Question

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, \( \mathbf{B} = B_0 \hat{j} \), where \( B_0 \) is a constant. The Hamiltonian of the system is \[ H = -\hbar \omega \sigma_y, \quad \text{where} \quad \omega = \frac{q B_0}{2m} < 0 \quad \text{and} \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}. \]
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:

• A. \( \frac{\pi}{8\omega} \)
• B. \( \frac{\pi}{4\omega} \)
• C. \( \frac{\pi}{2\omega} \)
• D. \( \frac{\pi}{\omega} \)

Hint:

For time evolution of quantum states under a Hamiltonian involving Pauli matrices, the time dependence follows a sinusoidal pattern, with the state oscillating between different spin orientations.

Answer 1

The Correct Option is B

Step 1: The Hamiltonian of the system is given as \( H = -\hbar \omega \sigma_y \), and the electron is initially in the spin-up state \[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \] at \( t = 0 \).

Step 2: The evolution of the spin state is governed by the time evolution operator: \[ |\psi(t)\rangle = e^{-\frac{i}{\hbar} H t} |\psi(0)\rangle. \] For the Hamiltonian \( H = -\hbar \omega \sigma_y \), this becomes: \[ |\psi(t)\rangle = e^{i \omega \sigma_y t} \begin{pmatrix} 1 \\ 0 \end{pmatrix}. \]

Step 3: The exponential of the Pauli matrix \( \sigma_y \) can be written as: \[ e^{i \omega \sigma_y t} = \cos(\omega t) I + i \sin(\omega t) \sigma_y, \] where \( I \) is the identity matrix and \( \sigma_y \) is the Pauli matrix. Applying this to the initial state: \[ |\psi(t)\rangle = \cos(\omega t) \begin{pmatrix} 1 \\ 0 \end{pmatrix} + i \sin(\omega t) \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}. \] This simplifies to: \[ |\psi(t)\rangle = \begin{pmatrix} \cos(\omega t) \\ -i \sin(\omega t) \end{pmatrix}. \]

Step 4: The electron will be in the spin-down state along the \( x \)-axis, i.e., \[ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}, \] when: \[ \cos(\omega t) = \frac{1}{\sqrt{2}}, \quad -i \sin(\omega t) = \frac{-1}{\sqrt{2}}. \] This occurs at \( \omega t = \frac{\pi}{4} \), so the minimum time is: \[ t = \frac{\pi}{4\omega}. \]