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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:)

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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.

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  • $\begingroup$ Thanks!! Could you explain a bit about how the two equations $𝑃_{12}𝑃_{34}=πΌβŠ—πœŽ_1$ and $𝑃_{13}𝑃_{24}=𝜎_1βŠ—I$ come from? $\endgroup$ – Zhengrong Oct 2 '20 at 19:10
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    $\begingroup$ You just write down the 4x4 matrices that do this job. They are symmetric, off-diagonal, and do just that: interchange the suitable entries of all 4-vectors. You then condense them into direct products of 2x2 ones, as always. $\endgroup$ – Cosmas Zachos Oct 2 '20 at 19:22
  • $\begingroup$ Thank you so much:) $\endgroup$ – Zhengrong Oct 2 '20 at 20:21

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