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Initial momentum amplitude to wave function

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    http://www.whoisntdavidrinaldi.com/Untitled.png [Broken]

    2. Relevant equations



    3. The attempt at a solution

    So I had to use this initial momentum and multiply it by a plane wave (sure you are aware of the exp[i(px+p^2(t)/2m)\hbar] )
    setting p-p_0 to q and then coupling all terms together to get an integral of the form where one could complete the square the problem here is the extra q, which seems to force me to to integration by parts. The problem is I can set u=q but if dv is my exponential term what is v? I've never seen how to

    \int_{-\infty}^\infty dq(q)exp[-q^2(\frac{\alpha^2}{2}+\frac{it}{2m\hbar})+ q(\frac{ix}{\hbar}+\frac{ipt}{2m\hbar})


    crud. I'm not sure how to post latex here. I guess this is my second question. To see the above the link is here www.whoisntdavidrinaldi.com\help1.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 30, 2013 #2

    Simon Bridge

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    ##\small \psi(x)## is the fourier transform of ##\small \phi(p)## - time-evolve - use the derivation for eq3.61 to guide your algebra.

    ##\small \LaTeX2e## is posted between double-hash marks for inline equations and double-dollar signs for display math. Use the "quote" button (bottom left of this post) to see the following example:
    $$\int \phi(p)=\sqrt{\frac{2\alpha^3}{\sqrt{\pi}}}(p-p_0)e^{-\alpha^2(p-p_0)^2/2}$$

    exp[i(px+p^2(t)/2m)\hbar] (a bit of confusion about brackets there?) comes out as: $$e^{-i(px+p^2t/2m)\hbar }$$

    Your actual question though...
    You are trying to do:
    $$\int_{-\infty}^\infty q\exp\left[-q^2(\frac{\alpha^2}{2}+\frac{it}{2m\hbar})+ q(\frac{ix}{\hbar}+\frac{ipt}{2m\hbar})\right ]\; dq$$ ... I tidied up your latex a bit.

    Integrals of form: ##\int \exp [x^2]dx## have no analytic solution.
     
    Last edited: Sep 30, 2013
  4. Sep 30, 2013 #3
    if we look at my first equation, and forget about the extra q lying around, we get something of the form

    ##\int_{-\infty}^\infty \exp [-x^2+x]dx##

    or $$\int_{-\infty}^\infty \exp [-x^2-x]dx$$

    do have solutions..it's just that this new q makes me need to use integration by parts as well.
     
  5. Sep 30, 2013 #4

    Simon Bridge

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    ... well $$\int\exp[ax-bx^2]dx = \frac{\sqrt{\pi}}{2\sqrt{b}}\exp \left[\frac{a^2}{4b}\right]\text{erf}\left[\frac{2bx-a}{2\sqrt{b}}\right]+c$$ ... but you have an integral of complex exponential so you could try to express is as a sum of sines and cosines.
     
  6. Sep 30, 2013 #5
    I am confused

    45a5a0_613ec63e9c15c66a8b6dd48bddee5596.png
     
  7. Sep 30, 2013 #6

    Simon Bridge

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    That differs from what you wrote in post #1... you have bx-ax^2 in the exponent.

    [edit]
    Assuming I misread your expression in post #1...
    Integrating your version by parts:

    $$\int xe^{-ax^2-bx} dx = \frac{b\sqrt{pi}}{4a^{3/2}}e^{\frac{b^2}{4a}}\text{erf}\bigg(\frac{2ax+b}{2\sqrt{a}}\bigg)-\frac{1}{2a}e^{-x(ax+b)}+c$$

    I've still got the nagging feeling you should also separate real and imaginary parts - especially considering you'll end up taking the complex conjugate after.
     
    Last edited: Sep 30, 2013
  8. Oct 1, 2013 #7
    Why don’t you complete the square in the integral?
    let a = α2/2 + ipt/2mhbar
    b = ix/hbar + ipt/2mhbar
    e-(a + b*b/4a*a)∫qe(b/2a-q)*(b/2a-q)dq

    and making the substitution q’ = b/2a-q
    I get
    e-(a + b*b/4a*a)[(b/2a)e-q’*q’-√(π/4)erf(√q’) ]

    --sorry about the notation. I don't know how to type the square of an exponential argument,
     
    Last edited: Oct 1, 2013
  9. Oct 1, 2013 #8

    Simon Bridge

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    the exponential of x^2 in LaTeX is \exp(x^2) or e^x^2
    using the bv code you have to write exp[x2].

    when you substitute back, you'll end up with the answer in post #6.
    That's as close as I want to get to doing the problem for OP - lets wait for feedback before providing the next clue OK?
     
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