Categories: Physics, Quantum mechanics.

Larmor precession

Consider a stationary spin-1/2 particle, placed in a magnetic field with magnitude BB pointing in the zz-direction. In that case, its Hamiltonian H^\hat{H} is given by:

H^=γBS^z=2γBσz^\begin{aligned} \hat{H} = - \gamma B \hat{S}_z = - \frac{\hbar}{2} \gamma B \hat{\sigma_z} \end{aligned}

Where γ=q/m\gamma = - q / m is the gyromagnetic ratio, and σ^z\hat{\sigma}_z is the Pauli spin matrix for the zz-direction. Since H^\hat{H} is proportional to σ^z\hat{\sigma}_z, they share eigenstates \Ket{\downarrow} and \Ket{\uparrow}. The respective eigenenergies EE_{\downarrow} and EE_{\uparrow} are as follows:

E=2γBE=2γB\begin{aligned} E_{\downarrow} = \frac{\hbar}{2} \gamma B \qquad E_{\uparrow} = - \frac{\hbar}{2} \gamma B \end{aligned}

Because H^\hat{H} is time-independent, the general time-dependent solution χ(t)\Ket{\chi(t)} is of the following form, where aa and bb are constants, and the exponentials are “twiddle factors”:

χ(t)=aexp(iEt/)+bexp(iEt/)\begin{aligned} \Ket{\chi(t)} = a \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow} \:+\: b \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow} \end{aligned}

For our purposes, we can safely assume that aa and bb are real, and then say that there exists an angle θ\theta satisfying a=sin(θ/2)a = \sin(\theta / 2) and b=cos(θ/2)b = \cos(\theta / 2), such that:

χ(t)=sin(θ/2)exp(iEt/)+cos(θ/2)exp(iEt/)\begin{aligned} \Ket{\chi(t)} = \sin(\theta / 2) \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow} \:+\: \cos(\theta / 2) \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow} \end{aligned}

Now, we find the expectation values of the spin operators S^x\expval{\hat{S}_x}, S^y\expval{\hat{S}_y}, and S^z\expval{\hat{S}_z}. The first is:

χS^xχ=2[aexp(iEt/)bexp(iEt/)]T[0110][aexp(iEt/)bexp(iEt/)]=2[aexp(iEt/)bexp(iEt/)]T[bexp(iEt/)aexp(iEt/)]=2(abexp(i(E ⁣ ⁣E)t/)+baexp(i(E ⁣ ⁣E)t/))=2cos(θ/2)sin(θ/2)(exp(iγBt)+exp(iγBt))=2cos(γBt)(cos(θ/2)sin(θ/2)+cos(θ/2)sin(θ/2))=2sin(θ)cos(γBt)\begin{aligned} \matrixel{\chi}{\hat{S}_x}{\chi} &= \frac{\hbar}{2} \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} a \exp(- i E_{\downarrow} t / \hbar) \\ b \exp(- i E_{\uparrow} t / \hbar) \end{bmatrix} \\ &= \frac{\hbar}{2} \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} \cdot \begin{bmatrix} b \exp(- i E_{\uparrow} t / \hbar) \\ a \exp(- i E_{\downarrow} t / \hbar) \end{bmatrix} \\ &= \frac{\hbar}{2} \Big( a b \exp(i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar) + b a \exp(i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar) \Big) \\ &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( \exp(i \gamma B t) + \exp(- i \gamma B t) \Big) \\ &= \frac{\hbar}{2} \cos(\gamma B t) \Big( \cos(\theta/2) \sin(\theta/2) + \cos(\theta/2) \sin(\theta/2) \Big) \\ &= \frac{\hbar}{2} \sin(\theta) \cos(\gamma B t) \end{aligned}

The other two are calculated in the same way, with the following results:

χS^yχ=2sin(θ)sin(γBt)χS^zχ=2cos(θ)\begin{aligned} \matrixel{\chi}{\hat{S}_y}{\chi} = - \frac{\hbar}{2} \sin(\theta) \sin(\gamma B t) \qquad \matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos(\theta) \end{aligned}

The result is that the spin axis is off by θ\theta from the zz-direction, and is rotating (or precessing) around the zz-axis at the Larmor frequency ω\omega:

ω=γB\begin{aligned} \boxed{ \omega = \gamma B } \end{aligned}


  1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.