Consider the quantum state: \[\left\vert \psi\right\rangle = \frac{1}{\sqrt{3}}\left\vert +\right\rangle+ i\frac{\sqrt{2}}{\sqrt{3}} \left\vert -\right\rangle\]

Find the normalized vector \(\vert \phi\rangle\) that is orthogonal to it.

You know that the normalized spatial eigenfunctions for a particle in a 1-D box of length \(L\) are \(\sqrt{\frac{2}{L}}\sin{\frac{n\pi x}{L}}\). If you want the eigenfunctions for a particle in a 2-D box, then you just multiply together the eigenfunctions for a 1-D box in each direction. (This is what the separation of variables procedure tells you to do.)

1. Find the normalized eigenfunctions for a particle in a 2-D box with sides of length \(L\) in the \(x\)-direction and length \(W\) in the \(y\)-direction.
2. Find the Hamiltonian for a 2-D box and show that your eigenstates are indeed eigenstates and find a formula for the possible energies

3. Any sufficiently smooth spatial wave function inside a 2-D box can be expanded in a double sum of the product wave functions, i.e. \begin{equation} \psi(x,y)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\, c_{nm}\; \hbox{eigenfunction}_n(x)\;\hbox{eigenfunction}_m(y) \end{equation} Using your expressions from part (a) above, write out all the terms in this sum out to \(n=3\), \(m=3\). Arrange the terms, conventionally, in terms of increasing energy.

   You may find it easier to work in bra/ket notation: \begin{align*} \left|{\psi}\right\rangle &=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\, c_{nm}\left|{n}\right\rangle \left|{m}\right\rangle \\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\, c_{nm}\left|{nm}\right\rangle \end{align*}

4. Find a formula for the \(c_{nm}\)s in part (c). Find the formula first in bra ket notation and then rewrite it in wave function notation.

Consider one particle confined to a cube of side \(L\); the concentration in effect is \(n=L^{-3}\). Find the kinetic energy of the particle when in the ground state. There will be a value of the concentration for which this zero-point quantum kinetic energy is equal to the temperature \(kT\). (At this concentration the occupancy of the lowest orbital is of the order of unity; the lowest orbital always has a higher occupancy than any other orbital.) Show that the concentration \(n_0\) thus defined is equal to the quantum concentration \(n_Q\) defined by (63): \begin{equation} n_Q \equiv \left(\frac{MkT}{2\pi\hbar^2}\right)^{\frac32} \end{equation} within a factor of the order of unity.

The following two problems ask you to make Fermi estimates. In a good Fermi estimate, you start from basic scientific facts you already know or quantities that you can reasonably estimate based on your life experiences and then reason your way to estimate a quantity that you would not be able guess. You may look up useful conversion factors or constants. Use words, pictures, and equations to explain your reasoning:

1. Imagine that you send a pea-sized bead of silver through a Stern-Gerlach device oriented to measure the z-component of intrinsic spin. Estimate the total z-component of the intrinsic spin of the ball you would measure in the HIGHLY improbable case that every atom is spin up.
2. Protons, neutrons, and electrons are all spin-1/2 particles. Give a (very crude) order of magnitude estimate of the number of these particles in your body.

First complete the problem Diagonalization. In that notation:

1. Find the matrix \(S\) whose columns are \(|\alpha\rangle\) and \(|\beta\rangle\). Show that \(S^{\dagger}=S^{-1}\) by calculating \(S^{\dagger}\) and multiplying it by \(S\). (Does the order of multiplication matter?)
2. Calculate \(B=S^{-1} C S\). How is the matrix \(E\) related to \(B\) and \(C\)? The transformation that you have just done is an example of a “change of basis”, sometimes called a “similarity transformation.” When the result of a change of basis is a diagonal matrix, the process is called diagonalization.

