What role does the non-commutativity of the tensor product play in experimental quantum computation?

Question

We know that $H_A\otimes H_B\neq H_B\otimes H_A$ (in general). Theoretically, we know the formalism and what observables to construct from the two compositions possible, but we never talk about both the possibilities. I wish to know that how experimentally the Measurements or Evolutions are done over such composite systems (let's just assume a bipartition as above). How does the experimentalist know whether he is working in the $A\otimes B$ or $B\otimes A$ composite Hilbert Space?

Answer 1

For many questions that appear on this site, and about quantum information and computation in general, it is possible to ask a completely classical version of the question, and often the (sometimes obvious) answer that one finds in the more familiar classical setting translates directly to the quantum setting. In this case, a reasonable classical version of the question asks what role the non-commutativity of the Cartesian product plays in experimental classical computing (or, let's say, in practical implementations of classical computation).

Suppose we have system $A$ that can be in any classical state drawn from a set $\mathcal{A}$, and a system $B$ that can be in any classical state drawn from the set $\mathcal{B}$. If we put system $A$ and system $B$ next to each other on the table, then we can represent the classical state of the two systems together as an element of the Cartesian product $\mathcal{A}\times\mathcal{B}$. Note that there is an implicit assumption here, which is that the two systems are distinguishable, and we're deciding more or less arbitrarily that when we talk about a state $(a,b)\in\mathcal{A}\times\mathcal{B}$ that the state $a$ of system $A$ is listed first and the state $b$ of system $B$ is listed second. We could just as easily have decided to represent the classical state of the two systems together as an element of the Cartesian product $\mathcal{B}\times\mathcal{A}$, with the understanding that the state of system $B$ now gets listed first.

As an aside, if the two systems were indistinguishable, implying that $\mathcal{A} = \mathcal{B}$, and further we placed the two systems in a bag rather than on the table, then I guess there would really be no difference between $(a,b)$ and $(b,a)$. For this reason we would probably not use the Cartesian product to represent states of the bagged systems -- maybe we would use the set of all multisets of size 2 instead -- but let us forget about this situation and assume $A$ and $B$ are distinguishable for simplicity.

Now, what role does this play in experiments or practical applications of classical computing? How does an experimenter or programmer know he or she is working in the $\mathcal{A}\times\mathcal{B}$ or $\mathcal{B}\times\mathcal{A}$ state space? When you think about the question this way, I believe it may come into focus. My answer, which is consistent with the other answers that concern the quantum setting, is that it really doesn't play any role at all, and the experimenter/programmer knows because it was his or her decision which order to use. We know the difference between the systems $A$ and $B$, and the decision to represent states of the two systems together by elements of $\mathcal{A}\times\mathcal{B}$ or $\mathcal{B}\times\mathcal{A}$ is totally arbitrary -- but once the decision is made we stick with it to avoid confusion. The decision will not affect any calculations we do, so long as the calculations are consistent with the decision of which order to use.

To my eye, at a fundamental level there is no difference between the classical version of this question and the quantum version. We decide whether to represent states of the compound quantum system using the space $H_A\otimes H_B$ or $H_B\otimes H_A$, and that's all there is to it. You'll get exactly the same results of any calculations you perform, so long as your calculations are consistent with the choice to use $H_A\otimes H_B$ or $H_B\otimes H_A$.

Answer 2

When you say $\neq$ I presume you are talking about the implied basis in usual ordering like (00, 01, 02, 10 etc). Otherwise you would have the isomorphism of Hilbert spaces vs an equality statement. That is, AB implies a certain ordered basis and BA a different one.

The experiment has it's observables on the combined system in a basis independent way. If the experimentalist wants to put their results down, they can choose whatever basis they like.

The distinction goes into the question being asked. What is the second entry of vector v in Hilbert space that combines A and B is not a well defined question. What is the second entry with respect to a given ordered basis is. The experimentalist has to ask the second in order to get an answer. You have to ask a sensible question if you want a sensible answer.

Answer 3

The order in the tensor product is a convention and has nothing to do with experiments.

As an example, if I have a cavity (with photons in it, $H_A$) and an atom (with internal states, $H_B$), it is clear which is the atom and which is the cavity, regardless of the order ones chooses for their Hilbert spaces in the tensor product when describing the setup theoretically.

Answer 4

The two spaces $A$ and $B$ are just labels, with arbitrary ordering. For distinguishable qubits (or more general), the experimentalist can just say "this one's $A$, and this other one's $B$". If you swap the labels, you need to swap the labels everywhere - in both the Hamiltonian and the state (including eigenvectors, density matrix etc).

In other words, if I define a swap operator $S$ such that $$ S(H_A\otimes H_B)S=H_B\otimes H_A, $$ then evolution of states can be calculated either using $$ e^{-i H_A\otimes H_B t}|\psi_{AB}\rangle \quad\text{or}\quad e^{-i H_B\otimes H_A t}|\psi_{BA}\rangle $$ where $|\psi_{BA}\rangle=S|\psi_{AB}\rangle$. Or, if you're working with a density matrix, you have $\rho_{BA}=S\rho_{AB}S$.