Use of traces in physics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:53:37Z http://mathoverflow.net/feeds/question/59824 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59824/use-of-traces-in-physics Use of traces in physics John R Ramsden 2011-03-28T12:29:36Z 2011-03-28T17:49:31Z <p>This is borderline physics, but I'll post here first as it relates to a mathematical concept.</p> <p>I was wondering if anyone could clarify what the physical significance is of traces, whether of matrices or more general operators, and are there more general notions of the same kind?</p> <p>For example, there is what seems (as far as I can follow it) quite a promising "precursor" quantum theory called Trace Dynamics, developed by Stephen Adler and summarized in <a href="http://www.worldscinet.com/ijmpd/19/1914/free-access/S0218271810018335.pdf" rel="nofollow">http://www.worldscinet.com/ijmpd/19/1914/free-access/S0218271810018335.pdf</a></p> <p>Also, I have seen traces used in many other physics contexts.</p> <p>I guess the first thing, more on topic here, would be to explain what mathematical significance this apparently slightly artificial (and coordinate dependent?) notion has.</p> <p>Feel free to waffle in generalities - The reply needn't be impeccably formal, as I am more interested in gaining intuition than results that could easily be looked up ;-)</p> http://mathoverflow.net/questions/59824/use-of-traces-in-physics/59825#59825 Answer by Pieter Naaijkens for Use of traces in physics Pieter Naaijkens 2011-03-28T12:38:47Z 2011-03-28T12:50:16Z <p>In quantum mechanics, states can (often) be described by density matrices. That is, if $A$ is an observable, it's expected value is given by $\langle A \rangle = Tr(\rho A)$, where $\rho$ is a density matrix (or operator in the infinite dimensional case). This formulation (as opposed to using state vectors for example) has the advantage that it's easy to described so-called mixed states (i.e. non-pure states). In this context it is in a sense more of a tool, especially useful when studying statistical mechanics.</p> http://mathoverflow.net/questions/59824/use-of-traces-in-physics/59826#59826 Answer by Jon Bannon for Use of traces in physics Jon Bannon 2011-03-28T12:42:21Z 2011-03-28T14:31:33Z <p>This is really just a long comment. I feel that the honor (naturality) of the trace has been called into question, and so it must be defended ;) I'm certain there are many other elegant things that can be said...but here are my two cents:</p> <p>First of all, one nice thing about a trace is that it is <em>not</em> coordinate dependent. This is because $Tr(S^{-1}TS)=Tr(SS^{-1}T)=Tr(T)$ for all $T$ in whatever algebra you are considering on which $Tr$ lives.</p> <p>As for the general significance of the trace, you may want to have a look at the <a href="http://en.wikipedia.org/wiki/Riesz_representation_theorem" rel="nofollow">Riesz representation theorem</a> for bounded linear functionals on the algebra of continuous functions with compact support with respect to pointwise multiplication. This theorem facilitates viewing a trace (or continuous functionals in general) as an integral in certain important abelian algebras. </p> <p>If you are looking at physics from a point of view that incorporates <a href="http://en.wikipedia.org/wiki/C%2a-algebra" rel="nofollow">$C^{\ast}$-algebras</a>, then traces play an important role. Say, for example, you are thinking about quantum mechanics from the operator algebra point of view. A good sort of $C^{\ast}$ -algebra in which to do this is a <a href="http://en.wikipedia.org/wiki/Von_Neumann_algebra" rel="nofollow">von Neumann algebra</a>, since these are generated by self-adjoint projections...which amount to 'yes' or 'no' questions (a nice description I first saw posted on John Baez's blog). A trace on the von Neumann algebra readily expresses a notion of dimension that generalizes the usual Hilbert space dimension. In (finite) matrices, two projections have ranges with the same dimension if and only if these projections have the same trace. So, equivalence classes of projections are given by the possible traces: ${0,1,2,...,dim(H)}$. There are von Neumann algebras that exhibit continous dimension (the equivalence classes range over $[0,1]$). All such algebras admit a (normal) trace state...which witnesses the dimension. In short, traces seem to be no more artificial than dimension...largely because of the coordinate independence.</p> <p>Also, traces are important in noncommutative geometry. For example, a non-normal trace called the <a href="http://en.wikipedia.org/wiki/Dixmier_trace" rel="nofollow">Dixmier trace</a> plays the role of the integral in the noncommutative differential calculus. For more on this, see <a href="http://www.alainconnes.org/en/" rel="nofollow">Connes's book</a>.</p> http://mathoverflow.net/questions/59824/use-of-traces-in-physics/59836#59836 Answer by JosÃ© Figueroa-O'Farrill for Use of traces in physics JosÃ© Figueroa-O'Farrill 2011-03-28T13:56:31Z 2011-03-28T13:56:31Z <p>In statistical mechanics, the trace $\mathrm{Tr} e^{-\beta H}$, where $\beta$ is the inverse temperature and $H$ is the hamiltonian, defines the partition function of a system at equilibrium.</p> <p>In supersymmetric quantum mechanical models, the supertrace (which is in particular also a trace) $\mathrm{Str} e^{-\beta H}$ computes the so-called <a href="http://en.wikipedia.org/wiki/Witten_index" rel="nofollow">Witten index</a> of the model. It is the index of an operator $D$ (the supercharge) such that $D^2 = H$, and there exist models (so-called supersymmetric sigma models) for which $D$ is the Dirac operator of one of the classical complexes. This observation then underlies the Physics proofs of the Atiyah-Singer index theorem. The Witten index also plays a rôle in less formal considerations: it can be used to probe the breaking of supersymmetry, which is a phenomenologically important question, since supersymmetry, if it exists at all, is broken at our energy scales. A nonvanishing Witten index indicates that supersymmetry is not spontaneously broken.</p> http://mathoverflow.net/questions/59824/use-of-traces-in-physics/59856#59856 Answer by Simon Wadsley for Use of traces in physics Simon Wadsley 2011-03-28T16:01:54Z 2011-03-28T16:01:54Z <p>For a more general notion of the same kind see <a href="http://ncatlab.org/nlab/show/span+trace" rel="nofollow">http://ncatlab.org/nlab/show/span+trace</a></p> http://mathoverflow.net/questions/59824/use-of-traces-in-physics/59872#59872 Answer by Russell May for Use of traces in physics Russell May 2011-03-28T17:49:31Z 2011-03-28T17:49:31Z <p>There was a similar question recently posed (and often answered) here:</p> <p><a href="http://mathoverflow.net/questions/13526/geometric-interpretation-of-trace" rel="nofollow">http://mathoverflow.net/questions/13526/geometric-interpretation-of-trace</a></p>