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[OL dx cos([k k'][x + L]) = sin(~k:k~'lL).
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17Bremmer's article is in English.
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Then notice from (9.12) that t 2 = r2 - r / r*, so that Ir - t exp( i2kL) .c( k Now use (9.16) to show that (9.9) yields (9.17). 9.3
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In section 9.1 (page 261) we have used Poisson's sum formula to show that
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derived in 2 by considering the Fourier expansion 18 of cos 7rX, sum the series in (9.20) and show that (9.20) is indeed equal to (9.7). Hint: Notice that as Irl < 1, (9.7) can be summed, its closed expression being given by.c(k) = t/ [1 + r exp (i2Lk)].
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Writing r as Irl exp[i arg(r)] in (9.7), show that
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- 7r]l}.
1=-00
Do the first sum and then use (9.12) to show that
1.c(k)1 2 =
{Irl ei [2kL+arg (r>-1r1}
{Irl e- i [2kL+ar g (r>-1r1} .
Now use Poisson's sum formula again in each of the two series on the right-hand side of the preceding equation and deduce (9.22). 9.S From (9.8) and (9.28) show that
[an,a~,]=
1 7r Lv'knkn, ,
dkksin([k-kn]L)sin([k-kn,]L)I.c(k)12 k - kn k - k n,
[an,a ,] = n
1 7rLv'kn k n,
dkk sin([k-kn ]L)sin([k+kn,]L)I.c(k)1 2 k - kn k + k n,
Solve the integrals Cnn' and Snn' by contour integration and show that the solutions are C nn' = (1/2) Re Jnn'- and Snn' = (1/2) Re Jnn'+' where
Jnn'- = 27rLI.c(kn)12kn8nn'
18See also [433].
PROBLEMS
where k n == n1r/ L, '"Y == -In Irl/2L, kn == k n - A, and A == [arg(r) - 1rl!2L. To sum the series Sand S', use the series summation result (see Chap. XII, Sec. 125, p. 370 of [77])
But be careful that it is valid only on the window 0 < x < 2L, as the series above is discontinuous at the boundaries of this window. To avoid the discontinuities and remain within the window, you can adopt a sort of regularization procedure writing Sas
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Remembering that sin([kn - knl]L)/(kn - knl)
8 = 2knlL 28nn l { 1 + i cotan ([k nl
L8nn l, show that
+ 6. - iT] L) }
and that
Conclude then that [an, a~/] that [an, anI] = o.
= 8nn l. Proceed in an analogous way with 8' to conclude
9.6 Using the canonical commutation relations for the radiation field inside the cavity,19
[Bin (x) , Ein(X')] = -i2h :x {8(x - x') - 8(x [Bin(x),Bin(x')] = [Ein(X),Ein(x')] =0,
show that [an,a~/] = 8nn , and [an, an'] = O.
+ x' + 2L)},
9.7 Using the canonical commutation relations for the radiation field outside the cavity,
A,] A [Bout(X), Eout(x)
= -~2h ax 8(x - x),
[Bout(x),Bout(x')] = [Eout(x),Eout(x')] = 0,
show that [b(k), b(k')t] = 8(k - k') and [b(k) , b(k')] = O.
9.8 Another way to test (9.39) is to substitute it in (9.2) and see if the resulting intracavity electric field agrees with (9.24). Substituting (9.39) in (9.2), we find that Ein(X) = Ein,D(X) + Ein,c(X), where
Ein,D(X) == -i2
2: an
Dn(x)
dk vac(k) sin ([x + L] k) { (k)eikLanl (k) - *(k)e- ikL a n2(k)} +H.c . .... '
&n,c(X) ==
-i21OO dk' b(k')
19The extra delta function is there because of the perfect mirror; see [448,4951.
PROBLEMS
" I(k',k)
+ H.C.
1. Using (9.40) and (9.41), show that Dn(x) is given by
Dn(x) =
J7r~~n
2 dk kl.c(k)1 sin ([x
+ LJ k)
(t=-
nJ k: L)
and that the integral above can be rewritten as
(_l)n 2
RejOO dk k 1.c(k)12P_l_ {e ikX _ ei(X+2L)k}.
k - kn
I,,(x)
Now noticing that inside the cavity x varies from - L to 0 only, do the integral In (x) by contour integration and verify that it is given by20
In(x) = -i27reik"xknl.c(kn)12
m=-oo
k m - h . ei(km-i"()x_ km+h. ei (X+2L)(km+i"()}. km - kn - ~"Y km - kn + ~"Y
S,,(x)
Show that Sn(x) can be rewritten as Sn(x) = Sln(X)
+ k nS 2n (x), where
Sln(X) =
m=-(X)
{ei(km-i"()X - e i(km+i-r)(x+2L)} ,
Now, using
m=-(X)
eikmu
m'=-(X)
8(<< - 2Lm'),
and noticing that - L
< x < 0, show that
Sln(X) = 2L {l- e- 2L -r} 8(x).
To sum S2n (x), use the Fourier series for cos([x + 7rJa) and its derivative with respect to x. Verify that
20The notation here is the same as in Problem 9.5.
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where 0 < u < 2L. Now using thafl IC(kn )12 = -Re{ i cotan( -i-yL)} [exp( -y) + exp( -L')')l![exp(L')') - exp( -L')')], finally show that
dk klC(kW sin ([x + L] k) sin (~k ~ k: ] L)
(-l)n1[ {I - e- 2 L-'Y} 8(x)
+ rrkn sin ([x + L] k n ) .