The Commutator
We study the precise reason why and when the order of measurements matter in Quantum Mechanics. Along the way we briefly review and codify what we learned in our study of the Stern-Gerlach experiment. In particular, we introduce some standard notation and language used by practicing physicists.
Spin and $\hbar$
We've learned that spin angular momentum is an intrinsic property of particles like protons, electrons and atoms. Spin is a vector $\overrightarrow{\mathbf{S}}$ that can point in any direction in three-dimensional space, $\mathbb{R}^{3}$, but measuring spin always amounts to measuring it along a specific direction.
The magnitude of $\overrightarrow{\mathbf{S}}$ is a fixed property of any species of particle. $|\overrightarrow{\mathbf{S}}|$ is constrained to be close to integral values of the quantity $h/4\pi$. More precisely, Quantum Mechanics says that the spin of any particle - of any thing really - must be given by
\begin{equation}\label{s2}|\overrightarrow{\mathbf{S}}|^{2} = \left( \frac{h}{2\pi}\right)^{2} \frac{n}{2}\left(\frac{n}{2}+1\right),\quad n = 0,1,2,3,\cdots.\end{equation}
The number $n$ is just a property of the particle, similar to its electric charge, which must be an integer number of the fundamental charge $e$.
It is conventional to define the quantity $\hbar$, pronounced ``h-bar'', as
$$\hbar = \frac{h}{2\pi},$$
so that we can simplify \eqref{s2} as
\begin{equation}\label{s22} |\overrightarrow{\mathbf{S}}|^{2} = \hbar^{2}\, \frac{n}{2}\left(\frac{n}{2}+1\right), n = 0, 1, 2,3, \cdots\end{equation}
Looking at \eqref{s22}, we see that spins are better parametrized by half-integers in this convention.
$$ |\overrightarrow{\mathbf{S}}|= \hbar\, \sqrt{ s\left(s+1\right)}, \quad s = 0,\frac{1}{2},1,\frac{3}{2},\cdots.$$
From that perspective we'd say that pions are spin 0, silver atoms are spin $\frac{1}{2}$, the $W$ and $Z$ bosons$^{1}$ are spin 1 and the $\Delta$ baryons have spin $\frac{3}{2}$. This is what physicists use in practice, so we'll use this convention from now on.
Components of Spin
We've also seen that $|\overrightarrow{\mathbf{S}}|$ determines the number of possible values for spin we can observe in nature. We can only measure spin in one direction.
To be precise, let $s$ parameterize a given particle's total spin:
$$|\overrightarrow{\mathbf{S}}| = \hbar\sqrt{s(s+1)}.$$
In our new language, we say that the values of spin in some direction, say $S_{i}$, are restricted to half integral values of $\hbar$, from $-s$ to $s$. That is, we can only observe $S_{i}$ as some value $k\hbar$, where $-s< k < s$, and $k$ is a half integer.
Exercise 1: For a particle with $|\overrightarrow{\mathbf{S}}|$ parameterized by the half-integer $s$, argue that there are $2s+1$ possible physical states states.
Spin $\frac{1}{2}$ revisited
We have studied the case where $s = \frac{1}{2}$ at length. This includes silver atoms and the famous Stern-Gerlach experiment. We parametrize the physical state of a spin $\frac{1}{2}$ particle by a two-dimensional vector
$$|\,\psi\,\rangle = \alpha | \uparrow\;\rangle + \beta |\downarrow\;\rangle\rightsquigarrow \left(\begin{array}{r} \alpha \\ \beta \end{array}\right).$$
To measure the spin of a particle described by $|\,\psi\,\rangle$, we act with a spin component operator like $S_{i} = \frac{\hbar}{2}\sigma_{i}$.
The $z$-direction
For the specific case of $S_{z}$,
$$S_{z}|\,\psi\,\rangle = \frac{\hbar}{2}\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right) \left(\begin{array}{r} \alpha \\ \beta \end{array}\right) = \left(\begin{array}{r}\alpha \\ -\beta \end{array}\right).$$
As we've seen, physical observation is restricted to a single eigenvector of $S_{z}$, which for us means that
$$S_{z}|\,\psi\,\rangle \rightarrow \frac{\hbar}{2} \left(\begin{array}{r} 1 \\ 0 \end{array}\right),\quad {\rm with\,probability\;}|\alpha |^{2},$$
and
$$S_{z}|\,\psi\,\rangle \rightarrow -\frac{\hbar}{2}\left(\begin{array}{r} 0 \\ 1 \end{array}\right),\quad {\rm with\,probability\;}|\beta |^{2},$$
because
$$|\alpha|^{2} + |\beta|^{2} = 1.$$
This should all be review. If it's not, you can go back and study our series on the Stern-Gerlach experiment to learn more.
