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Infinite potential well and linear superposition

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a particle in a potential ;
    [tex]V(x)=0\ for\ 0<=x<=L[/tex] and
    [tex]V(x)=\infty\ otherwise[/tex]

    . Its wave function is a linear superposition of the lowest two stationary states and given by
    [tex]\psi(x,t)=\frac{1}{2}*(\sqrt{3} \psi_{1}(x,t)-\psi_{2}(x,t)[/tex])

    (a) Use the results fromquestion 2 and find ψ(x,t). Then calculate [tex]|\psi(x,t)|^{2}[/tex].
    Make sure that your result for [tex]|\psi(x,t)|^{2}[/tex] is manifestly real (no “i” left in there). What is the frequency of the oscillation in [tex]|\psi(x,t)|^{2}[/tex]?

    (b) Calculate the expectation value of the Hamilton operator ˆH. Compare it with the energy eigenvalues
    E1 and E2.

    (c) What is the probability for the particle to be in state ψ1 and what is the probability for it to be in the state ψ2?

    2. Relevant equations

    Time dependent wave function for the nth stationary state in this potential well in this case is;
    [tex]\psi_{n}(x,t)=\sqrt{\frac{2}{L}}\sin(\frac{n\pix}{L})\exp{\frac{-iE_{n}t}{h/2\pi}\ n=1,2,3,...[/tex]

    3. The attempt at a solution

    Hi, I can get the solution to the first part of (a), I think,


    [tex]E_{1}=\frac{h^{2}}{8mL^{2}}[/tex] and

    But after that I have no idea of where to go. When I try and calculate the product of the conjugate times non-conjugate I end up with complex terms that I can't simplify, so my result isn't real. Also I have no idea how you find the frequancy of oscillation in [tex]|\psi(x,t)|^{2}[/tex], or how to calculate the expectation of the Hamiltonian operator. Any help would be a godsend to me, I feel completly lost now in QM.

    Thanks in advanc
    Last edited: Feb 2, 2009
  2. jcsd
  3. Feb 2, 2009 #2
    The complex conjugate of (a + ib) is (a - ib). So, the product of these two will give a real number. You stated your result has complex terms. Show how you took the product of the complex number and the conjugate.
  4. Feb 2, 2009 #3
    The conjugate of this,


    will be,


    where [tex]\phi_{n}(t)=exp{\frac{-iE_{n}t}{h/2\pi}[/tex]

    Now taking [tex][\psi_{1}(x)*\bar{\phi_{1}(t)}-\psi_{2}(x)*\bar{\phi_{2}(t)}]*[\psi_{1}(x)*{\phi_{1}(t)}-\psi_{2}(x)*{\phi_{2}(t)}][/tex]



    since [tex]\phi_{1}(x)\neq\phi_{2}(x)[/tex] in general I have two mixed terms in the middle. How do I get rid of them? What have I done wrong?
  5. Feb 2, 2009 #4
    One way of doing this is to express the compex exponential in terms of cosine and sine using the Euler relation. This will separate the real and imaginary parts then you can combine in the form a + ib. Then multiply by the conjugate giving a2 + b2.
  6. Feb 2, 2009 #5
    Okay. I'm going to simplify everything and call the arguments of both exponential theta(1) and theta(2).

    Expanding using Eulers notation gives,

    Thus, [tex]\bar\psi(x,t)*\psi(x,t)[/tex] is,


    where [tex]x=\psi_{1}(x)cos{\theta_{1}}-\psi_{2}(x)cos{\theta_{2}[/tex]

    Does this look alright now?
  7. Feb 2, 2009 #6
    Looks fine with the exception that I get a minus sign between the sine terms since both arguments are negative.
  8. Feb 2, 2009 #7
    Excellent thanks for the help, I will put the minus's in at the end since i always lose the damn things!

    Now please could you give me a hint as to how I find the frequency of oscillation in [tex]\bar\psi(x,t)*\psi(x,t)[/tex]?

    For part (b) I got the expectation of the Hamiltonian to be less than the sum of the energy eigenvalues. Is this right or have a made a mistake somewhere?

    Also for part (c) how do I find the probability for the particle to be in either of the states?
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