eigenket

PHY1520H Graduate Quantum Mechanics. Lecture 20: Perturbation theory. Taught by Prof. Arun Paramekanti

December 3, 2015 phy1520 , , , , , , , , , ,

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Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] ch. 5 content.

Perturbation theory

Given a \( 2 \times 2 \) Hamiltonian \( H = H_0 + V \), where

\begin{equation}\label{eqn:qmLecture20:20}
H =
\begin{bmatrix}
a & c \\
c^\conj & b
\end{bmatrix}
\end{equation}

which has eigenvalues

\begin{equation}\label{eqn:qmLecture20:40}
\lambda_\pm = \frac{a + b}{2} \pm \sqrt{ \lr{ \frac{a – b}{2}}^2 + \Abs{c}^2 }.
\end{equation}

If \( c = 0 \),

\begin{equation}\label{eqn:qmLecture20:60}
H_0 =
\begin{bmatrix}
a & 0 \\
0 & b
\end{bmatrix},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:80}
V =
\begin{bmatrix}
0 & c \\
c^\conj & 0
\end{bmatrix}.
\end{equation}

Suppose that \( \Abs{c} \ll \Abs{a – b} \), then

\begin{equation}\label{eqn:qmLecture20:100}
\lambda_\pm \approx \frac{a + b}{2} \pm \Abs{ \frac{a – b}{2} } \lr{ 1 + 2 \frac{\Abs{c}^2}{\Abs{a – b}^2} }.
\end{equation}

If \( a > b \), then

\begin{equation}\label{eqn:qmLecture20:120}
\lambda_\pm \approx \frac{a + b}{2} \pm \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} }.
\end{equation}

\begin{equation}\label{eqn:qmLecture20:140}
\begin{aligned}
\lambda_{+}
&= \frac{a + b}{2} + \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&= a + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= a + \frac{\Abs{c}^2}{a – b},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:qmLecture20:680}
\begin{aligned}
\lambda_{-}
&= \frac{a + b}{2} – \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&=
b + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= b + \frac{\Abs{c}^2}{a – b}.
\end{aligned}
\end{equation}

This adiabatic evolution displays a “level repulsion”, quadradic in \( \Abs{c} \) as sketched in fig. 1, and is described as a non-degenerate perbutation.

fig. 1. Adiabatic (non-degenerate) perturbation

fig. 1. Adiabatic (non-degenerate) perturbation

If \( \Abs{c} \gg \Abs{a -b} \), then

\begin{equation}\label{eqn:qmLecture20:160}
\begin{aligned}
\lambda_\pm
&= \frac{a + b}{2} \pm \Abs{c} \sqrt{ 1 + \inv{\Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&\approx \frac{a + b}{2} \pm \Abs{c} \lr{ 1 + \inv{2 \Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&= \frac{a + b}{2} \pm \Abs{c} \pm \frac{\lr{a – b}^2}{8 \Abs{c}}.
\end{aligned}
\end{equation}

Here we loose the adiabaticity, and have “level repulsion” that is linear in \( \Abs{c} \), as sketched in fig. 2. We no longer have the sign of \( a – b \) in the expansion. This is described as a degenerate perbutation.

fig. 2. Degenerate perbutation

fig. 2. Degenerate perbutation

General non-degenerate perturbation

Given an unperturbed system with solutions of the form

\begin{equation}\label{eqn:qmLecture20:180}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}},
\end{equation}

we want to solve the perturbed Hamiltonian equation

\begin{equation}\label{eqn:qmLecture20:200}
\lr{ H_0 + \lambda V } \ket{ n } = \lr{ E_n^{(0)} + \Delta n } \ket{n}.
\end{equation}

Here \( \Delta n \) is an energy shift as that goes to zero as \( \lambda \rightarrow 0 \). We can write this as

\begin{equation}\label{eqn:qmLecture20:220}
\lr{ E_n^{(0)} – H_0 } \ket{ n } = \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

We are hoping to iterate with application of the inverse to an initial estimate of \( \ket{n} \)

\begin{equation}\label{eqn:qmLecture20:240}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

This gets us into trouble if \( \lambda \rightarrow 0 \), which can be fixed by using

\begin{equation}\label{eqn:qmLecture20:260}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

which can be seen to be a solution to \ref{eqn:qmLecture20:220}. We want to ask if

\begin{equation}\label{eqn:qmLecture20:280}
\lr{ \lambda V – \Delta_n } \ket{n} ,
\end{equation}

contains a bit of \( \ket{ n^{(0)} } \)? To determine this act with \( \bra{n^{(0)}} \) on the left

\begin{equation}\label{eqn:qmLecture20:300}
\begin{aligned}
\bra{ n^{(0)} } \lr{ \lambda V – \Delta_n } \ket{n}
&=
\bra{ n^{(0)} } \lr{ E_n^{(0)} – H_0 } \ket{n} \\
&=
\lr{ E_n^{(0)} – E_n^{(0)} } \braket{n^{(0)}}{n} \\
&=
0.
\end{aligned}
\end{equation}

This shows that \( \ket{n} \) is entirely orthogonal to \( \ket{n^{(0)}} \).

