Measurements, probabilities and the S-G experiment
Contents:
# Measurements
“A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured”
P.A.M. Dirac
Before measurement:
$$ |\alpha\rangle = \sum_{a'}c_{a'}|a'\rangle = \sum_{a'}\left(\langle a'|\alpha\rangle\right)|a'\rangle $$
After measurement:
$$ |\alpha\rangle \xrightarrow{\text{measurement}} |a'\rangle $$
$\to$ some eigenstate, one of the many.
Measurement usually changes the state
$\to$ unless of course, the state is already in one of the eigenstates of the observable:
$$ |a'\rangle \xrightarrow{\text{measurement}} |a'\rangle $$
When a measurement causes $|\alpha\rangle$ to become $|a'\rangle$, it is said that $\hat A$ (Hermitian) is measured to be $a'$ (the eigenvalue).
# Probability
Which eigenket will $|\alpha\rangle$ collapse to?
$$ \boxed{\text{Prob. for } a' = |\langle a'|\alpha\rangle|^2} $$
↳ if $|\alpha\rangle$ is normalized.
To find $P(a')$ empirically, we need to perform measurements on a pure ensemble of $|\alpha\rangle$ states.
Note that if $P(a') = 1$ then $P(a'')$ or $P(a''') = 0$ with certainty. That is, if a state collapses to $|a'\rangle$ then it cannot collapse further to $|a''\rangle$ or $|a'''\rangle$, as $\langle a''|a'\rangle = 0$ $\forall a' \neq a''$.
We know $|\langle a'|\alpha\rangle|^2 \geq 0$ and $\sum_{a'}|\langle a'|\alpha\rangle|^2 = 1$ from earlier.
# Expectation value
$$ \langle\hat A\rangle = \langle\alpha|\hat A|\alpha\rangle $$
$\to$ the expectation value of an operator w.r.t. state $|\alpha\rangle$.
For example: $\langle\hat S_z\rangle \in \left[-\frac{\hbar}{2}, \frac{\hbar}{2}\right]$ ↳ expectation is a range, but $S_z = \pm\frac{\hbar}{2}$ ↳ eigenvalues are discrete.
# Filtration
Filtration $\to$ choose only one eigenket of $\hat A$, say $|a'\rangle$, and block the others.
Mathematically:
$$ \Lambda_{a'}|\alpha\rangle = |a'\rangle\langle a'|\alpha\rangle $$
This is like saying, “choose only $|a'\rangle$”.
# Sequential S-G experiments
Let’s talk about the results of the sequential S-G experiments.
┌──────────┐ Sz+
Sx+ ────────► │ SG │ ────────►
│ z-axis │ ────────►
└──────────┘ Sz-
$$ P(S_z+) = P(S_z-) = \frac{1}{2} $$
since it splits into equal intensities.
$$ P(|a'\rangle) = |\langle a'|\alpha\rangle|^2 $$
$\to$ Remember.
In our case $|\alpha\rangle = |S_x;+\rangle$ $\to$ the initial state. (We’re denoting the $S_z+$ and $S_z-$ states as $|+\rangle$ and $|-\rangle$ only.)
$$ P(|+\rangle) = P(|S_z;+\rangle) = \frac{1}{2} = P(|-\rangle) $$
$$ \boxed{|\langle+|S_x;+\rangle| = |\langle-|S_x;+\rangle| = \frac{1}{\sqrt{2}}} $$
So, what does $|\alpha\rangle$ or $|S_x;+\rangle$ look like before measurement?
$$ \boxed{|S_x;+\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}e^{i\delta_1}|-\rangle} $$
with $\delta_1$ as real. Why like this? $\delta_1 \to$ “phase” for the vector. Some angle. Helpful in differentiation.
What about $|S_x;-\rangle$ then? We know it must be orthogonal to $|S_x;+\rangle \implies \langle S_x;-|S_x;+\rangle = 0$
$$ \boxed{|S_x;-\rangle = \frac{1}{\sqrt{2}}|+\rangle - \frac{1}{\sqrt{2}}e^{i\delta_1}|-\rangle} $$
Why? Find
$$ \langle S_x;-| = \frac{1}{\sqrt{2}}\langle+| - \frac{1}{\sqrt{2}}e^{-i\delta_1}\langle-| $$
and
$$ \langle S_x;-|S_x;+\rangle = \frac{1}{2}\langle+|+\rangle - \frac{1}{2}\langle-|-\rangle = 0 $$
(cross terms die). Hence, they are indeed orthogonal.
We choose the coefficient of $|+\rangle$ to be positive by convention.
