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The role of representation theory in QM/QFT?

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I need help understanding the role of representation theory in QM/QFT. My understanding of representation theory in this context is as follows: there are physical symmetries of the system we are studying. This collection of of symmetries forms a group $G$. However, there are multiple Hilbert spaces that we can use to represent said system; so we would like to somehow represent $G$ as a collection of isomorphisms on whatever Hilbert space we choose to use. And to do so, we use a representation.

Firstly, is this a proper interpretation? Is there more to it?

Secondly, representations are typically group homomorphisms, not necessarily isomorphisms. Can we represent a group (in physics) with a representation that is not an isomorphism? And what does it mean physically if we can do so?

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user HeyJoe
asked Nov 13, 2012 in Theoretical Physics by HeyJoe (30 points) [ no revision ]

5 Answers

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Yes, in QM/QFT we always have some group $G$ that is required to have representation on a Hilbert space. This group in particular usually includes time translation and space translation operators as its generators which are interpreted as Hamiltonian and Momentum observables respectively. In fact all observables of the theory come from the generators of the group $G$.

In a classical field theory phase space is a space of fields. Observables are functions on this space, and hence are functionals of fields. A subalgebra of the Poisson algebra of these observables generates group $G$ (modulo some discrete symmetries). So upon quantization, $G$ gets represented on space of states through a representation of the (Poisson) algebra of fields. It is this thing which distinguishes a QFT from abstract representation theory of group $G$. We not only require that space of states $H$ be a representation space of $G$ but also that its action on $H$ be written in terms of action of some algebra of quantum fields on a given space time.

Two representations of algebra of quantum fields may not be equivalent to each other. Uniqueness of representation is not a problem with finitely many degrees of freedom. When number of degrees of freedom are infinite (which is usually the case in a typical QFT) then one usually works in Fock space representation.

In physics one usually looks for projective unitary representation of $G$ which are not true representations but representations modulo phase or equivalently unitary representation of some central extension of $G$. In some cases for the consistent definition of the theory it may be required that the representation of $G$ on space of states be a true representation.

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user user10001
answered Nov 13, 2012 by user10001 (635 points) [ no revision ]
Is the second paragraph more than the requirement that the Hilbert space has to have a cyclic vacuum w.r.t. the action of your field operator algebra? I guess once you have these you only need map from spacetime to these and if you know how to move along spacetime, you know how to move through your algebra and hence through your state space. Or do you describe a more general notion or what might be called a QFT? (And more far fetched: Aren't algebraists even trying to get rid of a vector space as primary object altogether?)

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user NikolajK
Hi @Nick i was not trying to be very rigorous in this answer. All i meant was that elements of group/algebra G of observables should be expressible in terms of elements of algebra A of fields. This of course follows from requirement of cyclicity of vacuum wrt A and as i think cyclicity of the vacuum wrt A is not a physical requirement but just one (though possibly only ?) way of implementing the requirement that observables be local expressions in terms of quantum fields. In algebraic approach too observables are required to be constructed out of elements of A in more abstract terms (right ?).

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user user10001
Using a Fock space in QFT is one of the reasons one meets infinities that must then be swept under the carpet. The Hilbert space of interacting field theories is not a Fock space (i.e., does not carry the Fock representation), by Haag's theorem.

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user Arnold Neumaier
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Representation theory is essential to define the relevant physical observables of a theory. The reason is that all separable Hilbert spaces of the same dimension are isomorphic (hence the same apart from labeling). Thus one needs extra structure to distinguish the interesting physics.

This is done by specifying a ''dynamical'' Lie algebra of important observables, whose physical interpretation is given by the form of the Lie bracket.

For example, the canonical commutation rules define the formal meaning of position and momentum in QM, and of free fields in QFT. The Lie algebra $so(3)$ defines angular momentum, the Poincare group defines 4-momentum and 4-angular momentum, $su(2)$ defines isospin, $su(3)$ defines color, etc.

The particular unitary (projective) representation on a Hilbert space then determines the quantum numbers and particle contents of a theory.

The Lie algebra of the symmetry group is usually much smaller than the dynamical Lie algebra, but plays another important role.

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user Arnold Neumaier
answered Nov 18, 2012 by Arnold Neumaier (12,385 points) [ no revision ]
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Have a look at the orbit method. The wikipedia article may provide a start. This philosophy provides a deep interpretation of quantization.

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user Jon Bannon
answered Nov 13, 2012 by Jon Bannon (20 points) [ no revision ]
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In quantum mechanics, if a system has a symmetry group $G$ representing physical operations we can perform on a system that will leave the evolution unchanged, then this group will act on the states of the system, which are state vectors in the system's Hilbert space $\mathcal{H}$. (The latter, by the way, is unique up to isomorphisms that preserve the operator algebra.)

This group action is equivalent to a homomorphism between $G$ and a group of linear operators on $\mathcal{H}$. (This is not completely trivial in that linearity follows from the physical postulate that the symmetries must also preserve superpositions.) After finding this group representation, then one usually decomposes it into a direct sum of irreducible representations, which give invariant subspaces under the symmetry group, usually labelled by the irrep that acts on them. These are probably the "multiple Hilbert spaces" you refer to.

Essentially, the group representations give a way to act with the symmetries on the actual states of the system. The different ways in which this is possible is studied by representation theory, and it can have a deep impact on the physics of the problem. This, of course, is simply saying that the symmetry of the problem has a deep impact on its solution, which should not be a surprise.

Finally, a group representation that is also an isomorphism is normally called a faithful representation, and it is not necessary for physically meaningful group representations to be faithful or, indeed, even nontrivial. If you look at the representations of the rotation group, $SO(3)$, you find at its lowest dimension the invariant subspace $j=0$ with the unit representation: completely trivial action, which does not stop us using it!

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user Emilio Pisanty
answered Nov 13, 2012 by Emilio Pisanty (520 points) [ no revision ]
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To your 2nd question "Secondly, representations are typically group homomorphisms, not necessarily isomorphisms. Can we represent a group (in physics) with a representation that is not an isomorphism? And what does it mean physically if we can do so?"

It can mean a lot of different things. Any time a representation has a kernel it means some subgroup acts trivially. For example, scalar field theory is concerned with the trivial representation of the Lorentz group (definitely not a group isomorphism). Vector fields are invariant under $2\pi$ rotations. The QFT vacuum is assumed to be Lorentz and Poincare invariant; it spans a (projectively) trivial 1d subrepresentation of the whole Hilbert space. Some kinds of dynamics are parity invariant; some aren't. The list of examples is practically endless.

This post imported from StackExchange Physics at 2015-03-30 13:55 (UTC), posted by SE-user user1504
answered Nov 13, 2012 by user1504 (1,100 points) [ no revision ]

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