Suppose I have a complete set of two commuting observables $\hat{A}, \hat{B}$ for which:

$$ \hat{A} |{a} \rangle = a |a\rangle $$

$$ \hat{B} |b \rangle = b |b\rangle $$

Now, I can find a common eigenstate in the Hilbert space where $\hat{A}$ and $\hat{B}$ operates, or I can define a new Hilbert space which is the direct product of the space spanned by $|{a} \rangle$ and $|{b} \rangle$:

$$\{ |a,b \rangle \}_{H} = \{ |a \rangle \}_{H_1} \otimes \{ |b \rangle \}_{H_2}$$

Where $H$ are Hilbert spaces, and with the operators extension $A \to A\ \otimes \ 1$ ; $B \to 1\ \otimes \ B$.

In the new Hilbert space I can then choose the common eigenbasis as $|{a} \rangle \otimes|{b} \rangle$.

Now there is a rationale in Quantum Mechanics to choose one method or another, or is a matter choice ? And if indeed is a matter of choice, what are the situation in which one may prefer a method over another?

  • $\begingroup$ The question is somewhat circular, since, in order to have two operators commuting, you need to define them on the same basis. Even though I do agree that asking under which conditions a basis can be represented as a product basis makes sense. $\endgroup$
    – Roger V.
    Oct 6, 2020 at 11:58
  • 2
    $\begingroup$ I agree with @Vadim that there is a problem with the question. Neither A nor B acts on a product space, so their common basis can't be written as a product. $\endgroup$
    – NDewolf
    Oct 6, 2020 at 12:02
  • $\begingroup$ Note that you don't need to explicitly reference edits to the post. There is an edit history available for those who are interested. $\endgroup$ Oct 6, 2020 at 12:46

1 Answer 1


That depends on the Hilbertspace:

$A$ is defined only on the first Hilbertspace spanned by the $|a\rangle$ vectors. $B$ is similarly defined on a second Hilbertspace spanned by the $|b\rangle$ vectors.

You ask about vectors in the tensor product of the two spaces, where neither $A$ nor $B$ is defined.

If you extend them by $A \mapsto A\otimes 1$, than your statement is of course true, since the $A$ only "cares" about the first part of the tensor product.

If you want $A$ and $B$ to akt on the same hilbertspace, the vectors $|a\rangle \otimes |b\rangle$ are not in the correct space and there is no logical answer.

  • $\begingroup$ Ok, I understand what you say from a mathematics point of view, but in general what is the ''common'' method in QM ? Suppose I have a quantum isotropic harmonic oscillator in two dimension. If I choose $ (n_x, n_y) $ as quantum numbers I say that the complete set of observables are the hamiltonians $(\hat{H}_x, \hat{H}_y)$. Now, there is an intrinsic reason for why do consider usually the direct product of the eigenbasis of the two operator, or is just a choice ? $\endgroup$ Oct 6, 2020 at 12:24

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