Question 12.5: Repeat Example 12.1 for an electron with spin down along the...

For an electron with spin-down along the y direction (Problem 4.32(a)):

|\Psi\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ -i \end{array}\right) ; \quad\langle\Psi|=\frac{1}{\sqrt{2}}\left(\begin{array}{ll} 1 & i \end{array}\right) .

So

\rho_{11}=\frac{1}{2}\left[\left(\begin{array}{ll} 1 & 0 \end{array}\right)\left(\begin{array}{c} 1 \\ -i \end{array}\right)\right]\left[\left(\begin{array}{ll} 1 & i \end{array}\right)\left(\begin{array}{l} 1 \\ 0 \end{array}\right)\right]=\frac{1}{2}(1)(1)=\frac{1}{2} .

\rho_{12}=\frac{1}{2}\left[\left(\begin{array}{ll} 1 & 0 \end{array}\right)\left(\begin{array}{c} 1 \\ -i \end{array}\right)\right]\left[\left(\begin{array}{ll} 1 & i \end{array}\right)\left(\begin{array}{l} 0 \\ 1 \end{array}\right)\right]=\frac{1}{2}(1)(i)=\frac{i}{2} .

\left.\rho_{21}=\frac{1}{2}\left[\left(\begin{array}{ll} 0 & 1 \end{array}\right)\left(\begin{array}{c} 1 \\ -i \end{array}\right)\right]\left[\begin{array}{ll} 1 & i\end{array}\right)\left(\begin{array}{l} 1 \\ 0 \end{array}\right)\right]=\frac{1}{2}(-i)(1)=-\frac{i}{2} .

\rho_{22}=\frac{1}{2}\left[\left(\begin{array}{ll} 0 & 1 \end{array}\right)\left(\begin{array}{c} 1 \\ -i \end{array}\right)\right]\left[\left(\begin{array}{ll} 1 & i \end{array}\right)\left(\begin{array}{l} 0 \\ 1 \end{array}\right)\right]=\frac{1}{2}(-i)(i)=\frac{1}{2} .

and

\rho=\frac{1}{2}\left(\begin{array}{cc} 1 & i \\ -i & 1 \end{array}\right) .

Or, more efficiently,

\rho=|\Psi\rangle\langle\Psi|=\frac{1}{2}\left(\begin{array}{c} 1 \\ -i \end{array}\right)\left(\begin{array}{ll} 1 & i \end{array}\right)=\frac{1}{2}\left(\begin{array}{cc} 1 & i \\ -i & 1 \end{array}\right) .