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Diffstat (limited to 'source/know/concept/larmor-precession/index.md')
| -rw-r--r-- | source/know/concept/larmor-precession/index.md | 24 |
1 files changed, 12 insertions, 12 deletions
diff --git a/source/know/concept/larmor-precession/index.md b/source/know/concept/larmor-precession/index.md index 774af7b..601dae7 100644 --- a/source/know/concept/larmor-precession/index.md +++ b/source/know/concept/larmor-precession/index.md @@ -36,8 +36,8 @@ and the exponentials are "twiddle factors": $$\begin{aligned} \Ket{\chi(t)} - = a \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow} - \:+\: b \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow} + = a e^{- i E_{\downarrow} t / \hbar} \Ket{\downarrow} + \:+\: b e^{- i E_{\uparrow} t / \hbar} \Ket{\uparrow} \end{aligned}$$ For our purposes, we can safely assume that $$a$$ and $$b$$ are real, @@ -45,8 +45,8 @@ and then say that there exists an angle $$\theta$$ satisfying $$a = \sin(\theta / 2)$$ and $$b = \cos(\theta / 2)$$, such that: $$\begin{aligned} - \Ket{\chi(t)} = \sin(\theta / 2) \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow} - \:+\: \cos(\theta / 2) \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow} + \Ket{\chi(t)} = \sin(\theta / 2) \: e^{- i E_{\downarrow} t / \hbar} \Ket{\downarrow} + \:+\: \cos(\theta / 2) \: e^{- i E_{\uparrow} t / \hbar} \Ket{\uparrow} \end{aligned}$$ Now, we find the expectation values of the spin operators @@ -56,23 +56,23 @@ The first is: $$\begin{aligned} \matrixel{\chi}{\hat{S}_x}{\chi} &= \frac{\hbar}{2} - \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} + \begin{bmatrix} a e^{i E_{\downarrow} t / \hbar} \\ b e^{i E_{\uparrow} t / \hbar} \end{bmatrix}^{\mathrm{T}} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \cdot - \begin{bmatrix} a \exp(- i E_{\downarrow} t / \hbar) \\ b \exp(- i E_{\uparrow} t / \hbar) \end{bmatrix} + \begin{bmatrix} a e^{- i E_{\downarrow} t / \hbar} \\ b e^{- i E_{\uparrow} t / \hbar} \end{bmatrix} \\ &= \frac{\hbar}{2} - \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}} + \begin{bmatrix} a e^{i E_{\downarrow} t / \hbar} \\ b e^{i E_{\uparrow} t / \hbar} \end{bmatrix}^{\mathrm{T}} \cdot - \begin{bmatrix} b \exp(- i E_{\uparrow} t / \hbar) \\ a \exp(- i E_{\downarrow} t / \hbar) \end{bmatrix} + \begin{bmatrix} b e^{- i E_{\uparrow} t / \hbar} \\ a e^{- i E_{\downarrow} t / \hbar} \end{bmatrix} \\ - &= \frac{\hbar}{2} \Big( a b \exp(i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar) - + b a \exp(i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar) \Big) + &= \frac{\hbar}{2} \Big( a b e^{i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar} + + b a e^{i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar} \Big) \\ - &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( \exp(i \gamma B t) + \exp(- i \gamma B t) \Big) + &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( e^{i \gamma B t} + e^{- i \gamma B t} \Big) \\ - &= \frac{\hbar}{2} \cos(\gamma B t) \Big( \cos(\theta/2) \sin(\theta/2) + \cos(\theta/2) \sin(\theta/2) \Big) + &= \frac{\hbar}{2} \cos(\gamma B t) \cdot 2 \cos(\theta/2) \sin(\theta/2) \\ &= \frac{\hbar}{2} \sin(\theta) \cos(\gamma B t) \end{aligned}$$ |