Rx(θ) =
∑∞
n=0, 2 , 4 ...(iθ 2 )n
n!∑∞
n=1, 3 , 5 ...(iθ 2 )n
n!
∑∞
n=1, 3 , 5 ...(iθ 2 )n
n!∑∞
n=0, 2 , 4 ...(iθ 2 )n
n!
=
(
cosθ 2 isinθ 2
isinθ 2 cosθ 2)
Note that all of these rotation matrices become the identity matrixfor rotations through 720 degrees
and are minus the identity for rotations through 360 degrees.
18.11.8NMR Transition Rate in a Oscillating B Field
We have the time dependent Schr ̈odinger equation for a proton in astatic field in the z direction
plus an oscillating field in the x direction.
i ̄h
dχ
dt= Hχi ̄h(
a ̇
b ̇)
= −
gp
2
μN(
Bz Bxcosωt
Bxcosωt −Bz)(
a
b)
(
a ̇
b ̇)
= i
gpμN
2 ̄h(
Bz Bxcosωt
Bxcosωt −Bz)(
a
b)
=i(
ω 0 ω 1 cosωt
ω 1 cosωt −ω 0)(
a
b)
So far all we have done is plugged things into the Schr ̈odinger equation. Now we have to solve this
system of two equations. This could be hard but we will do it only neart= 0, when the EM wave
starts. Assume that att= 0,a= 1 andb= 0, that is, the nucleus is in the lower energy state.
Then we have
a ̇ = iω 0 a
a = 1eiω^0 t
b ̇ = iω 1 cosωta−iω 0 b=iω 1 cosωteiω^0 t−iω 0 bb ̇ = iω^1
2(ei(ω+ω^0 )t+e−i(ω−ω^0 )t)−iω 0 bNow comes the one tricky part of the calculation. The diagonal terms in the Hamiltonian cause a
very rapid time dependence to the amplitudes. To get b to grow, we need to keep addingb ̇in phase
with b. To see that clearly, let’s compute the time derivative ofbeiω^0 t.
d
dt(beiω^0 t) =iω 1
2(ei(ω+2ω^0 )t+e−i(ω−^2 ω^0 )t)−iω 0 beiω^0 t+iω 0 beiω^0 t=iω 1
2(ei(ω+2ω^0 )t+e−i(ω−^2 ω^0 )t)Terms that oscillate rapidly will average to zero. To get a net changeinbeiω^0 t, we need to have
ω≈ 2 ω 0. Then the first term is important and we can neglect the second which oscillates with a
frequency of the order of 10^11. Note that this is exactly the condition that requires the energy of
the photons in the EM fieldE= ̄hωto be equal to the energy difference between the two spin states
∆E= 2 ̄hω 0.