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Singularities in the harmonic oscillator propagator

  1. Sep 21, 2014 #1
    Hi people!
    Today I was doing some QFT homework and in one of them they ask me to calculate the Harmonic Oscillator propagator, which, as you may know is:
    [tex]W(q_2,t_2 ; q_1,t_1) = \sqrt{\frac{m\omega}{2\pi i \hbar \sin \omega (t_2-t_1)}} \times \exp \left(\frac{im\omega}{2\hbar \sin \omega (t_2-t_1)}\left[ (q_1^2+q_2^2)\cos \omega (t_2-t_1)-2q_2q_1\right]\right)[/tex]

    So, the last question of the problem is explaining why do we have singularities at [itex]t_2-t_1=n\pi / \omega[/itex].
    I've been searching for it on books and I've found something about the "caustics", which I've no idea what is it...
    I understand that when [itex]t_2-t_1=0 [/itex] it has to diverge, because the particle hasn't moved from its place as the time hasn't passed, but why it happens every half period?
    It seems to me that if the function diverges at those points then we are sure of which is the position of the particle at those times, and I thought that there had to be an uncertainty for [itex]t_2-t_1 \neq 0 [/itex] (I don't know if I'm explaining well enought what I think hahahhaa....)

    Thanks in advance!
     
  2. jcsd
  3. Sep 22, 2014 #2

    vanhees71

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    Hm, that's a very good question. I've never thought about it, to be honest. Of course, any propagator must fulfill the boundary condition
    [tex]W(q_2,0;q_3,0)=\delta(q_2-q_3).[/tex]
    That this occurs periodically may be caused by the dynamical SU(2) symmetry of the simple harmonic oscillator. Mathematically it's easy to understand by deriving the propgator making use of the Heisenberg picture. Solving the equations of motion for the position and momentum operators is exactly as for c-numbers, because the Hamiltonian is quadratic not only in momentum but also in position, i.e., the equations of motion are linear.
     
  4. Sep 22, 2014 #3
    Where will the particle be after one period? Remember that for the harmonic oscillator every wave function is exactly periodic with period [itex]T = 2\pi/\omega[/itex].

    Now, where will the particle be after one half-period?

    You might find it fun to set up and play with a harmonic oscillator potential in this applet.
     
  5. Sep 22, 2014 #4

    stevendaryl

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    Actually, what's interesting about the harmonic oscillator propagator is that it is periodic:
    [itex]W(q_2,t_2 ; q_1,t_1) = W(q_2,t_2 + n T; q_1,t_1)[/itex]

    where [itex]T = \frac{2 \pi}{\omega}[/itex]

    Given that it is periodic, you would expect a singularity whenever [itex]t_2 - t_1 = n T[/itex], because every propagator has a singularity at [itex]t_2 - t_1 = 0[/itex]

    So the only mystery is why there are twice as many singularities as that. There is a singularity at [itex]t_2 - t_1 = \frac{T}{2}[/itex] as well as at [itex]t_2 - t_1 = T[/itex]
     
  6. Sep 24, 2014 #5
    Thanks for the answers!
     
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