Activities
Student explore the properties of an orthonormal basis using the Cartesian and \(S_z\) bases as examples.
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).
Students use a completeness relations to write hydrogen atoms states in the energy and position bases.
Students practice using inner products to find the components of the cartesian basis vectors in the polar basis and vice versa. Then, students use a completeness relation to change bases or cartesian/polar bases and for different spin bases.
Students each recall a representation of vectors that they have seen before and record it on an individual whiteboard. The instructor uses these responses to generate a whole class discussion that compares and contrasts the features of the representations. If appropriate for the class, the instructor introduces bra/ket notation as a new, but valuable representation.
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 \(\left|{-}\right\rangle _x\):
\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*}
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 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.
Problem
First complete the problem Diagonalization. In that notation:
- 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?)
- 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.
Students use completeness relations to write a matrix element of a spin component in a different basis.
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, 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.
Kinesthetic
30 min.
Students, working in pairs, represent two component complex vectors with their left arms. Through a short series of instructor led prompts, students move their left arms to show how various linear transformations affect each complex component.
Students learn how to express Angular Momentum as a vector quantity in polar coordinates, and then in Cylindrical and Spherical Coordinates
Problem
Consider the three quantum states: \[\left\vert \psi_1\right\rangle = \frac{1}{\sqrt{3}}\left\vert +\right\rangle+ i\frac{\sqrt{2}}{\sqrt{3}} \left\vert -\right\rangle\] \[\left\vert \psi_2\right\rangle = \frac{1}{\sqrt{5}}\left\vert +\right\rangle- \frac{2}{\sqrt{5}} \left\vert -\right\rangle\] \[\left\vert \psi_3\right\rangle = \frac{1}{\sqrt{2}}\left\vert +\right\rangle+ i\frac{e^{\frac{i\pi}{4}}}{\sqrt{2}} \left\vert -\right\rangle\]
- For each of the \(\vert \psi_i\rangle\) above, find the normalized vector \(\vert \phi_i\rangle\) that is orthogonal to it.
- Calculate the inner products \(\langle \psi_i\vert \psi_j\rangle\) for \(i\) and \(j=1\), \(2\), \(3\).
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.
Let \begin{align} \boldsymbol{\vec a} &= \boldsymbol{\hat x}-3\boldsymbol{\hat y}-\boldsymbol{\hat z}\\ \boldsymbol{\vec b} &= \boldsymbol{\hat x}+\boldsymbol{\hat y}+2\boldsymbol{\hat z}\\ {\boldsymbol{\vec c}} &= -2\boldsymbol{\hat x}-\boldsymbol{\hat y}+\boldsymbol{\hat z}\\ \boldsymbol{\vec d} &= -\boldsymbol{\hat x}-\boldsymbol{\hat y}+\boldsymbol{\hat z} \end{align}
Use the dot product to determine which pairs (if any) of these vectors
- Are perpendicular?
- Are parallel?
- Have an angle less than \(\pi/2\) between them?
- Have an angle of more than \(\pi/2\) between them?
Students consider the dimensions of spin-state kets and position-basis kets.
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 work in small groups to use completeness relations to change the basis of quantum states.
Students use the completeness relation for the position basis to re-express expressions in bra/ket notation in wavefunction notation.
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 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 take the inner product of vectors that lie on the spacetime axis to show that they are orthogonal. To do the inner product, students much use the Minkowski metric.
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. \[\vert\vec{r}-\vec{r}^\prime\vert=\sqrt{(x-x^\prime)^2+(y-y^\prime)^2-(z-z^\prime)^2}\]
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.
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.