title, topic, keyword
Small group, whiteboard, etc
Required in-class time for activities
Leave blank to search both

Activities

Small Group Activity

30 min.

\(|\pm\rangle\) Forms an Orthonormal Basis
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.

Kinesthetic

10 min.

Curvilinear Basis Vectors
Students use their arms to depict (sequentially) the different cylindrical and spherical basis vectors at the location of their shoulder (seen in relation to a specified origin of coordinates: either a set of axes hung from the ceiling of the room or perhaps a piece of furniture or a particular corner of the room).
  1. Given the polar basis kets written as a superposition of Cartesian kets \begin{eqnarray*} \left|{\hat{s}}\right\rangle &=& \cos\phi \left|{\hat{x}}\right\rangle + \sin\phi \left|{\hat{y}}\right\rangle \\ \left|{\hat{\phi}}\right\rangle &=& -\sin\phi \left|{\hat{x}}\right\rangle + \cos\phi \left|{\hat{y}}\right\rangle \end{eqnarray*}

    Find the following quantities: \[\left\langle {\hat{x}}\middle|{\hat{s}}\right\rangle ,\quad \left\langle {\hat{y}}\middle|{{\hat{s}}}\right\rangle ,\quad \left\langle {\hat{x}}\middle|{\hat{\phi}}\right\rangle ,\quad \left\langle {\hat{y}}\middle|{\hat{\phi}}\right\rangle \]

  2. Given a vector written in the polar basis \[\left|{\vec{v}}\right\rangle = a\left|{\hat{s}}\right\rangle + b\left|{\hat{\phi}}\right\rangle \] where \(a\) and \(b\) are known. Find coefficients \(c\) and \(d\) such that \[\left|{\vec{v}}\right\rangle = c\left|{\hat{x}}\right\rangle + d\left|{\hat{y}}\right\rangle \] Do this by using the completeness relation: \[\left|{\hat{x}}\right\rangle \left\langle {\hat{x}}\right| + \left|{\hat{y}}\right\rangle \left\langle {\hat{y}}\right| = 1\]
  3. Using a completeness relation, change the basis of the spin-1/2 state \[\left|{\Psi}\right\rangle = g\left|{+}\right\rangle + h\left|{-}\right\rangle \] into the \(S_y\) basis. In otherwords, find \(j\) and \(k\) such that \[\left|{\Psi}\right\rangle = j\left|{+}\right\rangle _y + k\left|{-}\right\rangle _y\]

Computational Activity

120 min.

Sinusoidal basis set
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.
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.

Small Group Activity

30 min.

Finding Matrix Elements
In this small group activity, students multiply a general 3x3 matrix with standard basis row/column vectors to pick out individual matrix elements. Students generate the expressions for the matrix elements in bra/ket notation.

Computational Activity

120 min.

Position operator
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.

Small Group Activity

30 min.

Outer Product of a Vector on Itself
Students compute the outer product of a vector on itself to product a projection operator. Students discover that projection operators are idempotent (square to themselves) and that a complete set of outer products of an orthonormal basis is the identity (a completeness relation).
Students work in small groups to use completeness relations to change the basis of quantum states.

Small Group Activity

60 min.

Going from Spin States to Wavefunctions
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.