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Physical Hilbert Space

  1. Jul 21, 2011 #1
    Hi,

    Could someone tell me, or refer me to a reference, about what the physical separable hilbert spaces are for the electroweak and strong forces. I'm looking for a defined inner product for the theories and a rigorous account of their hilbert spaces.

    Thanks a lot,
     
  2. jcsd
  3. Jul 21, 2011 #2

    tom.stoer

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    Think about quantizing QED in A°=0 gauge. You can construct a Hamiltonian H' acting on a kinematical Hilbert space. But this H contains three photon polarizations, namely two transversal and one longitudinal, A°=0 has already been eliminated.

    We know that there are only two physical polarizations, i.e. no longitudinal photon.

    We can fix this by implementing the Gauss law constraint, which is translated into an operator acting on the states. Instead of G(x) = 0 (which can be proven to be inconsistent with the comutation relations) we use G(x) |phys> = 0, i.e. the kernel of G(x) defines the physical Hilbert space.

    As G(x) is related to a gauge transformation this is something like "roting away" the unphysical states. The physical states w/o longitudinal photons are in |phys>, whereas the unphysical states are all perpendicular to these states. As [H, G(x)] = 0 the time evolution leaves |phys> invariant, i.e. time evolution does not mix physical and unpysical states, i.e. does not create longitudinal photons.

    The same mechanism works in QCD as well, but it is awefully complicated due to the non-abelian gauge groups and due to so-called Gribov copies. That means that there are different gauge field sectors which do not coincide in their definition of physical states; there is no global gauge fixing condition like in QED.

    If you like I can find a paper for you which discusses this non-perturbative gauge fixing = definition of the physical degrees of freedom of QCD.
     
  4. Jul 23, 2011 #3
    Is the kinematical state space, [itex]\mathcal{K}[/itex], for QCD the space of states that are SU(3) gauge invariant, diffeomorphism invariant, and satisfy the Hamiltonian for the dynamics of QCD?

    Similarly, is the kinematical state space for electroweak, the space of states that are SU(2)xU(1) gauge invariant, diffeomorphism invariant, and satisfy the Hamiltonian for the dynamics of the Electroweak theory?
     
  5. Jul 24, 2011 #4

    tom.stoer

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    In the kinematical Hilbert space you still have gauge degrees of freedom, so the states are not gauge invariant.

    The simplest way to quantize QCD canonically is to fix A°=0 (as A° has no conjugate momentum) and to derive the Hamiltonian H'. Then you have
    - three gluonic degrees of freedom
    - Gauss law acting as a generator of residual gauge transformations
    (these transformations are time-independent and do not violate the A°=0 gauge)

    Now you fix the residual gauge symmetry generated by the Gauss law; then you have
    - a physical Hamiltonian H
    - two gluonic degrees of freedom
    - Gauss law = identity operator in the physical Hilbert space
     
  6. Jul 24, 2011 #5
    Why does Wikipedia have that SU(3) is the gauge symmetry group for QCD?
     
  7. Jul 24, 2011 #6
    Because it is....
     
  8. Jul 25, 2011 #7

    tom.stoer

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    The idea was to introduce a new degree of freedom (color) which was required by the Pauli principle (there are some baryon states which would have totally symmetric wave functions w/o this extra degree of freedom). Then there was the idea to describe its dynamics using Yang-Mills equations, i.e. generalized Maxwell equations for a non-abelian gauge group.

    From collider experiments it became clear that there are three scatterers in a nucleon, so SU(3) was the most natural choice.
     
  9. Jul 25, 2011 #8
    I'm sorry, I am missing something from my understanding, I have another question, please help me. How is it that the kinematic state space has SU(3) as its gauge symmetry group but does not have SU(3) gauge invariant states in it?
     
  10. Jul 25, 2011 #9

    tom.stoer

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    I hope you are familiar with the Gauss law constraint. It must be implemented (b/c it is derived as a Lagrangian equation) on the physical states.

