91Ó°ÊÓ

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$\mathrm{μ}$B, where $\mathrm{μ}$ 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.

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

Where,

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

So,

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

Let $x=\mathrm{μ}B/kT$

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

Therefore the value of x is:

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

Substitute x into (1):

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

The probability of first state is:

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

Substitute x and z:

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

The probability of second state is:

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

Substitute x and z:

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

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:

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