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

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

A:

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

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

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

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

so

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

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

For $$\ket{-}$$ we must have

\label{eqn:someSpinProblems:120}
a – i b = 0,

so

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

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

The normalized eigenvectors are

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

For the probability question we are interested in

\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}

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

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

A:

\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}

The characteristic equation is
\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}

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

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

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

or

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

This squares to

\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 ),

so the normalized eigenkets are
\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}.
}

References

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