The Nonrelativistic Propagator

Our goal is to investigate the differential equation which defines $G$, and in particular to solve for $G_0$ explicitly, so that the expansion for $G$ can be explicitly carried out. From Huygens' principle,

$$\psi(\vec{x}^\prime,t^\prime) = i \int d^3x G(\vec{x}^\prime,t^\prime;\vec{x},t)\psi(\vec{x},t) \quad t^\prime > t .$$ (6.28)

A form valid for all time is

$$\theta(t^\prime-t)\psi(x^\prime) = i \int d^3x G(x^\prime;x) \psi(x) .$$ (6.29)

Using the chain rule and the fact that $\psi(x^\prime)$ satisfies the Schrödinger equation, we have

$$\left[ i\frac{\partial}{\partial t^\prime} -H(x^\prime) \rig... ...\partial t^\prime} \theta(t^\prime-t) \right] \psi(x^\prime) .$$ (6.30)

Since

$$\frac{d\theta(\tau)}{d\tau} = \delta(\tau) ,$$ (6.31)

we have

$$\left[ i\frac{\partial}{\partial t^\prime} - H(x^\prime) \ri... ...prime-t) \psi(x^\prime) = i\delta(t^\prime-t) \psi(x^\prime) .$$ (6.32)

Substituting equation 6.29 for $\theta(t^\prime-t)\psi(x^\prime)$, we have

$$i\int d^3x \left[ i\frac{\partial}{\partial t^\prime} - H(x^\prime) \right] G(x^\prime;x) \psi(x) = i \delta(t^\prime-t) \psi(x^\prime) ,$$

$$= i\int d^3x \delta(t^\prime-t) \delta^3(\vec{x}^\prime-\vec{x}) \psi(x) .$$ (6.33)

Therefore

$$\left[ i\frac{\partial}{\partial t^\prime} - H(x^\prime) \ri... ...me-t) \delta^3(\vec{x}^\prime-\vec{x}) = \delta^4(x^\prime-x)$$ (6.34)

is the Green's function equation in the Schrödinger theory. Along with the boundary conditions $G(x^\prime;x)=0$ for $t^\prime<t$ this defines the retarded Green's function.

For the free-particle propagator $H_0(x^\prime)=-\frac{1}{2m}\nabla^2_{x^\prime}$. $G_0(x^\prime;x)$ can depend only upon the difference of the coordinates $(x^\prime,t^\prime)$ and $(x,t)$. This is because the wave at $(\vec{x}^\prime,t^\prime)$ emerging from a unit source at $\vec{x}$ which is turned on at $t$ depends only on the interval $(\vec{x}^\prime-\vec{x},t^\prime-t)$, and $G_0(x^\prime;x)$ is precisely the amplitude of this wave.

Consider the Fourier transform

$$G_0(x^\prime;x) = G_0(x^\prime-x) = \int \frac{d^3pd\omega}{... ...rime-\vec{x})} e^{-i\omega(t^\prime-t)} G_0(\vec{p},\omega) .$$ (6.35)

The Schrödinger equation for the Green's function becomes

$$\left( i\frac{\partial}{\partial t^\prime} + \frac{1}{2m} \nabla^2_{x^\prime} \right) G_0(x^\prime;x) = \delta^4(x^\prime-x) ,$$

$$\int\frac{d^3pd\omega}{(2\pi)^4} \left( \omega - \frac{p^2}{2m} \... ... G_0(\vec{p},\omega) e^{-i\omega(t^\prime-t) + ip\cdot(\vec{x}^\prime-\vec{x})} = \int\frac{d^3pd\omega}{(2\pi)^4} e^{-i\omega(t^\prime-t) + ip\cdot(\vec{x}^\prime-\vec{x})}$$ (6.36)

and hence for $\omega\neq p^2/2m$,

$$G_0(\vec{p},\omega) = \frac{1}{\omega-p^2/2m} .$$ (6.37)

