Compute the eigenstates of the operator \(\mathbf{e}_{n} \cdot \mathbf{s}\), where
\(\mathbf{e}_{n}=(\sin \theta \cos \phi, \sin
\theta \sin \phi, \cos \theta)\) is a unit vector that points in
the \(\theta, \phi\) direction (we use
the physicist’s standard where \(\theta\) is the angle from the vertical and
\(\phi\) is the polar angle in the
\(x-y\) plane). Write \(\hat{\mathbf{S}}=\frac{\hbar}{2}
\mathbf{\sigma}\) 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 trigonometric identities if your answer
looks complicated; the half angle formulas will be helpful.)
Note that if we examine \[\left(\mathbf{e}_{n} \cdot
\hat{\mathbf{S}}\right)^{2}=\frac{\hbar^{2}}{4}\left(\mathbf{e}_{n}
\cdot \mathbf{\sigma}\right)^{2}=\frac{\hbar^{2}}{4} \mathbf{e}_{n}
\cdot \mathbf{e}_{n}=\frac{\hbar^{2}}{4}\] we see that the
eigenvalues of \(\mathbf{e}_{n} \cdot
\hat{\mathbf{S}}\) 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.
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 \(M_{x}=\left(\begin{array}{ccc}0 &
\frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 &
\frac{1}{\sqrt{2}} \\ 0 &
\frac{1}{\sqrt{2}}&0\end{array}\right), \quad
M_{y}=\left(\begin{array}{ccc}0 & -\frac{i}{\sqrt{2}} & 0 \\
\frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\ 0 &
\frac{i}{\sqrt{2}} & 0\end{array}\right),\quad \text{and }\\
M_{z}=\left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 0 & 0 \\
0 & 0 & -1\end{array}\right)\).
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 \(\left(\mathbf{e}_{n} \cdot
\mathbf{M}\right)^{3}=\left(\mathbf{e}_{n} \cdot
\mathbf{M}\right)\) with \(\mathbf{e}_{n}=\left(\sin \alpha \cos \beta , \sin
\alpha \sin \beta, \cos \alpha\right)\). You now will show
this.
First compute \[\mathbf{e}_{n} \cdot
\mathbf{M}=\left(\begin{array}{ccc}
\cos \alpha & \frac{1}{\sqrt{2}} \sin \alpha e^{-i \beta} & 0 \\
\frac{1}{\sqrt{2}} \sin \alpha e^{i \beta} & 0 &
\frac{1}{\sqrt{2}} \sin \alpha e^{i \beta} \\
0 & \frac{1}{\sqrt{2}} \sin \alpha e^{i \beta} & -\cos \alpha
\end{array}\right),\] Then compute \(\left(\mathbf{e}_{n},
\mathbf{M}\right)^{2}\) and \(\left(\mathbf{e}_{n} \cdot
\mathbf{M}\right)^{3}\) to verify
\[\left(\mathbf{e}_{n} \cdot \mathbf{M}\right)^{3}=\left(\mathbf{e}_{n} \cdot \mathbf{M}\right).\]
Use this result to show that \[\begin{aligned} & \exp [i \mathbf{v} \cdot \mathbf{M}]=\sum_{n=0}^{\infty} \frac{(i)^{n}}{n!}|v|^{n}\left(\mathbf{e}_{v} \cdot \mathbf{M}\right)^{n}\\ & \left.= \mathbb{I} + i \sin |v|\left(\mathbf{e}_{\sigma} \cdot \mathbf{M}\right)+(\cos |v|-1) ( \mathbf{e}_{v} \cdot \mathbf{M}\right)^{2}, \end{aligned}\] with \(\mathbf{e}_{v}=\frac{\mathbf{v}}{|v|} = (\sin \alpha \cos \beta , \sin \alpha \sin \beta , \cos \alpha)\).
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.
Using what you know about exponentials of operators \(e^{\hat{A}} e^{\hat{B}}\) show that, in general, we have
\[e^{i \mathbf{v} \cdot \mathbf{\sigma}}
e^{i \mathbf{v}^{\prime} \cdot \mathbf{\sigma}} \neq
e^{i\left(\mathbf{v}+\mathbf{v}^{\prime}\right) \cdot \sigma}\]
Under what circumstances are they equal (this will be a relation between
\(\mathbf{v}\) and \(\mathbf{v}^{\prime}\))?
Hint: Consider BCH for Pauli matrices; do not try to multiply
the matrices for \(e^{i \mathbf{v} \cdot
\mathbf{\sigma}}\) and \(e^{i
\mathbf{v}^{\prime} \cdot \mathbf{\sigma} }\).
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 trigonometric functions.
Consider the symplectic group algebra
\[\left[K_{0}, K_{ \pm}\right]= \pm K_{
\pm}\quad\text{and} \quad\left[K_{+}, K_{-}\right]=-2 K_{0}.\]
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 is given by \[\exp
\left[-\xi K_{+}+2 i \eta K_{0}+\xi^{*} K_{-}\right]
=e^{-\frac{\lambda}{K^{*}} K_{+}} e^{-2 \ln (K^{*}) K_{0}}
e^{\frac{\lambda^{*}}{K^{*}} K_{-}}.\]
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.