phy1520

Graduate Quantum Mechanics notes now available on paper from amazon

June 11, 2019 phy1520 ,

My notes for “Graduate Quantum Mechanics” (PHY1520H) taught by Prof. Arun Paramekanti, fall 2015. (435 pages), are now available on paper (black and white) through kindle-direct-publishing for $12 USD.

This book is dedicated to my siblings.

Kindle-direct-publishing is a print on demand service, and allows me to make the notes available for pretty close to cost (in this case, about $6 printing cost, $5 to amazon, and about $1 to me as a token royalty).  The notes are still available for free in PDF form, and the latex sources are also available should somebody feel motivated enough to submit a merge request with corrections or enhancements.

This grad quantum course was especially fun.  When I took this class, I had enjoyed the chance to revisit the subject.  Of my three round match against QM, I came out much less bloody this time than the first two rounds.

These notes are no longer redacted and include whatever portions of the problem I completed, errors and all.  In the event that any of the problem sets are recycled for future iterations of the course, students who are taking the course (all mature grad students pursuing science for the love of it, not for grades) are expected to act responsibly, and produce their own solutions, within the constraints provided by the professor.

Changelog:

phy1520.V0.1.9-3 (June 10, 2019)

  • First version posted to kindle-direct successfully.
  • Lots of 6×9 formatting fixes made.
  • Add commas and periods to equations.
  • Remove blank lines that cause additional undesired indenting (implied latex \par’s).

Non-interacting particles in a box

December 17, 2015 phy1520 , ,

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Q: [1] pr 4.1

Calculate the three lowest energy levels and their degeneracies for equal mass distinguishable spin half particles in a box of length \( L \). Consider

(a) Two particles.
(b) Three particles.
(c) Four particles.

A: part (a)

The problem statement doesn’t include the dimensionality of the box. The simplest case is the one dimensional box, for which the wave function of one particle is

\begin{equation}\label{eqn:noninteractingParticlesInABox:20}
\psi_1(x) = \sqrt{\frac{2}{L}} \sin\lr{ \frac{n \pi x}{L} },
\end{equation}

and the energy of that particle is

\begin{equation}\label{eqn:noninteractingParticlesInABox:40}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 n^2.
\end{equation}

If the box is two dimensional the energy is

\begin{equation}\label{eqn:noninteractingParticlesInABox:60}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ n_1^2 + n_2^2 },
\end{equation}

and if it’s a 3D box, we have

\begin{equation}\label{eqn:noninteractingParticlesInABox:80}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ n_1^2 + n_2^2 + n_3^2}.
\end{equation}

Suppose we are considering the 3D box. In statistical mechanics when we are considering particles Fermions, they are indistinguishable, and thus not allowed to share the same spin state at a given energy level. However, for distinguishable particles, that restriction doesn’t exist, and we can have two (or more) such particles in the lowest order energy state. The lowest such energy is

\begin{equation}\label{eqn:noninteractingParticlesInABox:100}
\begin{aligned}
E_{1,1,1 ;1,1,1}
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 6 \times 1^2 } \\
&=
\frac{6}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}
\end{equation}

The particle spin states can be any of \( \ket{++}, \ket{+-}, \ket{-+}, \ket{–} \), so there is a four way degeneracy in the ground state.

the next lowest energy level is

\begin{equation}\label{eqn:noninteractingParticlesInABox:120}
\begin{aligned}
E_{1,1,2 ;1,1,1}
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 5 \times 1^2 + 2^2 } \\
&=
\frac{9}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

where there are \( \binom{6}{1} = 6 \) ways to pick such a state for each variation of spin, for a total \( 6 \times 4 = 24 \) way degeneracy. Finally, since \( 2^2 + 2^2 < 3^2 + 1^2 \), the next lowest energy level is \begin{equation}\label{eqn:noninteractingParticlesInABox:140} \begin{aligned} E_{1,2,2 ;1,1,1} &= \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 4 \times 1^2 + 2 \times 2^2 } \\ &= \frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2, \end{aligned} \end{equation} with a \( \binom{6}{2} \times 4 = 15 \times 4 = 60 \) way degeneracy for this energy level.

