4
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ I'm not sure what you mean by $\neq$. They are always isomorphic. Also, I don't see what the order in the tensor product has to do with experiments. If I have a cavity and an atom, it is clear which is which, regardless of the order I choose for their Hilbert spaces in the tensor product. $\endgroup$ – Norbert Schuch Nov 5 '18 at 13:30
  • $\begingroup$ I think this is a great question & have been trying to figure out something similar if I am indeed understanding you correctly. My current thought is that you could XOR & XNOR the inputs (see ven diagrams on en.wikipedia.org/wiki/Exclusive_or + en.wikipedia.org/wiki/Logical_equality). $\endgroup$ – meowzz Nov 6 '18 at 0:00
11
$\begingroup$

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 $H_B\otimes H_A$.

$\endgroup$
  • $\begingroup$ Thank you for another amazing answer! I am curious if you might be able to comment on how this question/answer relates to CPT symmetries & the complexity class of CPT related computations. $\endgroup$ – meowzz Nov 5 '18 at 22:43
  • 4
    $\begingroup$ I'm sorry, but I don't know anything at all about CPT symmetries or the complexity class of CPT related computations. If you post your question as a new question, possibly someone with expertise on these topics can give you an answer. $\endgroup$ – John Watrous Nov 5 '18 at 23:27
  • $\begingroup$ Prof. @John Watrous, I do understand that the eigenspectrum is the same for both cases, as I did a short calculation. However, the ordering of the two entries in the eigenfunction is flipped for the two cases, in the eigenvectors below. Does this mean that for the same observable (the eigenvalue), we have a different eigenfunction? (that is, another state in another basis)? The density matrices in the two cases will be clearly different and hence the projectors. $\endgroup$ – Siddhānt Singh Nov 6 '18 at 10:55
  • $\begingroup$ Prof. @JohnWatrous, I understand this very well and it is true for the observables as the eigenvalues and their ordering is the same for the two cases. But is it true for the eigenstates as well? Please look at the answer I posted in the same thread. Let me know about it. Sorry, this is some doubt which sounds trivial, but I always had it. $\endgroup$ – Siddhānt Singh Nov 6 '18 at 10:57
  • 4
    $\begingroup$ Hello Siddhānt. If you'd like to ask the entire community about this, you could post it as a separate question. (This is not a discussion forum.) $\text{---}$ Having said that: From John Watrous' answer, it should be clear that the operators $A \otimes B$ and $B \otimes A$ will be different, in the same way that the ordered pairs $(a,b)$ and $(b,a)$ will be different: they are different mathematical objects, which contain much the same information (as with the list of eigenvalues) but in a different order. $A \otimes B$ is how we represent the physical system, but the system is not the math. $\endgroup$ – Niel de Beaudrap Nov 6 '18 at 11:06
5
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thank you, I get the idea. However, $A\otimes B$ is very different from $B\otimes A$ by construction. The density operators (which are not separable) will be completely different in two composite spaces. This is exactly where my question is. Are the observables in both the ways of ordering of the product the exact same? $\endgroup$ – Siddhānt Singh Nov 5 '18 at 14:49
  • $\begingroup$ Again: How you order the tensor products has nothing to do with experiments. I think you are asking a pure theory question. But maybe it is best if you update your question by giving an example of what you mean. It is very hard to give a clear answer without anything to hold on. $\endgroup$ – Norbert Schuch Nov 5 '18 at 14:58
  • $\begingroup$ "Actually, even if R is commutative, then the tensor product is not always commutative!" -Source Thoughts? $\endgroup$ – meowzz Nov 5 '18 at 22:33
  • $\begingroup$ Do you maybe care to elaborate first? What are your thoughts? What kind of thoughts do you want to hear from me? What does this have to do with the question or my answer? $\endgroup$ – Norbert Schuch Nov 5 '18 at 23:12
  • 1
    $\begingroup$ @SiddhāntSingh The order of the tensor product is a notational choice and has no physical consequences. If you were from right to left, this would also not change the physics. $\endgroup$ – Norbert Schuch Nov 6 '18 at 11:19
1
$\begingroup$

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 $\rho_{BA}=S\rho_{AB}S$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.