Commutators, Traces & [x,p] = I(hbar) Explained

[TrickyDicky]

Well, this fellow would do something along the following lines: we look for a Hilbert space H equipped with the CCR. We can see this statement as some abstract "equation" for which we seek for solutions. The "solutions" are pairs composed by: i) concrete Hilbert spaces; ii) equipped with a concrete realization of the CCR in terms of operators in this space. The problem will be solved once we find all the possible concrete solutions. The equivalence of solutions is given by unitary equivalence of the pairs. By inspection, we can see that no such solution exists if the Hilbert space is taken to be finite dimensional (because of the trace argument mentioned earlier). On the other hand, the solution offered in post #31 is a valid solution, so at least we know that solutions exist. A much more serious issue is raised by the fact that one can show that at least one of the operators in the (abstract) CCR cannot be bounded. This forces us to take the domain of the operators in consideration, since two pairs can be made inequivalent just by changing the domain of the operators (i.e., the formal expression of the operators is still the same for both, we just change the domain). A possible solution to this problem is to restrict the kind of solutions we look for, i.e., rather than considering some general solution for the CCR, we look for a solution that arises as the infinitesimal generators of the "exponentiated CCR", i.e., the Weyl *-algebra (based on a finite dimensional symplectic vector space, since we are dealing with ordinary QM rather than bosonic QFT). There's actually a physical justification for this restriction, since the Weyl relations are equivalent to the Imprimitivity Condition, which captures the notion of the homogeneity of (physical) space.

So, we have posed the problem in a way that takes into consideration the domain difficulties. What's next? we could keep trying to find solutions by inspection or guessing. But that's useless, since there could be an infinite number of inequivalent solutions!

The best way to attack the problem is to formulate it in such a way so that we can apply some powerful theorem that will give us automatically all of the possible solutions of the problem at once (note that, since this process will give us all the solutions, it already has to contain the previous result we got by inspection, namely, that there's no solution in finite dimensions).

There are two important ways to do this. The first is group theoretical. One notices that asking for a realization of the Weyl relations is equivalent to asking for a (unitary, etc.) representation of certain Lie group (the Heisenberg group, which is the unique connected, simply connected Lie group with the CCR as Lie algebra) and that satistifies certain condition. So, the problem now has been reduced to the mathematical problem of finding all of the possible representations of the Heisenberg group that satisfy certain condition. The Heisenberg group is a non-compact group, so we cannot use the Peter-Weyl theorem here (and that's good, otherwise we would get that the irreducible representations are necessarily finite dimensional, in contradiction with the previous result by inspection). But the Heisenberg group is a semidirect product of a normal, abelian subgroup and a closed subgroup. And there's a very powerful theorem, called the Mackey Machine, that deals with the representation theory of these type of groups. The theorem states that there's a family of irreducible and inequivalent representations (called induced representations) and that all other abstract or generic irreducible representation is equivalent to some representation in this family. Thus, the problem is solved if we can build this family of induced representations. Fortunately, there's an algorithmic recipe for doing this. In the case of the Heisenberg group, the result is the following: there's only one non-trivial induced representation (given by the exponentiation of the representation of the CCR given in post #31, sometimes called the "Schrödinger representation"). So, up to unitary equivalence of solutions, there's essentially only one solution to the problem (as stated in terms of the Weyl relations). This result is called the Stone-von Neumann theorem. Evidently, since this solution is over an infinite dimensional space and all other solution is equivalent to this one, then no solution exists over a finite dimensional space. Also, since the solution is over a separable space and all other solution is equivalent to this one, then no solution exists over a non-separable space.

