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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.