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Students learn about the geometric meaning of the amplitude and period parameters in the sine function. They also practice sketching the sum of two functions by hand.

We will show that the components of the angular momentum operator \(\vec{L}\), written in differential operator form in rectangular components, satisfy the commutation relations: \begin{equation} \left[L_x,L_y\right]=+i\hbar L_z \qquad(\text{and cyclic permutations}) \end{equation}

First calculate the components of angular momentum classically: \begin{align} \vec{L}&=\vec{r}\times\vec{p}\\ &=(x\hat{x}+y\hat{y}+z\hat{z})\times(p_x\hat{x}+p_y\hat{y}+p_z\hat{z})\\ &=(yp_z-zp_y)\hat{x}+(zp_x-xp_z)\hat{y}+(xp_y-yp_x)\hat{z} \end{align}

Making the standard quantum substitutions, \begin{align} p_x&\rightarrow -i\hbar\partial_x\\ p_y&\rightarrow -i\hbar\partial_y\\ p_z&\rightarrow -i\hbar\partial_z\\ \end{align} we obtain the following operators for the components of angular momentum: \begin{align} \hat{L}_x&=-i\hbar(y\partial_z-z\partial_y)\\ \hat{L}_y&=-i\hbar(z\partial_x-x\partial_z)\\ \hat{L}_z&=-i\hbar(x\partial_y-y\partial_x)\\ \end{align}

To see the role of the product rule in the commutation relations, it is helpful to give the partial derivatives an arbitrary function \(\psi\) to act on. \begin{align} \left[\hat{L}_x,\hat{L}_y\right]\psi &=\left[-i\hbar(y\partial_z-z\partial_y), -i\hbar(z\partial_x-x\partial_z)\right]\psi\\ &=-\hbar^2\left\{(y\partial_z-z\partial_y)(z\partial_x-x\partial_z) -(z\partial_x-x\partial_z)(y\partial_z-z\partial_y)\right\}\psi \end{align} Now, foil-like-mad. Make sure that all of the partial derivatives act on EVERYTHING to their right. Two of the terms above of the form \begin{align} y\,\partial_z(z\,\partial_x \psi) \end{align} require a product rule: \begin{align} y\,\partial_z(z\,\partial_x \psi) &=y((\partial_z z)(\partial_x\psi)+z(\partial_x\partial_x\psi))\\ &=y\partial_x\psi+yz(\partial_x\partial_x\psi) \end{align}

Continuing the calculation above, we see that all of the second derivative terms will cancel because the order of differentiation doesn't matter, leaving only the first derivative terms from the product rule. \begin{align} \left[\hat{L}_x,\hat{L}_y\right]\psi &=\left[-i\hbar(y\partial_z-z\partial_y), -i\hbar(z\partial_x-x\partial_z)\right]\psi\\ &=-\hbar^2\left\{(y\partial_z-z\partial_y)(z\partial_x-x\partial_z) -(z\partial_x-x\partial_z)(y\partial_z-z\partial_y)\right\}\psi\\ &=-\hbar^2\left\{\left(y\,\partial_z(z\,\partial_x \psi) -y\,\partial_z(x\,\partial_z \psi) -z\,\partial_y(z\,\partial_x \psi) +z\,\partial_y(x\,\partial_z \psi)\right)\right.\\ &\;\;\;\quad\quad\left.-\left(z\,\partial_x(y\,\partial_z \psi) -z\,\partial_x(z\,\partial_y \psi) -x\,\partial_z(y\,\partial_z \psi) +x\,\partial_z(z\,\partial_y \psi)\right) \right\}\\ &=-\hbar^2\left\{\left(\cancel{yz(\partial_z\partial_x\psi)} +y\partial_x\psi -\cancel{yx(\partial_z^2\psi)} -\cancel{z^2(\partial_y\partial_x\psi)} +\cancel{zx(\partial_y\partial_z\psi)}\right)\right.\\ &\;\;\;\quad\quad\left.-\left(\cancel{zy(\partial_x\partial_z\psi)} -\cancel{z^2(\partial_x\partial_y\psi)} -\cancel{xy(\partial_z^2\psi)} +\cancel{xz(\partial_z\partial_y\psi)} +x\partial_y\psi\right) \right\}\\ &=i\hbar\left(-i\hbar(-y\partial_x+x\partial_y)\psi\right)\\ &=i\hbar\hat{L}_z\, \psi \end{align} The other components are cyclic permutations of this calculation.

In this activity, students apply the Stefan-Boltzmann equation and the principle of energy balance in steady state to find the steady state temperature of a black object in near-Earth orbit.

This activity follows Planck spectral distribution

Google “phet blackbody spectrum”' and open the simulation.

