# Quantum mechanics, are simultaneous eigenstates to be intended always as a tensorial product of two eigenstates?

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 $$|\epsilon_1,\epsilon_2\rangle=|\epsilon_1\rangle\otimes|\epsilon_2\rangle?$$

• Why do you think so? May 14, 2022 at 19:30

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 $$H_{1/2}\oplus H_1$$.

• 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: $\psi_1(x)\psi_2(x)$? May 14, 2022 at 20:22
• @Salmone It depends on the operators - in general it's simply not the case. May 14, 2022 at 20:42
• The parity operator commutes with $H$ for any symmetric potential and the solutions are not products. May 14, 2022 at 20:44
• @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? May 15, 2022 at 0:27
• @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. May 15, 2022 at 9:26

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.

• 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? May 14, 2022 at 19:49
• 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. May 14, 2022 at 20:41
• @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). May 14, 2022 at 22:13

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 $$|a\rangle \overset{?}=|a\rangle\otimes |a^2\rangle$$.