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Relation between Clifford algebra and Lorentz algebra

  1. Apr 21, 2009 #1
    If the generators of the Lorentz spinor transformation can be expressed in terms of the gamma matrices (which can be used ot build a Clifford algebra), why cant the generators of the Lorentz vector transformation similarly be expressed in terms of the gamma matrices? And why does there exists this relationship between the Clifford algebra and the Lorentz algebra in the first place?
     
  2. jcsd
  3. Apr 23, 2009 #2
    This is an issue of representations of a group. I'm afraid I don't know of any simple answer to the question. The essentials are these:

    1. The Lorentz group is a Lie group (not compact, however): SO(3,1).
    2. The Lie algebra of this group has an isomorphism: so(3,1) ~= sl(2,C)
    4. The Dirac spinors are actually not an irreducible representation.
    5. The vector representation is simply so(3,1).

    It is probably just necessary to go and find a book on representation theory of Lie algebras and read it.
     
  4. Apr 23, 2009 #3
    Lorentz transformations form a group. Dirac algebra (special case of Clifford algebras) is a specific representation of this group. Irreducible representations of Lorentz group are enumerated with a pair of half-integer numbers u and v; a (u,v) representation has the dimension (2u+1)*(2v+1). (0,0) is scalar. (0,1/2) and (1/2,0) are representations that describe transformations of 2d Weyl spinors. Dirac algebra is non-irreducible (0,1/2) x (1/2,0) (4d). Vector algebra is irreducible (1/2,1/2) (also 4d).

    The Dirac algebra is reducible, which means that it describes two non-interacting fields (left-handed and right-handed 2d spinors) which transform independently of each other under boosts and rotations. However, Dirac equation contains a mass term which couples left-handed with right-handed and vice versa. Therefore we're forced to deal with 4-component spinors.

    Does this help?
     
    Last edited: Apr 23, 2009
  5. Apr 23, 2009 #4
    Yeah, but actually in the meantime Ive been looking at this question a bit more closely. It seems the gamma matrices can be used to build representations of the spin(N) algebra, and this is what dirac spinors transform under. But as you say, The Dirac spinor comprises of two Weyl spinors (which might get mixed up by a mass term), each of which transforms in the (0,1/2) and (1/2,0) reps of the Lorentz group. Spin(N) is the double cover of SO(N), so it does 2 copies of the Lorentz transformation basically. And because spinors can be complex, a two component spinor has 4 dofs, which is the correct amount for it to be acted on by an SO(4). But now we have the question - how can we call the theory Lorentz invariant if the things which it is built from, dirac spinors, dont transform under the lorentz group but instead its double cover? I think the answer is that there is nothing stopping us from separating out the left and right handed 2 spinors even when there is a mass term - such a theory now would be invariant under Lorentz since its constructed from 2 spinors. But we run into problems with gauge invraiance (which eventually leads to higgs mechanism), and also for many purposes I think its easier to speak about 4 components spinors and say they transform under Lorentz when what you strictly mean is that the 2 spinors it is comprised of each transforms under different copies of the Lorentz group.
     
  6. Apr 23, 2009 #5
    Err, not quite. The mass term really does mess up the separate transforms. The reason double covers don't make any difference is that they still share the same Lie algebra, i.e. infinitesimal transforms. The global differences manifest as topological effects, e.g. the existence of spin-1/2 particles.
     
  7. Apr 23, 2009 #6
    I think you misunderstand the meaning of double cover. Spin(4) and SO(4) groups have identical algebras which means that their generators commute in the same way. However, the continuous mapping between the two would be 2-to-1. A classic example is SU(2) vs SO(3).

    The theory is Lorentz invariant because we can do a series of infinitesimal boosts or rotations and transform all dirac spinors using corresponding elements of (0,1/2)x(1,2,0) rep and the Lagrangian will be preserved. If you rotate the whole thing 360 degrees around any axis, you will discover that all your spinors change sign instead of going back to where you started. That's because the spacetime symmetry group is SO(3,1) and the group of Dirac spinors is Spin(3,1) which is a double cover of SO(3,1), so by the time you got back to 1 in SO(3,1), your path takes you to an element -1 of Spin(3,1). But that is irrelevant since that sign does not affect the Lagrangian directly anyway.

    Mass term only matters when we're dealing with propagation.
     
    Last edited: Apr 23, 2009
  8. Apr 23, 2009 #7
    So the definition of Lorentz invariance is broader than just invariance under the action of a rep of the Lorentz group - really it means invariance under a rep of any group whose generators obey the Lorentz algebra?
     
    Last edited: Apr 24, 2009
  9. Apr 24, 2009 #8
    Basically, yes.

    The definition of Lorentz invariance is that physics laws are the same in all inertial reference frames. For that to be true, it is sufficient (but not necessary) for the Lagrangian density of the field to be invariant under boosts and rotations. Since any (proper) Lorentz transformation can be constructed out of infinitesimal boosts and rotations, it is sufficient and necessary for the Lagrangian density to be invariant under same. In order to achieve that, it is not necessary for internal variables such as spinors to transform under a rep of the Lorentz group. Any group that has the right Lie algebra will do.
     
  10. Apr 24, 2009 #9
    Ah it all makes sense now. Thank you hamster and genneth!
     
  11. Apr 24, 2009 #10
    What do you have in mind when you say it is sufficient but not necessary for the Lagrangian density to be invariant under boosts and rotations for the same laws of physics to appear in different inertial frames? I thought it was both necessary and sufficient. Are you thinking about physics that cannot be derived from an action principle but which are nevertheless Lorentz invariant?
     
  12. Apr 24, 2009 #11
    I was merely being pedantic. :) QFT lagrangians are typically lorentz scalars. But two lagrangians that differ by a constant (or more generally, by a total derivative) correspond to the same physics. So, you could have some kind of weird transformation law that changes the lagrangian but preserves physics.
     
  13. Apr 24, 2009 #12
    I think you're selling yourself short there... :-p In general, changes to the Lagrangian by a total derivative are considered harmless, except when the boundary effects are important. Usually, we postulate that the fields fall to zero at infinity, but things like gauge potentials don't have to, even if the physical fields like E and B do.

    More generally though, a symmetry of a quantum system is some operator which commutes with the Hamiltonian. It is often insinuated in textbooks that this is equivalent to the Lagrangian remaining invariant under the desired transform. Apart from the total derivative issue above, there is another one, more subtle. The connection between the quantum Hamiltonian, which is an operator, and the Lagrangian, which is classical, is via the path integral. The path integral involves a measure over the space of paths, and the measure may not be preserved under the transform, even if the Lagrangian is. These lead to the so-called (and badly called) anomalies.
     
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