A spin-half particle is in a linear superposition $0.8|\uparrow\rangle + 0.6|\downarrow\rangle$ of its spin-up and spin-down states. If $|\uparrow\rangle$ and $|\downarrow\rangle$ are the eigenstates of $\sigma_z$ then what is the expectation value, up to one decimal place, of the operator $10\sigma_z + 5\sigma_x$ ? Here, symbols have their usual meanings. ____________
The expectation value of an operator $A$ for a quantum state $|\psi\rangle$ is given by $\langle A \rangle = \langle \psi | A | \psi \rangle$. The given state is $|\psi\rangle = 0.8|\uparrow\rangle + 0.6|\downarrow\rangle$. The operator is $A = 10\sigma_z + 5\sigma_x$. We need to calculate $\langle A \rangle$. Since the expectation value is linear, $\langle A \rangle = 10\langle \sigma_z \rangle + 5\langle \sigma_x \rangle$. The basis states $|\uparrow\rangle$ and $|\downarrow\rangle$ are eigenstates of $\sigma_z$ with eigenvalues $+1$ and $-1$, respectively. In this basis, the Pauli matrices are:
$\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
The state vector in this basis is $|\psi\rangle = \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$. The corresponding bra vector is $\langle\psi| = \begin{pmatrix} 0.8 & 0.6 \end{pmatrix}$.
Calculate $\sigma_z$ Expectation Value
The expectation value of $\sigma_z$ is:
$\langle \sigma_z \rangle = \langle\psi|\sigma_z|\psi\rangle = \begin{pmatrix} 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$
$= \begin{pmatrix} 0.8 & -0.6 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$
$= (0.8)(0.8) + (-0.6)(0.6) = 0.64 - 0.36 = 0.28$
Calculate $\sigma_x$ Expectation Value
The expectation value of $\sigma_x$ is:
$\langle \sigma_x \rangle = \langle\psi|\sigma_x|\psi\rangle = \begin{pmatrix} 0.8 & 0.6 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$
$= \begin{pmatrix} 0.6 & 0.8 \end{pmatrix} \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$
$= (0.6)(0.8) + (0.8)(0.6) = 0.48 + 0.48 = 0.96$
Combined Expectation Value Calculation
Now, we combine the expectation values:
$\langle 10\sigma_z + 5\sigma_x \rangle = 10\langle \sigma_z \rangle + 5\langle \sigma_x \rangle$
$= 10(0.28) + 5(0.96)$
$= 2.8 + 4.8 = 7.6$
The expectation value is $7.6$.
Important Questions from Spin Electron Spin Pauli Matrices
- Consider the Pauli matrices $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$, $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
The value of $\text{Tr}(\sigma_z [\sigma_x, \sigma_y])$ is - The Hamiltonian of two interacting spin-1/2 particles is $H = \frac{A}{\hbar^2} \vec{S}_1 \cdot \vec{S}_2$, where $\vec{S}_1$ and $\vec{S}_2$ are the spin angular momenta of particles 1 and 2, respectively. Here, $A = 10.56 \text{ eV}$. The energy in eV required to induce an excitation from the ground state to the excited state (rounded off to two decimal places) is _____
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
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?
