Spin Eigenvalue Calculation
The given operator is $\hat{A} = \lambda \vec{\sigma} \cdot \vec{B}$.
The vector is specified as $\vec{B} = \frac{B}{\sqrt{2}}(\hat{x} + \hat{y})$.
The Pauli matrices are:
- $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
- $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$
- $\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
The components of $\vec{B}$ are $B_x = \frac{B}{\sqrt{2}}$, $B_y = \frac{B}{\sqrt{2}}$, and $B_z = 0$.
Operator Matrix Form
We first compute the term $\vec{\sigma} \cdot \vec{B}$:
$ \vec{\sigma} \cdot \vec{B} = \sigma_x B_x + \sigma_y B_y + \sigma_z B_z $
$ \vec{\sigma} \cdot \vec{B} = \sigma_x \left(\frac{B}{\sqrt{2}}\right) + \sigma_y \left(\frac{B}{\sqrt{2}}\right) + \sigma_z (0) $
$ \vec{\sigma} \cdot \vec{B} = \frac{B}{\sqrt{2}} (\sigma_x + \sigma_y) $
$ \vec{\sigma} \cdot \vec{B} = \frac{B}{\sqrt{2}} \left( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \right) $
$ \vec{\sigma} \cdot \vec{B} = \frac{B}{\sqrt{2}} \begin{pmatrix} 0 & 1-i \\ 1+i & 0 \end{pmatrix} $
Now, the operator $\hat{A}$ is:
$ \hat{A} = \lambda \vec{\sigma} \cdot \vec{B} = \frac{\lambda B}{\sqrt{2}} \begin{pmatrix} 0 & 1-i \\ 1+i & 0 \end{pmatrix} $
Eigenvalue Determination
To find the eigenvalues $\epsilon$, we solve the characteristic equation $\det(\hat{A} - \epsilon I) = 0$.
$ \det \left( \frac{\lambda B}{\sqrt{2}} \begin{pmatrix} 0 & 1-i \\ 1+i & 0 \end{pmatrix} - \epsilon \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) = 0 $
$ \det \begin{pmatrix} -\epsilon & \frac{\lambda B (1-i)}{\sqrt{2}} \\ \frac{\lambda B (1+i)}{\sqrt{2}} & -\epsilon \end{pmatrix} = 0 $
$ (-\epsilon)(-\epsilon) - \left( \frac{\lambda B (1-i)}{\sqrt{2}} \right) \left( \frac{\lambda B (1+i)}{\sqrt{2}} \right) = 0 $
$ \epsilon^2 - \frac{\lambda^2 B^2}{2} (1-i)(1+i) = 0 $
Since $(1-i)(1+i) = 1 - i^2 = 1 - (-1) = 2$, the equation becomes:
$ \epsilon^2 - \frac{\lambda^2 B^2}{2} (2) = 0 $
$ \epsilon^2 - \lambda^2 B^2 = 0 $
$ \epsilon^2 = \lambda^2 B^2 $
$ \epsilon = \pm \lambda B $
The eigenvalues of $\hat{A}$ are $\pm \lambda B$.
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?
