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Electric Potential Normalization

  1. Sep 14, 2009 #1
    I was in my Electrodynamics lecture last week, still working the Laplacian and Poisson equations, when we discussed an infinite parallelpipid (infinite in the [tex]x[/tex] direction, length [tex]a[/tex] and [tex]b[/tex] in the [tex]y[/tex] and [tex]z[/tex] direction respectively) with a potential of [tex]\Phi=\Phi_0[/tex] at [tex]x=0[/tex] plane and every other face having [tex]\Phi=0[/tex]. Here he said that, due to boundary conditions we should expect the potential to have the form

    [tex]
    \Phi(x,y,z)=\sum_{n_{y},n_{z}}A_{n_{y}n_{z}}\eta_{n_{y}}\sin\left[\frac{n_{y}\pi y}{a}\right]\eta_{n_{z}}\sin\left[\frac{n_{z}\pi z}{b}\right]\exp\left[-\pi\sqrt{\frac{n_{y}^{2}}{a^{2}}+\frac{n_{z}^{2}}{b^{2}}}x\right]
    [/tex]

    where [tex]\eta[/tex] and [tex]A[/tex] are both normalization constants. I agree with this form (not sure about the normalization constants, but that's the question to come). We let the eta's equal [tex]\sqrt{2/a}[/tex] and [tex]\sqrt{2/b}[/tex] in order to coincide with QM. In looking at the case [tex]x=0[/tex], we find

    [tex]
    \Phi_{0}=\sum_{n_{y}n_{z}}A_{n_{y}n_{z}}\sqrt{\frac{2}{a}}\sin\left[\frac{n_{y}\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_{z}\pi z}{b}\right]
    [/tex]

    Then multiply both sides by

    [tex]
    \sum_{n_{y}'n_{z}'}A_{n_{y}'n_{z}'}\sqrt{\frac{2}{a}}\sin\left[\frac{n_{y}'\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_{z}^{'}\pi z}{b}\right]
    [/tex]

    Which after integrating over [tex]dy'[/tex] and [tex]dz'[/tex] from [tex]0[/tex] to [tex]a[/tex] and [tex]0[/tex] to [tex]b[/tex], respectively, we find that [tex]n_y,n_z[/tex] must be odd integers. So then the normalization constant is found to be

    [tex]
    A_{n_{y}'n_{z}'}=\frac{8\Phi_{0}\sqrt{ab}}{n_{y}'n_{z}'\pi^{2}}
    [/tex]

    All of this makes sense and does work out mathematically. The trouble I find is that my professor said we could do this without having the [tex]\eta[/tex] terms in there, that [tex]A[/tex] would absorb it and we should still come out with the same answer. But following the same math as above, and removing the [tex]\eta[/tex] normalizations, I find the normalization constant to be

    [tex]
    A_{n_{y}',n_{z}'}=\frac{4\Phi_{0}ab}{n_{y}'n_{z}'\pi^{2}}
    [/tex]

    My question is two-fold:
    (1) Is my professor wrong and we should not expect the same normalization constant between the two (my answer is off by the factor [tex]2/\sqrt{ab}[/tex]?
    (2) Is my professor right, and I screwed up somewhere?
     
  2. jcsd
  3. Sep 14, 2009 #2
    I think I found an error in what my professor did in class. We have a line that reads

    [tex]
    \int_0^ady\int_0^bdz\sqrt{\frac{2}{a}}\sin\left[\frac{n_y'\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_z'\pi z}{b}\right]\Phi_0=A_{n_y'n_z'}
    [/tex]

    And it should read

    [tex]
    \int_0^ady\int_0^bdz\sqrt{\frac{2}{a}}\sin\left[\frac{n_y'\pi y}{a}\right]\sqrt{\frac{2}{b}}\sin\left[\frac{n_z'\pi z}{b}\right]\Phi_0=\frac{2}{\sqrt{ab}}A_{n_y'n_z'}
    [/tex]

    This makes the normalization constant [tex]A_{n_y'n_z'}[/tex] equal to what I get:

    [tex]
    A_{n_y'n_z'}=\frac{4\Phi_0ab}{n_y'n_z'\pi^2}
    [/tex]

    Guess the professor forgot to carry some factors from one line to the next. Thanks for your help anyway!
     
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