The other method relies on the algebraic formulation of QM. By putting a norm, the Weyl *-algebra is transformed into a C*-algebra. The GNS construction theorem states that any representation of a C*-algebra that satisfies <f|R_af>=w(a) (where |f> is a cyclic vector, R is the representation of the algebra, and w is an algebraic state), is equivalent to the GNS construction based on the state w. Now, for the case of the finite dimensional Weyl C*-algebra, one can show that for any abstract (i.e., generic) realization R of the algebra on a Hilbert space, there always exists a cyclic vector |f> such that <f|R_af>=w(a), where w is an algebraic state that does not depend on the particular realization, i.e., it's always the same. So, any realization is equivalent to the GNS construction based on this w. It can be shown, of course, that the previous Schrödinger representation satisfies this and in this way both methods simply prove the same result.

Thanks for the elaborated answer. Could you maybe expand on the counterexamples to the Stone-von Neumann theorem, and how to avoid issues with the domain of the unbounded operators an exponentiated form of the canonical commutation relations(the Wey relations you mention) have to be used that are not rigorously the same but close enough. My understanding is that this representation of the Heisenberg group is isomorphic to $L^2(R)$ but only as modules.

 

[aleazk]

Thanks for the elaborated answer. Could you maybe expand on the counterexamples to the Stone-von Neumann theorem, and how to avoid issues with the domain of the unbounded operators an exponentiated form of the canonical commutation relations(the Wey relations you mention) have to be used that are not rigorously the same but close enough. My understanding is that this representation of the Heisenberg group is isomorphic to $L^2(R)$ but only as modules.

I don't have much time to write a detailed answer now. I refer you to the following references:

Quantum Field Theory, a tourist guide for mathematicians - G.B.Folland (pages 43 to 46 for an exposition on how to interpret the problem in terms of the Heisenberg group/Weyl relations; a counterexample is mentioned)

Foundations of Quantum Mechanics - J.Jauch (pages 197, 198 for the relation between the Weyl relations and the Imprimitivity Condition/quantum localization in homogeneous physical space)

 

[TrickyDicky]

I don't have much time to write a detailed answer now. I refer you to the following references:

Quantum Field Theory, a tourist guide for mathematicians - G.B.Folland (pages 43 to 46 for an exposition on how to interpret the problem in terms of the Heisenberg group/Weyl relations; a counterexample is mentioned)

Foundations of Quantum Mechanics - J.Jauch (pages 197, 198 for the relation between the Weyl relations and the Imprimitivity Condition/quantum localization in homogeneous physical space)

Thank you for the great references.
The key point seems to be that in order to avoid counterexamples to the Stone-von Neuman theorem one must use the CanonicalCommutationRelations in the form of the Weyl relations, but the CCR are only equivalent to the Weyl relations in a basis-dependent way due to the unavoidable fact that the position and momentum operators in the CCR cannot be both bounded and that determines their non-uniqueness (unlike the Weyl relations case that is indeed unique up to unitary equivalence) and their inescapable(even in abstract terms) dependence on position or momentum representation.

The absence of a basis-independent unique unitary equivalence between the canonical and Weyl relations triggers questions about the causes of the absence of interaction picture in QFT(Haag's theorem), in QFT there is no position operator and precisely difficulties with the position representation in quantum relativistic mechanics i.e. the strange behaviour of Newton-Wigner localization prescription is a motivation for having to introduce quantum field theory, is this not existence of position operator in the way it exists in QM what prompts the problem with the interaction picture in QFT?, that is however salvaged in QM by the obliged preferred basis?

 

[TrickyDicky]

I've just recalled this was already explained by strangerep in a recent thread:

What's happening is that any unitary irreducible representation ("unirrep") of the Lorentz group is necessarily infinite-dimensional. In such cases, the usual Stone von-Neumann uniqueness theorem (concerning unitary equivalence of the unirreps of the canonical commutation relations) does not hold (unlike the situation in nonrel QM).

It is the failure of the S-vN thm in the inf-dim case which gives rise to the ambiguity of representation of the canonical commutation relations in QFT.

I just stressed that in the NRQM situation uniqueness is basis-dependent due to unboundedness of the position and momentum operators, but this gets too far from the OP and deserves a new thread.