A: part (b)

For three particles (the two particle case wasn’t actually in the problem statement, but seemed an easier starting place), the lowest energy state for a 3D box is

\begin{equation}\label{eqn:noninteractingParticlesInABox:160}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 9 \times 1^2 } \\
&=
\frac{9}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}
\end{equation}

There are now \( 2^3 = 8 \) variations of spin \( \ket{+++}, \ket{++-}, \cdots \), so the ground state is 8-way degenerate. Next up is

\begin{equation}\label{eqn:noninteractingParticlesInABox:180}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 8 \times 1^2 + 2^2 } \\
&=
\frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

where there is a \( \binom{9}{1} \times 8 = 9 \times 8 = 72 \) way degeneracy in this energy level. Finally, the next lowest energy level is

\begin{equation}\label{eqn:noninteractingParticlesInABox:200}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 7 \times 1^2 + 2 \times 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

with a \( \binom{9}{2} \times 8 = 36 \times 8 = 288 \) way degeneracy for this energy level.

A: part (c)

For four particles the lowest energy state for a 3D box is

\begin{equation}\label{eqn:noninteractingParticlesInABox:220}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 12 \times 1^2 } \\
&=
\frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}
\end{equation}

There are now \( 2^4 = 16 \) variations of spin \( \ket{++++}, \ket{+++-}, \cdots \), so the ground state is 16-way degenerate. For the second level

\begin{equation}\label{eqn:noninteractingParticlesInABox:240}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 11 \times 1^2 + 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

where there is a \( \binom{12}{1} \times 16 = 12 \times 16 = 192 \) way degeneracy in this energy level. Finally, the next lowest energy level is

\begin{equation}\label{eqn:noninteractingParticlesInABox:260}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 7 \times 1^2 + 2 \times 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

with a \( \binom{12}{2} \times 16 = 66 \times 16 = 1056 \) way degeneracy for this energy level.

References

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

Totally asymmetric potential

December 16, 2015 phy1520 , , ,

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Q: [1] pr 4.11

(a) Given a time reversal invariant Hamiltonian, show that for any energy eigenket

\begin{equation}\label{eqn:totallyAsymmetricPotential:20}
\expectation{\BL} = 0.
\end{equation}

(b) If the wave function of such a state is expanded as

\begin{equation}\label{eqn:totallyAsymmetricPotential:40}
\sum_{l,m} F_{l m} Y_{l m}(\theta, \phi),
\end{equation}

what are the phase restrictions on \( F_{lm} \)?

A: part (a)

For a time reversal invariant Hamiltonian \( H \) we have

\begin{equation}\label{eqn:totallyAsymmetricPotential:60}
H \Theta = \Theta H.
\end{equation}

If \( \ket{\psi} \) is an energy eigenstate with eigenvalue \( E \), we have

\begin{equation}\label{eqn:totallyAsymmetricPotential:80}
\begin{aligned}
H \Theta \ket{\psi}
&= \Theta H \ket{\psi} \\
&= \lambda \Theta \ket{\psi},
\end{aligned}
\end{equation}

so \( \Theta \ket{\psi} \) is also an eigenvalue of \( H \), so can only differ from \( \ket{\psi} \) by a phase factor. That is

\begin{equation}\label{eqn:totallyAsymmetricPotential:100}
\begin{aligned}
\ket{\psi’}
&=
\Theta \ket{\psi} \\
&= e^{i\delta} \ket{\psi}.
\end{aligned}
\end{equation}

Now consider the expectation of \( \BL \) with respect to a time reversed state

\begin{equation}\label{eqn:totallyAsymmetricPotential:120}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\bra{ \psi} \Theta^{-1} \BL \Theta \ket{\psi} \\
&=
\bra{ \psi} (-\BL) \ket{\psi},
\end{aligned}
\end{equation}

however, we also have

\begin{equation}\label{eqn:totallyAsymmetricPotential:140}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\lr{ \bra{ \psi} e^{-i\delta} } \BL \lr{ e^{i\delta} \ket{\psi} } \\
&=
\bra{\psi} \BL \ket{\psi},
\end{aligned}
\end{equation}

so we have \( \bra{\psi} \BL \ket{\psi} = -\bra{\psi} \BL \ket{\psi} \) which is only possible if \( \expectation{\BL} = \bra{\psi} \BL \ket{\psi} = 0\).

