# When does the Hamiltonian spectrum depend on the operator algebra representation?

Usually when we specify a first-quantized Hamiltonian on a Hilbert space of itinerant particles, we assume that the canonical commutation relation $[x_i,p_j] = i\, \delta_{ij}$ completely specifies all physical observables, like the Hamiltonian's spectrum. (Sometimes, if we're feeling fancy, we appeal to the Stone-von Neumann theorem to justify this, even though strictly speaking the theorem only applies to the exponentiated canonical commutation relations.) In second-quantized theories of bosons or fermions, we similarly assume that the commutation relations $[a_i, a_j]_\pm = [a_i^\dagger, a_j^\dagger]_\pm = 0$, $[a_i, a^\dagger_j]_\pm = \delta_{ij}$ completely specify the physical observables, independent of the choice of representation.

But for spin systems, the spectrum certainly can depend on the choice of representation - most simply, Heisenberg chains are believed to have a Haldane gap for integer-spin representations but not for half-integer-spin representations. Is there a general rule that determines when physical observables depend on the representation of the operator algebra?

(I bet someone's already asked this question before on SE, but I couldn't find a duplicate.)

• I'm not sure what you mean by general rule here: For algebras not covered by Stone-von Neumann, you need to examine the algebra in question whether it admits unitarily inequivalent representations or not. For the standard QFT case, there's Haag's theorem that shows there are uncountably many different inequivalent representations, but I'm not sure what exactly you're asking for here. – ACuriousMind Nov 27 '16 at 2:15
• If we are using creation operators (i.e. second-quantization) --note that spins are a special case by applying a Holstein-Primakoff transformation-- then I can't think of a case where the physics is not fixed by knowing the commutation relations and the cap: i.e. the smallest $n$ for which $\left( a^\dagger \right)^n = 0$. Disclaimer: I'm coming from the condensed matter side, so I'm not well-acquainted with the subtleties arising from infinite degrees of freedom (cf ACuriousMind's comment). – Ruben Verresen Dec 9 '16 at 21:37

Is there a general rule that determines when physical observables depend on the representation of the operator algebra?

This happens in quantum mechanics whenever the Lie algebra defined by the commutation relations does not uniquely determine the corresponding Lie group. In this case the most complete spectrum comes from the unique universal covering group, and the others are partial spectra coming from all quotients that preserve the Lie algebra.

The case SO(3) with universal cover SU(2) is just the simplest case of it.

In quantum field theory, the situation is more complicated since one has to represent infinite-dimensional Lie algebras and groups.

I'll be answering the following question. Given a quantum system defined by a $C^*$ algebra, how do we know if this algebra has a unique irreducible representation (up to equivalence) or not.

The first case corresponds to your first example (canonical commutation relations in finite dimensions), while the second case corresponds to spin example that you gave, and also to the canonical commutation (anti-commutation) relations in infinite dimension.

As you may already know, there is no known complete representation theory of $C^*$ algebras, so I will not be able to give you a complete answer; it is not known. However, I can give you a partial answer based on a general philosophy advanced by Rieffel, that representations of $C^*$ algebras can be obtained as the output of a process of quantization. This is very natural for quantum physicists since their final aim is really a quantum theory.

(From now on , I am following Landsman ref-1 and ref-2 )

Thus a representation will be attached to some configuration space that will be quantized. Here I'll restrict myself to finite dimensional configuration spaces (the representations though need not be finite dimensional, e.g., in the case of the canonical commutation relations). This restriction will exclude quantum field theoretical examples, which I'll address briefly in the sequel.

Next, another restriction will be added by requiring the configuration space to be homogeneous $Q=G/H$. (Again, all the examples in the question belong to this particular case). Physically, this is not really a restriction, because, these spaces define elementary quantum systems constituting the building blocks of more complicated quantum systems.

The $C^*$ algebra attached to this configuration space will be taken a matrix algebra of smooth functions on $Q$ carrying a finite dimensional irreducible representation of $G$. For example in the spin case, we can take a two dimensional matrix valued functions on the two sphere $S^2$ and for the Heisenberg-Weyl algebra we can take the algebra of smooth functions on $\mathbb{R}$.

Landsman argues that the non-equivalent representations arise in the case of non-trivial topology of the configuration space. In the Heisenberg-Weyl case, the configuration space is $\mathbb{R}$, thus the topology is trivial and the representation is unique, while in the $S^2$ case, the topology is not trivial and there are inequivalent representations.

According to the above reasoning we should have expected that the infinite dimensional CCR or CAR should have a unique representation as well because the configuration space would be $\mathbb{R}^{\infty}$, but this is not so simple, since issues of convergence become important. For example, if we want representations with a well-defined number operator, then we are talking about the oscillator group which is a central extension of the Weyl-Heisenberg group by $S^1$, giving rise to nontrivial topology and inequivalent representations. In the cases of current algebras, Wess-Zumino terms have a similar effect.

• What are $Q$, $G$, and $H$? – tparker Aug 10 '17 at 18:12
• $Q$ is the configuration space. It has a structure of a coset space $G/H$ because it is homogeneous. After quantization $G$ becomes a dynamical group and $H$ the isotropy of the vacuum. As I argued in the answer, such spaces give rise to elementary quantum systems. Please, see a formulation of the Dirac's 4th axiom on page 64 of Oliveira' s thesis.fenix.tecnico.ulisboa.pt/downloadFile/395140774040/… – David Bar Moshe Aug 13 '17 at 6:34