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

Question

Prepp Team · Accepted Answer

Electron Spin Dynamics in a Magnetic Field

The problem asks for the minimum time required for an electron, initially in the spin up state, to transition to the spin down state along the x-axis when subjected to a magnetic field.

Initial Conditions and Hamiltonian

Initial state at $t=0$: $|\psi(0)\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. This represents the spin up state.
Hamiltonian: $H = -\hbar \omega \sigma_y$, where $\omega = \frac{q B_0}{2m} > 0$ and $\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$.
Target state: A state proportional to $\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. This state corresponds to spin down along the x-axis.

Time Evolution

The time evolution of the quantum state is governed by the Schrödinger equation. The state at time $t$ is given by:

$ |\psi(t)\rangle = e^{-iHt/\hbar} |\psi(0)\rangle $

Substitute the Hamiltonian $H = -\hbar \omega \sigma_y$:

$ |\psi(t)\rangle = e^{-i(-\hbar \omega \sigma_y)t/\hbar} |\psi(0)\rangle = e^{i\omega t \sigma_y} |\psi(0)\rangle $

Calculating the Evolution Operator

The evolution operator $e^{i\omega t \sigma_y}$ can be calculated using the matrix properties. For a Pauli matrix $\sigma_k$, we have $e^{i\theta \sigma_k} = \cos(\theta) I + i \sin(\theta) \sigma_k$. Here, $k=y$ and $\theta = \omega t$. Thus:

$ e^{i\omega t \sigma_y} = \cos(\omega t) I + i \sin(\omega t) \sigma_y $ $ e^{i\omega t \sigma_y} = \cos(\omega t) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + i \sin(\omega t) \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $ $ e^{i\omega t \sigma_y} = \begin{pmatrix} \cos(\omega t) & 0 \\ 0 & \cos(\omega t) \end{pmatrix} + \begin{pmatrix} 0 & \sin(\omega t) \\ -\sin(\omega t) & 0 \end{pmatrix} $ $ e^{i\omega t \sigma_y} = \begin{pmatrix} \cos(\omega t) & \sin(\omega t) \\ -\sin(\omega t) & \cos(\omega t) \end{pmatrix} $

Determining the State at Time t

Now, apply the evolution operator to the initial state $|\psi(0)\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$:

$ |\psi(t)\rangle = \begin{pmatrix} \cos(\omega t) & \sin(\omega t) \\ -\sin(\omega t) & \cos(\omega t) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} $ $ |\psi(t)\rangle = \begin{pmatrix} \cos(\omega t) \cdot 1 + \sin(\omega t) \cdot 0 \\ -\sin(\omega t) \cdot 1 + \cos(\omega t) \cdot 0 \end{pmatrix} = \begin{pmatrix} \cos(\omega t) \\ -\sin(\omega t) \end{pmatrix} $

Finding the Transition Time

We want the electron to be in the state $\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Therefore, we need to find the smallest time $t > 0$ such that:

$ \begin{pmatrix} \cos(\omega t) \\ -\sin(\omega t) \end{pmatrix} = c \begin{pmatrix} 1 \\ -1 \end{pmatrix} $

for some proportionality constant $c$. Comparing the components:

$\cos(\omega t) = c$
$-\sin(\omega t) = -c \implies \sin(\omega t) = c$

From these conditions, we must have $\cos(\omega t) = \sin(\omega t)$. This implies $\tan(\omega t) = 1$. The smallest positive value for $\omega t$ that satisfies this condition is:

$ \omega t = \frac{\pi}{4} $

Solving for $t$ gives:

$ t = \frac{\pi}{4\omega} $

At this time, $c = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, and the state is indeed $\begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.