A: part (b)

Consider the expansion of the wave function of a time reversed energy eigenstate

\begin{equation}\label{eqn:totallyAsymmetricPotential:160}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} e^{i\delta} \ket{\psi} \\
&=
e^{i\delta} \braket{\Bx}{\psi},
\end{aligned}
\end{equation}

and then consider the same state expanded in the position basis

\begin{equation}\label{eqn:totallyAsymmetricPotential:180}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \ket{\Bx’}\bra{\Bx’} } \ket{\psi} \\
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} } \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \Theta \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \ket{\Bx’} \\
&=
\int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \braket{\Bx}{\Bx’} \\
&=
\int d^3 \Bx’ \braket{\psi}{\Bx’} \delta(\Bx- \Bx’) \\
&=
\braket{\psi}{\Bx}.
\end{aligned}
\end{equation}

This demonstrates a relationship between the wave function and its complex conjugate

\begin{equation}\label{eqn:totallyAsymmetricPotential:200}
\braket{\Bx}{\psi} = e^{-i\delta} \braket{\psi}{\Bx}.
\end{equation}

Now expand the wave function in the spherical harmonic basis

\begin{equation}\label{eqn:totallyAsymmetricPotential:220}
\begin{aligned}
\braket{\Bx}{\psi}
&=
\int d\Omega \braket{\Bx}{\ncap}\braket{\ncap}{\psi} \\
&=
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) \\
&=
e^{-i\delta}
\lr{
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) }^\conj \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj Y_{lm}^\conj(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj (-1)^m Y_{l,-m}(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{l,-m}(r)}^\conj (-1)^m Y_{l,m}(\theta, \phi),
\end{aligned}
\end{equation}

so the \( F_{lm} \) functions are constrained by

\begin{equation}\label{eqn:totallyAsymmetricPotential:240}
F_{lm}(r) = e^{-i\delta} \lr{ F_{l,-m}(r)}^\conj (-1)^m.
\end{equation}

References

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

Commutators for some symmetry operators

December 16, 2015 phy1520 , , ,

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Q: [1] pr 4.2

If \( \mathcal{T}_\Bd \), \( \mathcal{D}(\ncap, \phi) \), and \( \pi \) denote the translation, rotation, and parity operators respectively. Which of the following commute and why

  • (a) \( \mathcal{T}_\Bd \) and \( \mathcal{T}_{\Bd’} \), translations in different directions.
  • (b) \( \mathcal{D}(\ncap, \phi) \) and \( \mathcal{D}(\ncap’, \phi’) \), rotations in different directions.
  • (c) \( \mathcal{T}_\Bd \) and \( \pi \).
  • (d) \( \mathcal{D}(\ncap,\phi)\) and \( \pi \).

A: (a)

Consider
\begin{equation}\label{eqn:symmetryOperatorCommutators:20}
\begin{aligned}
\mathcal{T}_\Bd \mathcal{T}_{\Bd’} \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{\Bx + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd},
\end{aligned}
\end{equation}

and the reverse application of the translation operators
\begin{equation}\label{eqn:symmetryOperatorCommutators:40}
\begin{aligned}
\mathcal{T}_{\Bd’} \mathcal{T}_{\Bd} \ket{\Bx}
&=
\mathcal{T}_{\Bd’} \ket{\Bx + \Bd} \\
&=
\ket{\Bx + \Bd + \Bd’} \\
&=
\ket{\Bx + \Bd’ + \Bd}.
\end{aligned}
\end{equation}

so we see that

\begin{equation}\label{eqn:symmetryOperatorCommutators:60}
\antisymmetric{\mathcal{T}_\Bd}{\mathcal{T}_{\Bd’}} \ket{\Bx} = 0,
\end{equation}

for any position state \( \ket{\Bx} \), and therefore in general they commute.

A: (b)

That rotations do not commute when they are in different directions (like any two orthogonal directions) need not be belaboured.

