Are tensored qubits commutative?

Question

I am given a solution to a problem, saying that

$$ |\psi\rangle_{ABC} = \frac{1}{\sqrt{2}}(|0\rangle_A \otimes |1\rangle_C + |1\rangle_A \otimes |0\rangle_C) \otimes |+\rangle_B $$

$$ = \frac{1}{2}(|001\rangle_{ABC} + |100\rangle_{ABC} + |011\rangle_{ABC} + |110\rangle_{ABC}) $$

However, simply performing out the tensors would give the same results but with the second and third qubits reversed, i.e.

$$ \frac{1}{2}(|010\rangle_{ACB} + |100\rangle_{ACB} + |011\rangle_{ACB} + |101\rangle_{ACB}) $$

since $B$ is on the right. Am I mistaken thinking that qubits cannot commute like that? How is the given answer correct?

Answer 1

It may be clearer if instead of all qubits, you put $A$ as a qudit with $d=d_A$ etc for B,C.

The first state is then in the Hilbert space $\mathbb{C}^{d_A} \otimes \mathbb{C}^{d_B} \otimes \mathbb{C}^{d_C}$

but the second is in $\mathbb{C}^{d_A} \otimes \mathbb{C}^{d_C} \otimes \mathbb{C}^{d_B}$

those two Hilbert spaces are isomorphic when you swap around the B and C positions. So the only difference between the two answers was because of this simple mismatch.

Now specialize to $d_A=d_B=d_C=2$. Now you should be more clear about which $2$ went where.