Define a projection operator

\begin{equation}\label{eqn:qmLecture20:320}
P_n = \ket{n^{(0)}}\bra{n^{(0)}},
\end{equation}

which has the idempotent property \( P_n^2 = P_n \) that we expect of a projection operator.

Define a rejection operator
\begin{equation}\label{eqn:qmLecture20:340}
\overline{{P}}_n
= 1 –
\ket{n^{(0)}}\bra{n^{(0)}}
= \sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}.
\end{equation}

Because \( \ket{n} \) has no component in the direction \( \ket{n^{(0)}} \), the rejection operator can be inserted much like we normally do with the identity operator, yielding

\begin{equation}\label{eqn:qmLecture20:360}
\ket{n}’ = \lr{ E_n^{(0)} – H_0 }^{-1} \overline{{P}}_n \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

valid for any initial \( \ket{n} \).

Power series perturbation expansion

Instead of iterating, suppose that the unknown state and unknown energy difference operator can be expanded in a \( \lambda \) power series, say

\begin{equation}\label{eqn:qmLecture20:380}
\ket{n}
=
\ket{n_0}
+ \lambda \ket{n_1}
+ \lambda^2 \ket{n_2}
+ \lambda^3 \ket{n_3} + \cdots
\end{equation}

and

\begin{equation}\label{eqn:qmLecture20:400}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots
\end{equation}

We usually interpret functions of operators in terms of power series expansions. In the case of \( \lr{ E_n^{(0)} – H_0 }^{-1} \), we have a concrete interpretation when acting on one of the unpertubed eigenstates

\begin{equation}\label{eqn:qmLecture20:420}
\inv{ E_n^{(0)} – H_0 } \ket{m^{(0)}} =
\inv{ E_n^{(0)} – E_m^0 } \ket{m^{(0)}}.
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture20:440}
\ket{n}
=
\inv{ E_n^{(0)} – H_0 }
\sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}
\lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:460}
\boxed{
\ket{n}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_n } \ket{n}.
}
\end{equation}

From \ref{eqn:qmLecture20:220}, note that

\begin{equation}\label{eqn:qmLecture20:500}
\Delta_n =
\frac{\bra{n^{(0)}} \lambda V \ket{n}}{\braket{n^0}{n}},
\end{equation}

however, we will normalize by setting \( \braket{n^0}{n} = 1 \), so

\begin{equation}\label{eqn:qmLecture20:521}
\boxed{
\Delta_n =
\bra{n^{(0)}} \lambda V \ket{n}.
}
\end{equation}

to \( O(\lambda^0) \)

If all \( \lambda^n, n > 0 \) are zero, then we have

\label{eqn:qmLecture20:780}
\begin{equation}\label{eqn:qmLecture20:740}
\ket{n_0}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ – \Delta_{n_0} } \ket{n_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture20:800}
\Delta_{n_0} \braket{n^{(0)}}{n_0} = 0
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:540}
\begin{aligned}
\ket{n_0} &= \ket{n^{(0)}} \\
\Delta_{n_0} &= 0.
\end{aligned}
\end{equation}

to \( O(\lambda^1) \)

Requiring identity for all \( \lambda^1 \) terms means

\begin{equation}\label{eqn:qmLecture20:760}
\ket{n_1} \lambda
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_{n_1} \lambda } \ket{n_0},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:560}
\ket{n_1}
=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}} \bra{ m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} } \ket{n_0}.
\end{equation}

With the assumption that \( \ket{n^{(0)}} \) is normalized, and with the shorthand

\begin{equation}\label{eqn:qmLecture20:600}
V_{m n} = \bra{ m^{(0)}} V \ket{n^{(0)}},
\end{equation}

that is

\begin{equation}\label{eqn:qmLecture20:580}
\begin{aligned}
\ket{n_1}
&=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}
\\
\Delta_{n_1} &= \bra{ n^{(0)} } V \ket{ n^0} = V_{nn}.
\end{aligned}
\end{equation}

to \( O(\lambda^2) \)

The second order perturbation states are found by selecting only the \( \lambda^2 \) contributions to

\begin{equation}\label{eqn:qmLecture20:820}
\lambda^2 \ket{n_2}
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – (\lambda \Delta_{n_1} + \lambda^2 \Delta_{n_2}) }
\lr{
\ket{n_0}
+ \lambda \ket{n_1}
}.
\end{equation}

