# What is the spectrum of the Hamiltonian of the universe?

Let us assume that there exists a Hamiltonian that (together with the initial state) describes the whole universe.

Then my question is : What is the spectrum of this Hamiltonian and what are the multiplicities of the spectrum ? More precisely, what is the representation of the Hamiltonian in Spectral multiplicity theory (see http://en.wikipedia.org/wiki/Self-adjoint_operator, section "Spectral multiplicity theory").

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This question is vague. The problem with saying "A hamiltonian that describes the universe" is that you are either asking for a QFT hamiltonian that describes a model of the universe, excluding gravity, and then the question is definite, or else you are asking for a quantum gravity Hamiltonian which describes the entire universe. For the latter case, you have to deal with the fact that we don't have a quantum gravity of our universe (deSitter-like), and further, the quantum gravity doesn't necessarily have a Hamiltonian formulation, since time slicing is not generally available in all cases. –  Ron Maimon Jun 23 '12 at 8:14
As you formulate it, the answer is very simple. The spectrum is the whole set of real non-negative numbers and the multiplicity of each eigenvalue is infinite. It's enough to realize that a single photon may move to all directions to get the infinite degeneracy for each value of the energy. The Hamiltonian effectively knows about all of dynamics in the Universe but one must go beyond the simple "spectrum and degeneracy". In particular, the spectrum of the Hamiltonian corresponds to multiparticle states and you want to know the spectum of $m^2$ of each isolated particle. –  Luboš Motl Jun 23 '12 at 8:41
@Lubos : The repesentation of a self adjoint operator in Spectral multiplicity theory contains also measures . How do they look like ? –  jjcale Jun 23 '12 at 14:38
If the universe is infinite then no Hamiltonian of the universe exists since the total energy of the universe is infinite and we need an other aproach (C*-algebras ?). Otherwise it shoud be possible to deduce the number of space dimensions by counting eigenvalues of the Hamiltonian. –  jjcale Jun 24 '12 at 6:37
@LubošMotl: There's a problem with the argument you give if you include the gravitational field--- it might end up that there are no allowed states of nonzero total energy in the quantum gravity Hamiltonian (if the concept makes sense, that is if gravity has a Hamitlonian formulation, which I don't see why it should outside of a space that admits a light-cone slicing) because of the constraint H=0. This sort of stupidity makes it impossible to answer definitively. Your answer would be correct in a field theory model not on a large torus, of course. –  Ron Maimon Jun 24 '12 at 7:53

Assume that the universe is asymptotically flat and has finite total energy. (These appear to be the minimal assumptions under which talking about the Hamiltonian of the universe make sense.)

Then the following description holds independent of any other unknown details about its modeling. The latter are relevant only if you want to predict the set of bound states and their masses, and want to know something about other observables than Poincare symmetries.

Because of translation invariance of space-time, the spectrum is the positve halfline, including zero.

The appropriate spectral representation is in terms of an asymptotic representation where the Hamiltonian and the momentum are diagonal. In such a representation, assuming asymptotic completeness and disregarding infrared complications, a basis is given by states $|C;p^1,...,p^{|C|}\rangle$, where $C$ runs over all possible sets of bound states, and $p^j=(p^j_0,{\bf p}^j)$ is the asymptotic 4-momentum of the $j$-th entry of $C$, with $p^j_0=\sqrt{({\bf p}^j)^2+(m^jc)^2}$, where $m^j$ is the mass of the $j$-th entry of $C$ and $c$ is the speed of light.

On these states, 4-momentum $p$ acts by multiplication with $\sum_{j=1}^{|C|} p^j$, and $H=p_0c$ is the Hamiltonian. The representation is the direct sum of the corresponding Fock representations of the Poincare group (with statistics corresponding to the spin of the bound states). This gives a complete description of the spectrum in the $C^*$-algebraic sense. Clearly, every eigenvalue has infinite multiplicity.

A correct infrared description also needs to account for clouds of massless particles accompanying these states, defined formally by very weak limits that are not mathematically well-defined in a separable Hilbert space setting. It leads to the formation of uncountably many superselection sectors modifying the above description, but lacking a clear mathematical basis, the details are poorly understood.

The above assumes that the universe is asymptotically flat and has finite total energy. (These appear to be the minimal assumptions under which talking about the Hamiltonian of the universe make sense.) The description holds independent of any other unknown details about its modeling. The latter are relevant only if you want to predict the set of bound states and their masses, and want to know something about other observables than Poincare symmetries.

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The Hamiltonian you are describing has absolutely continuous spectrum $[0,\infty)$ with infinite countable multiplicity and is therefore unitary equivalent to $-\Delta$ acting on $L^{2}(\mathbb{R}^{2})$ . So this Hamiltonian contains almost no information and all information about the universe must be encoded in the initial state. Or another mathematical framework is needed. –  jjcale Jun 24 '12 at 18:19
@jjcale: Of course, a Hamiltonian generally contains only very little information about a system, once it has one or more continuous degree of freedom. For example, all nonrelativistic particles in a repulsive potential have (apart from a zero-point shift) the same spectrum, and so the same spectrum as QED. Nevertheless, my description gives the answer that you had asked for. You didn't ask for all information about the universe but about the spectrum of its Hamiltonian. –  Arnold Neumaier Jun 25 '12 at 11:41