A: (c)

We have
\begin{equation}\label{eqn:symmetryOperatorCommutators:80}
\begin{aligned}
\mathcal{T}_\Bd \pi \ket{\Bx}
&=
\mathcal{T}_\Bd \ket{-\Bx} \\
&=
\ket{-\Bx + \Bd},
\end{aligned}
\end{equation}

yet
\begin{equation}\label{eqn:symmetryOperatorCommutators:100}
\begin{aligned}
\pi \mathcal{T}_\Bd \ket{\Bx}
&=
\pi \ket{\Bx + \Bd} \\
&=
\ket{-\Bx – \Bd} \\
&\ne
\ket{-\Bx + \Bd}.
\end{aligned}
\end{equation}

so, in general \( \antisymmetric{\mathcal{T}_\Bd}{\pi} \ne 0 \).

A: (d)

We have

\begin{equation}\label{eqn:symmetryOperatorCommutators:120}
\begin{aligned}
\pi \mathcal{D}(\ncap, \phi) \ket{\Bx}
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \mathcal{D}(\ncap, \phi) \pi^\dagger \pi \ket{\Bx} \\
&=
\pi \lr{ \sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} } \pi^\dagger \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i (\pi \BJ \pi^\dagger) \cdot (\pi \ncap \pi^\dagger) )^k}{k!} \pi \ket{\Bx} \\
&=
\sum_{k=0}^\infty \frac{(-i \BJ \cdot \ncap)^k}{k!} \pi \ket{\Bx} \\
&=
\mathcal{D}(\ncap, \phi) \pi \ket{\Bx},
\end{aligned}
\end{equation}

so \( \antisymmetric{\mathcal{D}(\ncap, \phi)}{\pi} \ket{\Bx} = 0 \), for any position state \( \ket{\Bx} \), and therefore these operators commute in general.

References

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

Plane wave and spinor under time reversal

December 16, 2015 phy1520 , , , , ,

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Q: [1] pr 4.7

  1. (a)
    Find the time reversed form of a spinless plane wave state in three dimensions.

  2. (b)
    For the eigenspinor of \( \Bsigma \cdot \ncap \) expressed in terms of polar and azimuthal angles \( \beta\) and \( \gamma \), show that \( -i \sigma_y \chi^\conj(\ncap) \) has the reversed spin direction.

A: part (a)

The Hamiltonian for a plane wave is

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:20}
H = \frac{\Bp^2}{2m} = i \PD{t}.
\end{equation}

Under time reversal the momentum side transforms as

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:40}
\begin{aligned}
\Theta \frac{\Bp^2}{2m} \Theta^{-1}
&=
\frac{\lr{ \Theta \Bp \Theta^{-1}} \cdot \lr{ \Theta \Bp \Theta^{-1}} }{2m} \\
&=
\frac{(-\Bp) \cdot (-\Bp)}{2m} \\
&=
\frac{\Bp^2}{2m}.
\end{aligned}
\end{equation}

The time derivative side of the equation is also time reversal invariant
\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:60}
\begin{aligned}
\Theta i \PD{t}{} \Theta^{-1}
&=
\Theta i \Theta^{-1} \Theta \PD{t}{} \Theta^{-1} \\
&=
-i \PD{(-t)}{} \\
&=
i \PD{t}{}.
\end{aligned}
\end{equation}

Solutions to this equation are linear combinations of

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:80}
\psi(\Bx, t) = e^{i \Bk \cdot \Bx – i E t/\Hbar},
\end{equation}

where \( \Hbar^2 \Bk^2/2m = E \), the energy of the particle. Under time reversal we have

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:100}
\begin{aligned}
\psi(\Bx, t)
\rightarrow e^{-i \Bk \cdot \Bx + i E (-t)/\Hbar}
&= \lr{ e^{i \Bk \cdot \Bx – i E (-t)/\Hbar} }^\conj \\
&=
\psi^\conj(\Bx, -t)
\end{aligned}
\end{equation}

A: part (b)

The text uses a requirement for time reversal of spin states to show that the Pauli matrix form of the time reversal operator is

