The value of $\text{Tr}(\sigma_z [\sigma_x, \sigma_y])$ is
Pauli Matrix Commutator Calculation
To find the value of $\text{Tr}(\sigma_z [\sigma_x, \sigma_y])$, we first calculate the commutator $[\sigma_x, \sigma_y]$.
- Compute the product $\sigma_x \sigma_y$: $ \sigma_x \sigma_y = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = \begin{pmatrix} (0)(0) + (1)(i) & (0)(-i) + (1)(0) \\ (1)(0) + (0)(i) & (1)(-i) + (0)(0) \end{pmatrix} = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} $
- Compute the product $\sigma_y \sigma_x$: $ \sigma_y \sigma_x = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} (0)(0) + (-i)(1) & (0)(1) + (-i)(0) \\ (i)(0) + (0)(1) & (i)(1) + (0)(0) \end{pmatrix} = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} $
- Calculate the commutator $[\sigma_x, \sigma_y]$: $ [\sigma_x, \sigma_y] = \sigma_x \sigma_y - \sigma_y \sigma_x = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} - \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} = \begin{pmatrix} i - (-i) & 0 - 0 \\ 0 - 0 & -i - i \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & -2i \end{pmatrix} $
- Note that $[\sigma_x, \sigma_y] = 2i \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = 2i \sigma_z$.
Trace Calculation for $\sigma_z [\sigma_x, \sigma_y]$
Next, we compute the matrix product $\sigma_z [\sigma_x, \sigma_y]$.
- Substitute the result for the commutator: $ \sigma_z [\sigma_x, \sigma_y] = \sigma_z (2i \sigma_z) $
- Use the property that $\sigma_z^2 = I$ (the $2 \times 2$ identity matrix): $ \sigma_z (2i \sigma_z) = 2i \sigma_z^2 = 2i I = 2i \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2i & 0 \\ 0 & 2i \end{pmatrix} $
- Calculate the trace, which is the sum of the diagonal elements: $ \text{Tr}(\sigma_z [\sigma_x, \sigma_y]) = \text{Tr}\begin{pmatrix} 2i & 0 \\ 0 & 2i \end{pmatrix} = 2i + 2i = 4i $
Therefore, the value of $\text{Tr}(\sigma_z [\sigma_x, \sigma_y])$ is $4i$.
Important Questions from Spin Electron Spin Pauli Matrices
- 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?
- 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?
