Energy eigenkets and spin precession
Contents:
# Energy eigenkets
Okay, say we have an operator $\hat A$ s.t. $[\hat A, \hat H] = 0$, and $|a'\rangle$ is the base ket of $\hat A$ and its eigenket too. Then, the eigenkets of $\hat A$ are also eigenkets of $\hat H$ (simultaneous), and these eigenkets are called “energy eigenkets”:
$$ \hat H|a'\rangle = E_{a'}|a'\rangle $$
Recall that we can write any operator as
$$ \hat B = \sum_{a''}\sum_{a'}|a''\rangle\langle a''|\hat B|a'\rangle\langle a'| $$
We do this for $\exp\left(-\frac{i\hat Ht}{\hbar}\right)$ (taking $t_0 = 0$):
$$ \exp\left(-\frac{i\hat Ht}{\hbar}\right) = \sum_{a''}\sum_{a'}|a''\rangle\langle a''|\exp\left(-\frac{i\hat Ht}{\hbar}\right)|a'\rangle\langle a'| $$
Now, how is
$$ \exp\left(-\frac{i\hat Ht}{\hbar}\right)|a'\rangle = \exp\left(-\frac{iE_{a'}t}{\hbar}\right)|a'\rangle $$
??
This is cause we know $\hat H|a'\rangle = E_{a'}|a'\rangle$. Any Taylor-expandable function of $\hat H$ acting on $|a'\rangle$ just replaces $\hat H$ with $E_{a'}$:
$$ \exp\left(-\frac{i\hat Ht}{\hbar}\right)|a'\rangle = \sum_n\frac{1}{n!}\left(-\frac{it}{\hbar}\right)^n\hat H^n|a'\rangle = \sum_n\frac{1}{n!}\left(-\frac{it}{\hbar}\right)^nE_{a'}^n|a'\rangle $$
So the sandwiched part is $\exp\left(-\frac{iE_{a'}t}{\hbar}\right)\delta_{a''a'}$, and the double sum collapses:
$$ \boxed{\exp\left(-\frac{i\hat Ht}{\hbar}\right) = \sum_{a'}|a'\rangle\exp\left(-\frac{iE_{a'}t}{\hbar}\right)\langle a'|} $$
# How the expansion coefficients evolve
This is useful. For example, suppose
$$ |\alpha, t_0 = 0\rangle = \sum_{a'}|a'\rangle\langle a'|\alpha\rangle = \sum_{a'}c_{a'}|a'\rangle $$
then
$$ \begin{aligned} |\alpha, t_0 = 0; t\rangle &= \exp\left(-\frac{i\hat Ht}{\hbar}\right)|\alpha, t_0 = 0\rangle \\ &= \sum_{a'}|a'\rangle\langle a'|\alpha\rangle\exp\left(-\frac{iE_{a'}t}{\hbar}\right) \end{aligned} $$
$\implies c_{a'}(t = 0) \to c_{a'}(t)$, the expansion coefficients change as
$$ \boxed{c_{a'}(t) = c_{a'}(t = 0)\exp\left(-\frac{iE_{a'}t}{\hbar}\right)} $$
Only the phase changes, so $|c_{a'}(t)| = |c_{a'}(0)|$. That’s the “if $[\hat H, \hat A] = 0$ then they’re equal” from the time evolution notes.
A special case, when $|\alpha\rangle = |a'\rangle$:
$$ \boxed{|a', t_0 = 0; t\rangle = |a', t = 0\rangle\exp\left(-\frac{iE_{a'}t}{\hbar}\right)} $$
If a state is a simultaneous eigenket of $\hat A$ and $\hat H$, it remains so, at all times.
because $\exp\left(-\frac{iE_{a'}t}{\hbar}\right)$ is only doing phase modulation.
