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Inifinite potential barrier problem

  1. Oct 24, 2003 #1
    (modified this to reflect a better understanding)

    Stationary solutions to the wave equation mean that if I calculate <x> I will get a value independent of time. For a potential defined as V(x) =0 from x = 0 to a and infinite outside that range of x, the lowest energy solution is Psi(x,0) = A sin (pi x / a). I can do the
    math and show that <x> = a/2 for this solution.

    Now classically we know that a particle of energy greater than 0 will oscillate between the two boundaries. So any physical solution to PSI(x,t) must give such a solution. I am trying to show this for the simple case of PSI(x,t) = A sin(pi x/a) exp(-iwt) + Asin(2pi x/a)exp(-i2wt). I have 2 questions.

    1. I am correct in expecting that |PSI(x,t)|^2 is still a stationary value? My book suggests it is time varying but I don't believe it.

    2. When I form <x> for this wave function, I multiply by its complex conjugate and do the integral. The 'crossterms' have the time dependence and I get <x> = a * (0.5 - 0.18 * cos (3wt) ). This was
    not exactly what I expected but when I graphed |PSI(x,0)|^2 it seems
    reasonable. <x> = 0.32 a at t =0, and peak of |PSI(x,0)|^2 is 0.30.

    3. When I calc. <H> = 5/2 * pi * pi * hbar *hbar / (2ma2

    When I add up E1 + E2 for the eigenstates i get twice the value I calculated above. That seems strange to me but I can't find any faults in the math. Does anyone know if <H> is *not* supposed to equal E1 + E2?
    Last edited: Oct 25, 2003
  2. jcsd
  3. Nov 2, 2003 #2
    Hi mmwave!

    Here are the answers:

    1. The new state is not stationary. The book is right. Actually, any time-dependent state can be written as a sum of stationary states.

    2. Your result confirms my answer to question 1.

    3. <H> is the average energy. Average is (E1+E2)/2, not E1 + E2.

  4. Nov 2, 2003 #3
    Thanks for the reply. I am not sure I agree that <x> being time varying supports the argument that |Psi|^2 is time varying. That extra factor of x complicates the integral a lot. I will write out the |Psi(x,t)|^2 integral again and see what I get now that I've had some more sleep.

    The average of E1 + E2 is interesting. The value <Q> is the mean value I would get from a whole bunch of identically prepared systems if I measured the observable Q. That is not necessarily the same thing I would get if I measure Q of a single system repeatedly over time. In the case of <H> = E they are the same?
  5. Nov 5, 2003 #4
    After you measure the energy of the system once, it's state becomes the eigenstate of energy corresponding to the value you obtained. If you measure it again and again you will always get the same value.
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