26 May 2026

Electron Spin Resonance

Electron Zeeman splitting, magnetic-dipole transitions, resonance, populations, hyperfine structure, and linewidth.

msc semester-ii molecular-spectra electron-spin-resonance zeeman-effect magnetic-resonance

Electron spin resonance (ESR), also called electron paramagnetic resonance, measures transitions between magnetic energy levels of an unpaired electron. The electron must be unpaired because a closed shell has zero resultant electronic magnetic moment.

Spin magnetic moment

For a spin-$1/2$ electron,

\[\hat{\mathbf S}=\frac{\hbar}{2}\boldsymbol{\sigma},\]

where $\boldsymbol{\sigma}$ denotes the three Pauli matrices. The magnetic moment associated with spin is

\[\boxed{ \hat{\boldsymbol{\mu}}_{\mathrm e} =-g_{\mathrm e}\mu_B\frac{\hat{\mathbf S}}{\hbar} },\]

with

\[\mu_B=\frac{e\hbar}{2m_{\mathrm e}}.\]

Here $e$ is the positive elementary-charge magnitude. The minus sign follows from the negative electron charge: the electron magnetic moment is antiparallel to its spin. For a free electron, $g_{\mathrm e}\simeq2.0023$; in an atom, molecule, or solid, spin–orbit coupling and the local electronic environment generally make the measured effective $g$ different from the free-electron value.

Apply a static magnetic field

\[\mathbf B_0=B_0\hat{\mathbf z}.\]

The magnetic interaction energy is

\[\hat H_0=-\hat{\boldsymbol{\mu}}_{\mathrm e}\cdot\mathbf B_0.\]

Substituting the magnetic moment,

\[\hat H_0 =g_{\mathrm e}\mu_BB_0\frac{\hat S_z}{\hbar}.\]

The spin eigenvalue equation is

\[\hat S_z\lvert m_s\rangle =m_s\hbar\lvert m_s\rangle, \qquad m_s=\pm\frac12.\]

Therefore

\[\hat H_0\lvert m_s\rangle =g_{\mathrm e}\mu_BB_0m_s\lvert m_s\rangle,\]

and the two energies are

\[E_{+1/2}=+\frac12g_{\mathrm e}\mu_BB_0, \qquad E_{-1/2}=-\frac12g_{\mathrm e}\mu_BB_0.\]

Their separation is

\[\boxed{ \Delta E =E_{+1/2}-E_{-1/2} =g_{\mathrm e}\mu_BB_0 }.\]

The $m_s=-1/2$ state is lower in energy for $B_0>0$ because its magnetic moment is parallel to the field.

Why the oscillating field must be transverse

Let a microwave magnetic field be applied along $x$:

\[\mathbf B_1(t)=B_1\cos\omega t\,\hat{\mathbf x}.\]

Its interaction with the electron is

\[\hat H_1(t) =-\hat{\boldsymbol{\mu}}_{\mathrm e}\cdot\mathbf B_1(t) =g_{\mathrm e}\mu_BB_1 \frac{\hat S_x}{\hbar}\cos\omega t.\]

In the $\hat S_z$ basis,

\[\hat S_x =\frac{\hbar}{2} \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}.\]

The diagonal matrix elements vanish:

\[\langle +\tfrac12\lvert\hat S_x\rvert+\tfrac12\rangle=0, \qquad \langle -\tfrac12\lvert\hat S_x\rvert-\tfrac12\rangle=0.\]

The off-diagonal matrix element is nonzero:

\[\langle +\tfrac12\lvert\hat S_x\rvert-\tfrac12\rangle =\frac{\hbar}{2}.\]

Thus the transverse field couples the two Zeeman states and gives the magnetic-dipole selection rule

\[\boxed{\Delta m_s=\pm1}.\]

A field parallel to $z$ would be proportional to $\hat S_z$, whose off-diagonal elements vanish, so it would shift the levels without driving this transition.

Resonance condition

Absorption is largest when the microwave photon energy equals the Zeeman separation:

\[\hbar\omega=\Delta E.\]

Using $\omega=2\pi\nu$ and $\hbar\omega=h\nu$,

\[h\nu=g_{\mathrm e}\mu_BB_0.\]

Hence

\[\boxed{ \nu=\frac{g_{\mathrm e}\mu_B}{h}B_0 }.\]

An ESR spectrometer commonly holds $\nu$ fixed and sweeps $B_0$. The resonance field is then

\[\boxed{ B_{\mathrm{res}}=\frac{h\nu}{g_{\mathrm e}\mu_B} }.\]

Consequently, a measured resonance field determines

\[g_{\mathrm e}=\frac{h\nu}{\mu_BB_{\mathrm{res}}}.\]

The electron levels separate linearly with \(B_0\); a transverse microwave drives the transition when \(h\nu=g_{\mathrm e}\mu_BB_0\).

