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Free particle time evolution

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    The eigenstates of the momentum operator with eigenvalue k are denoted by |k>, and the state of the system at t = 0 is given by the vector
    [tex]|{ψ}>=\int \frac {dk}{2π} g(k)|{k}>[/tex]
    Find the state of the system at t, |ψ(t)>.

    Compute the expectation value of [itex]\hat{P}[/itex].
    2. Relevant equations


    3. The attempt at a solution
    From what I learned from the lecture, I just have to introduce (multiply) [itex] \exp[\frac{-i}{\hbar}\hat{H}t][/itex] where in this free particle case, [itex]\hat{H}=\frac{\hat{P}^2}{2m}[/itex], to |ψ>.

    So [tex]|{ψ(t)}>=\exp[\frac{-i}{\hbar}\frac{\hat{P}^2}{2m}t]\int \frac {dk}{2π} g(k)|{k}>[/tex]

    When I compute for the expectation value using [itex]<ψ(t)|\hat{P}|ψ(t)>[/itex], I get [itex]\frac{1}{4\pi^2}\int |k|^2 \hat{P} dx[/itex].

    The exponentials cancel due to multiplying of its complex conjugate.
    I was confused on how to get rid of the two integrals with dk. I assumed (without reason so probably wrong) they become 1 because they are the product of complex conjugate and the total probability is 1.

    Any help will be appreciated!

    PS. How do I type ket in latex?
     
  2. jcsd
  3. Oct 19, 2016 #2

    BvU

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    Hi,

    You want to be a bit more accurate.

    For time development I think you want ##H## and not ##\hat H##. ##H## is an operator and yes, the ##|k>## are simultaneous eigenfunctions of H and p . See the link for the normalization: you don't have two integrals with ##k## but one with ##k## and one with ##k'##.

    Equally, for ##\hat P## you want the operator ##\ p\ ## and not the number ##\hat P## (that would be kind of circular ...)

    Ket in ##LaTeX## is just what you did. Or use \left \langle a\middle | b\right \rangle : ##\ \ \left \langle a\middle | b\right \rangle ##
     
  4. Oct 19, 2016 #3
    Hmm, I'm a bit confused. From my understanding, ##\hat{H}=\hat{T}+\hat{U}## where ##\hat{T}## is the kinetic operator and ##\hat{U}## the potential operator. And for free particle, U=0. Then I wrote the kinetic energy operator in terms of the momentum operator.
    Isn't ##\hat P## the operator ##-i \hbar \frac {d}{dx}## and not a number? And do you mean that when I take the complex conjugate of one of the ##Ψ(t)##, I should take the integral of k' and the other one in k, resulting in ##\frac{1}{4\pi^2}\int dk' g(k') \int dk g(k)## ? I don't know why I need to consider ##\hat{p}^2## because it gets canceled out when multiplying the complex conjugate of the exponential.
     
  5. Oct 19, 2016 #4

    BvU

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    Sorry, my bad -- I wrongly interpreted your notation . So ##\hat P|k> = k |k>## . That way you can describe the time development of ##|k>##.
    For the time development, the exponent therefore stays under the integral: each ##|k>## has its own time development.

    yes. see the link: the integration yields something with a delta function.
     
  6. Oct 19, 2016 #5
    Not quite sure how to follow the steps on the link. The integral I have for the expectation is ##\frac {1}{(2\pi)^2}\int dx \int dk \int g(k')g(k)k'k(dk')##. Have I done something wrong?
     
  7. Oct 19, 2016 #6

    BvU

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    Where are the ##\ \ \left \langle k'\middle | k \right \rangle \ \ ##? ##\quad## And the operator only works to the right, doesn't it ?
     
  8. Oct 20, 2016 #7
    Hmm, is it supposed to be ##<k'| \frac{1}{2\pi} \int dk' g(k') \exp{\frac{i\hat P ^2 t}{2m\hbar}} \int \hat P \frac{1}{2\pi}dk g(k) \exp{\frac{-i\hat P^2t}{2m\hbar}}|k>## then the exponentials cancel so ##<k'| (\frac{1}{2\pi})^2 \int dk' g(k') \ \int \hat P dk g(k)|k>##? Do I then compute for the expected value using the method of integrating the products of the complex conjugate of <k'|..., ##\hat P## and ...|k>?

    Becomes ##\frac{1}{4\pi^2}\int dx k'^* [\int \int dk'dk \hat P g(k')g(k)k]##
     
    Last edited: Oct 20, 2016
  9. Oct 20, 2016 #8

    BvU

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    I also wonder how you can end up with a triple integral
    Did you find the normalization eqn for the momentum eigenstates in the link ?
     
  10. Oct 21, 2016 #9

    BvU

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    Note that for ##<p>## (expectation value for ##\hat p## ), you want to end up somewhere at ##\ \int k\;|g(k)| ^2\;dk\ ## (according to my Merzbacher, QM, ##\ |g(k)| ^2 ## is the probability density in momentum space)
     
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