Students implement a finite-difference approximation for the kinetic energy operator as a matrix, and then use numpy to solve for eigenvalues and eigenstates, which they visualize.

Students consider the relation (1) between the angular momentum and magnetic moment for a current loop and (2) the force on a magnetic moment in an inhomogeneous magnetic field. Students make a (classical) prediction of the outcome of a Stern-Gerlach experiment.

Lecture about finding \(\left|{\pm}\right\rangle _x\) and then \(\left|{\pm}\right\rangle _y\). There are two conventional choices to make: relative phase for \(_x\left\langle {+}\middle|{-}\right\rangle _x\) and \(_y\left\langle {+}\middle|{+}\right\rangle _x\).

So far, we've talked about how to calculate measurement probabilities if you know the input and output quantum states using the probability postulate:

\[\mathcal{P} = | \left\langle {\psi_{out}}\middle|{\psi_{in}}\right\rangle |^2 \]

Now we're going to do this process in reverse.

I want to be able to relate the output states of Stern-Gerlach analyzers oriented in different directions to each other (like \(\left|{\pm}\right\rangle _x\) and \(\left|{\pm}\right\rangle _x\) to \(\left|{\pm}\right\rangle \)). Since \(\left|{\pm}\right\rangle \) forms a basis, I can write any state for a spin-1/2 system as a linear combination of those states, including these special states.

I'll start with \(\left|{+}\right\rangle _x\) written in the \(S_z\) basis with general coefficients:

\[\left|{+}\right\rangle _x = a \left|{+}\right\rangle + be^{i\phi} \left|{-}\right\rangle \]

Notice that:

(1) \(a\), \(b\), and \(\phi\) are all real numbers; (2) the relative phase is loaded onto the second coefficient only.

My job is to use measurement probabilities to determine \(a\), \(b\), and \(\phi\).

I'll prepare a state \(\left|{+}\right\rangle _x\) and then send it through \(x\), \(y\), and \(z\) analyzers. When I do that, I see the following probabilities:

Input = \(\left|{+}\right\rangle _x\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	1	1/2	1/2
\(P(-\hbar/2)\)	0	1/2	1/2

First, looking at the probability for the \(S_z\) components:

\[\mathcal(S_z = +\hbar/2) = | \left\langle {+}\middle|{+}\right\rangle _x |^2 = 1/2\]

Plugging in the \(\left|{+}\right\rangle _x\) written in the \(S_z\) basis:

\[1/2 = \Big| \left\langle {+}\right|\Big( a\left|{+}\right\rangle + be^{i\phi} \left|{-}\right\rangle \Big) \Big|^2\]

Distributing the \(\left\langle {+}\right|\) through the parentheses and use orthonormality: \begin{align*} 1/2 &= \Big| a\cancelto{1}{\left\langle {+}\middle|{+}\right\rangle } + be^{i\phi} \cancelto{0}{\left\langle {+}\middle|{-}\right\rangle } \Big|^2 \\ &= |a|^2\\[12pt] \rightarrow a &= \frac{1}{\sqrt{2}} \end{align*}

Similarly, looking at \(S_z = -\hbar/2\): \begin{align*} \mathcal(S_z = +\hbar/2) &= | \left\langle {-}\middle|{+}\right\rangle _x |^2 = 1/2 \\ 1/2 = \Big| \left\langle {-}\right|\Big( a\left|{+}\right\rangle + be^{i\phi} \left|{-}\right\rangle \Big) \Big|^2\\ 1/2 &= \Big| a\cancelto{0}{\left\langle {-}\middle|{+}\right\rangle } + be^{i\phi} \cancelto{1}{\left\langle {-}\middle|{-}\right\rangle } \Big|^2 \\ &= |be^{i\phi}|^2\\ &= |b|^2 \cancelto{1}{(e^{i\phi})(e^{-i\phi})}\\[12pt] \rightarrow b &= \frac{1}{\sqrt{2}} \end{align*}

I can't yet solve for \(\phi\) but I can do similar calculations for \(\left|{-}\right\rangle _x\):

Input = \(\left|{-}\right\rangle _x\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	0	1/2	1/2
\(P(-\hbar/2)\)	1	1/2	1/2

\begin{align*} \left|{-}\right\rangle _x &= c \left|{+}\right\rangle + de^{i\gamma} \left|{-}\right\rangle \\ \mathcal(S_z = +\hbar/2) &= | \left\langle {+}\middle|{-}\right\rangle _x |^2 = 1/2\\ \rightarrow c = \frac{1}{\sqrt{2}}\\ \mathcal(S_z = +\hbar/2) &= | \left\langle {-}\middle|{-}\right\rangle _x |^2 = 1/2\\ \rightarrow d = \frac{1}{\sqrt{2}}\\ \end{align*}

So now I have: \begin{align*} \left|{+}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\beta} \left|{-}\right\rangle \\ \left|{-}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\gamma} \left|{-}\right\rangle \\ \end{align*}

