

    RGV
     
  6. Jul 19, 2011 #5

    hunt_mat

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    As I said Ray, apply them both to a function f to find:
    [tex]
    xf-\frac{df}{dx}=g(x),\quad e^{-x^{2}/2}\frac{df}{dx} \left( e^{-x^{2}/2}f\right) =g(x)
    [/tex]
    Both are g(x) as when applied to f they should give the same thing. Use the integrating factor [itex]e^{-x^{2}/2}[/itex] and treat it as an ODE.
     
  7. Jul 19, 2011 #6
    thank you both for helping :)
     
  8. Jul 20, 2011 #7

    Ray Vickson

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    I assume that -exp (-x^2 / 2) d /dx exp (-x^2 / 2) is to be interpreted as an operator that applies to a function f(x), giving -exp(-x^2/2)*(d/dx)[exp(-x^2/2)*f(x)] = x*exp(-x^2)*f(x) - exp(-x^2)*df(x)/dx = exp(-x^2)*[x - d/dx] f(x). The exp(-x^2) arises from exp(-x^2/2)*exp(-x^2/2). Where is my error?

    On the other hand (as I suggested before), -exp(+x^2/2) d/dx exp(-x^2/2) applied to f(x) does, indeed, give x*f(x) - df(x)/dx, because we get cancellation: exp(x^2/2)*exp(-x^2/2) = 1.

    RGV
     
  9. Jul 20, 2011 #8

    hunt_mat

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    let's check this, take the first equation, the integrating factor for this is [itex]-e^{-x^{2}/2}[/itex] and the equation becomes:
    [tex]
    \frac{d}{dx}\left( -e^{-x^{2}/2}f(x) \right) =-e^{-x^{2}/2}g(x)
    [/tex]
    Then to isolate g(x), multiply by [itex]-e^{x^{2}/2}[/itex] to get:
    [tex]
    -e^{x^{2}/2} \frac{d}{dx}\left( -e^{-x^{2}/2}f(x) \right) =g(x)
    [/tex]
    So there is a difference in signs as Ray said.
     
  10. Jul 20, 2011 #9
    The problem is related to the Hermite polynomials and operators of creation and annihilation. The creation operator:
    [itex] a^{\dagger}\equiv \frac{1}{\sqrt{2}}\left(x- \frac{d}{dx}\right)[/itex]
    I played with the Rodrigues formula for Hermite poyinomials (for n=1?):
    [itex]H_n (x)=(-1)^n e^{x^2} \frac{d^n}{dx^n}e^{-x^2}[/itex]
    and
    [itex] \psi_n(x)=\frac{1}{\sqrt{2^n n!\sqrt{\pi}}}e^{-\frac{x^2}{2}}H_n(x)[/itex]
    but run into some problems.
     
    Last edited: Jul 20, 2011
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