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Problems from angular momentum chapter of [1].

Q: $S_y$ eigenvectors

Find the eigenvectors of $\sigma_y$, and then find the probability that a measurement of $S_y$ will be $\Hbar/2$ when the state is initially

$$\begin{equation}\label{eqn:someSpinProblems:20} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \end{equation}$$

A:

The eigenvalues should be $\pm 1$, which is easily checked

$$\begin{equation}\label{eqn:someSpinProblems:40} \begin{aligned} 0 &= \Abs{ \sigma_y – \lambda } \\ &= \begin{vmatrix} -\lambda & -i \\ i & -\lambda \end{vmatrix} \\ &= \lambda^2 – 1. \end{aligned} \end{equation}$$

For $\ket{+} = (a,b)^\T$ we must have

$$\begin{equation}\label{eqn:someSpinProblems:60} -1 a – i b = 0, \end{equation}$$

so

$$\begin{equation}\label{eqn:someSpinProblems:80} \ket{+} \propto \begin{bmatrix} -i \\ 1 \end{bmatrix}, \end{equation}$$

or
$$\begin{equation}\label{eqn:someSpinProblems:100} \ket{+} = \inv{\sqrt{2}} \begin{bmatrix} 1 \\ i \end{bmatrix}. \end{equation}$$

For $\ket{-}$ we must have

$$\begin{equation}\label{eqn:someSpinProblems:120} a – i b = 0, \end{equation}$$

so

$$\begin{equation}\label{eqn:someSpinProblems:140} \ket{+} \propto \begin{bmatrix} i \\ 1 \end{bmatrix}, \end{equation}$$

or
$$\begin{equation}\label{eqn:someSpinProblems:160} \ket{+} = \inv{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix}. \end{equation}$$

The normalized eigenvectors are

$$\begin{equation}\label{eqn:someSpinProblems:180} \boxed{ \ket{\pm} = \inv{\sqrt{2}} \begin{bmatrix} 1 \\ \pm i \end{bmatrix}. } \end{equation}$$

For the probability question we are interested in

$$\begin{equation}\label{eqn:someSpinProblems:200} \begin{aligned} \Abs{\bra{S_y; +} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} }^2 &= \inv{2} \Abs{ \begin{bmatrix} 1 & -i \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} }^2 \\ &= \inv{2} \lr{ \Abs{\alpha}^2 + \Abs{\beta}^2 } \\ &= \inv{2}. \end{aligned} \end{equation}$$

There is a 50 % chance of finding the particle in the $\ket{S_x;+}$ state, independent of the initial state.

Q: Magnetic Hamiltonian eigenvectors

Using Pauli matrices, find the eigenvectors for the magnetic spin interaction Hamiltonian

$$\begin{equation}\label{eqn:someSpinProblems:220} H = – \inv{\Hbar} 2 \mu \BS \cdot \BB. \end{equation}$$

A:

$$\begin{equation}\label{eqn:someSpinProblems:240} \begin{aligned} H &= – \mu \Bsigma \cdot \BB \\ &= – \mu \lr{ B_x \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + B_y \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + B_z \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} } \\ &= – \mu \begin{bmatrix} B_z & B_x – i B_y \\ B_x + i B_y & -B_z \end{bmatrix}. \end{aligned} \end{equation}$$

The characteristic equation is
$$\begin{equation}\label{eqn:someSpinProblems:260} \begin{aligned} 0 &= \begin{vmatrix} -\mu B_z -\lambda & -\mu(B_x – i B_y) \\ -\mu(B_x + i B_y) & \mu B_z – \lambda \end{vmatrix} \\ &= -\lr{ (\mu B_z)^2 – \lambda^2 } – \mu^2\lr{ B_x^2 – (iB_y)^2 } \\ &= \lambda^2 – \mu^2 \BB^2. \end{aligned} \end{equation}$$

That is
$$\begin{equation}\label{eqn:someSpinProblems:360} \boxed{ \lambda = \pm \mu B. } \end{equation}$$

Now for the eigenvectors. We are looking for $\ket{\pm} = (a,b)^\T$ such that

$$\begin{equation}\label{eqn:someSpinProblems:300} 0 = (-\mu B_z \mp \mu B) a -\mu(B_x – i B_y) b \end{equation}$$

or

$$\begin{equation}\label{eqn:someSpinProblems:320} \ket{\pm} \propto \begin{bmatrix} B_x – i B_y \\ B_z \pm B \end{bmatrix}. \end{equation}$$

This squares to

$$\begin{equation}\label{eqn:someSpinProblems:340} B_x^2 + B_y^2 + B_z^2 + B^2 \pm 2 B B_z = 2 B( B \pm B_z ), \end{equation}$$

so the normalized eigenkets are
$$\begin{equation}\label{eqn:someSpinProblems:380} \boxed{ \ket{\pm} = \inv{\sqrt{2 B( B \pm B_z )}} \begin{bmatrix} B_x – i B_y \\ B_z \pm B \end{bmatrix}. } \end{equation}$$

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.