Categories: Physics, Quantum mechanics.

Larmor precession

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

\[\begin{aligned} \hat{H} = - \gamma B \hat{S}_z = - \frac{\hbar}{2} \gamma B \hat{\sigma_z} \end{aligned}\]

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

\[\begin{aligned} E_{\downarrow} = \frac{\hbar}{2} \gamma B \qquad E_{\uparrow} = - \frac{\hbar}{2} \gamma B \end{aligned}\]

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

\[\begin{aligned} \ket{\chi(t)} = a \exp\!(- i E_{\downarrow} t / \hbar) \: \ket{\downarrow} \:+\: b \exp\!(- i E_{\uparrow} t / \hbar) \: \ket{\uparrow} \end{aligned}\]

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

\[\begin{aligned} \ket{\chi(t)} = \sin\!(\theta / 2) \exp\!(- i E_{\downarrow} t / \hbar) \: \ket{\downarrow} \:+\: \cos\!(\theta / 2) \exp\!(- i E_{\uparrow} t / \hbar) \: \ket{\uparrow} \end{aligned}\]

Now, we find the expectation values of the spin operators \(\expval*{\hat{S}_x}\), \(\expval*{\hat{S}_y}\), and \(\expval*{\hat{S}_z}\). The first is:

\[\begin{aligned} \matrixel{\chi}{\hat{S}_x}{\chi} &= \frac{\hbar}{2} \begin{bmatrix} a \exp\!(i E_{\downarrow} t / \hbar) \\ b \exp\!(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} a \exp\!(- i E_{\downarrow} t / \hbar) \\ b \exp\!(- i E_{\uparrow} t / \hbar) \end{bmatrix} \\ &= \frac{\hbar}{2} \begin{bmatrix} a \exp\!(i E_{\downarrow} t / \hbar) \\ b \exp\!(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} \cdot \begin{bmatrix} b \exp\!(- i E_{\uparrow} t / \hbar) \\ a \exp\!(- i E_{\downarrow} t / \hbar) \end{bmatrix} \\ &= \frac{\hbar}{2} \Big( a b \exp\!(i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar) + b a \exp\!(i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar) \Big) \\ &= \frac{\hbar}{2} \cos\!(\theta/2) \sin\!(\theta/2) \Big( \exp\!(i \gamma B t) + \exp\!(- i \gamma B t) \Big) \\ &= \frac{\hbar}{2} \cos\!(\gamma B t) \Big( \cos\!(\theta/2) \sin\!(\theta/2) + \cos\!(\theta/2) \sin\!(\theta/2) \Big) \\ &= \frac{\hbar}{2} \sin\!(\theta) \cos\!(\gamma B t) \end{aligned}\]

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

\[\begin{aligned} \matrixel{\chi}{\hat{S}_y}{\chi} = - \frac{\hbar}{2} \sin\!(\theta) \sin\!(\gamma B t) \qquad \matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos\!(\theta) \end{aligned}\]

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

\[\begin{aligned} \boxed{ \omega = \gamma B } \end{aligned}\]

References

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

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