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

Deriving a wave function

  1. Aug 18, 2011 #1
    I'd like to get some help and pointers on how to derive a wave function. Since I'm not really good at physics I'm just going to use brute force, if anyone spot a mistake please please point it out.

    If I look at an electron in an well with infinite potential walls. The well has the dimensions -a<x<a

    [itex]\displaystyle \frac{\hbar^2}{2m} \frac{\Psi d^2}{dx^2} = E \Psi [/itex]

    [itex]\displaystyle k^2 = \frac{2mE}{\hbar^2}[/itex]

    Gives
    [itex]\displaystyle \frac{\Psi d^2}{dx^2} = -k^2 \Psi[/itex]

    Solutions to diff equation

    [itex]\displaystyle \Psi(x) = A \cos kx + B \sin kx [/itex]

    Boundarys [itex]\displaystyle \Psi(\pm a) = 0 \rightarrow B = 0[/itex] and [itex]\displaystyle A = 0[/itex] or [itex]\displaystyle \cos ka = 0[/itex]

    [itex]\displaystyle k=\frac{\pi(2n - 1)}{2a}[/itex]

    Now
    [itex]\displaystyle \Psi(x) = A \cos \frac{\pi(2n - 1)}{2a} x [/itex]

    I also know that the integral from -a to a of the square should be equal to one.

    [itex]\displaystyle \int^{-a}_a \left|\Psi(x) \right|^2\,dx = 1[/itex]

    Can now be written like this

    [itex]\displaystyle \int^{-a}_a A^2 \cos^2 \left( \frac{\pi(2n - 1)}{2a} x \right) \,dx = 1[/itex]

    Some trigonometry

    [itex]\displaystyle A^2 \int^{-a}_a \frac{1}{2} \,dx + A^2 \int^{-a}_a \cos \left( \frac{\pi(2n - 1)}{a} x \right) \,dx = 1[/itex]


    [itex]\displaystyle A^2 a + A^2 \frac{\pi(2n - 1)}{a} \sin \left( \frac{\pi(2n - 1)}{a}(+a) \right) - A^2 \frac{\pi(2n - 1)}{a} \sin \left( \frac{\pi(2n - 1)}{a} (-a) \right) = 1[/itex]

    Since the sin parts becomes zero.

    [itex]\displaystyle A = \frac{1}{\sqrt{a}}[/itex]

    Finally

    [itex]\displaystyle \Psi(x) = \frac{1}{\sqrt{a}} \cos \frac{\pi(2n - 1)}{2a} x [/itex]

    I take it this solution is in the interval -a < x < a, in all other points the function is zero. If this is somewhat correct, is there a more elegant way of doing this, or are there some standard solutions that can be used to find a particular solution?

    Edit, how do I derive a wave function if a wall is at a certain potential, not infinite? For instance an electron in the third state in a 5 eV potential wall. I know that the function is exponential declining. But how do I determine where the nodes are and how high the wave function is at x = a?
     
    Last edited: Aug 18, 2011
  2. jcsd
  3. Aug 18, 2011 #2

    jtbell

    User Avatar

    Staff: Mentor

    Better: (A = 0 and sin ka = 0) or (B = 0 and cos ka = 0). This leads to a second set of solutions using sines, with different energies. The total set of solutions is the union of the two sets.

    This is pretty much the standard way to do it for the infinite square well, which you can find in any number of textbooks and Web pages. It doesn't actually derive the fact that the general solution consists of sines and cosines, but you can get that from standard methods for solving ordinary differential equations; see a textbook on differential equations for that.

    At x = a the wave function for -a < x < a must match the wave function for x > a, that is:

    [tex]A \cos ka = Ce^{-ka}[/tex] (and B = 0)

    or

    [tex]B \sin ka = Ce^{-ka}[/tex] (and A = 0)

    The functions must also match at x = -a (the other boundary) but this simply leads you to the same results as the above equations.

    Unfortunately you cannot solve for a (and therefore E) using only algebra. You have to do it numerically. After you've done that you can find A, B and C. This is also a standard topic. If you do a Google search for "finite square well" you'll probably find some methods for doing it.
     
  4. Aug 18, 2011 #3
    Mhmm... when I got the cos() function I thought, all right an even function. Since the well I presumed was symmetric I thought that was a good thing. Wouldn't the sin(some angle) variant just be a image of the cos(the angle I calculated) solution..? (Considering the boundary conditions.)

    C1 sin(some angle) = C2 cos(the angle a calculated)

    Edit
    I get the boundaries explanation a with a non infinite potential. By just looking at it: If I know the potential isn't it quite easy to determine [itex]C e^{-ka}[/itex]?
     
    Last edited: Aug 18, 2011
  5. Aug 18, 2011 #4
    I just found a nice series of videos on youtube and in Physics Handbook the solutions are stated for different potentials(Ep). Like you said:

    [itex]\displaystyle \Psi(x) A e^{ikx} + B e^{-ikx}[/itex] where [itex]\displaystyle k = \frac{1}{\hbar} \sqrt{2m(E-E_p)}[/itex]

    If the potential is higher than the particle energy.

    [itex]\displaystyle \Psi(x) C e^{lx} + B e^{-lx}[/itex] where [itex]\displaystyle l = \frac{1}{\hbar} \sqrt{2m(E_p-E)}[/itex]

    By the way, thanks a heap!
     
  6. Aug 18, 2011 #5

    jtbell

    User Avatar

    Staff: Mentor

    A symmetric potential has both even and odd eigenfunctions of energy. See these graphs:

    http://en.wikipedia.org/wiki/File:Particle_in_a_box_wavefunctions_2.svg

    n = 1 and 3 in this diagram correspond to your n = 1 and 2.

    n = 2 and 4 in this diagram are the first two energy eigenfunctions that use sines. Their energies and wavelengths fall in between those for the cosines.
     
  7. Aug 18, 2011 #6

    jtbell

    User Avatar

    Staff: Mentor

    These are for non-zero potential inside the well, and infinite outside. I thought you were asking about the case where the potential is zero inside and non-infinite outside.
     
  8. Aug 18, 2011 #7
    Thanks for the clarification!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook