Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinite square well solution

  1. Mar 28, 2014 #1

    kmm

    User Avatar
    Gold Member

    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

    kmm

    User Avatar
    Gold Member

    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

    jtbell

    User Avatar

    Staff: Mentor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Infinite square well solution
  1. Infinite square well (Replies: 3)

Loading...