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Free particle propagation: paradox?

  1. Dec 6, 2015 #1
    Shankar ("Principles of Quantum Mechanics", 2nd ed.) shows that the free particle propagator "matrix element" is given by (see p. 153):

    ## \qquad \langle x | U(t) | x' \rangle = U(x,t;x') = \left(\frac{m}{2\pi\hbar it}\right)^{1/2} e^{im(x-x')^2/2m\hbar} ##,

    which can be used to evaluate the evolution of the particle's wavefunction in time, starting from its initial wavefunction:

    ## \qquad \psi(x,t) = \int_{-\infty}^\infty U(x,t;x') \, \psi(x',0) \, dx'##

    But there is an alternate formulation:

    ## \qquad \psi(x,t) = \langle x | U(t) |\psi(0)\rangle = \langle x | e^{-iHt/\hbar} |\psi(0)\rangle = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\frac{i\hbar t}{2m} \right)^n \frac{d^{2n}}{dx^{2n}} \, \psi(x) \qquad\qquad (\star) ##

    He shows (p. 154-5) that if the initial wavefunction ##\psi(x,0)## is a Gaussian, then it spreads out in time, and explains why.

    Exercise 5.1.4 (p. 156) gives a somewhat unsatisfying example of an initial wavefunction supported on the interval ##[-L/2, L/2]## (i.e., identically zero outside this interval). Hence a "paradox" using the ##(\star)## form of the propagator, because clearly all derivatives are zero outside the interval, so the wavefunction cannot spread in time (paradox). But the example given is discontinuous at the endpoints of the interval, so the derivatives do not exist.

    But what if the initial wavefunction is the function

    ## \qquad \psi(x,0) =
    \begin{cases}
    e^{-1/(1-x^2)} & |x|<1 \\
    0 & |x| \geq 1
    \end{cases}
    ##

    (I have not normalized ##\psi##, but never mind.) This function is supported on ##[-1, 1]## (identically zero outside the interval) and has infinitely many continuous derivatives on ##(-\infty, \infty)##.

    I would not know how to tackle this scary-looking integral:

    ## \qquad \psi(x,t) = \int_{-\infty}^\infty U(x,t;x') \, \psi(x',0) \, dx'##

    ## \qquad = \left(\frac{m}{2\pi\hbar it}\right)^{1/2} \int_{-\infty}^\infty e^{im(x-x')^2/2m\hbar} \, \psi(x',0) \, dx' ##

    ## \qquad = \left(\frac{m}{2\pi\hbar it}\right)^{1/2} \int_{-1}^1 e^{im(x-x')^2/2m\hbar} e^{-1/(1-x'^2)} \, dx'##

    BUT it does seem clear that using the alternative form ##(\star)##

    ## \qquad \psi(x,t) = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\frac{i\hbar t}{2m} \right)^n \frac{d^{2n}}{dx^{2n}} \psi(x,0)##

    that ## \psi(x,t) ## continues to be supported within ##[-1, 1]##. In other words, ## \psi(x,t) ## never spreads outside the interval ##[-1, 1]##!

    Is this not a paradox? There must be uncertainty in the initial velocity (momentum), which should result in a growing uncertainty in the position with increasing time. Yet the particle remains inside ##[-1, 1]## forever!
     
    Last edited: Dec 7, 2015
  2. jcsd
  3. Dec 7, 2015 #2

    Orodruin

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    The Green's function you quote is for a free particle. Regardless of what the initial state is, you cannot use the free particle Green's function for an infinite square well because it is not a Green's function of the problem you are trying to solve.
     
  4. Dec 7, 2015 #3
    From what I can see he is not talking about a particle in a square well but a free particle with the initial wavepacket given by the function he wrote.
     
  5. Dec 7, 2015 #4
    andresb - exactly right.
     
  6. Dec 7, 2015 #5

    Orodruin

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    Ok, so instead we are dealing with a situation where you expect he wave function to be completely given by its value and its derivatives in one point. This is not true for the function you have given, the series will only be a solution to the Schrödinger equation if it converges uniformly, which I strongly suspect it does not.
     
  7. Dec 7, 2015 #6
    I'm sorry I don't understand your comment. Can you please rephrase with more clarity?
     
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