Physics Problems & Solutions: Quantum Mechanics

The state of a spin ½ particle can be represented using the eigenstate $\mid\uparrow\rangle$ and $\mid\downarrow\rangle$ of the $S_z$ operator.

$\\S_z \mid\uparrow\rangle = \frac{1}{2}\hbar \mid\uparrow\rangle \\S_z \mid\downarrow\rangle = -\frac{1}{2}\hbar \mid\downarrow\rangle$

Given the Pauli matrix $\sigma_x=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ which of the following is an eigenstate of $S_x$ with eigenvalue $-\frac{1}{2} \hbar$ ?

$\\$A.$ \hspace{5 mm} \mid\uparrow\rangle \\$B.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle+\mid\downarrow\rangle\right ) \\$C.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle-\mid\downarrow\rangle\right ) \\$D.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle+i\mid\downarrow\rangle\right ) \\$E.$ \hspace{5 mm} \frac{1}{\sqrt{2}} \left( \mid\uparrow\rangle-i\mid\downarrow\rangle\right )$

(GR0177 #83)

Solution:

Spin Operator:    $S_x=\frac{\hbar}{2}\sigma_x=\frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Eigenvalue:    $\alpha=-\frac{\hbar}{2}$

Eigenfunction:    $S_x\chi =\alpha \chi$

$\begin{aligned} \frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}&=\begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix} \\&=\begin{bmatrix} -\hbar/2 & 0 \\ 0 & -\hbar/2 \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix} \end{aligned}$

$\begin{aligned} \left(\begin{bmatrix} 0 & \hbar/2 \\ \hbar/2 & 0 \end{bmatrix}-\begin{bmatrix} -\hbar/2 & 0 \\ 0 & -\hbar/2 \end{bmatrix} \right ) \begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}&=0 \\\frac{\hbar}{2}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}&=0 \end{aligned}$

$\begin{aligned} \chi_1&=-\chi_2\\ \chi&=\begin{bmatrix} \chi_1 \\ \chi_2 \end{bmatrix}= \chi_1\begin{bmatrix} 1 \\ -1 \end{bmatrix} \end{aligned}$

Normalized $\chi$ :
$\chi_1^2+\chi_1^2=1\rightarrow 2\chi_1^2=1\rightarrow\chi_1=\frac{1}{\sqrt{2}}$
$\chi= \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Answer: C

Physics Problems & Solutions

Quantum Mechanics - Spin operator

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