Consider a spin $S = \hbar/2$ particle in the state $| \phi \rangle = \frac{1}{3} \begin{bmatrix}2 + i \\2 \end{bmatrix} $. The probability that a measurement finds the state with $S_x = + \hbar/2$ is
Spin State Vector
The given quantum state for the spin $S = \hbar/2$ particle is:
$|\phi\rangle = \frac{1}{3} \begin{pmatrix} 2+i \\ 2 \end{pmatrix}$
We verify the normalization of the state vector:
$ \langle \phi | \phi \rangle = \left( \frac{1}{3} \begin{pmatrix} 2-i & 2 \end{pmatrix} \right) \left( \frac{1}{3} \begin{pmatrix} 2+i \\ 2 \end{pmatrix} \right) $
$ = \frac{1}{9} \left( (2-i)(2+i) + (2)(2) \right) = \frac{1}{9} \left( (4 - i^2) + 4 \right) $
$ = \frac{1}{9} \left( (4 - (-1)) + 4 \right) = \frac{1}{9} (5 + 4) = 1 $
The state vector $|\phi\rangle$ is correctly normalized.
$S_x$ Eigenstate for $+\hbar/2$
We need the eigenstate of the $S_x$ operator corresponding to the eigenvalue $S_x = +\hbar/2$. This state is given by:
$|s_x = +\hbar/2\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$
The corresponding bra vector is required for the projection:
$\langle s_x = +\hbar/2 | = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \end{pmatrix}$
$S_x$ Measurement Probability
The probability $P$ of finding the system in the state with $S_x = +\hbar/2$ upon measurement is calculated as the squared magnitude of the projection of $|\phi\rangle$ onto $|s_x = +\hbar/2\rangle$:
$P = |\langle s_x = +\hbar/2 | \phi \rangle|^2$
First, compute the inner product $\langle s_x = +\hbar/2 | \phi \rangle$:
$ \langle s_x = +\hbar/2 | \phi \rangle = \left( \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \right) \left( \frac{1}{3} \begin{pmatrix} 2+i \\ 2 \end{pmatrix} \right) $
$ = \frac{1}{3\sqrt{2}} \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 2+i \\ 2 \end{pmatrix} = \frac{1}{3\sqrt{2}} (1 \cdot (2+i) + 1 \cdot 2) $
$ = \frac{1}{3\sqrt{2}} (2+i+2) = \frac{4+i}{3\sqrt{2}} $
Next, find the squared magnitude of this inner product to get the probability:
$ P = \left| \frac{4+i}{3\sqrt{2}} \right|^2 = \frac{|4+i|^2}{|3\sqrt{2}|^2} $
$ = \frac{4^2 + 1^2}{(3\sqrt{2})^2} = \frac{16 + 1}{9 \times 2} = \frac{17}{18} $
The probability that a measurement finds the state with $S_x = +\hbar/2$ is $\frac{17}{18}$.
Important Questions from Spin Electron Spin Pauli Matrices
- Atomic numbers of V, Cr, Fe and Zn are 23, 24, 26 and 30, respectively. Which one of the following materials does NOT show an electron spin resonance (ESR) spectra?
Consider two non-identical spin $\frac{1}{2}$ particles labelled $1$ and $2$ in the spin product state $|\frac{1}{2}, \frac{1}{2}\rangle_1 |\frac{1}{2}, -\frac{1}{2}\rangle$. The Hamiltonian of the system is
$H = \frac{4\lambda}{\hbar^2} \vec{S}_1 \cdot \vec{S}_2$,
where $\vec{S}_1$ and $\vec{S}_2$ are the spin operators of particles $1$ and $2$, respectively, and $\lambda$ is a constant with appropriate dimensions. What is the expectation value of $H$ in the above state?- A spin $\frac{1}{2}$ particle is in a spin up state along the $x$-axis (with unit vector $\hat{x}$) and is denoted as $|\frac{1}{2}, \frac{1}{2}\rangle_x$. What is the probability of finding the particle to be in a spin up state along the direction $\hat{x}'$, which lies in the $xy$-plane and makes an angle $\theta$ with respect to the positive $x$-axis, if such a measurement is made?
- Pauli spin matrices satisfy
An electron with mass $m$ and charge $q$ is in the spin up state $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ at time $t = 0$. A constant magnetic field is applied along the y-axis, $\vec{B} = B_0 \hat{j}$, where $B_0$ is a constant. The Hamiltonian of the system is $H = -\hbar \omega \sigma_y$, where $\omega = \frac{q B_0}{2m} > 0$ and $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$. The minimum time after which the electron will be in the spin down state along the x-axis, i.e., $\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$, is
