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Time evolution operator - Confusion

  1. Sep 6, 2011 #1
    Hi everyone. I am given a somewhat common potential well [itex] V(x)=0 [/itex] for [itex] ¦x¦<a [/itex] and infinite elsewhere. I am told that at t = 0 my particle is in a state represented by the wavefunction

    [itex] \psi(x,0)= A(\sin{(\frac{\pi x}{a})}+ \sqrt{2} \cos{(\frac{3 \pi x}{2 a})}) [/itex]

    where A is a constant use for normalization. It turns easily out that this constant equals [itex] \sqrt{\frac{1}{3a}} [/itex]. On the other hand, the eigenfunctions in this situations are very well known to be [itex]\psi_n= \sqrt{\frac{1}{a}} \sin(\frac{n \pi x}{2a}) [/itex] for n even and [itex]\psi_n= \sqrt{\frac{1}{a}} \cos(\frac{n \pi x}{2a}) [/itex] for n odd. So given wave function can easily be written as a linear combination of the second and third eigenstates.

    Now come the problems. At some point of the exercise I am asked to calculate the expected value of some operator but at some time t. This is how the professor writes the wavefunction at the generic time t:

    [itex] e^{- \frac{i}{\hbar}(E_2-E_3)t} \psi(x,0) [/itex]

    Now my question is: since the given state is a combination of 2 eigenstates, why in the world does he write the temporal evolution this way, with (E_2-E_3) in the exponent? Shouldn't it be something like
    [itex] e^{- \frac{i}{\hbar}E_2 t} \psi_2(x,0) - e^{- \frac{i}{\hbar}E_3 t} \psi_3(x,0) [/itex]?
    Maybe some details are wrong but what I mean is I expect to find the sum of 2 exponentials whereas there is the exponential of a sum.
    My idea is that (simplifying the notation) the ket [itex] |2> + \sqrt{2} |3> [/itex] is of course different from the sum of the kets [itex] |2> [/itex] and [itex] |3> [/itex] but then why isn't the right expression something like
    [itex] e^{- \frac{i}{\hbar}(E_2-\sqrt{2}E_3)t} \psi(x,0) [/itex].

    Am I somewhat clear?
    Last edited: Sep 6, 2011
  2. jcsd
  3. Sep 6, 2011 #2


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    Es you can easily check by pluging your solution of the Schrödinger eq. back into the equation, you are right and the prof. is wrong. The correct state is

    [tex]|\psi,t \rangle>=A [\exp(-\mathrm{i} E_2 t) |2 \rangle+\sqrt{2} \exp(-\mathrm{i} E_3 t) |3 \rangle].[/tex]

    The norm should be 1, so you find [itex]A=1/\sqrt(1+2)=1/\sqrt{3}[/itex]. Note that the [itex]1/\sqrt{a}[/itex] factor is already included in the norm of the state vectors, which are given in position representation.

    I don't understand, how you come to the last forumula, which definitely is wrong.
  4. Sep 6, 2011 #3
    Yeah, problem solved. The unexplainable factor came from a multiplication of exponentials. I must have been drunk when i first read the solution.
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