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Diffstat (limited to 'source/know/concept/larmor-precession')
| -rw-r--r-- | source/know/concept/larmor-precession/index.md | 26 |
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{ |