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:120}
\Theta = -i \sigma_y K,
\end{equation}

where \( K \) is a complex conjugating operator. The form of the spin up state used in that demonstration was

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:140}
\begin{aligned}
\ket{\ncap ; +}
&= e^{-i S_z \beta/\Hbar} e^{-i S_y \gamma/\Hbar} \ket{+} \\
&= e^{-i \sigma_z \beta/2} e^{-i \sigma_y \gamma/2} \ket{+} \\
&= \lr{ \cos(\beta/2) – i \sigma_z \sin(\beta/2) }
\lr{ \cos(\gamma/2) – i \sigma_y \sin(\gamma/2) } \ket{+} \\
&= \lr{ \cos(\beta/2) – i \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \sin(\beta/2) }
\lr{ \cos(\gamma/2) – i \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \sin(\gamma/2) } \ket{+} \\
&=
\begin{bmatrix}
e^{-i\beta/2} & 0 \\
0 & e^{i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2) & -\sin(\gamma/2) \\
\sin(\gamma/2) & \cos(\gamma/2)
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{-i\beta/2} & 0 \\
0 & e^{i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2) \\
\sin(\gamma/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\gamma/2)
e^{-i\beta/2}
\\
\sin(\gamma/2)
e^{i \beta/2}
\end{bmatrix}.
\end{aligned}
\end{equation}

The state orthogonal to this one is claimed to be

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:180}
\begin{aligned}
\ket{\ncap ; -}
&= e^{-i S_z \beta/\Hbar} e^{-i S_y (\gamma + \pi)/\Hbar} \ket{+} \\
&= e^{-i \sigma_z \beta/2} e^{-i \sigma_y (\gamma + \pi)/2} \ket{+}.
\end{aligned}
\end{equation}

We have

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:200}
\begin{aligned}
\cos((\gamma + \pi)/2)
&=
\textrm{Re} e^{i(\gamma + \pi)/2} \\
&=
\textrm{Re} i e^{i\gamma/2} \\
&=
-\sin(\gamma/2),
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:220}
\begin{aligned}
\sin((\gamma + \pi)/2)
&=
\textrm{Im} e^{i(\gamma + \pi)/2} \\
&=
\textrm{Im} i e^{i\gamma/2} \\
&=
\cos(\gamma/2),
\end{aligned}
\end{equation}

so we should have

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:240}
\ket{\ncap ; -}
=
\begin{bmatrix}
-\sin(\gamma/2)
e^{-i\beta/2}
\\
\cos(\gamma/2)
e^{i \beta/2}
\end{bmatrix}.
\end{equation}

This looks right, but we can sanity check orthogonality

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:260}
\begin{aligned}
\braket{\ncap ; -}{\ncap ; +}
&=
\begin{bmatrix}
-\sin(\gamma/2)
e^{i\beta/2}
&
\cos(\gamma/2)
e^{-i \beta/2}
\end{bmatrix}
\begin{bmatrix}
\cos(\gamma/2)
e^{-i\beta/2}
\\
\sin(\gamma/2)
e^{i \beta/2}
\end{bmatrix} \\
&=
0,
\end{aligned}
\end{equation}

as expected.

The task at hand appears to be the operation on the column representation of \( \ket{\ncap; +} \) using the Pauli representation of the time reversal operator. That is

\begin{equation}\label{eqn:timeReversalPlaneWaveAndSpinor:160}
\begin{aligned}
\Theta \ket{\ncap ; +}
&=
-i \sigma_y K
\begin{bmatrix}
e^{-i\beta/2} \cos(\gamma/2) \\
e^{i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
-i \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{i\beta/2} \cos(\gamma/2) \\
e^{-i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
e^{i\beta/2} \cos(\gamma/2) \\
e^{-i \beta/2} \sin(\gamma/2)
\end{bmatrix} \\
&=
\begin{bmatrix}
-e^{-i \beta/2} \sin(\gamma/2) \\
e^{i\beta/2} \cos(\gamma/2) \\
\end{bmatrix} \\
&= \ket{\ncap ; -},
\end{aligned}
\end{equation}

which is the result to be demononstrated.

References

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