Observables
The main idea of this section can be summarized as this: if a physical state is modeled by a vector, observable quantities - such as spin - are modeled by linear operators on that vector space. Linear operators are just functions that take vectors to vectors. So that if \(\psi\) and \(\phi\) are vectors, \(a\) and \(b\) are scalars, and \(O\) is a linear operator then,
$$O(a\psi + b\phi) = aO(\psi) + bO(\phi).$$
In other words:
The action of a linear operator on a linear combination of vectors, is a linear combination of actions of that operator.
In this module we will review
physical observables
observables as operators
the expectation value of an operator
Spin as an Operator
Last time we saw that silver atoms - like electrons or protons - have two spin states: up: \(|\nwarrow\,\rangle\) and down: \(|\searrow\,\rangle\). We modeled this two level system as a column vector, so that a generic spin state was given by
\begin{equation}\label{psi}|\psi\rangle = \alpha \,|\nwarrow\,\rangle + \beta \,|\searrow\,\rangle \rightsquigarrow \left(\begin{array}{c}\alpha \\ \beta\end{array}\right).\end{equation}
The spin of those silver atoms - measured in the vertical or \(\overrightarrow{\bf z}\) direction - is encoded in the linear operator \(S_{z}\). In the language of column vectors, \(S_{z}\) is a matrix:
$$S_{z} = \frac{\hbar}{2}\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right).$$
From this perspective, we see that the action of $S_{z}$ on $|\psi \rangle$ is
\begin{equation}\label{spinmatrix}S_{z}|\psi\rangle = \frac{\hbar}{2}\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{c}\alpha \\ \beta\end{array}\right) = \frac{\hbar}{2}\left(\begin{array}{r}\alpha \\ -\beta\end{array}\right).\end{equation}
Expressed in components, we immediately see what's peculiar about quantum mechanics:
\begin{equation}\label{components}S_{z}|\psi\rangle = \frac{\hbar \alpha}{2} \,|\nwarrow\,\rangle -\frac{\hbar\beta}{2} \,|\searrow\,\rangle. \end{equation}
The spin of the particle in the quantum state $|\psi\rangle$ is in a linear combination of two states, and so has a linear combination of measured values: \(\frac{\hbar}{2}\) and \(-\frac{\hbar}{2}\). This fact introduces some questions:
How can you possibly measure the linear combination of two distinct values, up and down?
What does that even mean?
Is it \((\alpha - \beta)\frac{\hbar}{2}\)?
The answers to these questions aren't very satisfying:
You can't.
Such a concept is meaningless.
It is not a linear combination of the answers.
You can only measure one of those two values, as we saw with the Stern-Gerlach experiment. So the answer amounts to another question:
Which spin value gets measured?
As we saw last time, the answer relates $|\psi\rangle$ to a probability distribution.
Interlude: on the Normalization of Quantum States
Moving forward, we will generically assume that quantum states are normalized, that is
$$\langle \psi | \psi \rangle = 1.$$
Morally, it means that we take the concept of $|\psi\rangle$ as a probability distribution seriously. Practically, it means we no longer have to divide by this constant factor. There are times when such a normalization is impossible\(^{\dagger}\) - but those times typically involve mathematical simplifications\(^{\clubsuit}\).
The Expectation Value of \(S_{z}\)
Last time we saw that the probability of the state $|\psi\rangle$ being found in the spin up state, $|\nwarrow\,\rangle$ was given\(^{\star}\) by
$$P_{\uparrow} = |\langle \,\nwarrow | \psi\rangle |^{2} = |\alpha|^{2}.$$
This probability of measurement is different than the actual measured value. What it means is that if we do a single measurement, we can expect to find the atom in a spin up state with a probability of $P_{\uparrow}$. To access the value of the thing being measured, we must turn to the observable operators. To discuss the measured value of spin in the $\overrightarrow{\bf z}$ direction, we must turn to $S_{z}$.
Statisticians have a concept of expectation value, which asks: given a large number of measurements, what's the typical value we might expect? As we will now see, it is kind of an awkward question for a two-level system, but it essentially amounts the mean of the probability distribution.
