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Instantaneous action in potential

  1. Oct 31, 2015 #1
    Suppose we have a particle trapped in a potential V(x), we calculate the bound states of the particle for t<0, so that our most general solution is [tex]\Psi(x,t)=\sum_{n}C_{n}\phi_{n}(x)e^{-iE_{n}t/\hbar}[/tex]
    Then, at t=0. we change the potential, so that the potential is now another function V'(x).

    How does the state of the particle change? The energy spectrum is changed, but the particle couldn't change its energy levels (see that because psi is a LC of the (old) eigenstates, the particle is at the same time at different energy levels, each of them with different probability) inmediatly, since no action can be taken in an amount of time 0. I would say that the Schrodinger equation needs to be solved after t=0 with the old wavefunction being the new initial condition, is that alright? Ideas please! Thanks!
     
  2. jcsd
  3. Oct 31, 2015 #2
    Also, because the potential has changed, we would expect the particle to change to a new state as before, but with new energy levels and eigenfunctions. Energy is not conserved here, but how does this change to the new state takes place? any ideas?
     
  4. Oct 31, 2015 #3

    DrClaude

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    Staff: Mentor

    That's correct.

    What do you mean by state? The state of the quantum system doesn't change (see above), by its description in term of the eigensates of the Hamiltonian changes since the Hamiltonian is no longer the same.
     
  5. Nov 1, 2015 #4
    Yes it was a bit confusing as I said it. So we solve the Schrodinger equation for t<0 and after t=0, our more general state will have the form of the equation I wrote above, but with different eigenvectors and different energy states (En). The point is, how does this happen, so that for that general solution I wrote there (but now for t>0), our initial condition is the general solution I wrote before.
     
  6. Nov 1, 2015 #5

    DrClaude

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    Staff: Mentor

    You need to find how the states ##\phi(x)## obtained with ##V(x)## are related to the states ##\phi'(x)## obtained with ##V'(x)##.

    This is related to the general prescription for the spectral decomposition of a state: given a set of basis functions ##\phi_n(x)## (such as the eigenstates of the Hamiltonian, but any basis set will do), any wave function ##\psi(x)## can be written as
    $$
    \psi(x) = \sum_n c_n \phi_n(x)
    $$
    with ##c_n## begin complex coefficients given by
    $$
    c_n = \int \phi_n(x)^* \psi(x) dx
    $$

    In your case, since you know ##\psi(x,t=0)##, you simply need to rewrite in terms of the energy eigenstates of the new Hamiltonian.
     
  7. Nov 2, 2015 #6
    Yes but my point is, at t=0, the second solution has the form: [tex]\Psi_{2}=\sum_{k} c_{k}\phi_{k}(x)[/tex], whereas from the first part, we know that the wavefunction (also at t=0) was:\\[tex]\Psi_{1}=\sum_{n} c_{n}\phi_{n}(x)[/tex] and then the condition is:\\ [tex]\Psi_{1}=\Psi_{2}[/tex] at t=0, is that correct? Otherwise, there is a discontinuity in time in the wave-function, im just interested to know how that happens
     
  8. Nov 2, 2015 #7
    Of course both [tex]\phi(x)[/tex] are different, they are just the bound states of each potential, the one at t<0 and the one at t>0
     
  9. Nov 2, 2015 #8

    DrClaude

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    Staff: Mentor

    What you say isn't incorrect, but I fear that there might some underlying misunderstanding, so I'll try to clarify things.

    First, there is a single wave function describing the quantum system. Therefore, saying ##\Psi_1 = \Psi_2## doesn't make much sense, but your idea that the wave function is single-valued is correct. At any given time, there is a definite ##\Psi(t)##.

    Second, it doesn't matter which basis set you use to describe the wave function: both basis sets are valid for any time ##t##. It is simply that in one basis, calculating the time evolution when ##t<0## is trivial, while in the other basis it is trivial when ##t>0##. That's why you want to change basis. But describing the wave function using the eigenfunctions of the Hamiltonian of ##t<0## will also work after the potential has changed, but instead of having ##c_n e^{-i E_n t/\hbar}##, you will have ##c_n(t)## with a complex time dependence.