1. 1. At what wavelength is the peak in spectral intensity
      • \(\lambda_{\text{peak}}\) for a black rock on the Earth's surface,
      • \(\lambda_{\text{peak}}\) for the black walls of a pizza oven,
      • \(\lambda_{\text{peak}}\) for a light bulb,
      • \(\lambda_{\text{peak}}\) for the sun.
   2. Check that the peak wavelength decreases with temperature following a \(1/T\) relationship.
2. 1. Use the numerical integration feature (the checkbox labelled “intensity” near the upper-right corner of the graph) to find the total intensity, in units of \(\text{W/m}^2\), emitted by
      • a black rock on the Earth's surface,
      • the black walls of a pizza oven,
      • the surface of a tungsten light bulb filament,
      • the surface of the sun.
   2. Check that these intensities are proportional to \(T^4\). Note, the quick way to check involves ratios: Does \(\frac{I_1}{I_2} = \left(\frac{T_1}{T_2}\right)^4\)?
3. How cold should you make an object if you want zero thermal radiation emitted?
4. (Extra---if your group has time)
   1. For an incandescent light bulb with a filament surface area of \(A\), estimate how efficiently it converts electrical energy into visible photons. Hint: you will need to estimate the following ratio: \begin{align*} \frac{\text{Electromagnetic radiation in visible wavelengths}}{ \text{Total electromagnetic radiation} } = \frac{ A\int_{400\text{ nm}}^{700\text{ nm}} S_\lambda(\lambda, T)d\lambda }{ A\int_{0}^{\infty} S_\lambda(\lambda, T)d\lambda } \end{align*}
   2. Estimate the filament surface area \(A\) for a 60 W light bulb.

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

This activity follows https://paradigms.oregonstate.edu/act/2525 and requires at minimum the end of the lecture which discusses this experiment.

In https://paradigms.oregonstate.edu/act/2525 you learned about an experiment in which rubidium atoms are dropped from a trap into an optical two-slit experiment. During this experiment the atoms fall a total of 1.5 meters. What is the de Broglie wavelength of an atom after falling from rest 1.5 m? \begin{align} \lambda &= \frac{2\pi\hbar}{p} \end{align}

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.

Students calculate probabilities for a particle on a ring using three different notations: Dirac bra-ket, matrix, and wave function. After calculating the angular momentum and energy measurement probabilities, students compare their calculation methods for notation.

This lecture is one step in motivating the form of the Planck distribution.

These lecture notes for the second week of https://paradigms.oregonstate.edu/courses/ph441 involve relating entropy and temperature in the microcanonical ensemble, using a paramagnet as an example. These notes include a few small group activities.

Commutation Relations for Spin Operators

A commutator of two observables is defined as:

\[[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\]

Determine the results of the following commutators:

1. \([\hat{S}_x,\hat{S}_y]\)
   
2. \([\hat{S}_y,\hat{S}_z]\)
   
3. \([\hat{S}_z,\hat{S}_x]\)
   
4. \([\hat{S}_y,\hat{S}_x]\)
   
5. \([\hat{S}_z,\hat{S}_y]\)
   
6. \([\hat{S}_x,\hat{S}_z]\)
   

Remember that the matrix representation of the spin operators written in the \(S_z\) basis is: \begin{align*} \hat{S}_x \doteq \frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \quad \hat{S}_y \doteq \frac{\hbar}{2}\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \quad \hat{S}_z \doteq \frac{\hbar}{2}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \end{align*}

Activity: Introduction

Divide students into groups to work out whether the spin operators commute.

Activity: Wrap-up

Groups should find that none of the quantum operators commute and therefore do not share the same basis for their respective eigenvectors. Because of this, it provides mathematical evidence for many properties that have so far been only observed. Since none of them commute, none of them have the same basis, nor can the spin operators be measured simultaneously.

Students review using the Arms representation to represent states for discrete quantum systems and connecting the Arms representation to histogram and matrix representation. The student then extend the Arms representation to begin exploring the continuous position basis.

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.”

In this activity students work out energy level transitions in hydrogen that lead to visible light.

This activity reinforces the strategies students have been practicing on each system by letting them create their own matrix operators and columns on the hydrogen atom and do some calculations with them.

These notes from week 6 of https://paradigms.oregonstate.edu/courses/ph441 cover the ideal gas from a grand canonical standpoint starting with the solutions to a particle in a three-dimensional box. They include a number of small group activities.

Students use their arms to act out two spin-1/2 quantum states and their inner product.

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.

• Physics Content Learning Objectives
  1. Explain the consequences of energy and angular momentum conservation in a system of two particles interacting via a central force, in both classical and quantum systems.
  2. Use effective potential diagrams to determine properties of classical orbits.
  3. Solve for the quantum properties of a particle confined to a ring, rigid rotor, and the hydrogen atom in several different representations.
  4. Relate the state of a quantum system (ring, rigid rotor, hydrogen atom) to graphs of a wave function.
  5. Apply Schrödinger time dependence to central force systems (ring, rigid rotor, H atom).
• Mathmatics Content Learning Objectives
  1. Solve ordinary differential equations using power series methods.
  2. Use eigen expansions as an orthonormal basis.
  3. Solve the initial value problem for partial differential equations with more than one spatial variable.
• Professional Learning Objectives
  1. Communicate scientific ideas in writing and with other representations (e.g. graphs, code), using good scientific language and practices, concisely and without ambiguity.
  2. Learn to work with and communicate productively and respectfully with peers and collaborators of different backgrounds.
  3. Cite the information and ideas obtained from or with others in a clear and professional manner.
  4. Communicate in a timely and professional manner with others in the work environment when things don't go to plan.

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 explore what linear transformation matrices do to vectors. The whole class discussion compares & contrasts several different types of transformations (rotation, flip, projections, “scrinch”, scale) and how the properties of the matrices (the determinant, symmetries, which vectors are unchanged) are related to these transformations.

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.

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.

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.