The expectation value of the spin $S_{z}$ in the state $|\psi\rangle$ given in \eqref{psi} is defined to be
\begin{equation}\label{expectation}\langle S_{z} \rangle = \langle \psi | S_{z} | \psi \rangle,\end{equation}
which we can compute:
$$\langle S_{z} \rangle = \frac{\hbar}{2}\left( \alpha \langle \psi | \nwarrow\,\rangle - \beta \langle \psi | \searrow\,\rangle\right).$$
Since $|\nwarrow\,\rangle$ and $|\searrow\,\rangle$ are orthonormal states, this simplifies to
\begin{equation}\label{ab}\langle S_{z} \rangle = \frac{\hbar}{2} \left(|\alpha|^{2} - |\beta|^{2}\right),\end{equation}
which is to say that
\begin{equation}\label{szp}\langle S_{z} \rangle = \frac{\hbar}{2} \left(P_{\uparrow} - P_{\downarrow}\right).\end{equation}
For the Stern-Gerlach experiment, where the silver atoms were uniformly distributed, $P_{\uparrow} = P_{\downarrow} = \frac{1}{2}$, so that
$$\langle S_{z} \rangle = 0.$$
Given that the only two allowed measurements are $\pm\frac{\hbar}{2}$, this makes sense, although it might give you the false impression that the expected value of spin is zero. All \eqref{szp} is telling you here is that the probability distribution of the spins - the quantum state $|\psi\rangle$ - is uniformly distributed.
The Variance of \(S_{z}\)
As we have just seen, the mean of a probability distribution is not always its most descriptive property. It is often better to view the mean in conjunction with the variance. Like the expectation value, the variance of $S_{z}$ in the state $|\psi\rangle$ is defined to be:
\begin{equation}\label{var} \Delta S_{z} = \langle S_{z}^{2} \rangle - \langle S_{z}\rangle^{2}.\end{equation}
We know $\langle S_{z}\rangle$, so we can compute
$$\langle S_{z}^{2}\rangle = \langle \psi | S_{z}^{2}|\psi \rangle.$$
As you can easily check, the square of the matrix $S_{z}$ is proportional to the identity matrix, $\mathbb{1}$:
\begin{equation}\label{szsquared}S_{z}^{2} = \frac{\hbar^{2}}{4}\mathbb{1},\end{equation}
so, given our normalization convention,
$$\langle \psi | S_{z}^{2}|\psi \rangle = \frac{\hbar^{2}}{4}\langle \psi | \psi \rangle = \frac{\hbar^{2}}{4}.$$
In other words,
$$\Delta S_{z} = \frac{\hbar^{2}}{4} \left(1 - (P_{\uparrow} - P_{\downarrow})^{2} \right).$$
For the case of the Stern-Gerlach experiment, where $\langle S_{z}\rangle = 0$, this reduces to
$$\Delta S_{z} = \frac{\hbar^{2}}{4}.$$
The variance might be more familiar in relation to the standard deviation, which for the silver atoms is hopefully familiar:
\begin{equation}\label{sigma}\sigma_{S_{z}} = \sqrt{\Delta S_{z}} = \frac{\hbar}{2}.\end{equation}
Together with the mean, we now have enough information to interpret the quantum state's probability distribution.
Epilogue
Today we dug into the underlying machinery of quantum mechanics. The difference between observables and the quantum state will continue to become clear as we formalize our discussions, but hopefully their outlines are becoming clear. The quantum state - sometimes called the wavefunction - has the interpretation of a probability distribution. Measurement samples that probability distribution.
The observables - the things we measure - are linear operators on the vector space of quantum states. They are not measurements in an of themselves, but they can be used to ask the probability distribution questions, like
What is the most probable value we might measure?
and
What is the expected variance of our observations?
It is crucial to differentiate between measurement and the observable operators that frame it. This is made clear by examining the effect of the operator $S_{z}$ in \eqref{spinmatrix} and \eqref{components}.
As will hopefully be made clear in the exercises and the next section, observables play a far more important role in quantum mechanics than merely querying the quantum state. They have additional properties that help explain the very nature of reality.
