Problem 1

1.) Compute the eigenstates of the operator $\vec{e}_{n} \cdot \vec{s}$, where
$\vec{e}_{n}=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$ is a unit vector that points in the $\theta, \phi$ direction (we use the physicists standard where $\theta$ is the angle from the vertical and $\phi$ is the polar angle in the $x-y$ plane). Write $\hat{\vec{S}}=\frac{\hbar}{2} \vec{\sigma}$ and solve the problem by diagonalizing the $2 \times 2$ matrix. (remember to normalize your final answer).
Your final answer is a 2 component spinor of the form $\binom{\alpha}{\beta}$. Use only $\cos \frac{\theta}{2}, \sin \frac{\theta}{2}, e^{i \phi}$ and numbers in your final answer. Make sure your final answer is in the form where $\alpha$, the top component of the spinor, is real. (Look up some trig identities if your answer looks complicated; the half angle formulas will be helpful.)
Note that if we examine $$\left(\vec{e}_{n} \cdot \hat{\vec{S}}\right)^{2}=\frac{\hbar^{2}}{4}\left(\vec{e}_{n} \cdot \vec{\sigma}\right)^{2}=\frac{\hbar^{2}}{4} \vec{e}_{n} \cdot \vec{e}_{n}=\frac{\hbar^{2}}{4}$$ we see that the eigenvalues of $\vec{e}_{n} \cdot \hat{\vec{S}}$ must be $\pm \frac{\hbar}{2}$ for any direction $\theta, \phi$ ! If you know about the Stern-Gerlach experiment, this explains why it gives the result it gives.

Problem 2

2.) Derive the matrices corresponding to the operators $\hat{L}_{x}, \hat{L}_{y}$, and $\hat{L}_{z}$ in the $l=1$ angular momentum representation. They satisfy $$\left(L_{i}\right)_{m m^{\prime}}=\hbar\langle l=1, m| \hat{L}_{i}\left|l=1, m^{\prime}\right\rangle=\hbar\left(M_{i}\right)_{m m^{\prime}}$$ with $M$ a dimensionless matrix.
You should find the computation of $L_{z}$ is easiest because the states $|l=1 , m\rangle$ are eigenstates of $\hat{L}_{z}$. You may find using the raising and lowering operators and the fact that $\hat{L}_{ \pm}=\hat{L}_{x} \pm i \hat{L}_{y}$ make your calculations easier. (Use the result for $\hat{L}_{+} |lm\rangle$ etc.) (i.e, $\hat{L}_{+}| 1 0 \rangle = \sqrt{2} \hbar|1 1 \rangle$, etc.)
You should find $M_{x}=\left(\begin{array}{ccc}0 & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}}\end{array}\right) \quad M_{y}=\left(\begin{array}{ccc}0 & -\frac{i}{\sqrt{2}} & 0 \\ \frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\ 0 & \frac{i}{\sqrt{2}} & 0\end{array}\right)\quad \text{and }\\ M_{z}=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{array}\right)$.
Note that because the $\hat{L}_{z}$ eigenvalues are $\hbar, 0$, and $-\hbar$, we have $$\begin{gathered} \left(M_{z}-1\right) M_{z}\left(M_{z}+1\right)=0 \\ \Rightarrow M_{z}\left(M_{z}^{2}-1\right)=0 \\ \text { or } M_{z}^{3}=M_{z} \end{gathered}$$ But since this is true for any direction, we have $M_{i}^{3}=M_{i}$.
Indeed, just like we argued about spin $\frac{1}{2}$ above, we should have $\left(\vec{e}_{n} \cdot \vec{M}\right)^{3}=\left(\vec{e}_{n} \cdot \vec{M}\right)$ with $\vec{e}_{n}=\left(\sin \alpha \cos \beta , \sin \alpha \sin \beta, \cos \alpha\right)$. You now will show this.
First compute $$\vec{e}_{n} \cdot \vec{M}=\left(\begin{array}{ccc} \cos \alpha & \frac{1}{\sqrt{2}} \sin \alpha e^{-i \beta} & 0 \\ \frac{1}{\sqrt{2}} \sin \alpha e^{i \beta} & 0 & \frac{1}{\sqrt{2}} \sin \alpha e^{i \beta} \\ 0 & \frac{1}{\sqrt{2}} \sin \alpha e^{i \beta} & -\cos \alpha \end{array}\right),$$ Then compute $\left(\vec{e}_{n}, \vec{M}\right)^{2}$ and $\left(\vec{e}_{n} \cdot \vec{M}\right)^{3}$ to verify

$$\left(\vec{e}_{n} \cdot \vec{M}\right)^{3}=\left(\vec{e}_{n} \cdot \vec{M}\right).$$

Use this result to show that $$\begin{aligned} & \exp [i \vec{v} \cdot \vec{M}]=\sum_{n=0}^{\infty} \frac{(i)^{n}}{n!}|v|^{n}\left(\vec{e}_{v} \cdot \vec{M}\right)^{n}\\ & \left.= \mathbb {1} + i \sin |v|\left(\vec{e}_{\sigma} \cdot \vec{M}\right)+(\cos |v|-1) \mid \vec{e}_{v} \cdot \vec{M}\right)^{2}, \end{aligned}$$ with $\vec{e}_{v}=\frac{\vec{v}}{|v|} = (\sin \alpha \cos \beta , \sin \alpha \sin \beta , \cos \alpha)$.

This is another case where we can explicitly compute the exponential of a matrix. If you wish to try, it does not work for any higher angular momentum.

