# Combination of 'transposition operators': do they commute?

Suppose I have the Hamiltonian defined as $$H =\hat A\hat B+\hat C\hat D$$, where the operator $$A,B,C and D$$ are square matrices. If I label the positions of $$A,B,C,D$$ as $$1,2,3,4$$. Now I want to apply the transposition operator on the Hamiltonian, for example:

$$P_{12}P_{34}H=\hat B\hat A+\hat D\hat C; \\ P_{13}P_{24}H=\hat C\hat D+\hat A\hat B$$

The question I'm considering is do the combination of transposition operators $$P_{12}P_{34}$$ and $$P_{13}P_{24}$$ commute? In a simple case, if $$H =\hat A\hat B\hat C\hat D$$, I think the answer if yes, since $$P_{12}P_{34}P_{13}P_{24}\hat A\hat B\hat C \hat D=P_{13}P_{24}P_{12}P_{34}ABCD = DCBA$$. However, if the Hamiltonian is composed of two parts (like in this case), does this relation still hold? Or whether $$[P_{12}P_{34},P_{13}P_{24}]$$ equal to $$0$$ does not depend on the Hamiltonian?

Thanks:)

Of course they commute, in general, manifestly, $$P_{12}P_{34} = I\otimes \sigma_1 ; \qquad P_{13}P_{24} = \sigma_1\otimes I,$$ so $$P_{12}P_{34} ~ P_{13}P_{24}= P_{13}P_{24}~P_{12}P_{34} .$$
Of course, $$\sigma_1=\begin{bmatrix} 0& 1\\1 &0\end{bmatrix}$$ permutes the two entries of 2-vectors/spinors it acts on.
• Thanks!! Could you explain a bit about how the two equations $𝑃_{12}𝑃_{34}=𝐼⊗𝜎_1$ and $𝑃_{13}𝑃_{24}=𝜎_1⊗I$ come from? – Zhengrong Oct 2 '20 at 19:10