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Hey guys. I'm trying to understand spinorial representations in relativistic QM/QFT. If I make any mistakes in my statements (drawing mostly from chapter 2 of Maggiore) please correct them.
In QM things are simple enough. We have a different representation of ##SU(2)## for each fixed eigenvalue ##j## of ##J^2## whilst letting the eigenvalues ##j_z## of ##J_z## vary for each fixed ##j##. For the case of ##j = \frac{1}{2}##, the matrix representations ##J_i = \frac{1}{2}\sigma_i## of the rotation generators determine the ##j = \frac{1}{2}## unitary representation of ##SU(2)## by ##e^{i\theta \hat{n}\cdot \sigma/2}##. The representation space is that spanned by the eigenbasis ##\{|\uparrow \rangle, |\downarrow \rangle \}## which is ##\mathbb{C}^2## since the ##\sigma_i## are complex ##2\times 2## matrices.
Coming to relativistic QM, the vector representations are the fundamental representations of ##SO(3,1)## but as we know this isn't enough because we need to allow for objects which transform under ##2\pi## spatial rotations in such a way that they flip sign since this is not physically observable.
To do this we define the quantities ##J^{\pm } = \frac{1}{2}J\pm iK## and rewrite the Lorentz algebra as ##[J^{+}_i, J^{+}_{j}] = i\epsilon_{ijk}J^{+}_k##,## [J^{-}_i, J^{-}_{j}] = i\epsilon_{ijk}J^{-}_k##, and ##[J^{+}_i, J^{-}_{j}] = 0 ##. Why is this the Lie algebra of ##SU(2) \times SU(2)##? Better yet, how is the Lie algebra of a Cartesian product of ##SU(2)## with itself actually defined? Furthermore, why does ##[J^{+}_i, J^{-}_{j}] = 0 ## imply we can label representations of the Lorentz algebra by ##(j_+,j_-)##? Normally the representations of ##SU(2)## are labeled by eigenvalues of ##J^2## and not those of ##J_z## because of the relationship between the eigenvalues of the former and those of the latter i.e. for each fixed ##j## we have the integer spaced eigenvalues ##-j,...,j## of ##J_z##. So I don't see how we can talk about the eigenvalues ##j_+## and ##j_-## of ##J_z^+## and ##J_z^-## without talking about those of ##J^2## which are what label the ##SU(2)## representations to start with.
The ##(\frac{1}{2},0)## representation of the Lorentz algebra is determined by the matrix representations ##J^+ = 0## and ##J_i^- = \frac{\sigma_i}{2}##. The representation space is again ##\mathbb{C}^2## and its elements we call the the left-handed Weyl spinors ##\psi_L##. Similarly for ##(0,\frac{1}{2})## we have the right-handed Weyl spinors ##\psi_R## also in ##\mathbb{C}^2##. But how do we get the ##(\frac{1}{2},0)## unitary representation of the Lorentz group from this representation of the Lorentz algebra? In other words, while the Lorentz algebra and that of ##SU(2) \times SU(2)## are the same, the two groups are not equivalent so how does one use this representation of the Lorentz algebra to generate a unitary representation of the Lorentz group as opposed to one of ##SU(2) \times SU(2)##?
Also, as a side question, in the QM case the spinorial representations are the ##j = \frac{1}{2}## unitary representations of ##SU(2)## which is the universal covering group of ##SO(3)##. So in the relativistic case shouldn't we have the ##(\frac{1}{2},0)## unitary representation coming from the ##(\frac{1}{2},0)## Lorentz algebra representation be that of ##SL(2,\mathbb{C})## which is the universal cover of ##SO(3,1)##?
Thanks in advance!
In QM things are simple enough. We have a different representation of ##SU(2)## for each fixed eigenvalue ##j## of ##J^2## whilst letting the eigenvalues ##j_z## of ##J_z## vary for each fixed ##j##. For the case of ##j = \frac{1}{2}##, the matrix representations ##J_i = \frac{1}{2}\sigma_i## of the rotation generators determine the ##j = \frac{1}{2}## unitary representation of ##SU(2)## by ##e^{i\theta \hat{n}\cdot \sigma/2}##. The representation space is that spanned by the eigenbasis ##\{|\uparrow \rangle, |\downarrow \rangle \}## which is ##\mathbb{C}^2## since the ##\sigma_i## are complex ##2\times 2## matrices.
Coming to relativistic QM, the vector representations are the fundamental representations of ##SO(3,1)## but as we know this isn't enough because we need to allow for objects which transform under ##2\pi## spatial rotations in such a way that they flip sign since this is not physically observable.
To do this we define the quantities ##J^{\pm } = \frac{1}{2}J\pm iK## and rewrite the Lorentz algebra as ##[J^{+}_i, J^{+}_{j}] = i\epsilon_{ijk}J^{+}_k##,## [J^{-}_i, J^{-}_{j}] = i\epsilon_{ijk}J^{-}_k##, and ##[J^{+}_i, J^{-}_{j}] = 0 ##. Why is this the Lie algebra of ##SU(2) \times SU(2)##? Better yet, how is the Lie algebra of a Cartesian product of ##SU(2)## with itself actually defined? Furthermore, why does ##[J^{+}_i, J^{-}_{j}] = 0 ## imply we can label representations of the Lorentz algebra by ##(j_+,j_-)##? Normally the representations of ##SU(2)## are labeled by eigenvalues of ##J^2## and not those of ##J_z## because of the relationship between the eigenvalues of the former and those of the latter i.e. for each fixed ##j## we have the integer spaced eigenvalues ##-j,...,j## of ##J_z##. So I don't see how we can talk about the eigenvalues ##j_+## and ##j_-## of ##J_z^+## and ##J_z^-## without talking about those of ##J^2## which are what label the ##SU(2)## representations to start with.
The ##(\frac{1}{2},0)## representation of the Lorentz algebra is determined by the matrix representations ##J^+ = 0## and ##J_i^- = \frac{\sigma_i}{2}##. The representation space is again ##\mathbb{C}^2## and its elements we call the the left-handed Weyl spinors ##\psi_L##. Similarly for ##(0,\frac{1}{2})## we have the right-handed Weyl spinors ##\psi_R## also in ##\mathbb{C}^2##. But how do we get the ##(\frac{1}{2},0)## unitary representation of the Lorentz group from this representation of the Lorentz algebra? In other words, while the Lorentz algebra and that of ##SU(2) \times SU(2)## are the same, the two groups are not equivalent so how does one use this representation of the Lorentz algebra to generate a unitary representation of the Lorentz group as opposed to one of ##SU(2) \times SU(2)##?
Also, as a side question, in the QM case the spinorial representations are the ##j = \frac{1}{2}## unitary representations of ##SU(2)## which is the universal covering group of ##SO(3)##. So in the relativistic case shouldn't we have the ##(\frac{1}{2},0)## unitary representation coming from the ##(\frac{1}{2},0)## Lorentz algebra representation be that of ##SL(2,\mathbb{C})## which is the universal cover of ##SO(3,1)##?
Thanks in advance!