    OK, the Gauss law acts as a generator of gauge transformations, i.e. a conserved charge density that commutes with the Hamiltonian, i.e. [H, G(x)] = 0; that means it represents a symmetry of the system.

    By d³x integration you can derive conserved charges. b/c G(x) annihilates physical states

    G(x) |phys> = 0

    the charges annihilate the physical states as well. That means the physical states are singulett states w.r.t. to the symmetry group generated by the Gauss law.

    Now let's make a simple example: instead of looking at the local constraint

    G(x) |phys> = 0

    we look at a very simple constraint

    Li |phys> = 0

    Here L is the angular momentum and the physical states are singulett states with angular momentum zero (this constraint usually does not follow as a Lagrangian equation, it is only used as an example to make clear what happens).

    Now back to your question:

    In the kinematical Hilbert space Li acts as a generator of spatial rotations: the "states are rotated"; you can visualize that when looking at rotations of the spherical harmonics Ylm.

    In the physical Hilbert space again Li acts as a generator of rotations, but b/c only the l=0 states are in the physical Hilbert space, the rotation reduces to the identity; again you can see that by looking at the spherical harmonics Y00 - nothing happens when you rotate it!

    That means that the kinematical Hilbert space with |lm> carries the irreducible representations of the SO(3) rotations, each of them with dimension

    dim rep l = 2l+1; m = -1, ... 0, 1, ... +l

    But for the physical = singulett states you have l=0, i.e. |phys> = |00> which means it is the trivial representation and the rotations reduce to the identity.


    Now coming back to gauge symmetry:

    The kinematical Hilbert space carries a representation of the local SU(3) color-gauge group. Gauge transformations act somehow like "local color rotations" within that space. This is the gauge symmetry. By fixing the gauge with A°(x) = 0 plus a second residual (time-independent) gauge symmetry via implementing the Gauss law

    G(x) | phys> = 0

    you restrict the Hilbert to the physical one which is the gauge singulett subspace in which the gauge symmetry is reduced to the identity.
     
  11. Jul 25, 2011 #10
    Thank you, that was helpful. I seem to remember something along those lines in my Maggiorre book on quantum field theory during the covariant quantization of electromagnetism. I'll go back and look over it in detail.

    I have another question that is still relevant. Can the kinematical state space of QCD and Electroweak be a [itex]L_2[/itex] space over a measure?
     
  12. Jul 26, 2011 #11

    tom.stoer

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    In the mathematical sense? I don't think so b/c usually you quantize these theories using plane waves and aleady these states are not L2 states. [afaik we do not yet have a mathematically sound definition of an interacting quantum field theory - except for some simple examples]
     
  13. Jul 28, 2011 #12
    Hi,

    This is what I was looking for, but for QCD and the Weak force also. Here is an excerpt from Carlo Rovelli's Zakopane lectures where he describes the Hilbert space of QED.
    After this he describes the Hilbert space of lattice QCD. What mathematically makes lattice QCD different from canonical QCD that it can have its Hilbert space uniquely described?

    Thanks,
     
  14. Jul 28, 2011 #13

    tom.stoer

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    Lattice QCD ist defined over a finite lattice with nearest neighbor interactions. So mathematically lattice QCD is nothing else but quantum mechanics.
     
  15. Jul 28, 2011 #14
    Rovelli is being mathematically sloppy. Those limits he mentions do not exist in the naive mathematical sense (e.g. separability does not go through). For a physicist this should not matter because (as Rovelli insightfully points out) we always work with a truncation of the degrees of freedom anyway, so they suffice for calculational purposes, e.g. perturbation theory, lattice gauge calculations, etc.

    In some sense, those limits exist, but in weaker senses; e.g. "physically meaningful" calculations performed on an element of that series converge to a definite value (possibly with running of coupling constants). Unfortunately "physically meaningful" in this context does not have a simple (or agreed upon) mathematical definition.

    You should be (made) aware that it is an open problem to properly define the Hilbert space of interacting quantum (gauge) field theories in a mathematically satisfactory fashion. This is related to one of the Clay Institute problems.
     
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