A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = -3/2, -1/2, 1/2, 3/2. In a magnetic field B, the energies of these four states are E = m$\mu$B, where $\mu$ the constant is 1.03 x 10^-7 eV/T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K. Calculate the probability of a lithium nucleus being in each of its four spin states under these conditions. Then show that, if the field is suddenly reversed, the probabilities of the four states obey the Boltzmann distribution for T =-300 K.

A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = -3/2, -1/2, 1/2, 3/2. In a magnetic field B, the energies of these four states are E = m$\mu$B, where $\mu$ the constant is 1.03 x 10^-7 eV/T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K.

The partition function is: because none of the levels are degenerate.

$Z=\sum _s{e}^{-E\left(s\right)/kT}$

Where,

$E\left(s\right)=m\mu Bm=-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$

So,

$Z=e^{3\mu B/2kT}+e^{\mu B/2kT}+e^{-\mu B/2kT}+e^{-3\mu B/2kT}$

Let $x=\mu B/kT$

$Z=e^{3x/2}+e^{x/2}+e^{-x/2}+e^{-3x/2}\left(1\right)$

Therefore the value of x is:

$x=\frac{\left(1.03\times {10}^{-7}\mathrm{eV}/\mathrm{T}\right)(0.63\mathrm{T})}{\left(8.617\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)(300\mathrm{K})}=2.51\times {10}^{-6}$

Substitute x into (1):

$$\begin{gathered} Z=e^{3\left(2.51\times {10}^{-6}\right)/2}+e^{\left(2.51\times {10}^{-6}\right)/2}+e^{-\left(2.51\times {10}^{-6}\right)/2}+e^{-3\left(2.51\times {10}^{-6}\right)/2} \\ Z=4 \end{gathered}$$

The probability of first state is:

$P=\frac{1}{Z}{e}^{3x/2}$

Substitute x and z:

$$\begin{gathered} P_1=\left(\frac{1}{4}\right)e^{3\left(2.51\times {10}^{-6}\right)/2} \\ {P}_1=0.2500009413 \end{gathered}$$

The probability of second state is:

$P=\frac{1}{Z}{e}^{x/2}$

Substitute x and z:

$$\begin{gathered} P_2=\left(\frac{1}{4}\right)e^{\left(2.51\times {10}^{-6}\right)/2} \\ {P}_2=0.2500003138 \end{gathered}$$

The probability of third state is:

$P=\frac{1}{Z}{e}^{-x/2}$

Substitute x and z:

$P_3=0.2499996862$

The probability of fourth state is:

$P=\frac{1}{Z}{e}^{-3x/2}$

Substitute x and z:

$$\begin{gathered} P_4=\left(\frac{1}{4}\right)e^{-3\left(2.51\times {10}^{-6}\right)/2} \\ {P}_4=0.2499990587 \end{gathered}$$