Because \( \ket{n_0} = \ket{n^{(0)}} \), the \( \lambda^2 \Delta_{n_2} \) is killed, leaving

\begin{equation}\label{eqn:qmLecture20:840}
\begin{aligned}
\ket{n_2}
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\ket{n_1} \\
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\sum_{l \ne n}
\frac{
\ket{l^{(0)}}
}
{
E_n^{(0)} – E_l^{(0)}
}
V_{l n},
\end{aligned}
\end{equation}

which can be written as

\begin{equation}\label{eqn:qmLecture20:620}
\ket{n_2}
=
\sum_{l,m \ne n}
\ket{m^{(0)}}
\frac{V_{m l} V_{l n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}
–
\sum_{m \ne n}
\ket{m^{(0)}}
\frac{V_{n n} V_{m n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

For the second energy perturbation we have

\begin{equation}\label{eqn:qmLecture20:860}
\lambda^2 \Delta_{n_2} =
\bra{n^{(0)}} \lambda V \lr{ \lambda \ket{n_1} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:880}
\begin{aligned}
\Delta_{n_2}
&=
\bra{n^{(0)}} V \ket{n_1} \\
&=
\bra{n^{(0)}} V
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}.
\end{aligned}
\end{equation}

That is

\begin{equation}\label{eqn:qmLecture20:900}
\Delta_{n_2}
=
\sum_{m \ne n} \frac{V_{n m} V_{m n} }{E_n^{(0)} – E_m^{(0)}}.
\end{equation}

to \( O(\lambda^3) \)

Similarily, it can be shown that

\begin{equation}\label{eqn:qmLecture20:640}
\Delta_{n_3} =
\sum_{l, m \ne n} \frac{V_{n m} V_{m l} V_{l n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}
–
\sum_{ m \ne n} \frac{V_{n m} V_{n n} V_{m n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

In general, the energy perturbation is given by

\begin{equation}\label{eqn:qmLecture20:660}
\Delta_n^{(l)} = \bra{n^{(0)}} V \ket{n^{(l-1)}}.
\end{equation}

References

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

Two spin time evolution

November 14, 2015 phy1520 , , , ,

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Motivation

Our midterm posed a (low mark “quick question”) that I didn’t complete (or at least not properly). This shouldn’t have been a difficult question, but I spend way too much time on it, costing me time that I needed for other questions.

It turns out that there isn’t anything fancy required for this question, just perseverance and careful work.

Guts

The question asked for the time evolution of a two particle state

\begin{equation}\label{eqn:twoSpinHamiltonian:20}
\psi = \inv{\sqrt{2}} \lr{ \ket{\uparrow \downarrow} – \ket{\downarrow \uparrow} }
\end{equation}

under the action of the Hamiltonian

\begin{equation}\label{eqn:twoSpinHamiltonian:40}
H = – B S_{z,1} + 2 B S_{x,2} = \frac{\Hbar B}{2}\lr{ -\sigma_{z,1} + 2 \sigma_{x,2} } .
\end{equation}

We have to know the action of the Hamiltonian on all the states

\begin{equation}\label{eqn:twoSpinHamiltonian:60}
\begin{aligned}
H \ket{\uparrow \uparrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \uparrow} + 2 \ket{\uparrow \downarrow} } \\
H \ket{\uparrow \downarrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \downarrow} + 2 \ket{\uparrow \uparrow} } \\
H \ket{\downarrow \uparrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \uparrow} + 2 \ket{\downarrow \downarrow} } \\
H \ket{\downarrow \downarrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \downarrow} + 2 \ket{\downarrow \uparrow} } \\
\end{aligned}
\end{equation}

With respect to the basis \( \setlr{ \ket{\uparrow \uparrow}, \ket{\uparrow \downarrow}, \ket{\downarrow \uparrow}, \ket{\downarrow \downarrow} } \), the matrix of the Hamiltonian is

\begin{equation}\label{eqn:twoSpinHamiltonian:80}
H =
\frac{ \Hbar B }{2}
\begin{bmatrix}
-1 & 2 & 0 & 0 \\
2 & -1 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}
\end{equation}

Utilizing the block diagonal form (and ignoring the \( \Hbar B/2 \) factor for now), the characteristic equation is

\begin{equation}\label{eqn:twoSpinHamiltonian:100}
0
=
\begin{vmatrix}
-1 -\lambda & 2 \\
2 & -1 – \lambda
\end{vmatrix}
\begin{vmatrix}
1 -\lambda & 2 \\
2 & 1 – \lambda
\end{vmatrix}
=
\lr{(1 + \lambda)^2 – 4}
\lr{(1 – \lambda)^2 – 4}.
\end{equation}

This has solutions

\begin{equation}\label{eqn:twoSpinHamiltonian:120}
1 \pm \lambda = \pm 2,
\end{equation}

or, with the \( \Hbar B/2 \) factors put back in

\begin{equation}\label{eqn:twoSpinHamiltonian:140}
\lambda = \pm \Hbar B/2 , \pm 3 \Hbar B/2.
\end{equation}

