这是一份ucl伦敦大学学院 PHAS0069作业代写的成功案
In the two-dimensional subspace of $\mathscr{H}$, and using the basis of eigenstates of $s_{3}=\sigma_{3} / 2$, define the density matrix
$$
\varrho^{(0)}=\left(\begin{array}{cc}
w_{+} & 0 \
0 & w_{-}
\end{array}\right)=\frac{1}{2}\left(\mathbb{1}+\zeta \sigma_{3}\right) \quad \text { with } \quad \zeta=w_{+}-w_{-}
$$
By definition $w_{+}+w_{-}=1$ (this relation was used above), both numbers being real. The expectation values of the components of the spin operator are found to be
$$
\left\langle\mathbf{s}{1}\right\rangle=0=\left\langle\mathbf{s}{2}\right\rangle, \quad\left\langle\mathbf{s}{3}\right\rangle=\operatorname{tr}\left(e^{(0)} \frac{\sigma{3}}{2}\right)=\frac{1}{2} \zeta .
$$
In calculating these traces use was made of the formulae $\operatorname{tr} \sigma_{i}=0, \quad \operatorname{tr}\left(\sigma_{i} \sigma_{k}\right)=2 \delta_{i k}$
PHAS0042 COURSE NOTES ：
$$
q(t)=\beta+\frac{\alpha}{m} t,
$$
so that the flunction $S$ and its time derivative are
$$
\begin{aligned}
&S(\boldsymbol{q}, \boldsymbol{\alpha}, t)=\frac{\alpha^{2}}{2 m} t+\boldsymbol{\alpha} \cdot \boldsymbol{\beta}+c, \
&\frac{\mathrm{d} S(\boldsymbol{q}, \boldsymbol{\alpha}, t)}{\mathrm{d} t}=\boldsymbol{\alpha} \cdot \dot{\boldsymbol{q}}-\frac{1}{2 m} \alpha^{2}=\frac{\alpha^{2}}{2 m} .
\end{aligned}
$$