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Interpretation of solution in QM

  1. Nov 28, 2007 #1
    I'm having a problem understanding solutions of differential equation in QM:

    [tex]\psi''(z)+\frac{p}{z}\psi(z)+k^2\psi(z)=0[/tex] (1)

    I usualy use Fourbenious method, and in this case I get a 3 coefficients recursion relation which is really messy.
    So I do it like this:
    for really large z second term expires, and I have a simple H.O.equation:

    [tex]\psi''(z)+k^2\psi(z)=0[/tex] (2)

    with a solution:

    [tex]\psi(z)=Aexp(ikz)[/tex] (3)

    Now, the solution of (1) must be:

    [tex]\psi(z)=Aexp(ikz)*f(z)[/tex] (4)

    Where f(z) is some function of z. Supstituting (4) in (1) I get the equation for f(z):

    [tex]f''(z)+2kif'(z)+\frac{p}{z}f(z)=0[/tex]

    If I use Fourbenious method here, I get a nice recursion relation and the solution is something like this (hypergeometric function):

    [tex]f(z)=Cz(1-\frac{2ki+p}{2!}z+\frac{(2ki+p)(4ki+p)}{2!3!}z^2-....(-1)^{n+1}\frac{(2ki+p)(4ki+p)...(2kin+p)}{n!(n+1)!}z^n)[/tex]

    There is also a second solution, but no need for me to write it down becoase the same problem arrises.
    I checked with mathematica 5.0 and the solution is OK. Is this OK? I mean, every usual problem in QM, HO, hydrogen atom, square well... have real space part and imaginary time part. But my space part has an imaginery and real part. Have I done something wrong? Or is it a coincidence that these standard problems have just real part of space part of wave function.
     
    Last edited: Nov 28, 2007
  2. jcsd
  3. Nov 28, 2007 #2

    Avodyne

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    The real and imaginary parts of your solution are each also solutions. Your original equation is real, so if psi is a solution so is psi^*. Note that e^(-ikx) is also a solution of (2).
     
  4. Nov 29, 2007 #3
    What if I wanted to plot psi(z)? I have never seen 3D plot for a wave function in one dimensional problem. I mean, when somewere wave function is large you can say that the particle is probably there, but if you have real and imaginery you can't plot just real part and talk about probability.
     
  5. Nov 29, 2007 #4

    Avodyne

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    Your 2nd order differential equation has two solutions, which can be chosen, without loss of generality, to be real functions. This is no different than your eq(2), which also has two real solutions, sin(kz) and cos(kz).

    Your psi is a particular complex linear combination of these two real solutions.
     
  6. Nov 29, 2007 #5
    Ok, but how do I write my solution down as a combination of two real solutions, just like you did for (2)?
     
  7. Nov 29, 2007 #6

    Avodyne

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    Take the real and imaginary parts of your solution, just like cos(kx) and sin(kx) are the real and imaginary parts of e^(ikx). Any linear combination of these two real functions, with real or complex coefficients, is also a solution.
     
  8. Nov 30, 2007 #7
    I understand now, thanks.
     
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