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Time dependence of energy & probability

  1. Apr 24, 2010 #1
    I have a few questions about the time dependence of energy and probability etc. of systems.


    Say I have a particle in an infinite 1-D well.

    I can work out the general wave function as a Fourier sum of the orthogonal sine functions.
    Hence, the average/expectation value of the energy is :

    < H > = SUM (C_n)^2 (E_n) where C_n are the Fourier coefficients and E_n is the energy of the particle in the nth state.

    To me, i cant see how the energy here is dependent on time. I have seen a question in a textbook asking for the average energy at say time t=t_0. How would one approach this?

    Am I missing something here?
  2. jcsd
  3. Apr 24, 2010 #2


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    Yes, you are missing the wavefunction and the probability, which is where the time dependence shows up. The result you have above is just the *average* value of the energy. Remember that the measurement postulate says that any single measurement must give an eigenvalue, and so from the Born interpretation, we know that those cn2 coefficients represent the probabilities of observing a particular energy eigenvalue (assuming the wavefunction is normalized .. otherwise they are just relative probabilities).

    Do you know how to write the wavefunction and/or probability density for the superposition state giving rise to the average energy expression you wrote above?
  4. Apr 24, 2010 #3
    Sorry Im not sure what you mean.

    Well, for the time evolution of the wavefunction, dont I just multiply it by exp[-iEt/h]??
    (h is h-bar)
  5. Apr 24, 2010 #4


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    In fact, <H> is conserved (i.e. it doesn't change in time). This is true. One can get the result by noting Ehrenfest's theorems which says that the expectation values of observables obey classical (looking) laws. Alternatively, one can see this from the fact that the c's are time-independent (they don't become time-dependent unless there is a time-dependent perturbation).

    This is conservation of energy in QM.
  6. Apr 24, 2010 #5
    But if <H> is indep. of time, why would the question ask for the average energy at t=0 and then t=t_0 (ie. an arbitrary time)?

    Just to clarify, here is the wave function F(x,t) for the particle at t=0:

    F(x,0) = A.[ 1 + cos(px/a)].sin(px/a)

    where p is pi (dont know where to insert Greek letters!)

    A is a normalization constant.

    First it asks for F(x,t), so I expressed it as the Fourier sum (infinite square well).

    Then, it asks for the average energy at those 2 times, and so I used the formula I had in the first post for <H>. But that has no dependence on time?
  7. Apr 24, 2010 #6


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    Your analysis looks correct to me ... are you sure they are asking for the expectation value of energy? You are correct that if the Hamiltonian is time-independent, then the expectation value of energy will also be time-independent. Is it possible that this is a "trick" question designed to help come to the realization that <H> is conserved?
  8. Apr 25, 2010 #7
    Thanks for the input. Just to clarify, I assumed average energy was the same as < H >, the expectation value of the Hamiltonian yes?

    I slightly related question on probabilities:

    The wave function for Hydrogen is given by, at t=0:

    f(r,0) = 1/SQRT10 [ 2.f_1,0,0 + f_2,1,0 + SQRT2.f_2,1,1 + SQRT3.f_2,1,-1 ]

    where the subscripts to each wavefunction are the quantum numbers n,l,m respectively.

    a) It asks for the probability as a function of time for finding it in state l = 1 and m = +1


    b) how does the wavefunction evolve in time, ie. f(r,t) ?

    For (a) I took the standard / < f_n,1,1 / f (r,0) > / ^2 for the probabilty ( Dirac notation), where the only one then that counts is f_2,1,1 .

    But it asks for it as a function of time....again my answer would have no time depedence. Is this just refering to the fact the fucntions are written as f (r,t) , even tho the answer may not depend on time??

    And for (b), do ou just multiply each function by exp(-i(E_n)/h) (h is h-bar), where E_n is -13.6 / n^2?
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