130_notes.dvi

∆Ef=1−∆Ef=0=

4

3

(

1

137

) 4 (

. 51

938

)

(. 51 × 106 )(5.56) = 5. 84 × 10 −^6 eV

Recall that at room temperature,kBtis about 401 eV, so the states have about equal population at
room temperature. Even at a few degrees Kelvin, the upper stateis populated so that transitions
are possible. The wavelength is well known.

λ= 2π

̄hc

E

= 2π

1973

5. 84 × 10 −^6

̊A= 2× 109 ̊A= 21.2 cm

This transition is seen in interstellar gas. Thef= 1 state is excited by collisions. Electromagnetic
transitions are slow because of the selection rule ∆ℓ=±1 we will learn later, and because of the
small energy difference. Thef= 1 state does emit a photon to de-excite and those photons have a
long mean free path in the gas.

24.3.2 Hyperfine Splitting in a Weak B Field

Since the field is weak we work in the states|fmf〉in which the hyperfine perturbation is diagonal
and compute the matrix elements forWz=μBBσz. But to do the computation, we will have to
write those states in terms of|msmi〉which we will abbreviate like|+−〉, which means the electron’s
spin is up and the proton’s spin is down.

σz| 11 〉=σz|++〉=| 11 〉

σz| 1 − 1 〉=σz|−−〉=−| 1 − 1 〉

σz| 10 〉=σz√^12 (|+−〉+|−+〉) =√^12 (|+−〉−|−+〉) =| 00 〉

σz| 00 〉=σz√^12 (|+−〉−|−+〉) =√^12 (|+−〉+|−+〉) =| 10 〉

Now since the three (f= 1) states are degenerate, we have to make sure all the matrix elements
between those states are zero, otherwise we should bite the bulletand do the full problem as in the
intermediate field case. Thef= 1 matrix is diagonal, as we could have guessed.

μBB





1 0 0

0 0 0

0 0 − 1





The only nonzero connection between states is betweenf= 1 andf= 0 and we are assuming the
hyperfine splitting between these states is large compared to the matrix element.

So the full answer is

Ez(1)=μBBmf

which is correct for bothfstates.