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The question is the one of the title, let $\hat{O}_1$ and $\hat{O}_2$ two commuting operators: $[\hat{O}_1,\hat{O}_2]=0$, there is an orthonormal basis formed by their simultaneous eigenstates. These eigenstates could be written as $|\epsilon_1,\epsilon_2\rangle$ so that $\hat{O}_1|\epsilon_1,\epsilon_2\rangle=\epsilon_1|\epsilon_1,\epsilon_2\rangle$ and $\hat{O}_2|\epsilon_1,\epsilon_2\rangle=\epsilon_2|\epsilon_1,\epsilon_2\rangle$. Are $|\epsilon_1,\epsilon_2\rangle=|\epsilon_1\rangle\otimes|\epsilon_2\rangle?$

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  • $\begingroup$ Why do you think so? $\endgroup$ May 14, 2022 at 19:30

3 Answers 3

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No. A simple counterexample to this is taking a spin-1/2 representation of the rotation group $H_{1/2}$ and a spin-1 representation $H_1$ - the direct sum $H_{1/2}\oplus H_1$ is a five-dimensional Hilbert space in which $S^2$ and e.g. $S_z$ commute, and so it has a basis of common eigenstates $\lvert 1/2,1/2\rangle, \lvert 1/2,-1/2\rangle,\lvert 1,1\rangle, \lvert 1,0\rangle, \lvert 1,-1\rangle$. But since it is five-dimensional and 5 is prime, it is not the tensor product of any other space except in the trivial way (tensoring with the one-dimensional Hilbert space).

More explicitly, if we had $\lvert 1/2,-1/2\rangle = \lvert 1/2\rangle_{S^2}\otimes\lvert -1/2\rangle_{S_z}$ and $\lvert 1,1\rangle = \lvert 1\rangle_{S^2}\otimes\lvert 1\rangle_{S_z}$, then we would also have to have a state like $\lvert 1\rangle_{S^2}\otimes \lvert 1/2\rangle_{S_z}$, i.e. with eigenvalue 1 for $S^2$ and $1/2$ for $S_z$, but of course there is no such state in $H_{1/2}\oplus H_1$.

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  • $\begingroup$ Ok, what about the eigenfunctions in Schroedinger representation? If I have the simultaneous eigenstates of the questions and I'm interested in simultaneous eigenfunctions, are these one factorizable, that is: $\psi_1(x)\psi_2(x)$? $\endgroup$
    – Salmon
    May 14, 2022 at 20:22
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    $\begingroup$ @Salmone It depends on the operators - in general it's simply not the case. $\endgroup$
    – ACuriousMind
    May 14, 2022 at 20:42
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    $\begingroup$ The parity operator commutes with $H$ for any symmetric potential and the solutions are not products. $\endgroup$ May 14, 2022 at 20:44
  • $\begingroup$ @ACuriousMind Sorry but, when we use separable Hamiltonian method, don't we say that if the Hamiltonian is separable then the total eigenfunction is the product of singles eigenfunctions? $\endgroup$
    – Salmon
    May 15, 2022 at 0:27
  • $\begingroup$ @Salmone Sure (see this answer of mine) but that's not what you asked about - you asked about some unspecified operators $O_i$, and as I said, in general you don't get separability for that. $\endgroup$
    – ACuriousMind
    May 15, 2022 at 9:26
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Of course not.

  1. Eigenstates of the harmonic oscillator are simultaneously eigenstates of $H$ and the parity operator,
  2. The spherical harmonics are eigenstates of $L^2$ and $L_z$,

Now the eigenstates can be written as $\vert E_n, \pm\rangle$ or $\vert \ell,m\rangle$ but this hardly denotes a tensor product.

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  • $\begingroup$ For example, for $|l,m\rangle$ example, if I want to consider the simultaneous eigenfunctions, (in this case speherical harmonics), must them be the product of two eigenfunctions? The eigenfunction of $L^2$ and the eigenfunction of $L_z$? In general, if I have $|\epsilon_1,\epsilon_2\rangle$ and I want to wirte down simultaneous eigenfunctions, are these one the product of single eigenfunctions? $\endgroup$
    – Salmon
    May 14, 2022 at 19:49
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    $\begingroup$ So you mean separation of variables leads to a product of solutions (which doesn’t imply tensor product)? Fair enough for the spherical harmonics, but not for h.o. and parity. $\endgroup$ May 14, 2022 at 20:41
  • $\begingroup$ @ZeroToHero When we talk about separable Hamiltonians, don't we say that the total eigenfunctions are equal to the product of single eigenfunctions of single Hamiltonians? Isn't this a general rule? Maybe I'm confusing (separable Hamiltonians) and (commuting operators with simultaneous eigenstates). $\endgroup$
    – Salmon
    May 14, 2022 at 22:13
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The operators $\hat{O}_1$ and $\hat{O}_2$ don't necessarily act on distinct Hilbert spaces. Take the (trivial) example of the operators $\hat{A}$ and $\hat{A}^2$. They obviously commute, but have different eigenvalues. $$ \hat{A} | a \rangle = a | a \rangle\\ \hat{A}^2 | a \rangle = a^2 | a \rangle $$

If they both act on the same Hilbert space, it doesn't make sense to write this state as $|a\rangle \overset{?}=|a\rangle\otimes |a^2\rangle$.

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