Let the Hamiltonian for two spin-$\frac{1}{2}$ particles of equal masses $m$, momenta $\vec{p}_1$ and $\vec{p}_2$ and positions $\vec{r}_1$ and $\vec{r}_2$ be $H = \frac{1}{2m}p_1^2 + \frac{1}{2m}p_2^2 + \frac{1}{2}m\omega^2 (r_1^2 + r_2^2)+ k\vec{\sigma}_1 \cdot \vec{\sigma}_2$, where $\vec{\sigma}_1$ and $\vec{\sigma}_2$ denote the corresponding Pauli matrices, $\hbar\omega = 0.1 \text{ eV}$ and $k = 0.2 \text{ eV}$. If the ground state has net spin zero, then the energy (in eV) is ________
Hamiltonian Components Analysis
The given Hamiltonian $H$ describes a system of two spin-$\frac{1}{2}$ particles. It can be separated into spatial and spin parts:
$H = H_{spatial} + H_{spin}$
- $H_{spatial} = \frac{p_1^2}{2m} + \frac{1}{2}m\omega^2 r_1^2 + \frac{p_2^2}{2m} + \frac{1}{2}m\omega^2 r_2^2$. This represents two independent 3D quantum harmonic oscillators.
- $H_{spin} = k\vec{\sigma}_1 \cdot \vec{\sigma}_2$. This is the spin-spin interaction term.
Spatial Ground State Energy Calculation
Each 3D quantum harmonic oscillator has a ground state energy of $E_0 = \frac{3}{2}\hbar\omega$.
For two independent oscillators, the total spatial ground state energy is the sum of their individual ground state energies:
$E_{spatial, ground} = \frac{3}{2}\hbar\omega + \frac{3}{2}\hbar\omega = 3\hbar\omega$.
Spin Interaction Energy Determination
The spin interaction term is $H_{spin} = k\vec{\sigma}_1 \cdot \vec{\sigma}_2$.
The eigenvalues of the operator $\vec{\sigma}_1 \cdot \vec{\sigma}_2$ for two spin-$\frac{1}{2}$ particles are:
- $+1$ for triplet states (total spin $S=1$).
- $-3$ for the singlet state (total spin $S=0$).
The question specifies that the ground state has net spin zero, which corresponds to the singlet state.
Therefore, the spin energy for the ground state is:
$E_{spin, ground} = k \times (-3) = -3k$.
Total Ground State Energy Calculation
The total energy of the ground state is the sum of the spatial ground state energy and the spin ground state energy:
$E_{total} = E_{spatial, ground} + E_{spin, ground}$
$E_{total} = 3\hbar\omega - 3k$.
Final Energy Value
Substitute the given values: $\hbar\omega = 0.1 \text{ eV}$ and $k = 0.2 \text{ eV}$.
$E_{total} = 3 \times (0.1 \text{ eV}) - 3 \times (0.2 \text{ eV})$
$E_{total} = 0.3 \text{ eV} - 0.6 \text{ eV}$
$E_{total} = -0.3 \text{ eV}$.
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?
