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Calculate the normalization of a Particle in a box

  1. Feb 8, 2008 #1
    [SOLVED] Particle in a box

    1. The problem statement, all variables and given/known data

    A particle is described by the wave function [tex]\Psi(x)=Ae^{(-bx^2)}[/tex]. Calculate the normalization factor A.

    2. Relevant equations & 3. The attempt at a solution

    So, to normalize [tex]\psi(x)[/tex], we integrate the probability ([tex]\Psi(x)^2[/tex]) over the entire area (this is in an infinite potential well, so let's say the particle is between x=0 and x=L) and set this equal to 1. Doing this, we have:


    I feel pretty comfortable with calculus but I cannot figure out how to evaluate this integral. I was able to find this in an integral table, but it wasn't much help:

    [tex]\int e^{ax^2} dx = -\frac{i\sqrt{\pi}}{2\sqrt{a}}\text{erf}\left(ix\sqrt{a}\right)[/tex]

    I do not recall learning anything about the error function (erf), and google wasn't much help. How can I evaluate this integral? Should I expand it in series? Any help would be appreciated. Thanks

    Last edited: Feb 8, 2008
  2. jcsd
  3. Feb 8, 2008 #2
    Is this the right equation or is there an 'i' on the exponential?

    If not, try integrating a power series expansion of 'e^(x)'.

    recall that if e^(-ix), then the normalization would be the integral of the wavefunction times its complex conjugate e^(ix), which would be much simpler.
  4. Feb 8, 2008 #3
    it is not imaginary, unfortunately...I will try the power series expansion now though
  5. Feb 8, 2008 #4


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    [itex](e^x)^2[/itex] is [itex]e^{2x}[/itex], not [itex]e^{x^2}[/itex]. And if you're in an infinite potential well, shouldn't the wavefunction vanish at the boundaries?
  6. Feb 8, 2008 #5
    I apologize, that was a typo - it's fixed now. Yes the wavefunction will vanish at the boundaries...but since we know that our particle will be somewhere between x=0 and x=L, those are the limits when finding the normalization factor, A. Alternatively, the probability that we will find the particle anywhere should be 1...since our particle can only be between x=0 and x=L, these are the limits of the normalization integral.
  7. Feb 9, 2008 #6


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    The point is that the wavefunction you gave can not represent a particle in a square well because it does not vanish at the boundaries. My bet is that the question is about a free particle in which case you must integrate from minus infinity to plus infinity and the integral is easy to do (or look it up).
  8. Feb 9, 2008 #7


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    I also say that the wave function that you posted in the original post [tex]\Psi(x)=Ae^{(-bx^2)}[/tex]
    Can't represent a particle in a infinite deep box,
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