12.6 Propagators in frequency-momentum space

12.6 Propagators in frequency-momentum space

The propagator defined by equation 12.4 will take a wavefunction at time 0 and give the wavefunction at any other time t, positive or negative. We will find it useful to define a version of the propagator that takes into account causality, only giving a non-zero result for $t>0:$

Definition (Retarded propagator)

The retarded propagator is given by

$$U_{+}(t,q_t-q_0)=\{\begin{array}{ll}0& t<0\\ U(t,q_t-q_0)& t>0\end{array}$$

This can also be written

$$U_{+}(t,q_t-q_0)=\theta (t)U(t,q_t-q_0)$$

where $\theta (t)$ is the step-function

$$\theta (t)=\{\begin{array}{ll}1& t>0\\ 0& t<0\end{array}$$

We will use an integral representation of $\theta (t)$ given by

$$\theta (t)=\underset{ϵ\to 0^{+}}{\lim }\frac{i}{2\pi }{\int }_{-\infty }^{+\infty }\frac{1}{\omega +iϵ}{e}^{-i\omega t}d\omega \text{\hspace{1em}(12.10)}$$

To derive this, note that as a distribution, $\theta (t)$ has a Fourier transform given by

$$\underset{ϵ\to 0^{+}}{\lim }\frac{i}{\sqrt{2\pi }}\frac{1}{\omega +iϵ}$$

since the calculation

$$\begin{array}{rl}\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }\theta (t)e^{i\omega t}dt& =\frac{1}{\sqrt{2\pi }}{\int }_0^{+\infty }{e}^{i\omega t}dt\\ & =\frac{1}{\sqrt{2\pi }}\left(-\frac{1}{i\omega }\right)\end{array}$$

makes sense for $\omega$ replaced by lim ${\rho }_{10^{+}}\left(\omega +iϵ\right)$ (or, for real boundary values of $\omega$ complex, taking values in the upper half-plane). Fourier inversion then gives equation 12.10.

Digression. The integral 12.10 can also be computed using methods of complex analysis in the variable $\omega$. Cauchy’s integral formula says that the integral about a closed curve of a meromorphic function with simple poles is given $by$ 2$\pi$i times the sum of the residues at the poles. For $t<0$ , since $e^{-i\omega t}$ falls of exponentially if $\omega$ has a non-zero positive imaginary part, the integral along the real $\omega$ axis will be the same as for the semi-circle $C_{+}$ closed in the upper half-plane (with the radius of the semi-circle taken to infinity). $C_{+}$ encloses no poles so the integral is 0.

Figure 12.2: Evaluating $\theta (t)$ via contour integration.

For $t>0$ , one instead closes the path using $C_{-}$ in the lower half-plane, and finds that the integral can be evaluated in terms of the residue $of$ the pole at $\omega =-iϵ$ (with the minus sign coming from orientation of the curve), giving

$$\theta (t)=-2\pi i\left(\frac{i}{2\pi }\right)=1$$

By similar arguments one can show that $\theta (-t)$ has (as a distribution) Fourier transform

$$\underset{ϵ\to 0^{+}}{\lim }-\frac{i}{\sqrt{2\pi }}\frac{1}{\omega -iϵ}$$

and the integral representation

$$\theta (-t)=\underset{ϵ\to 0^{+}}{\lim }-\frac{i}{2\pi }{\int }_{-\infty }^{+\infty }\frac{1}{\omega -iϵ}{e}^{-i\omega t}d\omega$$

Taking $1/\sqrt{2\pi }$ times the sum of the Fourier transforms for $\theta (t)$ and $\theta (-t)$ gives the distribution

$$\begin{array}{rl}\underset{ϵ\to 0^{+}}{\lim }\frac{i}{2\pi }\left(\frac{1}{\omega +iϵ}-\frac{1}{\omega -iϵ}\right)& =\underset{ϵ\to 0^{+}}{\lim }\frac{i}{2\pi }\frac{-2iϵ}{{\omega }^2+{ϵ}^2}\\ & =\underset{ϵ\to 0^{+}}{\lim }\frac{1}{\pi }\frac{ϵ}{{\omega }^2+{ϵ}^2}\\ & =\delta (\omega )\end{array}\text{\hspace{1em}(12.11)}$$

as one expects since the delta-function $\delta (\omega )$ is the Fourier transform of

$$\frac{1}{\sqrt{2\pi }}(\theta (t)+\theta (-t))=\frac{1}{\sqrt{2\pi }}$$

Returning to the propagator, as in section 11.5 one can Fourier transform with respect to time, and thus get a propagator that depends on the frequency $\omega$. The Fourier transform of equation 12.6 with respect to time is

$$\hat{U}(\omega ,k)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }\left(\frac{1}{\sqrt{2\pi }}{e}^{-i\frac{1}{2m}{k}^{2}t}\right)e^{i\omega t}dt=\delta (\omega -\frac{1}{2m}{k}^2)$$

Using equations 12.5 and 12.10 the retarded progagator in position space is given by

$$\begin{array}{rl}{U}_{+}(t,q_t-q_0)& =\underset{ϵ\to 0^{+}}{\lim }{\left(\frac{1}{2\pi }\right)}^2{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }\frac{i}{\omega +iϵ}{e}^{-i\omega t}{e}^{ik(q_t-q_0)}{e}^{-i\frac{1}{2m}{k}^{2}t}d\omega dk\\ & =\underset{ϵ\to 0^{+}}{\lim }{\left(\frac{1}{2\pi }\right)}^2{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }\frac{i}{\omega +iϵ}{e}^{-i(\omega +\frac{1}{2m}{k}^2)t}{e}^{ik(q_t-q_0)}d\omega dk\end{array}$$

Shifting the integration variable by

$$\omega \to {\omega }^{\prime }=\omega +\frac{1}{2m}{k}^2$$

one finds

$$U_{+}(t,q_t-q_0)=\underset{ϵ\to 0^{+}}{\lim }{\left(\frac{1}{2\pi }\right)}^2{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }\frac{i}{{\omega }^{\prime }-\frac{1}{2m}{k}^2+iϵ}{e}^{-i{\omega }^{\prime }t}{e}^{ik(q_t-q_0)}d{\omega }^{\prime }dk$$

but this is the Fourier transform

$$U_{+}(t,q_t-q_0)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }{\hat{U}}_{+}(\omega ,k)e^{-i\omega t}{e}^{ik(q_t-q_0)}d\omega dk\text{\hspace{1em}(12.12)}$$

where

$${\hat{U}}_{+}(\omega ,k)=\underset{ϵ\to 0^{+}}{\lim }\frac{i}{2\pi }\frac{1}{\omega -\frac{1}{2m}{k}^2+iϵ}\text{\hspace{1em}(12.13)}$$

Digression. By the same argument as the one above for the integral representation of $\theta (t)$ , but with pole now at

$$\omega =\frac{1}{2m}{k}^2-iϵ$$

the $\omega$ integral in equation 12.12 can be evaluated by the Cauchy integral formula, recovering formula 12.9 for $U(t,q_t-q_0)$

Previous section: 12.5 The propagator in position space · Next section: 12.7 Green’s functions and solutions to the Schr¨odinger equations · 中文版本

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