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
- 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
