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
- 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?