I was thinking that we needed to compute the time evolution operator

\begin{equation}\label{eqn:twoSpinHamiltonian:160}
U = e^{-i H t/\Hbar},
\end{equation}

but we actually only need the eigenvectors, and the inverse relations. We can find the eigenvectors by inspection in each case from

\begin{equation}\label{eqn:twoSpinHamiltonian:180}
\begin{aligned}
H – (1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-2 & 2 & 0 & 0 \\
2 & -2 & 0 & 0 \\
0 & 0 & 0 & 2 \\
0 & 0 & 2 & 0 \\
\end{bmatrix} \\
H – (-1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
0 & 2 & 0 & 0 \\
2 & 0 & 0 & 0 \\
0 & 0 & 2 & 2 \\
0 & 0 & 2 & 2 \\
\end{bmatrix} \\
H – (3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-4 & 2 & 0 & 0 \\
2 & -4 & 0 & 0 \\
0 & 0 &-2 & 2 \\
0 & 0 & 2 &-2 \\
\end{bmatrix} \\
H – (-3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
2 & 2 & 0 & 0 \\
2 & 2 & 0 & 0 \\
0 & 0 & 4 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The eigenkets are

\begin{equation}\label{eqn:twoSpinHamiltonian:280}
\begin{aligned}
\ket{1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1 \\
0 \\
0 \\
\end{bmatrix} \\
\ket{-1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
-1 \\
\end{bmatrix} \\
\ket{3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
1 \\
\end{bmatrix} \\
\ket{-3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-1 \\
0 \\
0 \\
\end{bmatrix},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:twoSpinHamiltonian:300}
\begin{aligned}
\sqrt{2} \ket{1} &= \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} \\
\sqrt{2} \ket{-1} &= \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{3} &= \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{-3} &= \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow}.
\end{aligned}
\end{equation}

We can invert these

\begin{equation}\label{eqn:twoSpinHamiltonian:220}
\begin{aligned}
\ket{\uparrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{1} + \ket{-3} } \\
\ket{\uparrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{1} – \ket{-3} } \\
\ket{\downarrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{3} + \ket{-1} } \\
\ket{\downarrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{3} – \ket{-1} } \\
\end{aligned}
\end{equation}

The original state of interest can now be expressed in terms of the eigenkets

\begin{equation}\label{eqn:twoSpinHamiltonian:240}
\psi
=
\inv{2} \lr{
\ket{1} – \ket{-3} –
\ket{3} – \ket{-1}
}
\end{equation}

The time evolution of this ket is

\begin{equation}\label{eqn:twoSpinHamiltonian:260}
\begin{aligned}
\psi(t)
&=
\inv{2}
\lr{
e^{-i B t/2} \ket{1}
– e^{3 i B t/2} \ket{-3}
– e^{-3 i B t/2} \ket{3}
– e^{i B t/2} \ket{-1}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{-i B t/2} \lr{ \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} }
– e^{3 i B t/2} \lr{ \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow} }
– e^{-3 i B t/2} \lr{ \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} }
– e^{i B t/2} \lr{ \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} }
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
\lr{ e^{-i B t/2} – e^{3 i B t/2} } \ket{\uparrow \uparrow}
+ \lr{ e^{-i B t/2} + e^{3 i B t/2} } \ket{\uparrow \downarrow}
– \lr{ e^{-3 i B t/2} + e^{i B t/2} } \ket{\downarrow \uparrow}
+ \lr{ e^{i B t/2} – e^{-3 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{i B t/2} \lr{ e^{-2 i B t/2} – e^{2 i B t/2} } \ket{\uparrow \uparrow}
+ e^{i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\uparrow \downarrow}
– e^{- i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\downarrow \uparrow}
+ e^{- i B t/2} \lr{ e^{2 i B t/2} – e^{-2 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{\sqrt{2}}
\lr{
i \sin( B t )
\lr{
e^{- i B t/2} \ket{\downarrow \downarrow} – e^{i B t/2} \ket{\uparrow \uparrow}
}
+ \cos( B t ) \lr{
e^{i B t/2} \ket{\uparrow \downarrow}
– e^{- i B t/2} \ket{\downarrow \uparrow}
}
}
\end{aligned}
\end{equation}

Note that this returns to the original state when \( t = \frac{2 \pi n}{B}, n \in \mathbb{Z} \). I think I’ve got it right this time (although I got a slightly different answer on paper before typing it up.)

This doesn’t exactly seem like a quick answer question, at least to me. Is there some easier way to do it?

Bra-ket and spin one-half problems

July 27, 2015 phy1520 , , , , , , , , ,

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Question: Operator matrix representation ([1] pr. 1.5)

(a)

Determine the matrix representation of \( \ket{\alpha}\bra{\beta} \) given a complete set of eigenvectors \( \ket{a^r} \).

(b)

Verify with \( \ket{\alpha} = \ket{s_z = \Hbar/2}, \ket{s_x = \Hbar/2} \).

