The potential in this problem can be replaced by the boundary condition G(r, r', t) = 0 and z = 0. The boundary problem can then be solved by the method of images. Suppose at r” is the image of the electron at r’ about z = 0. Then
\((ih \partial _t - H)G(r, r', t) = \delta (t) |\delta (r - r' - \delta (r - r''))|\) (1)
The Green's function is zero for x ≤ 0 and for x ≥ 0 is equal to the x > 0 part of the solution of (1). Let
We have \(ih\partial _TG = h \omega G\) and \(H = \frac{h^2 k^2}{2m}\), and the substitution of (2) in (1) gives
We first integrate with respect to \(\omega\). The path r is chosen to satisfy the causality condition.
Causality requires that when t < 0,G(r, r', t) = 0. First let the polar point of w shift a little, say by \(-i\varepsilon\), where \(\varepsilon\) is a small positive number. Finally letting \(\varepsilon\) → 0, we get
Hence when both x and t are greater than zero, the Green's function is given by (5); otherwise, it is zero. When x > 0 and t > 0,
If the potential V(x, y, z) were absent, the Green's function for the free space, IG(r, r', t)|2, would be proportional to t-3. But because of the presence of the reflection wall the interference term occurs.
