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QED in 3D -> QED in 1D, what changes?

  1. May 14, 2010 #1
    QED in 3D --> QED in 1D, what changes?

    Can we have massive charged and massless fields which interact in one space dimension? What might the interaction term look like?

    Thanks for any help!
     
  2. jcsd
  3. May 15, 2010 #2
    Re: QED in 3D --> QED in 1D, what changes?

    you have a guess?
     
  4. May 15, 2010 #3
    Re: QED in 3D --> QED in 1D, what changes?

    Can I have a go, since I'm trying to teach myself QFT now ? (not got very far yet)

    I would start by asking what sort of fields you can have in 1 space dimension. In 3, your interaction term has spinors, vectors and gamma matrices, but in 1 dimension you may be restricted to just (perhaps complex) scalar fields ?


    I know in 1+1 dimension, the propagator is different too (was one of the early questions in Zee's book)
     
  5. May 15, 2010 #4
    Re: QED in 3D --> QED in 1D, what changes?

    I would guess we could have a massive complex field in 1D, say phi(x,t), as well as massless complex fields in 1D, say A(x,t). What is the simplest interaction that might make sense between these fields?

    Thanks for any help!
     
  6. May 16, 2010 #5
    Re: QED in 3D --> QED in 1D, what changes?

    will there be scattering angles?
     
  7. May 16, 2010 #6

    tom.stoer

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    Re: QED in 3D --> QED in 1D, what changes?

    You can couple a massless gauge boson ("photon") to scalar or to fermion fields ("electrons"). The interaction formally looks exactly like QED, the indices are 0,1 instead of 0..3. The structure of the gamma-matrices reduces to two Pauli-matrices. QCD with SU(N) symmetry is possible as well.

    In 1D the only non-vanishing component of the photon field is due to the F01 component of the el.-mag field strength. In 1D one can eliminiate A0 as usual due to gauge invariance (A0 is not a dynamical degree of freedom because it comes w/o time derivative in the Larangian; therefore it acts both as a gauge field and a Lagrange multiplier; setting it to zero leaves behind a constraint, namely the Gauss law; this is rather similar to 3D); that means the only dynamical variable is A1. But A1 is subject to the Gauss law constraint. It is rather subtle, but the Gauss law does not eliminate all dynamical photon field completely, but it leaves a zero mode, that means a constant photon field = one single quantum mechanical degree of freedom a(t) instead of A1(x,t), together with its conjugate momentum e(t) instead of E(x,t) = F01(x,t). (the difference between the Gauss law and the generator of gauge transformations is just an integration by parts; it is exactly this derivative which kills a constant zero mode; that means the zero mode is unconstraint)

    The interaction of the photons with the electrons is due to the zero mode plus a 1D Coulomb potential which is basically V(x) ~|x|. Its Fourier transform is 1/k² as usual.

    In 1D QCD it's more complicated due to the non-abelian nature of the gluons. Nevertheless the reasoning just explained for QED goes through: the gluon field reduces to a set of quantum mechanical degrees of freedom a(t) with su(N) algebra. One finds (motivated by a large-N limit) a theory of weakly interacting mesonic fluctuations. This can be proven e.g. via "bosonization" of the fundamental quark fields. In addition (in the large-N limit) one can calculate a non-vanishing quark condensate which breaks the chiral symmetry. The colour-Coulomb potential is something like a shifted 1/k² where the shift is due to the gluonic zero mode a(t), that means something like 1/(k+a)². The colour-Coulomb potential is an inverted D² operator with D=d+a(t) being the covariant derivative after gauge fixing.

    http://adsabs.harvard.edu/abs/1993NuPhB.397..705S
    http://www.springerlink.com/content/k5l726883w437101/

    http://theorie3.physik.uni-erlangen.de/theses/Dip-1981-1994.html (unfortunately mostly in German)
     
    Last edited by a moderator: Apr 25, 2017
  8. May 17, 2010 #7
    Re: QED in 3D --> QED in 1D, what changes?

    Interesting. I'm a bit confused as to how you model the fermions though. The source of my confusion is this:

    Fermi statistics would require half integer spin fields. In 1 space dimension though, do you have enough geometrical structure to define spin 1/2 ? I'm thinking of the case in 3 space dimensions where we describe spin 1/2 behaviour by the requirement for two 360 degree rotations to restore an object to its original relationship with its environment. Is there an analog of this in one space dimension ?

    I guess that it could be phrased in terms of some generalisation of the 3 dimensional SU(2)->SO(3) double cover...
     
  9. May 17, 2010 #8

    tom.stoer

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    Re: QED in 3D --> QED in 1D, what changes?

    I have never thought about the geometry and the Clifford algebra, but you can look up the relevant spinor reps. in Wikipedia. http://en.wikipedia.org/wiki/Spinor. I guess that in 1+1 dim. one has Spin(2) ~ "double cover of SO(2)" and therefore 2-component spinors.

