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Calculated the expectation of the energy

  1. Apr 9, 2009 #1
    http://img23.imageshack.us/img23/1649/93412460.th.png [Broken]

    For question 2 in the above link,

    I calculated the expectation of the energy by

    [itex]E=<\hat{H}>=\int_0^a \psi^* \hat{H} \psi dx[/itex]
    where [itex]\psi=\psi^*=x(a-x)[/itex]

    this gave [itex]E=0[/itex]. this answer confused me for two reasons:
    (i) is it ok for the energy to be 0?
    (ii) i haven't made any use whatsoever of the eigenfunctions [itex]u_n(x)[/itex]

    secondly for the bit about repeated measurements, I said that repeated measurements are where the same measurement is made on each member of an enemble of identically prepared systems and so the outcome should just be the expectation obtained above???

    clearly ive missed something as i haven't used all the information given...
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 9, 2009 #2

    Matterwave

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    Re: Quantum

    E=0 is definitely NOT okay! Actually, <H> has to be greater than the ground state energy. One way of doing this problem is to expand the wave function into your set of eigenstates which can be done since eigenstates are complete. You get something like:

    [tex]\psi = \sum_{n=1}^{\infty} c_n u_n(x)[/tex]

    Each eigenstate has a corresponding energy, and you can get E by summing up all the energies multiplied by the absolute square of the coefficients for that state:

    [tex] E = \sum_{1}^{\infty} |c_n|^2 E_n[/tex]

    As you can see, you can never get an answer below the ground state energy since you're always adding and the absolute square of the coefficients must sum to 1.

    You can certainly do it your way, but first you have to normalize the wave function. Can you post some of your work on what you did for the integration? The integration should come out to be a positive answer that is greater than the ground state energy. You must have made a mistake somewhere in your integration.
     
  4. Apr 10, 2009 #3
    Re: Quantum

    so i would get the [itex]E_n[/itex] by [itex]\int_0^a \psi^* u_n(x) \psi dx[/itex]?
    i can do that easyily enough,
    i get [itex]E_n=\frac{\hbar^2 \pi^2}{2ma^2}n^2[/itex]

    i don't know how to find the [late]c_n[/itex] though? i know its complete and so theoretically we should be able to expand [itex]\psi[/itex] in terms of the [itex]u_n(x)[/itex] but how?

    and then how does the infinite sum work to give me the energy or do i just leave the final answer as the sum of ....
     
  5. Apr 10, 2009 #4

    Matterwave

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    Re: Quantum

    No, you certainly can't do that integral you just posted, it won't give you anything (notice how you have an "n" in your expression when you shouldn't). To obtain the [tex]c_n[/tex] you need to know a little bit about Fourier analysis. In this case the way I presented of finding E by finding the coefficiencts is much harder than doing the first integral you posted. I just said that to show you E>ground state energy.

    Do the integral you set up in the first place, but after you normalize the wave function. You should get a positive answer.

    Do this integral again:

    [itex]E=<\hat{H}>=\int_0^a \psi^* \hat{H} \psi dx[/itex]

    If you get 0 again, then show me your work, so I can see where you go wrong.
     
  6. Apr 10, 2009 #5
    Re: Quantum

    ok. why am i not allowed an n in that expression for the expectation of the hamiltonian? wasn't i trying to get En (basically what's wrong with [itex]\int u_n^* \hat{H} u_n dx[/itex]

    secondly how do i normalise this wavefunction
    [itex]|\psi|^2=1 \Rightarrow \psi \psi^*=1 \Rightarrow x^2(a-x)^2=1[/itex]
    i don't know how to proceed...
     
  7. Apr 10, 2009 #6

    Matterwave

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    Re: Quantum

    You want to find the energy of that wave function, not the energy of the eigenstates, correct? What you have found was the energy of the eigenstates (which, can be found in easier ways).

    The wave function you presented has 1 expectation energy associated with it, not infinitely many.

    To normalize the wave function you have to go reread the question and see that the wave functions are not EQUAL to x(a-x) but are PROPORTIONAL to x(a-x)

    So in general

    [tex]\psi = Ax(a-x)[/tex] for some constant A. To normalize the expression is not:

    [tex]\psi^*\psi =1[/tex] but rather:

    [tex]\int_{0}^{a} \psi^*\psi dx = 1[/tex]

    After you integrate that you should be able to find A. Then put that back into the original integration you had.
     
  8. Apr 10, 2009 #7
    Re: Quantum

    ok so i get [itex]A=\sqrt{\frac{17}{60}} a^{\frac{5}{2}}[/itex]

    then i do [itex]E=\int_0^a \psi^* \hat{H} \psi dx[/itex] to get the energy - that's all ok now thanks.

    wot about the last bit of the question as to how it would differ for repeated measurements...
     
  9. Apr 12, 2009 #8

    Matterwave

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    Re: Quantum

    Once you measure the energy of the particle, you collapse it into that energy eigenstate. All subsequent measurements (if there are no perturbations) will yield the same energy that you saw the first time.

    So the answer is no. If you make repeated measurements on your system, you'd just keep getting what you measured the first time back, over and over again.

    The expectation value of the energy (as with any other observable) refers to the average you'd get if you observed the energy over many identically prepared wavefunctions. NOT on one wavefunction over and over again.
     
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