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Substitution I don't get

  1. Sep 9, 2005 #1

    quasar987

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    Yes, it's me and the wave packets... again!

    This is taken from the text of Gasiorowicz's Quantum Physics 3rd ed. pp.26.

    We have a gaussian wave packet at t=0 that is is described by

    [tex]\psi(x,0)=\int_{-\infty}^{\infty}dke^{-\alpha (k-k_0)^2/2}e^{ikx}[/tex]

    and we apply the change of variable [itex]q' = k-k_0[/itex]. Suposedly the wave packet becomes

    [tex]\psi(x,0)=e^{ik_0x}e^{-x^2/2\alpha}\int_{-\infty}^{\infty}dq'e^{-\alpha q'^2/2}[/tex]

    How does one gets to that? When I make the substitution k = q' + k_0, I get

    [tex]e^{ik_0x}\int_{-\infty}^{\infty}dq' e^{-\alpha q'^2/2}e^{iq'x}[/tex]
     
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  3. Sep 9, 2005 #2

    George Jones

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    Combine the exponentials under the integral and complete the square (for q') in the resulting exponent.

    Regards,
    George
     
  4. Sep 9, 2005 #3

    AKG

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    I really don't like that Gasiorowicz book. It's the one that's purple and yellow on the front right? When substituting I get the same thing you do. It seems that he's "pulling out" the [itex]e^{iq'x}[/itex] from the integral and it "becomes" [itex]e^{-x^2/2\alpha }[/itex]. Let's simplify some thing. Let b = -a/2, and let y be used where x was being used, and let x be used where q' was being used. The integrand is then:

    dx ebx2 + ixy

    We want to see why:

    [tex]\int _{-\infty } ^{\infty } dx\, \exp (bx^2 + ixy) = \exp \left (\frac{y^2}{4b}\right ) \int _{-\infty } ^{\infty }dx\, \exp (bx^2)[/tex]

    Using Wolfram's "The Integrator" we get that the left side of the equation in indefinite form is the top thing in the attached image, and the right side is the bottom thing.

    This is exactly what we want if

    [tex]\lim _{x \to \infty} \left [\mbox{Erfi}\left (\frac{2bx + iy}{2\sqrt{b}}\right ) - \mbox{Erfi}\left (\frac{2b(-x) + iy}{2\sqrt{b}}\right )\right ] = \lim _{x \to \infty} \left [\mbox{Erfi}(\sqrt{b}x ) - \mbox{Erfi} (\sqrt{b}(-x))\right ][/tex]

    [tex]\lim _{x \to \infty} \left [\mbox{Erfi}\left (\sqrt{b}x + \frac{iy}{2\sqrt{b}}\right ) - \mbox{Erfi}\left (\sqrt{b}(-x) + \frac{iy}{2\sqrt{b}}\right )\right ] = \lim _{x \to \infty} \left [\mbox{Erfi}(\sqrt{b}x ) - \mbox{Erfi} (\sqrt{b}(-x))\right ][/tex]

    This would seem plausible because iy/2√b would just be a constant, but it's a complex number, and from what I could tell, Erfi takes real arguments only. I have no idea why the Integrator would put the "i" in the argument, I don't always trust that application.
     

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    Last edited: Sep 9, 2005
  5. Sep 9, 2005 #4

    AKG

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    Ouch, yes, completing the square is much less involved. However, I still run into a problem, namely that I end up with:

    [tex]\psi (x, 0) = e^{ik_0x}e^{-x^2/2\alpha }\int _{-\infty } ^{\infty} dq' \, e^{-\alpha \mathbf{(q' - xi/\alpha )}^2/2}[/tex]

    instead of:

    [tex]\psi (x, 0) = e^{ik_0x}e^{-x^2/2\alpha }\int _{-\infty } ^{\infty} dq' \, e^{-\alpha \mathbf{q}'^2/2}[/tex]

    If [itex]i[/itex] weren't imaginary, then a simple change of variables from q' to q' - xi/a (or something like that) would make everything okay. I haven't done any complex analysis - is such a substitution valid? If q' goes from [itex]-\infty [/itex] to [itex]\infty [/itex], would it make sense to say that q' - xi/a ranges over the line in the complex plane {c + di : d = -x/a}?
     
  6. Sep 9, 2005 #5

    quasar987

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    With reason, see post #13 of https://www.physicsforums.com/showthread.php?t=88019.
     