    What you can do is to write down a Dirac equation formally. Than you can try to find a rep. for the gamma matrices; one possible choice is

    [tex]\gamma^0 = \left( \array{cc}0 & 1 \\ 1 & 0 \end{array} \right) [/tex]

    [tex]\gamma^1 = \left( \array{cc}0 & -1 \\ 1 & 0 \end{array} \right) [/tex]

    [tex]\gamma^5 = \left( \array{cc}1 & 0 \\ 0 & -1 \end{array} \right) [/tex]

    why does the LaTeX command \right) not work properly???

    One can explicitly check for the algebra

    [tex]\{\gamma^\mu, \gamma^\nu\} = 2 g^{\mu\nu}[/tex]

    This proves the existence of 2-component spinors by explicit construction.

    Btw.: there is another major difference in 1+1 dim. QED and QCD: both are super-renormalizable which means that there are only a finite number of divergent Feynman graphs. In the canonical formalism one can derive the Hamiltonian; its regularization is done via gauge-invariant point splitting + subtraction of infinite constant from Dirac sea. This procedure is sufficient to get an explicitly finite operator - no infinitues anymore! The reason for this is due to the fact that the coupling constant is no longer dimensionless in 1+1 dim. Looking at momentum space integrals it can be understood as follows

    [tex]A_4 \sim \int d^4p \, a(p) = \int dp \, p^3 \, a(p) \to A_2 \sim \int d^2p \, a(p) = \int dp \, p \, a(p)[/tex]

    The function a(p) is roughly the same in 1+1 dim., but the integral itself contributes less powers in the momentum than in 3+1 dim. Therefore all integrals except for a finite number become explicitly finite.
     
    Last edited: May 17, 2010
  10. May 17, 2010 #9
    Re: QED in 3D --> QED in 1D, what changes?

    Thanks Tom, there does indeed appear to be a spinor algebra in (1+1) dimensions. I'll have to do a bit of digging to try to understand the geometrical interpretation.
     
  11. May 17, 2010 #10

    tom.stoer

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    Re: QED in 3D --> QED in 1D, what changes?

    Think about a complexification of SO(1,1).

    And check the Poincare generators; they are [tex]P^{\mu}[/tex] with the [tex]P^0 = H[/tex] the energy and [tex]P^1[/tex] the momentum plus [tex]M^{\mu\nu}[/tex] where [tex]M^{00}=0[/tex] and [tex]M^{01}[/tex] is just one boost generator (no rotations, no angular momentum).
     
  12. May 17, 2010 #11
    Re: QED in 3D --> QED in 1D, what changes?

    If you included the time axis we could consider angles?

    There would be Feynman diagrams?

    Thanks for the help!
     
  13. May 17, 2010 #12
    Re: QED in 3D --> QED in 1D, what changes?

    That is so cool! Thank you.
     
    Last edited by a moderator: Apr 25, 2017
  14. May 17, 2010 #13
    Re: QED in 3D --> QED in 1D, what changes?

    Thanks Tom! Much to digest.

    Seems like much stays the same from 4D space-time to 2D space-time.
     
    Last edited: May 17, 2010
  15. May 17, 2010 #14
    Re: QED in 3D --> QED in 1D, what changes?

    If you have access:

    http://iopscience.iop.org/0143-0807/29/4/014

    A toy model of quantum electrodynamics in (1 + 1) dimensions.

    From the abstract,

    We present a toy model of quantum electrodynamics (QED) in (1 + 1) dimensions. The QED model is much simpler than QED in (3 + 1) dimensions but exhibits many of the same physical phenomena, and serves as a pedagogical introduction to both QED and quantum field theory in general. We show how the QED model can be derived by quantizing a toy model of classical electrodynamics, and we discuss the connections between the classical and quantum models. In addition, we use the QED model to discuss the radiation of atoms and the Lamb shift.


    Or free from the web:

    http://www.its.caltech.edu/~boozer/electrodynamics/electrodynamics.html [Broken]

    "Here I describe a toy model of electrodynamics in (1+1) dimensions that I developed to try to help clarify various problems in the foundations of electrodynamics. The toy model shares much of the conceptual structure of ordinary electrodynamics but is mathematically much simpler. "
     
    Last edited by a moderator: May 4, 2017
  16. May 18, 2010 #15
    Re: QED in 3D --> QED in 1D, what changes?

    feynman diagrams are just a pictorical way to keep track on where in pertrubation calculation you are.
     
  17. May 18, 2010 #16

    tom.stoer

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    Re: QED in 3D --> QED in 1D, what changes?

    I have never seen Feynman diagrams in 1+1 QED as almost all calculations can be done w/o perturbation theory.

    And of course there are no angles in one spatial dimension :-)
     
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