# Building Sx and Sy
Now, we know
$$ \boxed{\hat A = \sum_{a'}a'|a'\rangle\langle a'| = \sum_{a'}a'\Lambda_{a'}} $$
In our case:
$$ \hat S_x = \frac{\hbar}{2}\left(|S_x;+\rangle\langle S_x;+| - |S_x;-\rangle\langle S_x;-|\right) $$
cause now we know the ket $|S_x\rangle$ exactly,
$$ \boxed{\hat S_x = \frac{\hbar}{2}\left[e^{-i\delta_1}\left(|+\rangle\langle-|\right) + e^{i\delta_1}\left(|-\rangle\langle+|\right)\right]} $$
To tell if it’s Hermitian ($\hat S_x = \hat S_x^\dagger$), use
$$ \left(|a\rangle\langle b|\right)^\dagger = |b\rangle\langle a|, \qquad (cA)^\dagger = c^{\ast}A^\dagger, \qquad (A+B)^\dagger = A^\dagger + B^\dagger $$
We can compute
$$ \hat S_x^\dagger = \frac{\hbar}{2}\left(e^{i\delta_1}|-\rangle\langle+| + e^{-i\delta_1}|+\rangle\langle-|\right) = \hat S_x $$
$\therefore$
$\hat S_x$ is indeed Hermitian.
Similarly, we can find
$$ |S_y;\pm\rangle = \frac{1}{\sqrt{2}}|+\rangle \pm \frac{1}{\sqrt{2}}e^{i\delta_2}|-\rangle $$
and
$$ \hat S_y = \frac{\hbar}{2}\left[e^{-i\delta_2}\left(|+\rangle\langle-|\right) + e^{i\delta_2}\left(|-\rangle\langle+|\right)\right] $$
# Finding the phases
But what are $\delta_1$ and $\delta_2$? $\to$ Can we find them?
If we have
┌──────────┐ Sy+
Sx± ────────► │ SG │ ────────►
│ y-axis │ ────────►
└──────────┘ Sy-
$$ \implies |\langle S_y;\pm|S_x;\pm\rangle| = \frac{1}{\sqrt{2}} $$
Same as before. (This also hints at the invariance of the system under rotation.)
But,
$$ |S_y;+\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{e^{i\delta_2}}{\sqrt{2}}|-\rangle, \qquad |S_y;-\rangle = \frac{1}{\sqrt{2}}|+\rangle - \frac{e^{i\delta_2}}{\sqrt{2}}|-\rangle $$
and
$$ |S_x;+\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{e^{i\delta_1}}{\sqrt{2}}|-\rangle, \qquad \langle S_y;+| = \frac{1}{\sqrt{2}}\langle+| + \frac{e^{-i\delta_2}}{\sqrt{2}}\langle-| $$
$$ \implies |\langle S_y;+|S_x;+\rangle| = \left|\frac{1}{2} + \frac{1}{2}e^{i(\delta_1-\delta_2)}\right| = \frac{1}{\sqrt{2}} $$
$$ \implies \frac{1}{2}\left|1 + e^{i(\delta_1-\delta_2)}\right| = \frac{1}{\sqrt{2}} $$
only possible when
$$ \boxed{\delta_2 - \delta_1 = \pm\frac{\pi}{2}} $$
Since the coefficients of $\hat S_x$ and $\hat S_y$ are $e^{i\delta}$, their matrix elements are not all real!!
$\to$ Convenient to take the $S_x$ matrix to be real and set $\delta_1 = 0$.
$$ \implies |S_x;+\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle \quad\text{and}\quad |S_y;+\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{e^{i\pi/2}}{\sqrt{2}}|-\rangle $$
where $|+\rangle$ and $|-\rangle$ are shorthand for $|S_z;+\rangle$ and $|S_z;-\rangle$, which form our basis vectors.
Note that with $\delta_1 = 0$, $\delta_2 = \pm\frac{\pi}{2}$ is still ambiguous. Choose $\frac{\pi}{2}$, we’ll see why later.
$$ \boxed{ \begin{aligned} |S_x;\pm\rangle &= \frac{1}{\sqrt{2}}|+\rangle \pm \frac{1}{\sqrt{2}}|-\rangle \\ |S_y;\pm\rangle &= \frac{1}{\sqrt{2}}|+\rangle \pm \frac{i}{\sqrt{2}}|-\rangle \end{aligned} } $$
Which is the same guess we made from polarized light in the Stern-Gerlach notes (the $S_x$ one up to an overall sign, which doesn’t change the state).
# Spin flipping operators
And the two non-Hermitian operators of spin flipping:
$$ \boxed{S_\pm = S_x \pm iS_y} $$
$$ S_+ = \hbar|+\rangle\langle-|, \qquad S_- = \hbar|-\rangle\langle+| $$
Quick check: with $\delta_1 = 0$ and $\delta_2 = \frac{\pi}{2}$ (so $e^{\mp i\pi/2} = \mp i$),
$$ \hat S_x = \frac{\hbar}{2}\left(|+\rangle\langle-| + |-\rangle\langle+|\right), \qquad \hat S_y = \frac{\hbar}{2}\left(-i|+\rangle\langle-| + i|-\rangle\langle+|\right) $$
$$ \implies S_x + iS_y = \frac{\hbar}{2}\left(|+\rangle\langle-| + |-\rangle\langle+| + |+\rangle\langle-| - |-\rangle\langle+|\right) = \hbar|+\rangle\langle-| = S_+ $$
and the same way $S_x - iS_y = \hbar|-\rangle\langle+| = S_-$. These are the increment / decrement operators from the spin ½ notes: they raise (or lower) the spin component by $\hbar$. They’re not Hermitian since $S_+^\dagger = S_-$.
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