10. Appendix

which can be found in standard textbooks on mathematical analysis:
°(1 n) n°(n);
°(n)°(1 n) apple/ sin(napple) provided 0 n 1;
and the Weierstrass definition of °(n):
∞
1
zeÁz 1
°(z) z
z
e m .
m
m1
Solution to Problem 6.11 635
In this definition Á is the Euler constant, equal to 0.5772157...
Next, we use analytic continuation to extend these results to complex arguments.
From the above results one can show that:
°(1 i·) (i·)°(i·)
and hence:
°(i·)°(i·)
apple
(i·) sin(i·apple)
From the Weierstrass definition we can show that the complex conjugate of
1/°(i·) is1/°(i·). Combining these results together we find the magnitude
of the Gamma function with an imaginary argument:
|°(1 i·)| 2 |·| 2 |°(i·)| 2 ·apple/ sinh(·apple).
Substituting this result back into the relation between °(i·) and °(1 i·)
we obtain:
|°(1 i·)| 2 |·| 2 |°(i·)| 2 ·apple/ sinh(·apple).
Finally, when we substitute this expression for |°(1 i·)| 2 into |R(0)| 2 we
obtain:
|R(0)| 2 ·apple exp(apple·)/[V sinh(·apple)]
For a given photon energy ˆ bigger than the energy gap Eg, energy conservation
leads to the relation: E ˆEg and · [R ∗ /(ˆEg)] 1/2 . We obtain
(6.153) when we replace ·apple by Ù and use the atomic units for which 1.
Solution to Problem 6.11
(a) Let us start with the wave equation for the electric field: ∇ 2 E
(Â/c 2 )( 2 E/t 2 ) 0 in a medium with dielectric constant Â. Without loss of
generality we can assume that the interface corresponds to z 0 and the vacuum
is given by z 0 and the solid fills z 0. The wave equation applies to