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- Thread starter Bobhawke
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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.

- #3

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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?

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?

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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.

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.

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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?

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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.

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Ah it all makes sense now. Thank you hamster and genneth!

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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.

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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.

I think you're selling yourself short there... 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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