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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.

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