None

1-D Particle-in-a-box

Hamiltonian: \begin{align} \hat{H} &\doteq -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \end{align} Eigenstates: \begin{align} \left|{n}\right\rangle &\doteq\sqrt{\frac{2}{L}}\, \sin\frac{n\pi x}{L}\\ n&=\left\{1, 2, 3, \dots\right\} \end{align}

Eigenvalue Equations: \begin{align} \hat{H}\left|{n}\right\rangle &=\frac{\pi^2\hbar^2}{2\mu L^2}\, n^2 \left|{n}\right\rangle \\ \end{align}

Particle-on-a-Ring

Hamiltonian: \begin{align} \hat{H} &\doteq -\frac{\hbar^2}{2I}\frac{\partial^2}{\partial \phi^2} \end{align} Eigenstates: \begin{align} \left|{m}\right\rangle &\doteq\frac{1}{\sqrt{2\pi r_0}}\, e^{im\phi}\\ m&=\left\{\dots 2, 1, 0, -1, -2, \dots\right\} \end{align} Eigenvalue Equations: \begin{align} \hat{H}\left|{m}\right\rangle &=\frac{\hbar^2}{2I}\, m^2 \left|{m}\right\rangle \\ \hat{L}^2\left|{m}\right\rangle &=\hbar^2\, m^2 \left|{m}\right\rangle \\ \hat{L}_z\left|{m}\right\rangle &=\hbar\, m \left|{m}\right\rangle \end{align}

2-D Particle-in-a-Box

Hamiltonian: \begin{align} \hat{H} &\doteq -\frac{\hbar^2}{2m}\Big(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}) \end{align} Eigenstates: \begin{align} \left|{mn}\right\rangle &\doteq\sqrt{\frac{2}{L_x}}\sqrt{\frac{2}{L_y}}\, \sin\frac{m\pi x}{L_x}\sin\frac{n\pi y}{L_y}\\ m&=\left\{1, 2, 3, \dots\right\}\\ n&=\left\{1, 2, 3, \dots\right\} \end{align} Eigenvalue Equations: \begin{align} \hat{H}\left|{mn}\right\rangle &=\frac{\pi^2\hbar^2}{2\mu}\, \left(\frac{m^2}{L_x^2}+\frac{n^2}{L_y^2}\right) \left|{mn}\right\rangle \\ \end{align}

Particle-on-a-Sphere

Hamiltonian: \begin{align} \hat{H} &\doteq -\frac{\hbar^2}{2I} \Big[\frac{1}{\sin\theta}\frac{\partial}{\partial \theta} \Big(\sin\theta \frac{\partial}{\partial \theta} \big) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \Big] \end{align} Eigenstates: \begin{align} \left|{\ell m}\right\rangle &\doteq Y_{\ell}^m(\theta, \phi)\\ &=(-1)^{\frac{m+|m|}{2}}\sqrt{\frac{2\ell+1}{4\pi}\frac{(\ell-m)!}{(\ell+m)!}} \,P_{\ell}^m(\cos\theta)\, e^{im\phi}\\ \ell&=\left\{0, 1, 2, \dots\right\}\\ m&=\left\{\ell, \dots , 0, \dots,-\ell\right\} \end{align} Eigenvalue Equations: \begin{align} \hat{H}\left|{\ell m}\right\rangle &=\frac{\hbar^2}{2I}\, \ell(\ell+1) \left|{\ell m}\right\rangle \\ \hat{L}^2\left|{\ell m}\right\rangle &=\hbar^2\, \ell(\ell+1) \left|{\ell m}\right\rangle \\ \hat{L}_z\left|{\ell m}\right\rangle &=\hbar\, m \left|{\ell m}\right\rangle \end{align}