The $x$-direction
For the specific case of $S_{x}$,
$$S_{x}|\,\psi\,\rangle = \frac{\hbar}{2}\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \left(\begin{array}{r} \alpha \\ \beta \end{array}\right) = \left(\begin{array}{r} \beta \\ \alpha \end{array}\right).$$
As we've seen, physical observation is restricted to a single eigenvector of $S_{x}$, which for us means that
\begin{equation}\label{x1}S_{x}|\,\psi\,\rangle \rightarrow \frac{\hbar}{2} \frac{1}{\sqrt{2}}\left(\begin{array}{r} 1 \\ 1 \end{array}\right),\quad {\rm with\,probability\;} \frac{1}{2}|\alpha + \beta |^{2},\end{equation}
and
\begin{equation}\label{x2}S_{x}|\,\psi\,\rangle \rightarrow -\frac{\hbar}{2} \frac{1}{\sqrt{2}}\left(\begin{array}{r} 1 \\ -1 \end{array}\right),\quad {\rm with\,probability\;} \frac{1}{2}|\alpha - \beta |^{2} ,\end{equation}
again because
$$|\alpha|^{2} + |\beta|^{2} = 1.$$
Exercise 2 : Verify by explicit computation the probabilities for observing $S_{x}$ in both up an down states in equations \eqref{x1} and \eqref{x2}. Hence verify that their probabilities sum to one.
The Order of Observations
We cannot simultaneously measure $S_{x}$ and $S_{z}$. Quantum Mechanics says we can only measure spin in one linearly-independent direction at a time. You might ask, how exactly does Quantum Mechanics say this? How do we know what other things cannot be simultaneously measured?
Because measurement changes the physical state of a quantum system, the order in which we do the measurements can affect the final outcome. This point we explored in detail by studying circuits of Stern-Gerlach devices. If we measure $S_{z}$ first, and then $S_{x}$, we will end up with a physical state that is an eigenvector of $S_{x}$:
$$\frac{1}{\sqrt{2}}\left(\begin{array}{r} 1 \\ 2 \end{array}\right)\quad \mathrm{or}\quad \frac{1}{\sqrt{2}}\left(\begin{array}{r} 1 \\ -1 \end{array}\right).$$
If we measure $S_{x}$ first and then $S_{z}$, we'll end up with a physical state that is an eigenvector of $S_{z}$:
$$\left(\begin{array}{r} 1 \\ 0 \end{array}\right)\quad \mathrm{or\quad } \left(\begin{array}{r} 0 \\ 1 \end{array}\right).$$
Because the eigenvectors are different, the final states - the final measurement - will be different. The only way the order of two operations would not matter is if they operators shared the same eigenvectors.
Exercise 3 : Define the linear operator $M$ by
$$M = \left(\begin{array}{cc} \mu & 0 \\ 0 & \nu \end{array}\right),$$
where $\mu$ and $\nu$ are two constants. Argue that $|\uparrow\;\rangle$ and $|\downarrow\;\rangle$ are eigenvectors of $M$. Hence show that the order in which we measure $M$ and $S_{z}$ does not matter. For what values of $\mu$ and $\nu$ is this true for $M$ and $S_{x}$?
The Commutator
There is an easy way to check whether two operators have the same eigenvectors. Let $M$ and $N$ be two matrix operators. They have can the same eigenvectors if
$$MN = NM.$$
That is to say, they have simultaneous eigenvectors if they commute. This is such an important concept that we formalize it with some notation
The commutator of two linear operators $A$ and $B$ is defined to be $[A,B]$, where
$$[A,B] = AB - BA.$$
The commutator captures the failure of $A$ and $B$ to commute. If it is nonzero, these operators cannot have the same eigenvectors.
Exercise 4 : Compute the commutator of $S_{z}$ and $S_{x}$.
The commutator of many physical observables is nonzero. Famously, position and momentum do not commute in Quantum Mechanics:
$$[x,p] \neq 0.$$
We shall learn all about that in due course. But first, we leave you with some instructive exercises:
Exercise 5 : Let $Y = [S_{z},S_{x}]$. Compute the commutator of $Y$ with $S_{z}$ and $S_{x}$.
Exercise 6 : Let $Y$ be as above, and let $Z = [S_{x}, Y]$. Compute $[Z,S_{z}]$.
$^{1}$ : Photons, having two polarization states, are an exception to this rule. Their exception derives from the fact that photons have zero mass, and so special relativity is required to give a proper description of their angular momentum states
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