Question 12.7: (a) Construct the density matrix for an electron that is eit...

(a) From Example 12.1, the density matrix for an electron in the state spin up along x is

\rho_{x+}=\left(\begin{array}{ll} 1 / 2 & 1 / 2 \\ 1 / 2 & 1 / 2 \end{array}\right) ;

from Problem 12.5 the density matrix for an electron in the state spin down along y is

\rho_{y-}=\left(\begin{array}{cc} 1 / 2 & i / 2 \\ -i / 2 & 1 / 2 \end{array}\right) .

Therefore the density matrix for the state in question is

\rho=\frac{1}{3} \rho_{x+}+\frac{2}{3} \rho_{y-}=\left(\begin{array}{cc} 1 / 2 & 1 / 6+i / 3 \\ 1 / 6-i / 3 & 1 / 2 \end{array}\right) .

(b) From Equation 4.147,

S _{x}=\frac{\hbar}{2}\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right), \quad S _{y}=\frac{\hbar}{2}\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)                (4.147).

S _{y}=\frac{\hbar}{2}\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right) .

so

\left\langle S_{y}\right\rangle=\operatorname{Tr}\left(\rho S _{y}\right)=\frac{\hbar}{2} \operatorname{Tr}\left[\left(\begin{array}{cc} 1 / 2 & 1 / 6+i / 3 \\ 1 / 6-i / 3 & 1 / 2 \end{array}\right)\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)\right]=\frac{\hbar}{2} \operatorname{Tr}\left(\begin{array}{cc} i / 6-1 / 3 & -i / 2 \\ i / 2 & -i / 6-1 / 3 \end{array}\right) .