Propagator for the Harmonic Oscillator

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We will now evaluate the propagator for the harmonic oscillator. The Lagrangian for this system is


Before we begin, let us prove that the propagator will separate into two factors; one of these comes entirely from the classical motion of the system, and the other comes entirely from quantum fluctuations about said trajectory. To this end, let us write x=x_c+y,\! where x_c\! is the classical trajectory and y\! is the fluctuation, which will be a new integration variable for the path integral. If we take t_i\! and t_f\! to be the initial and final times, respectively, then y(t_i)=y(t_f)=0.\! Substituting this into the action, we get

S=\int_{t_i}^{t_f} dt\,[\tfrac{1}{2}m(\dot{x}_c+\dot{y})^2-\tfrac{1}{2}m\omega^2(x_c+y)^2].

We now expand out the squares, obtaining

S=\int_{t_i}^{t_f} dt\,(\tfrac{1}{2}m\dot{x}_c^2-\tfrac{1}{2}m\omega^2x_c^2)+\int_{t_i}^{t_f} dt\,(\tfrac{1}{2}m\dot{y}^2-\tfrac{1}{2}m\omega^2y^2)+\int_{t_i}^{t_f} dt\,(m\dot{x}_c\dot{y}-m\omega^2x_cy).

If we integrate by parts in the third term, we get

S=\int_{t_i}^{t_f} dt\,(\tfrac{1}{2}m\dot{x}_c^2-\tfrac{1}{2}m\omega^2x_c^2)+\int_{t_i}^{t_f} dt\,(\tfrac{1}{2}m\dot{y}^2-\tfrac{1}{2}m\omega^2y^2)-\int_{t_i}^{t_f} dt\,m(\ddot{x}_c+\omega^2x_c)y.

We know, however, that the classical motion obeys the equation, \ddot{x}_c+\omega^2x_c=0.\! As a result, the third term is zero, and the action separates into two contributions, one coming entirely from the classical motion, and the other coming entirely from quantum fluctuations. Denoting these two contributions as

S_c(x_c)=\int_{t_i}^{t_f} dt\,(\tfrac{1}{2}m\dot{x}_c^2-\tfrac{1}{2}m\omega^2x_c^2)


S_q(y)=\int_{t_i}^{t_f} dt\,(\tfrac{1}{2}m\dot{y}^2-\tfrac{1}{2}m\omega^2y^2),\!

the propagator may now be written as

K(x_f,t_f;x_i,t_i)=e^{iS_c/\hbar}\int D[y(t)]\,e^{iS_q/\hbar}.

We will now evaluate each of these contributions.

Contribution from Classical Path

We will begin by evaluating the "classical" contribution to the propagator. This is essentially just a problem of classical mechanics; we begin by solving for the classical motion of the particle. The equation of motion is, as stated earlier,


We impose the boundary conditions, x(t_i)=x_i\! and x(t_f)=x_f.\! The solution of the equation of motion that satisfies these boundary conditions is


and the corresponding velocity is

x_c(t)=-\omega x_i\frac{\cos{\omega(t_f-t)}}{\sin{\omega(t_f-t_i)}}+\omega x_f\frac{\cos{\omega(t-t_i)}}{\sin{\omega(t_f-t_i)}}.

If we now substitute these expressions into the Lagrangian and simplify, we obtain


If we now substitute this into the action, we finally obtain


Contribution From Fluctuations

We now turn our attention to the "quantum" contribution to the propagator. It is given by

K_q=\int D[y(t')]\,\exp\left [\frac{i}{\hbar}\int_{0}^{\delta t}(\tfrac{1}{2}m\dot{y}^2-\tfrac{1}{2}m\omega^2y^2)\,dt'\right ].

Note that we changed variables in the time integral to t'=t-t_i;\! this will simplify the subsequent calculations. To further simplify our notation, we introduce the quantity, \delta t=t_f-t_i.\!

Because y(t'=0)=y(t'=\delta t)=0,\! we may expand y(t')\! in a Fourier sine series:

y(t')=\sum_{n=1}^{\infty} a_n \sin\left(\frac{n\pi t'}{\delta t}\right)

We now re-express the path integral in terms of the coefficients a_n\! of this series. One may verify, with the aid of the fact that

\int_0^{\delta t}dt'\,\sin\left (\frac{m\pi t'}{\delta t}\right )\cos\left (\frac{n\pi t'}{\delta t}\right )=0

for all m\! and n\! and

\int_0^{\delta t}dt'\,\sin\left (\frac{m\pi t'}{\delta t}\right )\sin\left (\frac{n\pi t'}{\delta t}\right )=\int_0^{\delta t}dt'\,\cos\left (\frac{m\pi t'}{\delta t}\right )\cos\left (\frac{n\pi t'}{\delta t}\right )=0

if m\neq n,\! that

K_q=C\int_{-\infty}^{\infty} da_1\,\int_{-\infty}^{\infty} da_2\,\ldots\,\exp{\left[\sum_{n=1}^{\infty}\frac{im}{2\hbar}\left(\left(\frac{n\pi}{\delta t}\right)^2-\omega^2\right)a^2_n\right]},

where C\! is a constant that is independent of the frequency that comes from the Jacobian of the transformation. The important point is that it does not depend on the frequency \omega\!. The integrals over the a_n\! are just Gaussians; evaluating them gives us

K_q=C'\prod_{n=1}^{\infty}\left[\left(\frac{n\pi}{\delta t}\right)^2-\omega^2\right]^{-\frac{1}{2}}=
C'\prod_{n=1}^{\infty}\left(\frac{\delta t}{n\pi}\right)
\prod_{n=1}^{\infty}\left[1-\left(\frac{\omega\,\delta t}{n\pi}\right)^2\right]^{-\frac{1}{2}},

where C'\! is a new constant directly related to C\! that is also independent of the frequency of motion. Since the first product in this expression is also independent of the frequency of motion, we will absorb it into C'\!, thus defining yet another constant, C''.\! The remaining, frequency-dependent, product then evaluates to

K_q=C''\sqrt{\frac{\omega\,\delta t}{\sin(\omega\,\delta t)}}.

In the limit \omega\rightarrow 0, we should recover the propagator for a free particle that propagates back to its initial position; using this fact, we find that

C''=\sqrt{\frac{m}{2\pi i \hbar\,\delta t}}.

We have thus determined the full "quantum" contribution to the propagator,

K_q=\sqrt{\frac{m}{2\pi i \hbar\,\delta t}}\sqrt{\frac{\omega\,\delta t}{\sin(\omega\,\delta t)}}=\sqrt{\frac{m\omega}{2\pi i \hbar \sin(\omega\,\delta t)}},

and therefore the full propagator,

K(x_f,t_f;x_i,t_i)=\sqrt{\frac{m\omega}{2\pi i \hbar \sin{\omega(t_f-t_i)}}}\exp\left \{\frac{im\omega}{2\hbar\sin{\omega(t_f-t_i)}}[(x_i^2+x_f^2)\cos{\omega(t_f-t_i)}-2x_ix_f]\right \}.


Our evaluation of the "quantum" contribution to the propagator uses the method presented here: File:FeynmanHibbs H O Amplitude.pdf

For a more detailed evaluation of this problem, please see Barone, F. A.; Boschi-Filho, H.; Farina, C. 2002. "Three methods for calculating the Feynman propagator". American Association of Physics Teachers, 2003. Am. J. Phys. 71 (5), May 2003. pp 483-491.

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