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Infinite square well solution

  1. Mar 28, 2014 #1


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    In finding solutions to the time independent Schrodinger equation we have to normalize [itex] \psi [/itex] to find the constant A. So we get [tex] \int_{0}^{a} |A|^{2} sin^{2}(kx) dx = |A|^2 \frac{a}{2}=1 [/tex]

    For A we then get [itex] |A|^2 = \frac{2}{a} [/itex]. Griffiths says that this only determines the magnitude of A but it's simplest to pick the positive real root. I know how to work with complex numbers generally but I'm a little confused as to what the imaginary root would be anyway. It looks to me like [itex] A= \pm \sqrt{ \frac{2}{a} } [/itex]. Since we aren't square rooting a negative I don't see where the [itex] i [/itex] comes in. Also, I know the magnitude of a complex number is real, so are we saying that A itself could be complex? I don't want to make any assumptions about this. Thanks!
  2. jcsd
  3. Mar 28, 2014 #2
    You have thought it right, 'A' can be complex indeed. In fact, [tex]A=\sqrt\frac 2 a\,e^{i\phi}[/tex] satisfies the the normalization constraint for any real \phi. Mind that the phase of the wave function is arbitrary.
  4. Mar 28, 2014 #3


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    Cool, thanks!
  5. Oct 26, 2014 #4
    I was passing here by coincidence, and after reading the post I had few questions indeed:
    If A is just the magnitude of the normalization constant, then will it differ if we took the complex complete part ? I mean, why do we always take the positive part?

    Also why does A only represent the magnitude? Where does this appear from?
    I always solve the particle in a box case taking the solution of A as granted without noticing that it is just a magnitude !!
  6. Oct 26, 2014 #5
    A is a constant that we have added after integration Sorry I do not mean added in the sense of + but more in the sense of incoperated in our equation!
  7. Oct 26, 2014 #6
    Here is the equation:

    We start of with:

    1=∫ abs(psi(x))^2 dx

    substitute psi(a)= A sin((npix)/a))

    1=abs(A)^2∫ abs( sin((npix)/a)))^2 dx

    so A=(2/a)^1/2

    where I have simply replaced the k of in the top most post with sinnpi/a
  8. Oct 26, 2014 #7


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    Strictly speaking, the normalization constant can be as csopi noted, letting ##\sqrt{2/a}## be real and using any value for φ. However, when you use the wave function to calculate anything that is physically observable (e.g. a probability or an expectation value), φ always cancels out. Therefore, for simplicity, we usually choose φ = 0.
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