The oscillator path integral

Intuition

The oscillator’s action is quadratic too, so its path integral is again exact: the classical action times a factor that depends only on the time. The classical path between the ends is a combination of $\sin\omega t$ and $\cos\omega t$ , and its action works out to $\tfrac{m\omega}{2\sin\omega T}[(x_{a}^{2}+x_{b}^{2})\cos\omega T-2x_{a}x_{b}]$ . The prefactor is $\sqrt{m\omega/2\pi i\hbar\sin\omega T}$ , which becomes the free one for short times. Something striking happens at $\omega T=\pi$ : every classical path that leaves a point comes back to a single point, as the figure shows, and the prefactor blows up — a focus of paths. Best of all, the propagator knows the spectrum. Set $x_{b}=x_{a}$ and integrate over $x$ : the result is a geometric series in $e^{-i\omega T}$ , and reading off its frequencies gives the energies $\hbar\omega(n+\tfrac12)$ , zero-point energy included, without solving any differential equation.

Pendulums set swinging from the same spot with different pushes all pass back through that spot together after half a swing, however hard they were pushed. The oscillator’s paths refocus the same way, and its propagator has a focus there.

Classical oscillator paths leaving $x=0$ at $t=0$ with different speeds, $x=A\sin\omega t$ , with time across in units of . Every one returns to at : the paths refocus there, and the propagator’s prefactor diverges.

For $V=\tfrac12m\omega^{2}x^{2}$ , with :

\begin{aligned}K&=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega T}}\;e^{iS_{\text{cl}}/\hbar}\\ S_{\text{cl}}&=\frac{m\omega}{2\sin\omega T}\bigl[(x_{a}^{2}+x_{b}^{2})\cos\omega T-2x_{a}x_{b}\bigr]\end{aligned}

Properties

As $\omega T\to0$ it becomes the free propagator: $\sin\omega T\approx\omega T$ and $S_{\text{cl}}\to m(x_{b}-x_{a})^{2}/2T$ .

The oscillator’s energies from its propagator

\begin{aligned}\int K(x,x;T)\,dx&=\frac{1}{2i\sin(\omega T/2)}\\ &=\sum_{n=0}^{\infty}e^{-i\omega(n+\frac12)T}\end{aligned}

Put $x_{a}=x_{b}=x$ : the classical action becomes a single Gaussian exponent, $x^{2}$ times over with a minus sign. The Gaussian integral and the prefactor combine into , which expands as a geometric series in . Comparing with the spectral form of the trace, , reads off the energies.

Proof steps

\begin{aligned}S_{\text{cl}}(x,x)&=\frac{m\omega x^{2}(\cos\omega T-1)}{\sin\omega T}\\ &=-m\omega x^{2}\tan\frac{\omega T}{2}\end{aligned}

Put $x_{a}=x_{b}=x$ ; use $\frac{1 - \cos θ}{\sin θ} = \tan$ .

\int e^{-\frac{i}{\hbar}m\omega x^{2}\tan\frac{\omega T}{2}}\,dx=\sqrt{\frac{\pi\hbar}{i\,m\omega\tan\frac{\omega T}{2}}}

A Gaussian integral.

\begin{aligned}\int K(x,x;T)\,dx&=\sqrt{\frac{1}{-2\sin\omega T\,\tan\frac{\omega T}{2}}}\\ &=\frac{1}{2i\sin\frac{\omega T}{2}}\end{aligned}

Multiply by the prefactor; $\sin\omega T\,\tan\frac{\omega T}{2}=2\sin^{2}\frac{\omega T}{2}$ .

\frac{1}{2i\sin\frac{\omega T}{2}}=\frac{e^{-i\omega T/2}}{1-e^{-i\omega T}}=\sum_{n=0}^{\infty}e^{-i\omega(n+\frac12)T}

A geometric series, convergent when $T$ has a small negative imaginary part.

\begin{aligned}&\int K(x,x;T)\,dx=\sum_{n}e^{-iE_{n}T/\hbar}\\ &\Longrightarrow\ E_{n}=\hbar\omega\Bigl(n+\tfrac12\Bigr)\end{aligned}

The spectral form of $K$ , integrated over $x$ with normalised $\psi_{n}$ .

Applications

Practice

Energies From the Trace

The integral of $K(x,x;T)$ over $x$ is a sum of $e^{-iE_{n}T/\hbar}$ ; for the oscillator it is a geometric series with the frequencies .

\int K(x,x;T)\,dx=\sum_{n}e^{-iE_{n}T/\hbar}

Try it

From the geometric series, what is the energy of the level $n=2$ , in units of $\hbar\omega$ ?