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### Q: [1] pr 4.7

- (a)

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