Hamiltonian Matrix Representation
The given Hamiltonian is $H = a_0 I + \vec{b} \cdot \vec{\sigma}$.
Substituting the Pauli matrices ($\sigma_x, \sigma_y, \sigma_z$) and the identity matrix ($I$), the Hamiltonian can be written as a $2 \times 2$ matrix:
$ H = a_0 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + b_x \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + b_y \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} + b_z \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $ $ H = \begin{pmatrix} a_0 + b_z & b_x - i b_y \\ b_x + i b_y & a_0 - b_z \end{pmatrix} $Let $|b| = \sqrt{b_x^2 + b_y^2 + b_z^2}$. The matrix can also be written as:
$ H = \begin{pmatrix} a_0 & 0 \\ 0 & a_0 \end{pmatrix} + \begin{pmatrix} b_z & b_+ \\ b_+^* & -b_z \end{pmatrix} = \begin{pmatrix} a_0 + b_z & b_+ \\ b_+^* & a_0 - b_z \end{pmatrix} $where $b_+ = b_x + i b_y$.
Eigenvalue Calculation
The energy eigenvalues ($E$) are found by solving the characteristic equation $\det(H - E I) = 0$.
$ \det \begin{pmatrix} a_0 + b_z - E & b_x - i b_y \\ b_x + i b_y & a_0 - b_z - E \end{pmatrix} = 0 $Expanding the determinant:
$ (a_0 + b_z - E)(a_0 - b_z - E) - (b_x - i b_y)(b_x + i b_y) = 0 $Simplify the terms:
$ ((a_0 - E) + b_z)((a_0 - E) - b_z) - (b_x^2 - (i b_y)^2) = 0 $ $ (a_0 - E)^2 - b_z^2 - (b_x^2 + b_y^2) = 0 $Combine the vector components:
$ (a_0 - E)^2 - (b_x^2 + b_y^2 + b_z^2) = 0 $ $ (a_0 - E)^2 - |b|^2 = 0 $Solve for $E$:
$ (a_0 - E)^2 = |b|^2 $ $ a_0 - E = \pm |b| $ $ E = a_0 \mp |b| $The two possible energy eigenvalues are $E_1 = a_0 - |b|$ and $E_2 = a_0 + |b|$.
Ground State Energy
The ground state energy corresponds to the lowest eigenvalue.
Since $|b| \ge 0$, we have $a_0 - |b| \le a_0 + |b|$.
Therefore, the ground state energy is $E_{ground} = a_0 - |b|$.
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