Answer

(a)

Forming the matrix element

\begin{equation}\label{eqn:moreBraKetProblems:20}
\begin{aligned}
\bra{a^r} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^s}
&=
\braket{a^r}{\alpha}\braket{\beta}{a^s} \\
&=
\braket{a^r}{\alpha}
\braket{a^s}{\beta}^\conj,
\end{aligned}
\end{equation}

the matrix representation is seen to be

\begin{equation}\label{eqn:moreBraKetProblems:40}
\ket{\alpha}\bra{\beta}
\sim
\begin{bmatrix}
\bra{a^1} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^1} & \bra{a^1} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^2} & \cdots \\
\bra{a^2} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^1} & \bra{a^2} \lr{ \ket{\alpha}\bra{\beta} } \ket{a^2} & \cdots \\
\vdots & \vdots & \ddots \\
\end{bmatrix}
=
\begin{bmatrix}
\braket{a^1}{\alpha} \braket{a^1}{\beta}^\conj & \braket{a^1}{\alpha} \braket{a^2}{\beta}^\conj & \cdots \\
\braket{a^2}{\alpha} \braket{a^1}{\beta}^\conj & \braket{a^2}{\alpha} \braket{a^2}{\beta}^\conj & \cdots \\
\vdots & \vdots & \ddots \\
\end{bmatrix}.
\end{equation}

(b)

First compute the spin-z representation of \( \ket{s_x = \Hbar/2 } \).

\begin{equation}\label{eqn:moreBraKetProblems:60}
\begin{aligned}
\lr{ S_x – \Hbar/2 I }
\begin{bmatrix}
a \\
b
\end{bmatrix}
&=
\lr{
\begin{bmatrix}
0 & \Hbar/2 \\
\Hbar/2 & 0 \\
\end{bmatrix}
–
\begin{bmatrix}
\Hbar/2 & 0 \\
0 & \Hbar/2 \\
\end{bmatrix}
} \\
&=
\begin{bmatrix}
a \\
b
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
-1 & 1 \\
1 & -1 \\
\end{bmatrix}
\begin{bmatrix}
a \\
b
\end{bmatrix},
\end{aligned}
\end{equation}

so \( \ket{s_x = \Hbar/2 } \propto (1,1) \).

Normalized we have

\begin{equation}\label{eqn:moreBraKetProblems:80}
\begin{aligned}
\ket{\alpha} &= \ket{s_z = \Hbar/2 } =
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
\ket{\beta} &= \ket{s_z = \Hbar/2 }
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1
\end{bmatrix}.
\end{aligned}
\end{equation}

Using \ref{eqn:moreBraKetProblems:40} the matrix representation is

\begin{equation}\label{eqn:moreBraKetProblems:100}
\ket{\alpha}\bra{\beta}
\sim
\begin{bmatrix}
(1) (1/\sqrt{2})^\conj & (1) (1/\sqrt{2})^\conj \\
(0) (1/\sqrt{2})^\conj & (0) (1/\sqrt{2})^\conj \\
\end{bmatrix}
=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}.
\end{equation}

This can be confirmed with direct computation
\begin{equation}\label{eqn:moreBraKetProblems:120}
\begin{aligned}
\ket{\alpha}\bra{\beta}
&=
\begin{bmatrix}
1 \\
0
\end{bmatrix}
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
0 & 0
\end{bmatrix}.
\end{aligned}
\end{equation}

Question: eigenvalue of sum of kets ([1] pr. 1.6)

Given eigenkets \( \ket{i}, \ket{j} \) of an operator \( A \), what are the conditions that \( \ket{i} + \ket{j} \) is also an eigenvector?

Answer

Let \( A \ket{i} = i \ket{i}, A \ket{j} = j \ket{j} \), and suppose that the sum is an eigenket. Then there must be a value \( a \) such that

\begin{equation}\label{eqn:moreBraKetProblems:140}
A \lr{ \ket{i} + \ket{j} } = a \lr{ \ket{i} + \ket{j} },
\end{equation}

so

\begin{equation}\label{eqn:moreBraKetProblems:160}
i \ket{i} + j \ket{j} = a \lr{ \ket{i} + \ket{j} }.
\end{equation}

Operating with \( \bra{i}, \bra{j} \) respectively, gives

\begin{equation}\label{eqn:moreBraKetProblems:180}
\begin{aligned}
i &= a \\
j &= a,
\end{aligned}
\end{equation}

so for the sum to be an eigenket, both of the corresponding energy eigenvalues must be identical (i.e. linear combinations of degenerate eigenkets are also eigenkets).

Question: Null operator ([1] pr. 1.7)

Given eigenkets \( \ket{a’} \) of operator \( A \)

(a)

show that

\begin{equation}\label{eqn:moreBraKetProblems:200}
\prod_{a’} \lr{ A – a’ }
\end{equation}

is the null operator.

(b)

\begin{equation}\label{eqn:moreBraKetProblems:220}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”}
\end{equation}

(c)

Illustrate using \( S_z \) for a spin 1/2 system.

