6
$\begingroup$

Say we have two representations of the observables from an abstract $C^*$-algebra $\mathcal A$ on two Hilbert spaces $H_1$ and $H_2$, i.e. consider the maps $\pi_1,\pi_2: \mathcal A \longrightarrow \mathcal B(H_1), \mathcal B(H_2)$.

If these representations are unitary equivalent, this means that there exists a unitary $U$ such that $\pi_2(A) = U\pi_1(A)U^{-1}$ for all $A\in \mathcal A$. Similarly, if an abstract state is normal with respect to both of these representations, i.e. can be represented by a density matrix on the respective Hilbert space, we find $\rho_2 = U\rho_1 U^{-1}$. And so all predictions we make are the same, irrespective of the Hilbert space on which we represent these observables and states.

However, even for unitary inequivalent representations we can find a unitary $U$ (as the Hilbert spaces are isomorphic) such that $\rho^\prime=U\rho_1 U^{-1}$ and $A^\prime = U\pi_1(A)U^{-1}$ on $H_2$, but with $\rho_2\neq \rho^\prime$ and $\pi_2(A)\neq A^\prime$.

But all expectation values and hence all predictions agree with our first representation. And isn't this the only important thing? So what is the significance of unitary (in-)equivalent representations? It seems that I am missing a crucial thing here.

$\endgroup$
3
  • $\begingroup$ See Haag's Theorem $\endgroup$ Jan 27, 2023 at 17:55
  • $\begingroup$ 1. "(as the Hilbert spaces are isomorphic)" You must assume infinite-dimensional separable Hilbert spaces here, might be worth mentioning. 2. "But all expectation values and hence all predictions agree with our first representation" - yes, by construction! You just constructed $U\circ \pi_1$ as a representation unitarily equivalent to $\pi_1$, and this representation has nothing to do with $\pi_2$. I don't understand what this construction is supposed to have to do with the fact that $\pi_1$ and $\pi_2$ are unitarily inequivalent and what the question actually is. $\endgroup$
    – ACuriousMind
    Jan 27, 2023 at 21:24
  • $\begingroup$ @ACuriousMind Thanks for your feedback. To 1.: Or finite-dimensional (complex) Hilbert spaces of the same dimension, no? 2. What confused me was the fact that I can represent the state and observable on $H_2$ via a unitary transformation (you're right, this has nothing to do with $\pi_2$ at all, tho), and so I did not understand why unitary inequivalent representations are a problem, because I thought that as long as we can represent all observables on a Hilbert space, we are fine. But as the answer points out, the problem is to find the correct representation (up to unitary eq.) first. $\endgroup$ Jan 27, 2023 at 21:33

1 Answer 1

6
$\begingroup$

When the algebra has a non trivial center, then it is easy to understand how inequivalent representations pop out and what is the physical difference between them.

Consider a $C^*$ (or simply $*$) algebra $\cal A$ with non trivial center and two unitary representations $\pi_i: {\cal A} \to \mathfrak{B}(H)$, $i=1,2$, on the same Hilbert space $H$. Finally assume that the two representations are irreducible. If $a=a^* \in \cal A$ is an element of the center (e.g. a Casimir operator in the universal enveloping algebra associated to a Lie group), the operators $\pi_i(a)$ must be proportional to the unit element $\pi_i(a)= c_iI$ for a pair of reals $c_i$ (Schur's lemma). These values are physical observables and can be measured. The choice of these values distinguishes between the two representations which are unitarily inequivalent: Every unitary intertwiner operator should transform $c_1I$ to $c_2I$ and this is not possible if the two constants are different. However both operators represent the same central element $a$. Both representations are physically admissible, in principle, but physics chooses the value of $a$ and declares which representation is the right one. If $\pi_1$ produces the wrong value of $c_1$, every unitary map from $H$ to $H$ gives rise to a new representation $\pi'_1= U\pi_1U^{-1}$ as you say, but it is not able to change the wrong value of $c_1$. The new representation is physically admissible, but it is not the right one.

The values of the constants representing the center are fixed by the algebraic pure states. (Pure states gives rise to irreducible GNS representations.) Indeed, if $\omega: {\cal A}\to \mathbb{C}$ is an algebraic pure state and $\pi_\omega$ is its GNS representation and $a$ stays in the center, then $$\omega(a)= \langle \Psi_\omega, \pi_\omega(a)\Psi_\omega\rangle= \langle \Psi_\omega, c_aI \Psi_\omega\rangle= c_a.$$

The nature of the difference of inequivalent unitary representations of algebras with trivial center (as an infinite dimensions Weyl algebra) is more subtle. A recent book on that is the one by Schmudgen.

$\endgroup$
8
  • $\begingroup$ Dear Valter, thank you for your wonderful answer! So two unitary non-equivalent representations can (or must?) associate to an abstract observable two self-adjoint operators which differ in their spectrum and thus can be experimentally distinguished, correct? Could you elaborate "The values of the constants representing the center are fixed by the algebraic states, so are the states which decide which (GNS) representation is the right one (for a given state)." a bit more, I am a bit confused? Thanks in advance! $\endgroup$ Jan 27, 2023 at 20:57
  • 1
    $\begingroup$ Yes you are right. I however focused on special elements whose spectrum is made of a single point. $\endgroup$ Jan 27, 2023 at 21:20
  • 1
    $\begingroup$ @Tobias Fünke There are subtleties however. A C*-representation is injective if and only if it preserves the spectrum of each element. So a pair of faithful representations of the same C*-algebra have corresponding elements with identical spectrum. This does not mean that the representations are unitarily equivalent. However the converse implication is true (as I pointed out is in a special case): if the spectra are not preserved they are not unitarily equivalent $\endgroup$ Jan 27, 2023 at 22:02
  • 1
    $\begingroup$ For instance, you may have two representations with the same point spectrum of a certain element, but the eigenspaces of a given eigenvalue have different dimensions. In that case unitary equivalence is not possible. $\endgroup$ Jan 27, 2023 at 22:07
  • 1
    $\begingroup$ @LucasBaldo Yes, generally they make different experimental predictions, Valter gave a nice example. See also e.g. my question here. The answer there shows that if two representations (+some important mathematical assumptions, though) make the same predictions, then they are unitarily equivalent. By modus tollens it follows that two unitary inequivalent representations make different predictions, which in principle can be experimentally verified. $\endgroup$ Jan 5 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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