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To show that two operators $\hat{A}$ and $\hat{B}$ commute, we can check whether $\hat{A}\hat{B}f(x)$ = $\hat{B}\hat{A}f(x)$.

My question is regarding the function $f(x)$. To check that $\hat{A}$ and $\hat{B}$ commute, can we use any function? Or are there requirements on the function we "test" commutativity with?

This relates to an exercise in a textbook which asks me to show why two operators in quantum chemistry commute. Both operators act on functions of multiple electron coordinates, e.g. $\Psi(\mathbf{x}_{1},\mathbf{x}_{2},\ldots,\mathbf{x}_{N})$, where $\mathbf{x}_{i}$ is the vector coordinate of electron $i$. One of the operators, $\hat{P}_{n}$ permutes the coordinates of the electrons (swapping the positions of two electrons, for $N$ electrons, there are $N!$ permutations), the other, $\hat{H}_{0}$, is a sum over one-electron operators, $\sum_{i}^{N} \hat{f}(i)$, where $\hat{f}(i)$ only acts on electron $i$. Whilst I know how $\hat{P}_{n}$ will act on any general function over multiple electron coordinates, I only know the outcome of the action of $\hat{H}_{0}$ on $\Psi$ which are made up of one-electron eigenfunctions, $\chi_{j}(\mathbf{x}_{i})$, of the operators $\hat{f}(i)$, e.g. determinants or products of these one-electron eigenfunctions.

If I can show that the operators commute using a test function $\Psi$ of this specific form, does this imply the operators have the general property that they commute?

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Yes, you can conclude that in this case, the reason being that the determinants (or, in the bosonic case, the symmetrized products) of single-particle functions form a basis of the Fock space; more precisely, the set of all determinants or symmetrized products of all possible combinations of $n$ basis elements of a single-particle basis form a basis of the $n$-particle sector of the Fock space.

Generally, if the test functions form a basis, commutativity for the entire space follows by linearity. The conclusion is generally not valid if the test functions don't form a basis; the operators might theoretically commute on the entire space except for just a two-dimensional subspace.

By the way, why do you know the action of $\hat H_0$ only for determinants/products? You should be able to show that $\hat H_0$ commutes with $\hat P_n$ quite generally, without applying it to any test functions.

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Without introducing a test function, I do not know how I would demonstrate commutativity. I say that I only know the action of $\hat{H}_{0}$ on determinants/products of the basis functions $\chi_{a}$ because I know that $\hat{f}_{i}\chi_{a}(\mathbf{x}_{i}) = \epsilon_{a} \chi_{a}(\mathbf{x}_{i})$ (an eigenvalue problem) but not the outcome of $\hat{f}_{i}$ acting on any arbitrary function of $\mathbf{x}_{i}$ (not necessarily an eigenfunction of $\hat{f}_{i}$). What do you mean by "commutativity for the entire space follows by linearity"? – James Womack Aug 12 '12 at 14:27

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