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+---
+title: "Larmor precession"
+firstLetter: "L"
+publishDate: 2021-07-02
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-07-02T15:48:41+02:00
+draft: false
+markup: pandoc
+---
+
+# Larmor precession
+
+Consider a stationary spin-1/2 particle,
+placed in a [magnetic field](/know/concept/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.