  7. Sep 9, 2005 #6

    quasar987

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    I was gonna post that.

    Edit: Actually, no. Instead of

    [tex]\psi (x, 0) = e^{ik_0x}e^{-x^2/2\alpha }\int _{-\infty } ^{\infty} dq' \, e^{-\alpha \mathbf{(q' - xi/\alpha )}^2/2}[/tex]

    I got

    [tex]\psi (x, 0) = e^{ik_0x}e^{-x^2/2\alpha }\int _{-\infty } ^{\infty} dq' \, e^{-\alpha \mathbf{(iq' + x/\alpha )}^2/2}[/tex]
     
    Last edited: Sep 9, 2005
  8. Sep 9, 2005 #7

    AKG

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    -a(iq + x/a)²/2 = -a(iq + ix/ia)²/2 = -a(i(q - ix/a))²/2 = -ai²(q - ix/a)²/2 = a(q - ix/a)²/2, so your exponent is just the negative of mine. However, I'm pretty sure I completed the square correctly. Also, when you complete the square, say, in the variable q, then you want to end up with something of the form A(q - B)² + C, you shouldn't have iq or anything like that (you should have factored the i out, but there was no i to factor). Actually, you should be able to see that yours is wrong, because when you square it, the q² term will be -a(iq)² = aq², but we know that the original thing we were working with had -aq² as the q² term.
     
  9. Sep 9, 2005 #8

    AKG

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  10. Sep 9, 2005 #9

    quasar987

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    About the trust part.
     
  11. Sep 9, 2005 #10

    quasar987

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    I wrote it wrong, sorry!! Forget the negative sign.

    The way I completed the square is

    [tex]\frac{-\alpha q'^2}{2}+iq'x = \left(i\sqrt{\frac{\alpha}{2}}q'+\frac{x}{\sqrt{2\alpha}}\right)^2-\frac{x^2}{2\alpha}[/tex]
     
  12. Sep 10, 2005 #11

    AKG

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    You normally don't see people complete the square as you have. Normally they do this:

    Aq² + Bq + C
    = A(q² + B/Aq) + C
    = A[q² + B/Aq + (B/2A)² - (B/2A)²] + C
    = A(q + B/2A)² + (C - B²/4A)

    so inside the bracket, the "q" is not multiplied by any factor (the factor appears outside the bracket). However, what you have now is technically correct, but now what do we do with it? Since it's the same thing I have in post 4, how do we get the "-xi/a" to disappear?
     
  13. Sep 10, 2005 #12

    George Jones

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    Sorry - from eyeballing the problem I thought completing the square would do the trick. AKG's substitution can be justified by using the residue theorem and an appropriate contour of integration.

    Consider

    [tex]
    \int dz e^{-\frac{\alpha}{2} \left( z - ib \right)^2}
    [/tex]

    for complex values [itex]z = x + iy[/itex].

    Because the integrand in analytic, the residue theorem give that this integral is zero around any closed path (contour) in the complex plane. Consider the closed contour that consists of a horizontal path [itex]z = -x + 0i[/itex] for [itex]x = -a[/itex] to [itex]x = a[/itex], a vertical path [itex]z = a + iy[/itex] for [itex]y = 0[/itex] to [itex]y = b[/itex], a horizontal path [itex]z = x + ib[/itex] for [itex]x = a[/itex] to [itex]x = -a[/itex], and a vertical path [itex]z = a + iy[/itex] for [itex]y = b[/itex] to [itex]y = 0[/itex].

    Along the bottom horizontal path, [itex]dz = dx[/itex], and if we let [itex]a \rightarrow[/itex] \infty[/itex], we get the required integral. Along the to horizontal path, [itex]dz = dx[/itex], and if we let [itex]a \rightarrow[/itex] \infty[/itex], we get AKG's substitution. If it can be shown that the integrals along the 2 vertical paths die as [itex]a \rightarrow[/itex] \infty[/itex], then the "substitution" is valid.

    Along the right vertical path, [itex]dz = dy[/itex], and the integral is

    [tex]
    \int_{0}^{b} dy e^{-\frac{\alpha}{2} \left( a + iy - ib \right)^2}.
    [/tex]

    Square the brackets and factor out the

    [tex]
    e^{-\frac{\alpha}{2} a^2}.
    [/tex]

    This dies fast enough to kill the integral. Same thing on the left.

    Regards,
    George
     
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