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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.