Hydrogen Atom

Hamiltonian: \begin{align} \hat{H} &\doteq -\frac{\hbar^2}{2\mu} \nabla^2 - \frac{ke^2}{r} \\ &\doteq -\frac{\hbar^2}{2\mu r^2} \Big[\frac{\partial}{\partial r} \Big( r^2 \frac{\partial}{\partial r} \Big) + \frac{1}{\sin\theta}\frac{\partial}{\partial \theta} \Big(\sin\theta \frac{\partial}{\partial \theta} \big) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \Big] - \frac{ke^2}{r} \end{align} Eigenstates: \begin{align} \left|{n\ell m}\right\rangle &\doteq R_{n\ell}(r)\, Y_{\ell}^m(\theta, \phi)\\ &=-\sqrt{\left(\frac{2Z}{na_0}\right)^3 \frac{(n-\ell-1)!}{2n[(n+\ell)!]^3}} \left(\frac{2\rho}{n}\right)^{\ell}\, e^{-\frac{\rho}{n}}\, L_{n+\ell}^{2\ell+1}{\scriptstyle{\left(\frac{2\rho}{n}\right)}} (-1)^{\frac{m+|m|}{2}} \sqrt{\frac{2\ell+1}{4\pi}\frac{(\ell-m)!}{(\ell+m)!}} \,P_{\ell}^m(\cos\theta)\, e^{im\phi}\\ \rho&=\frac{Zr}{a_0}\\ n&=\left\{1, 2, 3,\dots\right\}\\ \ell&=\left\{0, 1, 2, \dots, n-1\right\}\\ m&=\left\{\ell, \dots , 0, \dots,-\ell\right\} \end{align} Eigenvalue Equations: \begin{align} \hat{H}\left|{n\ell m}\right\rangle &=-\frac{1}{2}\left(\frac{Ze^2}{4\pi\epsilon_0}\right)^2 \frac{\mu}{\hbar^2}\,\frac{1}{n^2}\, \left|{n \ell m}\right\rangle \\ &=-13.6 \text{eV}\,\frac{1}{n^2}\, \left|{n \ell m}\right\rangle \\ \hat{L}^2\left|{n \ell m}\right\rangle &=\hbar^2\, \ell(\ell+1) \left|{n \ell m}\right\rangle \\ \hat{L}_z\left|{n \ell m}\right\rangle &=\hbar\, m \left|{n \ell m}\right\rangle \end{align}

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

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 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.

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 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.

Eigenvalues and Eigenvectors

Each group will be assigned one of the following matrices.

\[ A_1\doteq \begin{pmatrix} 0&-1\\ 1&0\\ \end{pmatrix} \hspace{2em} A_2\doteq \begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \hspace{2em} A_3\doteq \begin{pmatrix} -1&0\\ 0&-1\\ \end{pmatrix} \]
\[ A_4\doteq \begin{pmatrix} a&0\\ 0&d\\ \end{pmatrix} \hspace{2em} A_5\doteq \begin{pmatrix} 3&-i\\ i&3\\ \end{pmatrix} \hspace{2em} A_6\doteq \begin{pmatrix} 0&0\\ 0&1\\ \end{pmatrix} \hspace{2em} A_7\doteq \begin{pmatrix} 1&2\\ 1&2\\ \end{pmatrix} \]
\[ A_8\doteq \begin{pmatrix} -1&0&0\\ 0&-1&0\\ 0&0&-1\\ \end{pmatrix} \hspace{2em} A_9\doteq \begin{pmatrix} -1&0&0\\ 0&-1&0\\ 0&0&1\\ \end{pmatrix} \]
\[ S_x\doteq \frac{\hbar}{2}\begin{pmatrix} 0&1\\ 1&0\\ \end{pmatrix} \hspace{2em} S_y\doteq \frac{\hbar}{2}\begin{pmatrix} 0&-i\\ i&0\\ \end{pmatrix} \hspace{2em} S_z\doteq \frac{\hbar}{2}\begin{pmatrix} 1&0\\ 0&-1\\ \end{pmatrix} \]

For your matrix:

1. Find the eigenvalues.
2. Find the (unnormalized) eigenvectors.
3. Describe what this transformation does.
4. Normalize your eigenstates.

If you finish early, try another matrix with a different structure, i.e. real vs. complex entries, diagonal vs. non-diagonal, \(2\times 2\) vs. \(3\times 3\), with vs. without explicit dimensions.

Instructor's Guide

Main Ideas

This is a small group activity for groups of 3-4. The students will be given one of 10 matrices. The students are then instructed to find the eigenvectors and eigenvalues for this matrix and record their calculations on their medium-sized whiteboards. In the class discussion that follows students report their finding and compare and contrast the properties of the eigenvalues and eigenvectors they find. Two topics that should specifically discussed are the case of repeated eigenvalues (degeneracy) and complex eigenvectors, e.g., in the case of some pure rotations, special properties of the eigenvectors and eigenvalues of hermitian matrices, common eigenvectors of commuting operators.