Answer

(a)

Application of \( \ket{a} \), the eigenket of \( A \) with eigenvalue \( a \) to any term \( A – a’ \) scales \( \ket{a} \) by \( a – a’ \), so the product operating on \( \ket{a} \) is

\begin{equation}\label{eqn:moreBraKetProblems:240}
\prod_{a’} \lr{ A – a’ } \ket{a} = \prod_{a’} \lr{ a – a’ } \ket{a}.
\end{equation}

Since \( \ket{a} \) is one of the \( \setlr{\ket{a’}} \) eigenkets of \( A \), one of these terms must be zero.

(b)

Again, consider the action of the operator on \( \ket{a} \),

\begin{equation}\label{eqn:moreBraKetProblems:260}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a}
=
\prod_{a” \ne a’} \frac{\lr{ a – a” }}{a’ – a”} \ket{a}.
\end{equation}

If \( \ket{a} = \ket{a’} \), then \( \prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a} = \ket{a} \), whereas if it does not, then it equals one of the \( a” \) energy eigenvalues. This is a representation of the Kronecker delta function

\begin{equation}\label{eqn:moreBraKetProblems:300}
\prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a} \equiv \delta_{a’, a} \ket{a}
\end{equation}

(c)

For operator \( S_z \) the eigenvalues are \( \setlr{ \Hbar/2, -\Hbar/2 } \), so the null operator must be

\begin{equation}\label{eqn:moreBraKetProblems:280}
\begin{aligned}
\prod_{a’} \lr{ A – a’ }
&=
\lr{ \frac{\Hbar}{2} }^2 \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} – \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\begin{bmatrix}
0 & 0 \\
0 & -2
\end{bmatrix}
\begin{bmatrix}
2 & 0 \\
0 & 0 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 0 \\
0 & 0 \\
\end{bmatrix}
\end{aligned}
\end{equation}

For the delta representation, consider the \( \ket{\pm} \) states and their eigenvalue. The delta operators are

\begin{equation}\label{eqn:moreBraKetProblems:320}
\begin{aligned}
\prod_{a” \ne \Hbar/2} \frac{\lr{ A – a” }}{\Hbar/2 – a”}
&=
\frac{S_z – (-\Hbar/2) I}{\Hbar/2 – (-\Hbar/2)} \\
&=
\inv{2} \lr{ \sigma_z + I } \\
&=
\inv{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\inv{2}
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix}
\\
&=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:moreBraKetProblems:340}
\begin{aligned}
\prod_{a” \ne -\Hbar/2} \frac{\lr{ A – a” }}{-\Hbar/2 – a”}
&=
\frac{S_z – (\Hbar/2) I}{-\Hbar/2 – \Hbar/2} \\
&=
\inv{2} \lr{ \sigma_z – I } \\
&=
\inv{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} – \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} } \\
&=
\inv{2}
\begin{bmatrix}
0 & 0 \\
0 & -2
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.
\end{aligned}
\end{equation}

These clearly have the expected delta function property acting on kets \( \ket{+} = (1,0), \ket{-} = (0, 1) \).

Question: Spin half general normal ([1] pr. 1.9)

Construct \( \ket{\BS \cdot \ncap ; + } \), where \( \ncap = ( \cos\alpha \sin\beta, \sin\alpha \sin\beta, \cos\beta ) \) such that

\begin{equation}\label{eqn:moreBraKetProblems:360}
\BS \cdot \ncap \ket{\BS \cdot \ncap ; + } =
\frac{\Hbar}{2} \ket{\BS \cdot \ncap ; + },
\end{equation}

Solve this as an eigenvalue problem.

Answer

The spin operator for this direction is

\begin{equation}\label{eqn:moreBraKetProblems:380}
\begin{aligned}
\BS \cdot \ncap
&= \frac{\Hbar}{2} \Bsigma \cdot \ncap \\
&= \frac{\Hbar}{2}
\lr{
\cos\alpha \sin\beta \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} + \sin\alpha \sin\beta \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} + \cos\beta \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta &
e^{-i\alpha}
\sin\beta
\\
e^{i\alpha}
\sin\beta
& -\cos\beta
\end{bmatrix}.
\end{aligned}
\end{equation}

Observed that this is traceless and has a \( -\Hbar/2 \) determinant like any of the \( x,y,z \) spin operators.