    To summarize, you can always write
    $$
    \Psi(x;t) = \sum_n c_n(t) \phi_n(x)
    $$
    with the ##\phi_n## forming a complete basis set of functions. Choosing these functions to be the eigenstates of the Hamiltonian, ##\hat{H} \phi_n = E_n \phi_n##, then the time evolution becomes
    $$
    \begin{align*}
    \Psi(x;t) &= e^{-i \hat{H} t/\hbar} \Psi(x;0) \\
    &= e^{-i \hat{H} t/\hbar} \sum_n c_n(0) \phi_n(x) \\
    &= \sum_n c_n(0) e^{-i E_n t/\hbar} \phi_n(x)
    \end{align*}
    $$
    The coefficients ##c_n(0)## are obtained from ##\Psi(x;0)## following the prescription I give in post #5.
     
  10. Nov 2, 2015 #9
    I see, is much clearer now, many thanks!
     
  11. Nov 4, 2015 #10
    Hi, sorry but I have to come back again one more time. Im going to describe exactly the example Im working in. Imagine our particle in a box of size L/2 at t<0. Exactly at t=0, we remove the barrier and the well is now of length L. Now lets prepate our initial state being a linear combination of the first eigenstates (that is, the states where the size of the well was L/2) This is:[tex]
    \Psi(x,t<0)=\sum_{n}C_{n}\sin(k_{n}x)e^{-iE_{n}t/\hbar}
    [/tex]
    where explicitly I am choosing a very particular case, where I want my initial wavefunction to be a sum of all possible eigenstates, and the coefficients I choose are:
    [tex]
    C_{n}=\frac{\sqrt{6}}{n\pi}
    [/tex]
    It is clear that with that choose
    [tex]
    \sum_{n}|C_{n}|^{2}=1
    [/tex]
    and this is is going to be my initial state. Of course, [tex]
    k_{n}=\frac{2\pi n }{L}
    [/tex]
    since the size of the box is L/2. The initial wavefunction is then:[tex]
    \Psi(x,t=0)=\sum_{n}C_{n}\frac{2}{\sqrt{L}}\sin(k_{n}x)e^{-iE_{n}t/\hbar}
    [/tex]
    where the factor of 2/sqrt(L) comes from the normalization of the eigenstates for the L/2 well.


    Now, following our previous discussion, we work out the eigenstates for the well of size L (that is when we remove the barrier at t=0), and we want exactly to express the total wavefunction as a linear combination of the eigenstates for the well of size L. With that [tex]
    \Psi(x,t>0)=\sum_{m}A_{m}\sqrt{\frac{2}{L}}\sin(k_{m}x)e^{-iE_{m}t/\hbar}
    [/tex]
    where now the [tex]
    k_{m}=\frac{\pi m }{L}
    [/tex]
    following our discussion, calculate the Am coefficients:[tex]
    A_{m}=\int_{0}^{L}dx\sum_{n=1}^{+\infty}|C_{n}|\frac{2\sqrt{2}}{L}\sin(k_{n}x)\sin(k_{m}x)
    [/tex]
    This reads[tex]
    A_{m}=\frac{2\sqrt{12}}{\pi L}\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{L}\sin(\frac{2n\pi x}{L})\sin(\frac{m\pi x}{L})
    [/tex]
    The integral gives [tex]
    \frac{L}{2\pi}\bigg(\frac{\sin(\pi(m-2n))}{m-2n}-\frac{\sin(\pi(m+2n))}{m+2n}\bigg)
    [/tex]
    This integral is non zero only for a single value of n [tex]n=\frac{m}{2}[/tex] in which the left side of the integral becomes Pi. Our coefficients [tex]
    A_{m}=\frac{2\sqrt{12}}{(m\pi)}
    [/tex]
    But now, with that, the sum [tex]\sum_{m}|A_{m}|^{2}\neq 1[/tex]
    What is going wrong here? This is really annoying me because I cant see what going on, I would be grateful if you can help or see where am I doing it wrong, thanks!
     
  12. Nov 4, 2015 #11

    DrClaude

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    Staff: Mentor

    Since ##t=0##, you actually have
    [tex]
    \Psi(x,t=0)=\sum_{n}C_{n}\frac{2}{\sqrt{L}}\sin(k_{n}x)
    [/tex]
    but that's not very important.

    This is where your problem lies. ##\Psi(x,t=0) = 0## for ##x> L/2##, so the integral actually only runs from ##0## to ##L/2##.
     
  13. Nov 4, 2015 #12
    Fixed, many thanks!
     
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