Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Time evolution operator - Confusion
Loading...