Assuming that this has an \( \Hbar/2 \) eigenvalue (to be verified later), the eigenvalue problem is

\begin{equation}\label{eqn:moreBraKetProblems:400}
\begin{aligned}
0
&=
\BS \cdot \ncap – \Hbar/2 I \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta -1 &
e^{-i\alpha}
\sin\beta
\\
e^{i\alpha}
\sin\beta
& -\cos\beta -1
\end{bmatrix} \\
&=
\Hbar
\begin{bmatrix}
– \sin^2 \frac{\beta}{2} &
e^{-i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
\\
e^{i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
& -\cos^2 \frac{\beta}{2}
\end{bmatrix}
\end{aligned}
\end{equation}

This has a zero determinant as expected, and the eigenvector \( (a,b) \) will satisfy

\begin{equation}\label{eqn:moreBraKetProblems:420}
\begin{aligned}
0
&= – \sin^2 \frac{\beta}{2} a +
e^{-i\alpha}
\sin\frac{\beta}{2} \cos\frac{\beta}{2}
b \\
&= \sin\frac{\beta}{2} \lr{ – \sin \frac{\beta}{2} a +
e^{-i\alpha} b
\cos\frac{\beta}{2}
}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:moreBraKetProblems:440}
\begin{bmatrix}
a \\
b
\end{bmatrix}
\propto
\begin{bmatrix}
\cos\frac{\beta}{2} \\
e^{i\alpha}
\sin\frac{\beta}{2}
\end{bmatrix}.
\end{equation}

This is appropriately normalized, so the ket for \( \BS \cdot \ncap \) is

\begin{equation}\label{eqn:moreBraKetProblems:460}
\ket{ \BS \cdot \ncap ; + } =
\cos\frac{\beta}{2} \ket{+} +
e^{i\alpha}
\sin\frac{\beta}{2}
\ket{-}.
\end{equation}

Note that the other eigenvalue is

\begin{equation}\label{eqn:moreBraKetProblems:480}
\ket{ \BS \cdot \ncap ; – } =
-\sin\frac{\beta}{2} \ket{+} +
e^{i\alpha}
\cos\frac{\beta}{2}
\ket{-}.
\end{equation}

It is straightforward to show that these are orthogonal and that this has the \( -\Hbar/2 \) eigenvalue.

Question: Two state Hamiltonian ([1] pr. 1.10)

Solve the eigenproblem for

\begin{equation}\label{eqn:moreBraKetProblems:500}
H = a \biglr{
\ket{1}\bra{1}
-\ket{2}\bra{2}
+\ket{1}\bra{2}
+\ket{2}\bra{1}
}
\end{equation}

Answer

In matrix form the Hamiltonian is

\begin{equation}\label{eqn:moreBraKetProblems:520}
H = a
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.
\end{equation}

The eigenvalue problem is

\begin{equation}\label{eqn:moreBraKetProblems:540}
\begin{aligned}
0
&= \Abs{ H – \lambda I } \\
&= (a – \lambda)(-a – \lambda) – a^2 \\
&= (-a + \lambda)(a + \lambda) – a^2 \\
&= \lambda^2 – a^2 – a^2,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:moreBraKetProblems:560}
\lambda = \pm \sqrt{2} a.
\end{equation}

An eigenket proportional to \( (\alpha,\beta) \) must satisfy

\begin{equation}\label{eqn:moreBraKetProblems:580}
0
= ( 1 \mp \sqrt{2} ) \alpha + \beta,
\end{equation}

so

\begin{equation}\label{eqn:moreBraKetProblems:600}
\ket{\pm} \propto
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix},
\end{equation}

or

\begin{equation}\label{eqn:moreBraKetProblems:620}
\begin{aligned}
\ket{\pm}
&=
\inv{2(2 – \sqrt{2})}
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix} \\
&=
\frac{2 + \sqrt{2}}{4}
\begin{bmatrix}
-1 \\
1 \mp \sqrt{2}
\end{bmatrix}.
\end{aligned}
\end{equation}

That is
\begin{equation}\label{eqn:moreBraKetProblems:640}
\ket{\pm} =
\frac{2 + \sqrt{2}}{4} \lr{
-\ket{1} + (1 \mp \sqrt{2}) \ket{2}
}.
\end{equation}

Question: Spin half probability and dispersion ([1] pr. 1.12)

A spin \( 1/2 \) system \( \BS \cdot \ncap \), with \( \ncap = \sin \gamma \xcap + \cos\gamma \zcap \), is in state with eigenvalue \( \Hbar/2 \).

(a)

If \( S_x \) is measured. What is the probability of getting \( + \Hbar/2 \)?

(b)

Evaluate the dispersion in \( S_x \), that is,

\begin{equation}\label{eqn:moreBraKetProblems:660}
\expectation{\lr{ S_x – \expectation{S_x}}^2}.
\end{equation}

Answer

(a)

In matrix form the spin operator for the system is

\begin{equation}\label{eqn:moreBraKetProblems:680}
\begin{aligned}
\BS \cdot \ncap
&= \frac{\Hbar}{2} \lr{ \cos\gamma \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} + \sin\gamma \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}} \\
&= \frac{\Hbar}{2}
\begin{bmatrix}
\cos\gamma & \sin\gamma \\
\sin\gamma & -\cos\gamma \\
\end{bmatrix}
\end{aligned}
\end{equation}

