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Probability of particle

  1. Mar 29, 2006 #1
    I am to find if the following wave function can predict the probability that a particle is somewhere (anywhere) in the box to be t-dependent. And whether it has time-dependent average energy.

    [tex] \Psi_A (x,t) =K (-\Psi_6 (x,t) + \Psi_4 (x,t)- \Psi_2 (x,t))[/tex]

    here's what I did to find if this finction predicts that the particle is somehere in the box:

    [tex] \int \Psi_A (x,t) * \Psi_A (x,t) dx =K (-\Psi_6 (x,t) + \Psi_4 (x,t)- \Psi_2 (x,t)) K (-*\Psi_6 (x,t) + *\Psi_4 (x,t)- *\Psi_2 (x,t))=3k^2[/tex]

    this means that this finction does predict that a particle is somehere in the box

    if this function has time dependent average energy:
    [tex]\frac{E_6 + E_4 + E_2}{3}[/tex]

    right? my friend said that the energy cant be found, but isnt this the average energy?
  2. jcsd
  3. Mar 30, 2006 #2
    This is not the average energy. Each energy in the numerator needs to be weighted to the relative probability of each state, ie.
    [tex]\frac{E_6\int dx \Psi_6^*\Psi_6 + E_4\int dx \Psi_4^*\Psi_4 + E_2\int dx \Psi_2^*\Psi_2}{\int dx \Psi_A^*\Psi_A}[/tex]

  4. Mar 30, 2006 #3


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    I think it's safe to say (altough the OP didn't mention it) that the [itex]\Psi_i[/itex] are normalized eigenstates of the Hamiltonian, so their absolute square integrates to 1.

    So, that's the correct expression for the energy. If you are asked to prove that it is independent of time, it's trivial once you know that if [itex]\Psi_i[/itex] is normalized at some instant, it stays normalized. The relative difficulty is in showing that.

    In general, the average of any time-independent observable that commutes with the Hamiltonian is always constant.
  5. Mar 30, 2006 #4


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    Others have given many useful hints but let me add a few details. First, the question is not about whether the wavefunction predicts that the particle is somewhere in the box! Of course it does (it would have to be identically zero to predict that the particle is nowhere in the box).

    Strictly speaking, K may be complex, so you should be getting 3 KK* =3 |K|^2 (assuming that Psi_1, Psi_2, etc are normalized and orthogonal..which they are if you mean the standard normalized energy eigenstates)

    Again, this just follows from the fact that the wavefunction is not zero!
    This is the correct expression but how did you get it? (just curious).
    But your conclusion is wrong!! This result clearly does not depend on time (there is no time dependence!!)
    Of course the average energy can be found since the wavefunction is known. (however, asking "what is the energy of that state" is not a well defined question in quantum physics, as opposed to classical mechanics. Measuring the energy can produced either of the three energies with a probability of 33.3%...Maybe that's what you friend was referring to. But the *average energy* is clearly well-defined).

    Last edited: Mar 30, 2006
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