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Finite square potential

  1. May 26, 2009 #1
    Hi, I am in the process of learning QM.
    I am looking at this problem regarding to a finite square potential well.
    I have derived psi(x) and the k's for the 3 domains,

    psi1(x) = Ae^kx => k = sqrt(2m(V-E)/h^2)
    psi2(x) = Ce^jkx + De^-jkx => k=sqrt(2mE)/h
    psi3(x) = He^-kx => same k as in domain 1

    and then what I did was to take the boundry conditions and substitue into the equations and make a system of 4 equations with so that I can solve the unknowns.

    What I don't understand is the way I've been told to do this from the above.

    I've been told to calculate the determinant of the system of the 4 equations, and by scanning E. When the determinant goes to 0, I will get the energy level for E.
    Any one can explain to me why this will give the answer?

    And also, it seems like I will need to get the answer by ploting the determinant from its imaginary component. Why?

    And when I have the 3 different energy levels that satisfy the conditions, how do I find the unknowns?

    Thanks for any help.
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2
    Re: Qm101

    The boundary conditions on the wavefunction and its first derivative give you the system of equations you are referring to.

    This is a system of equations linear in the unknown coefficients. So the condition you stated is the condition for the existence of a nontrivial solution for this system.

    Not sure I understand this. The determinant is a complex number. Setting it equivalent to zero is equivalent to setting its real and imaginary parts to zero separately. Perhaps what you want to do is treat the determinant as a function of the parameter E and plot it as a function of E, to determine the zero crossings, which will give you the eigenvalues.
  4. May 27, 2009 #3
    Re: Qm101

    And when I get the energy states Es, how do I find out the unknowns?
  5. May 27, 2009 #4
    Re: Qm101

    You must have obtained a transcendental equation..just solve for the eigenvalues numerically.
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