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Expected energy and viability of a linear combination of wave functions

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the following wave function eigenfunction decomposition is viable and find it's expected energy.

    ψ(x, t) = [1/(1+i)]ψ1(x)e^(-iw1t) - (1/√2)ψ2(x)e^(-iw2t)

    2. Relevant equations
    ∫ψ(x)*ψ(x)dx = 1 probability

    ∫ψm(x)*ψn(x)dx = 0 orthogonality

    ∫ψm(x)*ψn(x)dx = δmn

    Ʃ|C[itex]_{n}[/itex]|[itex]^{2}[/itex] = 1

    <H> = Ʃ|C[itex]_{n}[/itex]|[itex]^{2}[/itex]*E[itex]_{n}[/itex]

    3. The attempt at a solution

    From my understanding so far, δmn is value given when you combine the condition for both probability and orthogonality. However I dont think I fully understand why you can just then said that the sum of the constants squared of the wave functions is equal to 1. Using the equation I can find that the first part of wave function is 1/2 and the second part is 1/2 so the sum of that is 1 so the wave function is viable. However I'm not sure what to do with the expected energy. I think that ω = En/h but im unsure as to how the linear combination effects the expression of expected energy because shouldn't <H> =En since the sum of the constants is just 1?

    Thanks for everyone that helps.
     
  2. jcsd
  3. Feb 5, 2013 #2

    Simon Bridge

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    Well - if you made a measurement of the eigenvalue, what is the probability of finding the particle in one of the eigenstates?
    How do you find the expectation value of a discrete random variable in normal statistics?
    But each value of the energy does not appear with equal probability does it?
     
  4. Feb 5, 2013 #3
    Alright, so the first part just requires some understanding of linear algebra. We are still dealing with time independent wave functions but I think we will be learning that notation very soon.

    Well, I could see showing the different probability for each value of energy but in this case the expected energy would be <H> = (1/2)E1 + (1/2)E2 where En = ω[itex]_{n}[/itex]h
     
  5. Feb 5, 2013 #4

    Simon Bridge

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    That last equation doesn't work - what is the relationship between frequency and energy?

    ... no idea what you are talking about.
     
  6. Feb 5, 2013 #5
    Wouldn't the relationship just be based off planck's equation for quantized energy? That would show that En = hv where "h" is planck's constant and "v" is the frequency. As far as I can see, the last equation is in that form unless I'm not understanding it correctly.
     
  7. Feb 5, 2013 #6

    Simon Bridge

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    ##\omega = 2\pi\nu## ... for some reason I left the h off the end when I quoted you...

    I'm sorry, the way you wrote
    I inferred that you had some issue with the resulting relation.
    I was trying to guess what it was instead of just asking. Now I'm asking :)
     
    Last edited: Feb 5, 2013
  8. Feb 5, 2013 #7
    That was just because I wasn't sure about using the Cn probabilities before where both E1 and E2 have an equivalent probability. I think the professor was just using ω as a substitute for En/h because he never brought up the term of 2*pi. The ω variable is taking the place of the wave function ψn(x,t) = Cn*ψn(x)e[itex]^{-iEnt/h}[/itex]. I was just uncertain if I was correctly using the formula for the expected energy <H>.
     
  9. Feb 5, 2013 #8

    Simon Bridge

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    Oh OK - well that was exactly how to use them.
    ##\langle H \rangle = c_1^\star c_1 E_1 + c_2^\star c_2 E_2##

    But in the wave-function he(?) writes ##e^{-i\omega_n t}## as the time-dependency ... in that relation, ##\omega_n = E_n/\hbar##
    If he is using ##\omega=E_n/h## then that would be out by a factor of ##2\pi##.

    If he uses units of ##\hbar=1## then you can write ##\omega_n = E_n##.
     
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