Gyromagnetic ratio of the electron

The gyromagnetic ratio of the electron $\small \gamma_e$ can be evaluated by the following electron spin resonance (ESR) experiment:

In the above diagram, a sample of hydrogen atoms is placed in a uniform magnetic field (1 T), which gives rise to the Zeeman effect, in which degenerate energy states are split. The sample is then irradiated with microwaves at different frequencies. When the electromagnetic source is turned off, no absorption is detected. However, as the frequency of the electromagnetic radiation is varied in the microwave range, absorption is observed at $\small v=2.8\times 10^{10}Hz$ , indicating a transition between two energy states of the atom (c.f.: in a proton NMR experiment, transition frequencies are typically in the $\small 10^{6}Hz$ range, while usual electronic transitions in hydrogen occur on the order of $10^{14}Hz$ ).

This behaviour can be understood by noting that the electron in a hydrogen atom occupies the 1s orbital ( $\small l=0$ ), so the atom has no orbital angular momentum contribution. The observed transition in the external magnetic field therefore arises from the electron’s spin angular momentum $\boldsymbol{\mathit{S}}$ , while the nuclear contribution is comparatively weak.

From eq67, the classical relation between the energy of a charged particle $\small U$ in a magnetic field $\small \boldsymbol{\mathit{B}}$ and the particle’s angular momentum $\small \boldsymbol{\mathit{L}}$ is $\small U=-\gamma\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{L}}$ , where $\gamma=\frac{q}{2m}$ is the classical gyromagnetic ratio. It follows that:

$\small U=-\gamma_e\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{S}}=-\gamma_e[B_x(x,y,z)\boldsymbol{\mathit{i}}+B_y(x,y,z)\boldsymbol{\mathit{j}}+B_z(x,y,z)\boldsymbol{\mathit{k}}]\cdot(S_x\boldsymbol{\mathit{i}}+S_y\boldsymbol{\mathit{j}}+S_z\boldsymbol{\mathit{k}})$

Analysing the effect of the uniform magnetic field (with magnitude $\small B_0$ ) on the energy states of hydrogen in the $\small z$ -direction, the above equation becomes:

$\small U=-\gamma_eB_0S_z\; \; \; \; \; \; \; \; 163$

From eq75, each of the eigenvalues of $\small \hat{L}^{2}$ is the square of the magnitude of the orbital angular momentum of an electron, which makes each of the eigenvalues of $\small \hat{L}_z$ the $\small z$ -component of the magnitude of the orbital angular momentum of an electron. Since $\small \hat{S}_z$ is the analogue of $\small \hat{L}_z$ , we postulate that the eigenvalues of $\small \hat{S}_z$ is the $\small z$ -component of the magnitude of the spin angular momentum of an electron, which from eq168, is $\small m_s\hbar=\pm\frac{\hbar}{2}$ . Therefore, we have

$h\nu=U_f-U_i=-\gamma_eB_0\left ( +\frac{\hbar}{2} \right )+\gamma_eB_0\left ( -\frac{\hbar}{2} \right )=-\gamma_eB_0\hbar$

$\gamma_e=-\frac{2\pi \nu}{B_0}=-1.759292\times 10^{11}Ckg^{-1}$

Question

Why is the spin magnetic momentum quantum number $\small m_s\hbar=-\frac{\hbar}{2}$ associated with the lower energy state $\small U_i$ ?

Answer

The experiment involves a transition from a lower initial energy state $U_i$ to a higher final state $U_f$ . Since $\gamma_e<0$ and $h\nu>0$ , the quantum number $m_s=-\frac{1}{2}$ must correspond to $U_i$ .

If we replace $\small q$ with $\small -e$ and $\small m$ with $\small m_e$ in $\small \gamma$ , we have $\small \gamma=-8.7941\times 10^{10}Ckg^{-1}$ . So, $\small \gamma_e$ is about twice the value of $\small \gamma$ . Due to this difference, the classical notion of the electron spinning on its own axis (which is equivalent to a current loop) has no physical reality. Therefore, the gyromagnetic ratio of the electron is formerly defined as:

$\small \gamma_e=-g_e\frac{e}{2m_e}\; \; \; \; \; \; \; \; 164$

where $\small g_e$ is the g-value of the electron, which is measured in a recent experiment to be 2.00231930436256 with an uncertainty of 1.7×10^-13.

This experiment also provides evidence that the spin angular momentum quantum number of the electron is $\small \frac{1}{2}$ . Since the spectrum consists of only two spin states connected by a single allowed magnetic dipole transition, and a particle with spin quantum number $s$ has $2s+1$ possible spin projections $m_s=-s,-s+1,\cdots,s$ , the observation of two spin states implies $2s+1=2$ , giving $s=\frac{1}{2}$ .

Question

Show that the ESR selection rule is $\Delta m_s=\pm 1$ .

Answer

According to time-dependent perturbation theory, the perturbation Hamiltonian for a hydrogen atom in the experiment is:

$H'(t)=-\gamma_e\boldsymbol{\mathit{B}}(t)\cdot\boldsymbol{\mathit{S}}$

where $\boldsymbol{\mathit{B}}(t)$ is the oscillating magnetic field of the incident radiation.

If the incident radiation lies in the $xy$ -plane, then $H'(t)\propto \hat{S}_x$ or $H'(t)\propto \hat{S}_y$ . Since the ladder operators of spin angular momentum are defined analogously to the corresponding orbital angular momentum ladder operators, we have $\hat{S}_+=\hat{S}_x+i\hat{S}_y$ and $\hat{S}_-=\hat{S}_x-i\hat{S}_y$ , with $\hat{S}_x=\frac{1}{2}(\hat{S}_++\hat{S}_-)$ and $\hat{S}_y=\frac{1}{2i}(\hat{S}_+-\hat{S}_-)$ . Therefore, the perturbation contains only the raising and lowering operators $\hat{S}_+$ and $\hat{S}_-$ , whose matrix elements are given by: