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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.