Students' Task

Introduction

Give a mini-lecture on how to calculate eigenvalues and eigenvectors. It is often easiest to do this with an example. We like to use the matrix \[A_7\doteq\begin{pmatrix}1&2\cr 9&4\cr\end{pmatrix}\] from the https://paradigms.oregonstate.edu/activities/2179 https://paradigms.oregonstate.edu/activities/2179 Finding Eigenvectors and Eigenvalues since the students have already seen this matrix and know what it's eigenvectors are. Then every group is given a handout, assigned a matrix, and then asked to: - Find the eigenvalues - Find the (unnormalized) eigenvectors - Normalize the eigenvectors - Describe what this transformation does

Student Conversations

• Typically, students can find the eigenvalues without too much problem. Eigenvectors are a different story. To find the eigenvectors, they will have two equations with two unknowns. They expect to be able to find a unique solution. But, since any scalar multiple of an eigenvector is also an eigenvector, their two equations will be redundant. Typically, they must choose any convenient value for one of the components (e.g. \(x=1\)) and solve for the other one. Later, they can use this scale freedom to normalize their vector.
• The examples in this activity were chosen to include many of the special cases that can trip students up. A common example is when the two equations for the eigenvector amount to something like \(x=x\) and \(y=-y\). For the first equation, they may need help to realize that \(x=\) “anything” is the solution. And for the second equation, sadly, many students need to be helped to the realization that the only solution is \(y=0\).

Wrap-up

The majority of the this activity is in the wrap-up conversation.

The [[whitepapers:narratives:eigenvectorslong|Eigenvalues and Eigenvectors Narrative]] provides a detailed narrative interpretation of this activity, focusing on the wrap-up conversation.

• Complex eigenvectors: connect to discussion of rotations in the Linear Transformations activity where there did not appear to be any vectors that stayed the same.
• Degeneracy: Define degeneracy as the case when there are repeated eigenvalues. Make sure the students see that, in the case of degeneracy, an entire subspace of vectors are all eigenvectors.
• Diagonal Matrices: Discuss that diagonal matrices are trivial. Their eigenvalues are just their diagonal elements and their eigenvectors are just the standard basis.
• Matrices with dimensions: Students should see from these examples that when you multiply a transformation by a real scalar, its eigenvalues are multiplied by that scalar and its eigenvectors are unchanges. If the scalar has dimensions (e.g. \(\hbar/2\)), then the eigenvalues have the same dimensions.

Students use Mathematica to visualize the probability density distribution for the hydrogen atom orbitals with the option to vary the values of \(n\), \(\ell\), and \(m\).

This activity lets students explore translating a wavefunction that isn't obviously made up of eigenstates at first glance into ket and matrix form. Then students explore wave functions, probabilities in a region, expectation values, and what wavefunctions can tell you about measurements of \(L_z\).

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.

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 are asked to find eigenvalues, probabilities, and expectation values for \(H\), \(L^2\), and \(L_z\) for a superposition of \(\vert n \ell m \rangle\) states. This can be done on small whiteboards or with the students working in groups on large whiteboards.

Students then work together in small groups to find the matrices that correspond to \(H\), \(L^2\), and \(L_z\) and to redo \(\langle E\rangle\) in matrix notation.

First, students are shown diagrams of cylindrical and spherical coordinates. Common notation systems are discussed, especially that physicists and mathematicians use opposite conventions for the angles \(\theta\) and \(\phi\). Then students are asked to check their understanding by sketching several coordinate equals constant surfaces on their small whiteboards.

• Students evaluate two given partial derivatives from a system of equations.
• Students learn/review generalized Leibniz notation.
• Students may find it helpful to use a chain rule diagram.

A short improvisational role-playing skit based on the Star Trek series in which students explore the definition and notation for position vectors, the importance of choosing an origin, and the geometric nature of the distance formula. \[\vert\vec{r}-\vec{r}^\prime\vert=\sqrt{(x-x^\prime)^2+(y-y^\prime)^2-(z-z^\prime)^2}\]

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