# Example: spin ½ in a magnetic field
The Hamiltonian of a spin $\frac{1}{2}$ system, with magnetic moment $\mu = \frac{e\hbar}{2mc}$, in an external field $\vec B$:
$$ \boxed{\hat H = -\left(\frac{e}{mc}\right)\vec S\cdot\vec B} $$
($e < 0$ for the electron). We let $\vec B$ be static and uniform in $\hat z$. Evaluating the dot:
$$ \boxed{\hat H = -\frac{eB}{mc}\hat S_z} $$
Clearly $[\hat H, \hat S_z] = 0$, since there’s just a constant multiplier. So they have simultaneous energy eigenkets ($|\pm\rangle$), with energy eigenvalues
$$ \boxed{E_\pm = \mp\frac{e\hbar B}{2mc}} $$
as $\hat S_z|\pm\rangle = \pm\frac{\hbar}{2}|\pm\rangle$. We define $\omega$ as
$$ \boxed{\omega = \frac{|e|B}{mc}} $$
and then $\hat H = \omega\hat S_z$, so $\hat H|\pm\rangle = \pm\frac{\hbar\omega}{2}|\pm\rangle$.
# Time development
Now if we consider the time development, it’s captured in
$$ \mathcal U(t, 0) = \exp\left(-\frac{i\omega\hat S_zt}{\hbar}\right) $$
Suppose at $t = 0$, the system is
$$ |\alpha\rangle = c_+|+\rangle + c_-|-\rangle $$
Apply the time evolution to it:
$$ \begin{aligned} |\alpha, t_0 = 0; t\rangle &= \exp\left(-\frac{i\omega\hat S_zt}{\hbar}\right)\left(c_+|+\rangle + c_-|-\rangle\right) \\ &= c_+\exp\left(-\frac{i\omega t}{2}\right)|+\rangle + c_-\exp\left(\frac{i\omega t}{2}\right)|-\rangle \end{aligned} $$
If $|\alpha\rangle$ at $t = 0$ was $|+\rangle$ ($c_+ = 1$, $c_- = 0$), then we see that
$$ \boxed{|\alpha, 0; t\rangle = \exp\left(-\frac{i\omega t}{2}\right)|+\rangle} $$
$\to$ it forever stays spin up.
If $|\alpha\rangle$ was in the $S_x+$ state,
$$ |\alpha, 0\rangle = \frac{1}{\sqrt 2}|+\rangle + \frac{1}{\sqrt 2}|-\rangle $$
then we can work out the probabilities…
$$ \begin{aligned} |\langle S_x;\pm|\alpha, t_0 = 0; t\rangle|^2 &= \left|\underbrace{\left[\frac{1}{\sqrt 2}\langle+| \pm \frac{1}{\sqrt 2}\langle-|\right]}_{\langle S_x;\pm|}\cdot\underbrace{\left[\frac{1}{\sqrt 2}e^{-i\omega t/2}|+\rangle + \frac{1}{\sqrt 2}e^{i\omega t/2}|-\rangle\right]}_{|\alpha, t_0 = 0; t\rangle}\right|^2 \\ &= \left|\frac{1}{2}e^{-i\omega t/2} \pm \frac{1}{2}e^{i\omega t/2}\right|^2 \end{aligned} $$
$$ \boxed{ |\langle S_x;+|\alpha, t_0 = 0; t\rangle|^2 = \cos^2\frac{\omega t}{2}, \qquad |\langle S_x;-|\alpha, t_0 = 0; t\rangle|^2 = \sin^2\frac{\omega t}{2} } $$
(as $\frac{1}{2}(e^{-i\theta} + e^{i\theta}) = \cos\theta$ and $\frac{1}{2}(e^{-i\theta} - e^{i\theta}) = -i\sin\theta$).
$\to$ This means the spin can flip directions after some time!!
# Spin precession
And we can write $\langle\hat S_x\rangle$ using
$$ \langle A\rangle = \sum_{a'}a'|\langle a'|\alpha\rangle|^2 $$
(measured value $\times$ the prob. of finding it):
$$ \langle\hat S_x\rangle = \frac{\hbar}{2}\cos^2\frac{\omega t}{2} + \left(-\frac{\hbar}{2}\right)\sin^2\frac{\omega t}{2} $$
$$ \boxed{\langle\hat S_x\rangle = \frac{\hbar}{2}\cos\omega t} $$
and
$$ \boxed{\langle\hat S_y\rangle = \frac{\hbar}{2}\sin\omega t, \qquad \langle\hat S_z\rangle = 0} $$
↳ we expect $\hat S_z$ to not change at all (it commutes with $\hat H$). It’ll be $\pm\frac{\hbar}{2}$, with the same probabilities, always.
So the spin’s expectation value rotates in the $xy$-plane with angular frequency $\omega$: spin precession.
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