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. The conservation of energy condition must be satisfied
well enough to get a significant transition rate. Actually we will find later that for rapid transitions,
energy conservation does not have to be exact.
So we have proven that we should set the frequencyωof our EM wave according to the energy
difference between the two spin states. This allows us to cause transitions to the higher energy
state. In NMR, we observe the transitions back to the lower energy state. These emit EM radiation
at the same frequency and we can detect it after the stronger input pulse ends (or by more complex
methods). We don’t yet know why the higher energy state will spontaneously decay to the lower
energy state. To calculate this, we will have to quantize the field. But we already see that the energy
termse−iEt/ ̄hof standard wave mechanics will require energy conservation with photon energies of
E= ̄hω.
NMR is a powerful tool in chemical analysis because the molecular fieldadds to the external B field
so that the resonant frequency depends on the molecule as well asthe nucleus. We can learn about
molecular fields or just use NMR to see what molecules are present in asample.
In MRI, we typically concentrate on one nucleus like hydrogen. We can put a gradient inBzso that
only a thin slice of the material hasωtuned to the resonant frequency. Therefore we can excite
transitions to the higher energy state in only a slice of the sample. Ifwe vary (in the orthogonal
direction!) the B field during the decay of the excited state, we can get a two dimensional picture.
If we vary B as a function of time during the decay, we can get to 3D.While there are more complex
methods used in MRI, we now understand the basis of the technique. MRIs are a very safe way to
examine the inside of the body. All the field variation takes some time though. Ultimately, a very
powerful tool for scanning materials (a la Star Trek) is possible.
18.11Derivations and Computations
18.11.1Theℓ= 1 Angular Momentum Operators*
We will use states of definiteLz, theY 1 m.
〈ℓm′|Lz|ℓm〉=m ̄hδm′mLz= ̄h
1 0 0
0 0 0
0 0 − 1
〈ℓm′|L±|ℓm〉 =√
ℓ(ℓ+ 1)−m(m±1) ̄hδm′(m±1)L+ = ̄h
0
√
2 0
0 0
√
2
0 0 0
L− = ̄h