Imagine that an electron with an initial spin state
$$|\psi\rangle=\left[\begin{array}{l}
\sqrt{1 / 2} \
\sqrt{1 / 2}
\end{array}\right]$$
Goes into an SGz device and comes out the - channel of that device. The electron's spin state is now
A) $\left[\begin{array}{l}\sqrt{1 / 2} \ \sqrt{1 / 2}\end{array}\right]$
B) $\left[\begin{array}{l}-\sqrt{1 / 2} \ -\sqrt{1 / 2}\end{array}\right]$
C) $\left[\begin{array}{l}\sqrt{1 / 2} \ -\sqrt{1 / 2}\end{array}\right]$
D) $\left[\begin{array}{l}1 \ 0\end{array}\right]$
E) $\left[\begin{array}{l}0 \ 1\end{array}\right]$

Question

Quizplus · Accepted Answer

The SGz device separates electrons based on their spin along the z-axis. The - channel corresponds to spin down, which is represented by the state vector $\left[\begin{array}{l}0 \ 1\end{array}\right]$, indicating the electron's spin is now in the down state.

Imagine That an Electron with an Initial Spin State ∣ψ⟩=[1/21/2]|\psi\rangle=\left[\begin{array}{l}\sqrt{1 / 2} \\\sqrt{1 / 2}\end{array}\right]∣ψ⟩=[1/2​1/2​​]

Imagine That an Electron with an Initial Spin State $|\psi\rangle=\left[\begin{array}{l}\sqrt{1 / 2} \\\sqrt{1 / 2}\end{array}\right]$