Thermal population and net absorption

Write the two energies as

\[E_{\mathrm{upper}}=+\frac{\Delta E}{2}, \qquad E_{\mathrm{lower}}=-\frac{\Delta E}{2}.\]

The Boltzmann populations obey

\[\frac{N_{\mathrm{upper}}}{N_{\mathrm{lower}}} =\exp\left[ -\frac{E_{\mathrm{upper}}-E_{\mathrm{lower}}}{k_BT} \right] =e^{-\Delta E/(k_BT)}.\]

If $N=N_{\mathrm{lower}}+N_{\mathrm{upper}}$, direct substitution of the two Boltzmann weights gives

\[\frac{N_{\mathrm{lower}}-N_{\mathrm{upper}}}{N} = \frac{e^{\Delta E/(2k_BT)}-e^{-\Delta E/(2k_BT)}} {e^{\Delta E/(2k_BT)}+e^{-\Delta E/(2k_BT)}} =\tanh\left(\frac{\Delta E}{2k_BT}\right).\]

For $\Delta E\ll k_BT$, $\tanh x\simeq x$, so

\[N_{\mathrm{lower}}-N_{\mathrm{upper}} \simeq N\frac{g_{\mathrm e}\mu_BB_0}{2k_BT}.\]

There are therefore slightly more absorptive upward transitions than stimulated downward transitions. This small population excess produces the observed net ESR absorption.

Hyperfine splitting

If the electron spin interacts with a nucleus of spin $I$, an isotropic hyperfine Hamiltonian can be written

\[\hat H_{\mathrm{hf}} =\frac{A}{\hbar^2}\hat{\mathbf I}\cdot\hat{\mathbf S},\]

where $A$ has dimensions of energy. In a strong field, the Zeeman direction defines the quantization axis and the nonsecular terms are neglected to first order:

\[\hat H_{\mathrm{hf}} \simeq\frac{A}{\hbar^2}\hat I_z\hat S_z.\]

Since

\[\hat I_z\lvert m_I,m_s\rangle =m_I\hbar\lvert m_I,m_s\rangle, \qquad \hat S_z\lvert m_I,m_s\rangle =m_s\hbar\lvert m_I,m_s\rangle,\]

the approximate energy is

\[E(m_s,m_I) =g_{\mathrm e}\mu_BB_0m_s+Am_Im_s.\]

For an allowed ESR transition, $m_s$ changes from $-1/2$ to $+1/2$ while $m_I$ is unchanged. Therefore

\[\Delta E(m_I) =g_{\mathrm e}\mu_BB_0+Am_I.\]

The resonance field at fixed frequency becomes

\[\boxed{ B_{\mathrm{res}}(m_I) =\frac{h\nu-Am_I}{g_{\mathrm e}\mu_B} }.\]

Because $m_I=-I,-I+1,\ldots,I$, one equivalent nucleus produces $2I+1$ hyperfine lines.

Linewidth and spin relaxation

After excitation, transverse spin coherence decays approximately as

\[M_\perp(t)=M_\perp(0)e^{-t/T_2}e^{-i\omega_0t}.\]

Its frequency response is obtained from

\[\int_0^\infty e^{-t/T_2}e^{i(\omega-\omega_0)t}\,dt = \left[ \frac{-e^{-[1/T_2-i(\omega-\omega_0)]t}} {1/T_2-i(\omega-\omega_0)} \right]_0^\infty,\]

which gives

\[\frac{1}{1/T_2-i(\omega-\omega_0)}.\]

The absorbed intensity is therefore Lorentzian:

\[\mathcal I(\omega)\propto \frac{1/T_2} {(\omega-\omega_0)^2+(1/T_2)^2}.\]

At half maximum,

\[\lvert\omega-\omega_0\rvert=\frac1{T_2}.\]

Since the electron resonance angular frequency satisfies

\[\omega_0=\frac{g_{\mathrm e}\mu_B}{\hbar}B_{\mathrm{res}},\]

the field half-width at half maximum is

\[\boxed{ \Delta B_{\mathrm{HWHM}} =\frac{\hbar}{g_{\mathrm e}\mu_BT_2} }.\]

Thus a shorter transverse relaxation time $T_2$ produces a broader resonance line. Longitudinal relaxation, characterized by $T_1$, restores the equilibrium population difference required for continued absorption.

The spin-matrix, resonance, hyperfine, and linewidth identities are checked in the Maxima worksheet.