Next Time
the complex numbers
the Pauli matrices
the algebra of spin operators
Further Reading
The late J.J. Sakurai’s textbook Modern Quantum Mechanics is an excellent, thorough pass through the material. Gordon Baym’s Lectures on Quantum Mechanics takes a similar path. For more context on the history of quantum theory, you might try Helge Kragh’s book, Dirac: A scientific biography. Despite its age, Dirac’s book Principles of Quantum Mechanics is also a great read..
If you’re serious about learning statistics, then check out An Introduction to Statistical Learning: with Applications in R by Gareth, Witten, Hastie and Tibshirani. A simpler reference is Statistics by Barlow.
Exercises
Verify \eqref{ab} and \eqref{szsquared} by direct computation. Hence verify \eqref{sigma}.
Last time you verified that the probability of finding $k$ spin up measurements in a mean of $N$ silver atoms was given by the binomial distribution. For the case of $\alpha = \beta = \frac{1}{\sqrt{2}}$, and building on your prior programming work, argue by numeric approximation that the binomial distribution can be approximated by the normal distribution with a mean value $N/2$ and variance $N/4$.
How would the results from the previous problem generalize for a generic state, $\psi\rangle$? You may assume it is normalized.
The operator $S_{z}$ quantifies spin angular momentum in the $\overrightarrow{\bf z}$ direction. One of the peculiar things about quantum mechanics is that there are actually three such operators, one for each direction of space:
$$S_{x},\;S_{y},\; S_{z}.$$
In particular, $S_{x}$ is modeled by the matrix
$$ S_{x} = \frac{\hbar}{2}\left( \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right).$$
Supposed a Stern-Gerlach apparatus is set up to measure $S_{z}$. What is the expectation value of $S_{x}$ just before the Stern-Gerlach measurement? What is it just after? (You may pick either of the two beams emitted from the Stern-Gerlach device).
For a general quantum state $|\psi\rangle$, as in \eqref{psi}, compute $\langle S_{x}\rangle$ and $\Delta S_{x}$. Show that $S_{x}^{2} = S_{z}^{2}$.
Compute $S_{x} |\nwarrow\,\rangle$ and $S_{x} |\searrow\,\rangle$.
Find $\alpha$ and $\beta$ in \eqref{psi} such that $|\psi\rangle$ satisfies:
$$S_{x}|\psi\rangle = \frac{\hbar}{2}|\psi\rangle.$$
Call this quantum state vector $|\nwarrow x\, \rangle$. Find also $|\searrow x\, \rangle$, where
$$S_{x}|\searrow x\,\rangle = -\frac{\hbar}{2}|\searrow x \,\rangle.$$
Verify that
$$\langle \, \nwarrow x|\nwarrow x\,\rangle = \langle \, \searrow x|\searrow x\,\rangle = 1,\quad \langle \, \nwarrow x|\searrow x\,\rangle = 0.$$
Compute
$$\langle \searrow |\searrow x\,\rangle \quad \mathrm{and}\quad \langle \nwarrow |\searrow x\,\rangle.$$
Compare
$$\langle \, \nwarrow | S_{z} | \nwarrow\,\rangle \quad \mathrm{with} \quad \langle \, \nwarrow | S_{x} | \nwarrow\,\rangle .$$
Finally, compare
$$\langle \, \nwarrow x| S_{z} | \nwarrow x\,\rangle \quad \mathrm{with} \quad \langle \, \nwarrow x| S_{x} | \nwarrow x\,\rangle .$$
Explain the difference between $| \nwarrow x\,\rangle $ and $| \nwarrow \,\rangle $.
Solutions available upon request. But first, please do send me your attempts at the problems!
\(^{\dagger}\) Its worth pointing out that some computational algorithms, like the famous Metropolis Algorithm, do not even require us to know the normalization factor. We merely have to assume that the state is normalizable.
\(^{\clubsuit}\) The plane wave vs. a Gaussian wave packet is a typical example of this scenario.
\(^{\star}\) Note that our normalization convention is now in effect.
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