To complete the expression, we must have a rule for handling the singularity. We will see that the retarded boundary condition

$$G(x^\prime;x) = 0 \quad\textrm{for}\quad t^\prime < t$$ (6.38)

requires us to add a positive infinitesimal imaginary part to the denominator. Lets integrate

$$G_0(x^\prime-x) = \int\frac{d^3p}{(2\pi)^3} e^{i\vec{p}\cdot(\vec{x}^\prime-\vec{x}... ...rac{d\omega}{2\pi} \frac{e^{-i\omega(t^\prime-t)}}{\omega - p^2/2m + i\epsilon}$$

$$= \int\frac{d^3p}{(2\pi)^3} e^{i\vec{p}\cdot(\vec{x}^\prime-\vec{x}... ...c{d\omega}{2\pi} \frac{e^{-i(\omega + p^2/2m)(t^\prime-t)}}{\omega + i\epsilon}$$

$$= \int\frac{d^3p}{(2\pi)^3} e^{i\vec{p}\cdot(\vec{x}^\prime-\vec{x}... ...)} \int\frac{d\omega}{2\pi} \frac{e^{-i\omega(t^\prime-t)}}{\omega + i\epsilon}$$ (6.39)

The integral over $\omega$ is evaluated as a contour integral in the complex $\omega$-plane as shown in figure 6.3. For $t^\prime > t$, the contour may be closed along an infinite semicircle below the real axis in order to ensure exponential damping of the integrand, and the value of the integral is $-i$ by Cauchy's theorem. For $t > t^\prime$, the contour is closed above and the integral vanishes because the pole at $-i\epsilon$ lies outside the contour. The integral is thus just $-i$ times the step function. Continuing our previous integration, we have

Figure 6.3: Contour in the complex $\omega$-plane for integrating the unit step function.

$$G_0(x^\prime-x) = \int\frac{d^3p}{(2\pi)^3} e^{i\vec{p}\cdot(\vec{x}^\prime-\vec{x})} e^{-i(p^2/2m)(t^\prime-t)} (-i) \theta(t^\prime-t)$$

$$= -i\theta(t^\prime-t) \int d^3p \frac{e^{i\vec{p}\cdot\vec{x}^\pri... ...ime}}{(2\pi)^{3/2}} \frac{e^{-i\vec{p}\cdot\vec{x} + i(p^2/2m)t}}{(2\pi)^{3/2}}$$

$$= -i\theta(t^\prime-t) \int d^3p \frac{e^{i(\vec{p}\cdot\vec{x}^\pr... ...me)}}{(2\pi)^{3/2}} \frac{e^{-i(\vec{p}\cdot\vec{x} - \omega t)}}{(2\pi)^{3/2}}$$

$$= -i\theta(t^\prime-t) \int d^3p \phi_p(\vec{x}^\prime,t^\prime) \phi_p^*(\vec{x},t) ,$$ (6.40)

where the subscript $p$ stands for plane waves. This is a special case for plane waves. In general

$$G(x^\prime;x) = -i\theta(t^\prime-t) \sum_n \psi_n(\vec{x}^\prime,t) \psi_n^*(\vec{x},t) ,$$ (6.41)

where $\sum_n$ is a generalized sum and integral over the spectrum of quantum numbers $n$, $\psi_n(x)$ is a complete set of normalized solutions to the Schrödinger equation which satisfy a completeness statement

$$\sum_n \psi_n(\vec{x}^\prime,t) \psi_n^*(\vec{x},t) = \delta^3(\vec{x}-\vec{x}^\prime) .$$ (6.42)

There is an enormous amount of information contained in $G(x^\prime;x)$. All the solutions of the Schrödinger equation, including the bound states, as required in the completeness relation, appear with equal weight. It is no wonder that $G$ is difficult to compute.

The same Green's function which propagates a solution of the Schrödinger equation forward in time propagates its complex conjugate backward in time. From (6.41) we have

$$i\int d^3x G(x^\prime;x) \psi_n(x) = \theta(t^\prime-t) \sum_m \psi_m(x^\prime) \int d^3x \psi_m^*(x) \psi_m(x) ,$$

$$= \theta(t^\prime-t) \psi_n(x^\prime)$$ (6.43)

and

$$i\int d^3x^\prime\psi_m^*(x^\prime)G(x^\prime;x) = \theta(t^\prime-t) \sum_n \psi_n^*(x) \int d^3x^\prime \psi_n(x^\prime) \psi_m^*(x^\prime) ,$$

$$= \theta(t^\prime-t) \psi_m^*(x) .$$ (6.44)