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Fill in a(t) and b(t) in (4.38) with the expressions given in (4.40):
(cid:32) (cid:33)(cid:32) (cid:33) (cid:32) (cid:33)
¯h ω0 ω1cosωt c(t)e−iω0t/2 = i¯h ddt(c(t)e−iω0t/2)
2 ω1cosωt −ω0 d(t)eiω0t/2 ddt(d(t)eiω0t/2)
Dividing ¯h from both sides and multiplying out the matrix on the left:
(cid:32) (cid:33) (cid:32) (cid:33)
1 ω0c(t)e−iω0t/2+ω1cos(ωt)d(t)eiω0t/2 = i ddt(c(t)e−iω0t/2)
2 ω1cos(ωt)c(t)e−iω0t/2−ω0d(t)eiω0t/2 ddt(d(t)eiω0t/2)
Taking the derivative of the right side using the product rule:
(cid:32) (cid:33) (cid:32) (cid:33)
1 ω0c(t)e−iω0t/2+ω1cos(ωt)d(t)eiω0t/2 = i ddcte−iω0t/2+c(t)(−i2ω0)(e−iω0t/2)
2 ω1cos(ωt)c(t)e−iω0t/2−ω0d(t)eiω0t/2 dddteiω0t/2+d(t)(iω20)(eiω0t/2)
(cid:32) (cid:33) (cid:32) (cid:33) (cid:32) (cid:33)
1 ω0c(t)e−iω0t/2+ω1cos(ωt)d(t)eiω0t/2 = i ddcte−iω0t/2 +i c(t)(−i2ω0)(e−iω0t/2)
2 ω1cos(ωt)c(t)e−iω0t/2−ω0d(t)eiω0t/2 dddteiω0t/2 d(t)(iω20)(eiω0t/2)
(cid:32) (cid:33) (cid:32) (cid:33) (cid:32) (cid:33)
1 ω0c(t)e−iω0t/2+ω1cos(ωt)d(t)eiω0t/2 = i ddcte−iω0t/2 +ω0 c(t)(e−iω0t/2)
2 ω1cos(ωt)c(t)e−iω0t/2−ω0d(t)eiω0t/2 dddteiω0t/2 2 −d(t)(eiω0t/2)
Subtracting the matrix on the far right on both sides:
(cid:32) (cid:33) (cid:32) (cid:33)
1 ω1cos(ωt)d(t)eiω0t/2 = i ddcte−iω0t/2
2 ω1cos(ωt)c(t)e−iω0t/2 dddteiω0t/2
(cid:32) (cid:33) (cid:32) (cid:33)
ω1cos(ωt) d(t)eiω0t/2 = i ddcte−iω0t/2
2 c(t)e−iω0t/2 ddeiω0t/2
dt
Now the right side of the equation can be rewritten as a matrix times a
column vector:
(cid:32) (cid:33) (cid:32) (cid:33)(cid:32) (cid:33)
ω1cos(ωt) d(t)eiω0t/2 = i e−iω0t/2 0 ddct
2 c(t)e−iω0t/2 0 eiω0t/2 dd
dt
Taking the inverse of the matrix on both sides:
1
(cid:32) (cid:33)(cid:32) (cid:33) (cid:32) (cid:33)
ω1cos(ωt) eiω0t/2 0 d(t)eiω0t/2 = i ddct
2 0 e−iω0t/2 c(t)e−iω0t/2 dd
dt
(cid:32) (cid:33) (cid:32) (cid:33)
ω1cos(ωt) d(t)eiω0t = i ddct
2 c(t)e−iω0t dd
dt
2