Problem 3

3.) Using what you know about exponentials of operators $e^{\hat{A}} e^{\hat{B}}$ show that, in general, we have

$$e^{i \vec{v} \cdot \vec{\sigma}} e^{i \vec{v}^{\prime} \cdot \vec{\sigma}} \neq e^{i\left(\vec{v}+\vec{v}^{\prime}\right) \cdot \sigma}$$ Under what circumstances are they equal (this will be a relation between $\vec{v}$ and $\vec{v}^{\prime}$)?
Hint: Consider BCH for Pauli matrices; do not try to multiply the matrices for $e^{i \vec{v} \cdot \vec{\sigma}}$ and $e^{i \vec{v}^{\prime} \cdot \vec{\sigma} }$.

Problem 4

4.) Working with the $l=1$ angular momentum matrices, compute $e^{-i \theta M_{z}} M_{i} e^{i \theta M_z}$. Use the Hadamard relation (which holds for matrices). Note that the commutators never terminate, but they do eventually repeat in a pattern. Determine what the pattern yields in terms of trig functions.

Problem 5

5.) Consider the symplectic group algebra

$$\left[K_{0}, K_{ \pm}\right]= \pm K_{ \pm} \quad\left[K_{+}, K_{-}\right]=-2 K_{0}$$ This is the same as the SU(2) algebra, but there is a minus sign on the $K_{0}$ operator.
Verify that $K_{+}=\left(\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right) \quad K_{-}=\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right)$ and $K_{0}=\frac{1}{2}\left(\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right)$ satisfy the above algebra.
Compute $\exp \left(-\xi K_{+}+2 i \eta K_{0}+\xi^{*} K_{-}\right)$, where $\xi$ and $\eta$ are complex numbers.
Hint: First compute $\left(-\xi K_{+}+2 i \eta K_{0}+\xi^{*} K_{-}\right)^{2}$ and use that result to simplify your work. Review hyperbolic functions if the power series are unfamiliar. Your final result will have the form

$$\left(\begin{array}{ll} K^{*} & \lambda^{*} \\ \lambda & K \end{array}\right) \quad \text { ($K$ and } \lambda \text { are functions of } \xi \text { and } \eta)$$

Factorize this to show the exponential disentangling identity for the symplectic group given by $$\exp \left[-\xi K_{+}+2 i \eta K_{0}+\xi^{*} K_{-}\right] =e^{-\frac{\lambda}{K^{*}} K_{+}} e^{-2 \ln K^{*} K_{0}} e^{\frac{\lambda^{*}}{K^{*}} K_{-}}.$$

Problem 6

6.) In lecture 2, we derived the the following simplified BCH formula $$e^{\hat{A}} e^{\hat{B}}=e^{\hat{A}+\hat{B}+\frac{1}{2}[\hat{A}, \hat{B}]+\frac{1}{12}[\hat{A}, [\hat{A}, \hat{B}]]+\frac{1}{12}[\hat{B},[\hat{B}, \hat{A}]]},$$ which is exact if $$\begin{aligned} & [\hat{A}, [\hat{A},[\hat{A}, \hat{B}]]] = 0 \\ & [\hat{B}, [\hat{A},[\hat{A}, \hat{B}]]] = 0 \\ & [\hat{A}, [\hat{B},[\hat{B}, \hat{A}]]] = 0 \\ & [\hat{B}, [\hat{B},[\hat{B}, \hat{A}]]] = 0. \end{aligned}$$

We want to re-express this in a different form.
Let $\hat{X}=\hat{A}$ and $\hat{Y}=\hat{B}+\frac{1}{2}[\hat{A}, \hat{B}]$
Then $$\begin{aligned} \hat{A}=\hat{X} \text { and } \hat{B} & =\hat{Y}-\frac{1}{2}[\hat{A}, \hat{B}] \\ & =\hat{Y}-\frac{1}{2}\left[\hat{X}, \hat{Y}-\frac{1}{2}[\hat{A}, \hat{B}]\right] \\ & =\hat{Y}-\frac{1}{2}[\hat{X}, \hat{Y}]+\frac{1}{4}[\hat{X},[\hat{X}, \hat{Y}]] \end{aligned}$$ since higher-order terms vanish.
Rearrange the BCH formula to its equivalent form $$\begin{array}{r} e^{\hat{X}} e^{\hat{Y}} e^{-\frac{1}{2}[\hat{X}, \hat{Y}]} e^{-\frac{1}{3}[\hat{Y},[\hat{Y}, \hat{X}]]} e^{\frac{1}{6}[\hat{X},[\hat{X}, \hat{Y}]]} \\ =e^{\hat{X}+\hat{Y}} \end{array}$$ (show your work, and recall $[\hat{X}[\hat{X}, \hat{Y}]]$ and $[\hat{Y},[\hat{Y}, \dot{X}]]$ commute with everything.)
Now consider the time evolution of a particle mowing in a linear potential with

$$\hat{H}=\frac{\hat{p}^{2}}{2 m}+F \hat{x} \quad \text { (a gravitational potential) }$$ The time evolution operator is $e^{-i \hat{H} t}=e^{-it[\frac{\hat{p}^{2}}{2 m}+F \hat{x}]}$. Using the notation from earlier in the problem, pick $\hat{X}=-i t \frac{\hat{p}^{2}}{2 m} \quad \hat{Y}=-i t F \hat{x} \quad$, with $[\hat{x}, \hat{p}]=i \hbar$. Use the BCH formula you derived above to compute a factorized form of $e^{-i \hat{H} t}$. Your answer will have four factors in it. Be careful. The order of the factors matters.