Activities
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: \[\left\langle {\hat{x}}\middle|{\hat{s}}\right\rangle ,\quad \left\langle {\hat{y}}\middle|{{\hat{s}}}\right\rangle ,\quad \left\langle {\hat{x}}\middle|{\hat{\phi}}\right\rangle ,\quad \left\langle {\hat{y}}\middle|{\hat{\phi}}\right\rangle \]
- 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: \[\left|{\hat{x}}\right\rangle \left\langle {\hat{x}}\right| + \left|{\hat{y}}\right\rangle \left\langle {\hat{y}}\right| = 1\]
- 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 \[\left|{\Psi}\right\rangle = j\left|{+}\right\rangle _y + k\left|{-}\right\rangle _y\]
Student explore the properties of an orthonormal basis using the Cartesian and \(S_z\) bases as examples.
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 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).
Consider the region \(D\) in the \(xy\)-plane shown below, which is bounded by \[u=9 \qquad u=36 \qquad v=1 \qquad v=4\] where \[u=xy \qquad v={y\over x}\] If you want to determine \(x\) and \(y\) as functions of \(u\) and \(v\), consider \(uv\) and \(u/v\).
List as many methods as you can think of for finding the area of the given region.
It is enough to refer to the methods by name or describe them briefly.
For at least 3 of these methods, give explicitly the formulas you would use to find the area.
You must put limits on your integrals, but you do not need to evaluate them.
Using any 2 of these methods, find the area.
One of these should be a method we have learned recently.
Now consider the following integral over the same region \(D\):
\(\int\!\!\int_D {y\over x} \>\>dA\)
- Which of the above methods can you use to do this integral?
- Do the integral.
Main ideas
- There are many ways to solve this problem!
- Using Jacobians (and inverse Jacobians)
Prerequisites
- Surface integrals
- Jacobians
- Green's/Stokes' Theorem
Warmup
Perhaps a discussion of single and double integral techniques for solving this problem.
Props
- whiteboards and pens
Wrapup
This is a good conclusion to the course, as it reviews many integration techniques. We emphasize that (2-dimensional) change-of-variable problems are a special case of surface integrals.
Here are some of the methods one could use to do these integrals:
- change of variables (at least 2 ways)
- Area Corollary to Green's Theorem (at least 2 ways)
- ordinary single integral (at least 2 ways)
- ordinary double integral (at least 2 ways)
- surface integral
Details
In the Classroom
- Some students will want to simply use Jacobian formulas; encourage such students to try to solve this problem both by computing \(\frac{\partial(x,y)}{\partial(u,v)}\) and by computing \(\frac{\partial(u,v)}{\partial(x,y)}\).
- Other students will want to work directly with \(d\boldsymbol{\vec{r}}_1\) and \(d\boldsymbol{\vec{r}}_2\). This works fine if one first solves for \(x\) and \(y\) in terms of \(u\) and \(v\).
Students who compute \(d\boldsymbol{\vec{r}}_1\) and \(d\boldsymbol{\vec{r}}_2\) directly can easily get confused, since they may try to eliminate \(x\) or \(y\), rather than \(u\) or \(v\).
Along the curve \(v=\hbox{constant}\), one has \(dy=v\,dx\), so that \(d\boldsymbol{\vec{r}}_1 = dx\,\boldsymbol{\hat{x}} + dy\,\boldsymbol{\hat{y}} = (\boldsymbol{\hat{x}} + v\,\boldsymbol{\hat{y}})\,dx\), which some students will want to write in terms of \(x\) alone. But one needs to express this in terms of \(du\)! This can be done using \(du = x\,dy + y\,dx = x (v\,dx) + y\,dx = 2y\,dx\), so that \(d\boldsymbol{\vec{r}}_1 = (\boldsymbol{\hat{x}} + v\,\boldsymbol{\hat{y}}) \,\frac{du}{2y}\). A similar argument leads to \(d\boldsymbol{\vec{r}}_2 = (-\frac{1}{v}\,\boldsymbol{\hat{x}}+\boldsymbol{\hat{y}})\,\frac{x\,dv}{2}\) for \(u=\hbox{constant}\), so that \(d\boldsymbol{\vec{S}} = d\boldsymbol{\vec{r}}_1\times d\boldsymbol{\vec{r}}_2 = \boldsymbol{\hat{z}} \,\frac{x}{2y}\,du\,dv = \boldsymbol{\hat{z}} \,{du\,dv\over2v}\). This calculation can be done without solving for \(x\) and \(y\), provided one recognizes \(v\) in the penultimate expression.
Emphasize that one must choose parameters, both on the region, and on each curve, and that \(u\) and \(v\) are chosen to make the limits easy.
- Take time before the activity to gauge students' recollection of single variable techniques and the Jacobian. After the activity, be sure to set up more than one approach. People will be fine after the first couple of steps but shouldn't leave class feeling stuck.
Subsidiary ideas
- Review of Green's Theorem
- Review of single integral techniques
- Review of double integral techniques
Enrichment
- Discuss the 3-dimensional case, perhaps relating it to volume integrals.
In this activity, students will explore how to calculate a derivative from measured data. Students should have prior exposure to differential calculus. At the start of the activity, orient the students to the contour plot - it's busy.
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.
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 use prepared Sage code or a prepared Mathematica notebook to plot \(\sin\theta\) simultaneously with several terms of a power series expansion to judge how well the approximation fits. Students can alter the worksheet to change the number of terms in the expansion and even to change the function that is being considered. Students should have already calculated the coefficients for the power series expansion in a previous activity, Calculating Coefficients for a Power Series.
Students use Tinker Toys to represent each component in a two-state quantum spin system in all three standard bases (\(x\), \(y\), and \(z\)). Through a short series of instructor-led prompts, students explore the difference between overall phase (which does NOT change the state of the system) and relative phase (which does change the state of the system). This activity is optional in the Arms Sequence Arms Sequence for Complex Numbers and Quantum States.
Students consider the change in internal energy during three different processes involving a container of water vapor on a stove. Using the 1st Law of Thermodynamics, students reason about how the internal energy would change and then compare this prediction with data from NIST presented as a contour plot.
This small group activity is designed to provide practice with the multivariable chain rule. Students determine a particular rate of change using given information involving other rates of change. The discussion emphasizes the equivalence of a variety of approaches, including the use of differentials. Good “review” problem; can also be used as a homework problem.
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 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 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 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.
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 will determine the change in entropy (positive, negative, or none) for both the system and surroundings in three different cases. This is followed by an active whole-class discussion about where the entropy comes from during an irreversible process.
Students work out heat and work for rectangular paths on \(pV\) and \(TS\) plots. This gives with computing heat and work, applying the First Law, and recognizing that internal energy is a state function, which cannot change after a cyclic process.
In this small group activity, students determine various aspects of local points on an elliptic hill which is a function of two variables. The gradient is emphasized as a local quantity which points in the direction of greatest change at a point in the scalar field.
Students work in small groups to use completeness relations to change the basis of quantum states.
In this small group activity, students determine various aspects of local points on an elliptic hill which is a function of two variables. The gradient is emphasized as a local quantity which points in the direction of greatest change at a point in the scalar field.