Students calculate probabilities for a particle on a ring whose wavefunction is not easily separated into eigenstates by inspection. To find the energy, angular momentum, and position probabilities, students perform integrations with the wavefunction or decompose the wavefunction into a superposition of eigenfunctions.

Students calculate the expectation value of energy and angular momentum as a function of time for an initial state for a particle on a ring. This state is a linear combination of energy/angular momentum eigenstates written in bra-ket notation.

In this small group activity, students solve for the time dependence of two quantum spin 1/2 particles under the influence of a Hamiltonian. Students determine, given a Hamiltonian, which states are stationary and under what circumstances measurement probabilities do change with time.

Students, working in pairs, use the Arms representations to represent states of spin 1/2 system. Through a short series of instructor-led prompts, students explore the difference between overall phase (which does NOT distinguish quantum states) and relative phase (which does distinguish quantum states).

Students observe the motion of a puck tethered to the center of the airtable. Then they plot the potential energy for the puck on their small whiteboards. A class discussion follows based on what students have written on their whiteboards.

Students, working in pairs, use their left arms to represent each component in a two-state quantum spin 1/2 system. Reinforces the idea that quantum states are complex valued vectors. Students make connections between Dirac, matrix, and Arms representation.

Students examine a plastic "surface" graph of the gravitational potential energy of a Earth-satellite system to make connections between gravitational force and gravitational potential energy.

Students examine a plastic “surface” graph of the gravitational potential energy of an Earth-satellite system to explore the properties of gravitational potential energy for a spherically symmetric system.

Students compute probabilities and averages given a probability density in one dimension. This activity serves as a soft introduction to the particle in a box, introducing all the concepts that are needed.

Students compute inner products to expand a wave function in a sinusoidal basis set. This activity introduces the inner product for wave functions, and the idea of approximating a wave function using a finite set of basis functions.

Students find matrix elements of the position operator \(\hat x\) in a sinusoidal basis. This allows them to express this operator as a matrix, which they can then numerically diagonalize and visualize the eigenfunctions.

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 solve for the equations of motion of a box sliding down (frictionlessly) a wedge, which itself slides on a horizontal surface, in order to answer the question "how much time does it take for the box to slide a distance \(d\) down the wedge?". This activities highlights finding kinetic energies when the coordinate system is not orthonormal and checking special cases, functional behavior, and dimensions.

Students become acquainted with the Spins Simulations of Stern-Gerlach Experiments and record measurement probabilities of spin components for a spin-1/2 system. Students start developing intuitions for the results of quantum measurements for this system.

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.

This activity allows students to puzzle through indexing, the from of operators in quantum mechanics, and working with the new quantum numbers on the sphere in an applied context.

This short small group activity introduces students to the Leibniz notation used for partial derivatives in thermodynamics, in which the variables being held constant are given explicitly. Students are guided to associate variables to their proper categories.

This lab gives students a chance to take data on the first day of class (or later, but I prefer to do it the first day of class). It provides an immediate context for thermodynamics, and also gives them a chance to experimentally measure a change in entropy. Students are required to measure the energy required to melt ice and raise the temperature of water, and measure the change in entropy by integrating the heat capacity.