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1 files changed, 14 insertions, 12 deletions

diff --git a/source/know/concept/larmor-precession/index.md b/source/know/concept/larmor-precession/index.md
index 601dae7..e29432b 100644
--- a/source/know/concept/larmor-precession/index.md
+++ b/source/know/concept/larmor-precession/index.md
@@ -11,28 +11,29 @@ layout: "concept"
 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:
+In that case, the 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,
+Where $$\gamma = q g / (2 m)$$ is the so-called *gyromagnetic ratio*
+for a particle with charge $$q$$, mass $$m$$,
+and a system-dependent *$$g$$-factor* (for electrons $$g \approx 2$$),
 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:
+Because $$\hat{H}$$ is proportional to $$\hat{\sigma}_z$$,
+they share eigenstates $$\Ket{\downarrow}$$ and $$\Ket{\uparrow}$$,
+so the respective eigenenergies $$E_{\downarrow}$$ and $$E_{\uparrow}$$ are as follows:
 
 $$\begin{aligned}
     E_{\downarrow} = \frac{\hbar}{2} \gamma B
-    \qquad
+    \qquad \qquad
     E_{\uparrow} = - \frac{\hbar}{2} \gamma B
 \end{aligned}$$
 
 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":
+where $$a$$ and $$b$$ are constants:
 
 $$\begin{aligned}
     \Ket{\chi(t)}
@@ -72,7 +73,7 @@ $$\begin{aligned}
     \\
     &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( e^{i \gamma B t} + e^{- i \gamma B t} \Big)
     \\
-    &= \frac{\hbar}{2} \cos(\gamma B t) \cdot 2 \cos(\theta/2) \sin(\theta/2)
+    &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \cdot 2 \cos(\gamma B t)
     \\
     &= \frac{\hbar}{2} \sin(\theta) \cos(\gamma B t)
 \end{aligned}$$
@@ -82,12 +83,13 @@ with the following results:
 
 $$\begin{aligned}
     \matrixel{\chi}{\hat{S}_y}{\chi} = - \frac{\hbar}{2} \sin(\theta) \sin(\gamma B t)
-    \qquad
+    \qquad \qquad
     \matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos(\theta)
 \end{aligned}$$
 
-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** $$\omega$$:
+The result is that, if the spin axis is off by $$\theta$$ from the $$z$$-direction,
+then it rotates (or **precesses**) around the $$z$$-axis
+at the **Larmor frequency** $$\omega$$:
 
 $$\begin{aligned}
     \boxed{