I know \(\beta \neq \gamma\) because these are not the same state - they are orthogonal to each other: \begin{align*} 0 &= \,_x\left\langle {+}\middle|{-}\right\rangle _x \\ &= \Big(\frac{1}{\sqrt{2}} \left\langle {+}\right| + \frac{1}{\sqrt{2}}e^{i\beta} \left\langle {-}\right| \Big)\Big( \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\gamma} \left|{-}\right\rangle \Big)\\ \end{align*}

Now FOIL like mad and use orthonormality: \begin{align*} 0 &= \frac{1}{2}\Big(\cancelto{1}{\left\langle {+}\middle|{+}\right\rangle } + e^{i\gamma} \cancelto{0}{\left\langle {+}\middle|{-}\right\rangle } + e^{i\beta} \cancelto{0}{\left\langle {-}\middle|{+}\right\rangle } + e^{i(\gamma - \beta)}\cancelto{1}{\left\langle {-}\middle|{-}\right\rangle } \Big)\\ &= \frac{1}{2}\Big(1 + e^{i(\gamma - \beta} \Big) \\ \rightarrow & \quad e^{i(\gamma-\beta)} = -1 \end{align*}

This means that \(\gamma-\beta = \pi\). I don't have enough information to solve for \(\beta\) and \(\gamma\), but there is a one-time conventional choice made that \(\beta = 0\) and \(\gamma = 1\), so that: \begin{align*} \left|{+}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{1}{e^{i0}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{-1}{e^{i\pi}} \left|{-}\right\rangle \\[12pt] \rightarrow \left|{+}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{+} \frac{1}{\sqrt{2}}\left|{-}\right\rangle \\ \left|{-}\right\rangle _x &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{-} \frac{1}{\sqrt{2}}\left|{-}\right\rangle \\[12pt] \end{align*}

When \(\left|{\pm}\right\rangle _y\) is the input state:

Input = \(\left|{+}\right\rangle _y\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	1/2	1	1/2
\(P(-\hbar/2)\)	1/2	0	1/2

Input = \(\left|{-}\right\rangle _y\)	\(S_x\)	\(S_y\)	\(S_z\)
\(P(\hbar/2)\)	1/2	0	1/2
\(P(-\hbar/2)\)	1/2	1	1/2

The calculations proceed in the same way. The \(S_z\) probabilities give me: \begin{align*} \left|{+}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{1}{e^{i\alpha}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{-1}{e^{i\theta}} \left|{-}\right\rangle \\ \end{align*}

The orthongality between \(\left|{+}\right\rangle _y\) and \(\left|{-}\right\rangle _y\) mean that \(\theta - \alpha = \pi\).

But I also know the \(S_x\) probabilities and how to write \(|ket{\pm}_x\) in the \(S_z\) basis. For an input of \(\left|{+}\right\rangle _y\): \begin{align*} \mathcal(S_x = +\hbar/2) &= | \,_x\left\langle {+}\middle|{+}\right\rangle _y |^2 = 1/2 \\ 1/2 &= \Big| \Big(\frac{1}{\sqrt{2}} \left\langle {+}\right| + \frac{1}{\sqrt{2}}\left\langle {-}\right|\Big) \Big( \frac{1}{\sqrt{2}}\left|{+}\right\rangle + \frac{1}{\sqrt{2}}e^{i\alpha} \left|{-}\right\rangle \Big) \Big|^2\\ 1/2 &= \Big| \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \cancelto{1}{\left\langle {+}\middle|{+}\right\rangle } + \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}e^{i\alpha} \cancelto{1}{\left\langle {-}\middle|{-}\right\rangle } \Big|^2 \\ &= \frac{1}{4}|1+e^{i\alpha}|^2\\ &= \frac{1}{4} \Big( 1+e^{i\alpha}\Big) \Big( 1+e^{-i\alpha}\Big)\\ &= \frac{1}{4} \Big( 2+e^{i\alpha} + e^{-i\alpha}\Big)\\ &= \frac{1}{4} \Big( 2+2\cos\alpha\Big)\\ \frac{1}{2} &= \frac{1}{2} + \frac{1}{2}\cos\alpha \\ 0 &= \cos\alpha\\ \rightarrow \alpha = \pm \frac{\pi}{2} \end{align*}

Here, again, I can't solve exactly for alpha (or \(\theta\)), but the convention is to choose \(alpha = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\), making \begin{align*} \left|{+}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{i}{e^{i\pi/2}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle + \frac{1}{\sqrt{2}}\cancelto{-i}{e^{i3\pi/2}} \left|{-}\right\rangle \\ \rightarrow \left|{+}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{+} \frac{\color{red}{i}}{\sqrt{2}} \left|{-}\right\rangle \\ \left|{-}\right\rangle _y &= \frac{1}{\sqrt{2}} \left|{+}\right\rangle \color{red}{-} \frac{\color{red}{i}}{\sqrt{2}} \left|{-}\right\rangle \\ \end{align*}

