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Forced Harmonic Oscillator with Path Integral

  1. Jan 31, 2009 #1
    how do I compute the transition amplitude of the forced harmonic oscillator with the method of path integration?

    Mr. Fogg
  2. jcsd
  3. Jan 31, 2009 #2
    Do you mean the vacuum to vacuum transition amplitude? You take the Fourier transformation of the path, plug the path into the functional, complete the square to separate out a path-independent term and a path-dependent term, shift the path so that the path-dependent term is no longer a function of the source, then set the whole expression equal to 1 when the source is zero to correctly normalize it, and this will show you that the transition amplitude is the path-independent term.
  4. Feb 1, 2009 #3
    Make the ansatz: [tex]x=\bar{x}+y[/tex], where [tex]\bar{x}[/tex] is the classical solution. Then we can write the "variation" y as a Fourier expansion in time. Thus the kernel K[b,a] is


    and normalization constant in F(T) could be found by the limit w->0 (free particle case). The action of the classical trajectory could be obtained using the Green function. We have the system: x''+w^2x=f(t)/m and BC x(ta)=Xa and x(tb)=Xb. The Lagrangian could then be obtained if we know x(t) which is given as:

    [tex]\bar{x}(t)=\int_{ta}^{tb}G(t,\chi)f(t)/m\cdot d\chi[/tex]

    Hope this helps you a bit...
  5. Feb 3, 2009 #4
    Thanks for your help.

    The Lagrangian is
    [tex] L = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} m \omega^2 x^2 + J(t) x [/tex]

    I started with:

    [tex]x = x_{cl} + y[/tex]

    and showed that the transition amplitude can be written as

    [tex] U(x_a,x_b,t_b) = \exp(-\frac{i}{\hbar}S(x_{cl})) + \int_{y(0)=0}^{y(t_{b})=0} [dy]\exp(-\frac{i}{\hbar} \int_0^{t_{b}} dt \frac{1}{2}m (\dot{y}^2 - \omega^2 y^2)[/tex]

    Now I wan't to show with Path Integration that the transition amplitude is then

    [tex] U(x_a, x_b, t_b) = \lim\limits_{N \rightarrow \infty}\frac{m}{2 \pi i \hbar \epsilon Q_{N-1}} e^{\frac{i}{\hbar} S(x_{cl})} \quad \text{(1.1)}[/tex]

    with [tex] \epsilon = \frac{t_b}{N} \quad Q_{N-1} = det(A)[/tex]

    A is a quadratic Matrix.

    I don't know, how to get the Matrix-Expression (1.1) of the transition amplitude.

    Last edited: Feb 3, 2009
  6. Feb 5, 2009 #5
    Maybe this link will help:


    The det(A) doesn't really matter.

    As I think already has been mentioned, for your particular problem (no more than quadratic in q), you can also find the classical solution to your Lagrangian, and plug that solution into your Lagrangian. The classical solution should be:

    [tex] q(t)=\int \frac{d^4n}{(2\pi)^4} \frac{f(n)}{m(\omega^2-n^2)}e^{-int}[/tex]

    where f(n) is the Fourier-Transform of f(t).
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