An eigenket \( \ket{\BS \cdot \ncap ; + } = (a,b) \) must satisfy

\begin{equation}\label{eqn:moreBraKetProblems:700}
\begin{aligned}
0
&= \lr{ \cos \gamma – 1 } a + \sin\gamma b \\
&= \lr{ -2 \sin^2 \frac{\gamma}{2} } a + 2 \sin\frac{\gamma}{2} \cos\frac{\gamma}{2} b \\
&= -\sin \frac{\gamma}{2} a + \cos\frac{\gamma}{2} b,
\end{aligned}
\end{equation}

so the eigenstate is
\begin{equation}\label{eqn:moreBraKetProblems:720}
\ket{\BS \cdot \ncap ; + }
=
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}.
\end{equation}

Pick \( \ket{S_x ; \pm } = \inv{\sqrt{2}}
\begin{bmatrix}
1 \\ \pm 1
\end{bmatrix} \) as the basis for the \( S_x \) operator. Then, for the probability that the system will end up in the \( + \Hbar/2 \) state of \( S_x \), we have

\begin{equation}\label{eqn:moreBraKetProblems:740}
\begin{aligned}
P
&= \Abs{\braket{ S_x ; + }{ \BS \cdot \ncap ; + } }^2 \\
&= \Abs{ \inv{\sqrt{2} }
{
\begin{bmatrix}
1 \\
1
\end{bmatrix}}^\dagger
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}
}^2 \\
&=\inv{2}
\Abs{
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix}
}^2 \\
&=
\inv{2}
\lr{
\cos\frac{\gamma}{2} +
\sin\frac{\gamma}{2}
}^2 \\
&=
\inv{2}
\lr{ 1 + 2 \cos\frac{\gamma}{2} \sin\frac{\gamma}{2} } \\
&=
\inv{2}
\lr{ 1 + \sin\gamma }.
\end{aligned}
\end{equation}

This is a reasonable seeming result, with \( P \in [0, 1] \). Some special values also further validate this

\begin{equation}\label{eqn:moreBraKetProblems:760}
\begin{aligned}
\gamma &= 0, \ket{\BS \cdot \ncap ; + } =
\begin{bmatrix}
1 \\
0
\end{bmatrix}
=
\ket{S_z ; +}
=
\inv{\sqrt{2}} \ket{S_x;+}
+\inv{\sqrt{2}} \ket{S_x;-}
\\
\gamma &= \pi/2, \ket{\BS \cdot \ncap ; + } =
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
=
\ket{S_x ; +}
\\
\gamma &= \pi, \ket{\BS \cdot \ncap ; + } =
\begin{bmatrix}
0 \\
1
\end{bmatrix}
=
\ket{S_z ; -}
=
\inv{\sqrt{2}} \ket{S_x;+}
-\inv{\sqrt{2}} \ket{S_x;-},
\end{aligned}
\end{equation}

where we see that the probabilites are in proportion to the projection of the initial state onto the measured state \( \ket{S_x ; +} \).

(b)

The \( S_x \) expectation is

\begin{equation}\label{eqn:moreBraKetProblems:780}
\begin{aligned}
\expectation{S_x}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix}
\sin\frac{\gamma}{2} \\
\cos\frac{\gamma}{2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2} 2 \sin\frac{\gamma}{2} \cos\frac{\gamma}{2} \\
&=
\frac{\Hbar}{2} \sin\gamma.
\end{aligned}
\end{equation}

Note that \( S_x^2 = (\Hbar/2)^2I \), so

\begin{equation}\label{eqn:moreBraKetProblems:800}
\begin{aligned}
\expectation{S_x^2}
&=
\lr{\frac{\Hbar}{2}}^2
\begin{bmatrix}
\cos\frac{\gamma}{2} & \sin\frac{\gamma}{2}
\end{bmatrix}
\begin{bmatrix}
\cos\frac{\gamma}{2} \\
\sin\frac{\gamma}{2}
\end{bmatrix} \\
&=
\lr{ \frac{\Hbar}{2} }^2
\cos^2\frac{\gamma}{2} + \sin^2 \frac{\gamma}{2} \\
&=
\lr{ \frac{\Hbar}{2} }^2.
\end{aligned}
\end{equation}

The dispersion is

\begin{equation}\label{eqn:moreBraKetProblems:820}
\begin{aligned}
\expectation{\lr{ S_x – \expectation{S_x}}^2}
&=
\expectation{S_x^2} – \expectation{S_x}^2 \\
&=
\lr{ \frac{\Hbar}{2} }^2
\lr{1 – \sin^2 \gamma} \\
&=
\lr{ \frac{\Hbar}{2} }^2
\cos^2 \gamma.
\end{aligned}
\end{equation}

At \( \gamma = \pi/2 \) the dispersion is 0, which is expected since \( \ket{\BS \cdot \ncap ; + } = \ket{ S_x ; + } \) at that point. Similarily, the dispersion is maximized at \( \gamma = 0,\pi \) where the \( \ket{\BS \cdot \ncap ; + } \) component in the \( \ket{S_x ; + } \) direction is minimized.

References

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