If I use these two convenctions for the relative phases, then I can write down \(\left|{\pm}\right\rangle _n\) in an arbitrary direction described by the spherical coordinates \(\theta\) and \(\phi\) as:

Discuss the generalize eigenstates: \begin{align*}\ \left|{+}\right\rangle _n &= \cos \frac{\theta}{2} \left|{+}\right\rangle + \sin \frac{\theta}{2} e^{i\phi} \left|{-}\right\rangle \\ \left|{-}\right\rangle _n &= \sin \frac{\theta}{2} \left|{+}\right\rangle - \cos \frac{\theta}{2} e^{i\phi} \left|{-}\right\rangle \end{align*}

And how the \(\left|{\pm}\right\rangle _x\) and \(\left|{\pm}\right\rangle _y\) are consistent.

Student explore the properties of an orthonormal basis using the Cartesian and \(S_z\) bases as examples.

Expectation Values and Uncertainty

You have a system that consists of quantum particles with spin. On this system, you will perform a Stern-Gerlach experiment with an analyzer oriented in the \(z\)-direction.

Consider one of the different initial spin states described below:

A spin 1/2 particle described by:

1. \(\left|{+}\right\rangle \)
2. \(\frac{i}{2}\left|{+}\right\rangle -\frac{\sqrt{3}}{2}\left|{-}\right\rangle \)

3. \(\left|{+}\right\rangle _x\)

   A spin 1 particle described by:

4. \(\left|{0}\right\rangle \)
5. \(\left|{-1}\right\rangle _x\)
6. \(\frac{2}{3}\left|{1}\right\rangle +\frac{i}{3}\left|{0}\right\rangle -\frac{2}{3}\left|{-1}\right\rangle \)

• List the possible values of spin you could measure and determine the probability associated with each value of the z-component of spin.

  
  

• Plot a histogram of the probabilities.

  
  

• Find the expectation value of the z-component of spin.

  
  

• Find the uncertainty of the z-component of spin.

Introduction

I like to break this activity into two parts:

(1) Calculating expectation values and relating them to the associated distributions of the probabilities of results, and

(2) Calculating the quantum uncertainty of the state and relating the uncertainty to distributions of the probabilities of results.

Therefore, I have my students do the first part of the activity before I introduce quantum uncertainty.

I introduce the activity by reminding students about two ways of calculating the expectation value. Given a quantum state \(\left|{\psi}\right\rangle \), for a measurement of an observable represented by an operator \(\hat{A}\) with eigenstates \(\left|{a_i}\right\rangle \) and eigenvalues \(a_i\): \begin{align*} \langle \hat{A} \rangle &= \sum_{i} a_i\mathcal{P}(a_i) \\ &= \left\langle {\psi}\right|\hat{A}\left|{\psi}\right\rangle \end{align*}

After the students calculate expectation values and we have a whole class discussion about 1 of the examples, then I do a lecture introducing the quantity of quantum uncertainty (relating it to the standard deviation of the distribution of probabilities by spin component value) and deriving the simplified equation:

\[\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle ^2}\]

Student Conversations

1. One could have each group report out, or the instructor could discuss a few key examples.

   For expectation value, I like to talk about Case 2: \(\frac{i}{2}\left|{+}\right\rangle -\frac{\sqrt{3}}{2}\left|{-}\right\rangle \), where the probabilities of the two outcomes are not equal to show how the weighting plays out. Also, the expectation value is not a possible measurement value, and I like to talk about that. “Expectation” value is a misleading name for this quantity - it characterizes the distribution and is not necessarily a result of an individual measurement.

   I also like to discuss an example like Case 5: \(\left|{1}\right\rangle _x\) where the distribution is symmetric around \(0\hbar\).

2. I think it's important to encourage students to calculate expectation values both ways (with probabilities and as a bracket with matrix notation) while the teaching team is available to help them.

3. For quantum uncertainty, I like to talk about an example like Case 3: \(\left|{+}\right\rangle _x\) where all the individual measurements are the same ”distance” away from the expectation value as a sensemaking exercise to connect to a conceptual interpretation of physics.

   I also like to discuss an example like Case 5: \(\left|{-1}\right\rangle _x\), where the fact that we're taking an rms average is apparent: half the measurements are \(\hbar\) away from the expectation value and the other half are \(0\hbar\) away, but the uncertainty is \(\hbar/\sqrt{2}\).

Students sketch the temperature-dependent heat capacity of molecular nitrogen. They apply the equipartition theorem and compute the temperatures at which degrees of freedom “freeze out.”

Students use a completeness relations to write hydrogen atoms states in the energy and position bases.

In this lecture, the instructor guides a discussion about translating between bra-ket notation and wavefunction notation for quantum systems.