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=−ℏ2γBσz^\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 E↓E_{\downarrow} and E↑E_{\uparrow} are as follows:

E↓=ℏ2γBE↑=−ℏ2γB\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⁡(−iE↓t/ℏ) ∣↓⟩ + bexp⁡(−iE↑t/ℏ) ∣↑⟩\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⁡(−iE↓t/ℏ) ∣↓⟩ + cos⁡(θ/2)exp⁡(−iE↑t/ℏ) ∣↑⟩\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:

⟨χ∣S^x∣χ⟩=ℏ2[aexp⁡(iE↓t/ℏ)bexp⁡(iE↑t/ℏ)]T⋅[0110]⋅[aexp⁡(−iE↓t/ℏ)bexp⁡(−iE↑t/ℏ)]=ℏ2[aexp⁡(iE↓t/ℏ)bexp⁡(iE↑t/ℏ)]T⋅[bexp⁡(−iE↑t/ℏ)aexp⁡(−iE↓t/ℏ)]=ℏ2(abexp⁡(i(E↓ ⁣− ⁣E↑)t/ℏ)+baexp⁡(i(E↑ ⁣− ⁣E↓)t/ℏ))=ℏ2cos⁡(θ/2)sin⁡(θ/2)(exp⁡(iγBt)+exp⁡(−iγBt))=ℏ2cos⁡(γBt)(cos⁡(θ/2)sin⁡(θ/2)+cos⁡(θ/2)sin⁡(θ/2))=ℏ2sin⁡(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:

⟨χ∣S^y∣χ⟩=−ℏ2sin⁡(θ)sin⁡(γBt)⟨χ∣S^z∣χ⟩=ℏ2cos⁡(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}

References

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