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Spinorial representations

  1. Jan 31, 2014 #1

    WannabeNewton

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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!
     
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  3. Jan 31, 2014 #2

    Avodyne

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    I'm not going to try to give a complete answer here, but there are a couple of simple points that can be stated easily.

    It's defined by precisely that set of commutators.

    You've misunderstood the notation. The eigenvalue of ##(J^+_1)^2+(J^+_2)^2+(J^+_3)^2## is ##j_+(j_++1)##, and similarly for ##-##.

    Also, these finite-dimensional representations of the Lorentz algebra are not unitary. These reps do not label states, but rather fields. The Lorentz transformation matrix that acts on (say) a vector field is not unitary.
     
  4. Jan 31, 2014 #3
    With regard to the SO(4) being isomorphic to SU(2)xSU(2) Tom S made a good post here:

    https://www.physicsforums.com/showthread.php?t=446309

    My observation on the second part of your first questioning paragraph would be j(j+1) =j^2 still applies,
     
  5. Jan 31, 2014 #4

    WannabeNewton

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    So the statement that ##so(3,1) \simeq su(2) \oplus su(2)## simply amounts to the algebraic manipulation of the usual Lie bracket relations for ##so(3,1)## into an identical pair of ##su(2)## Lie bracket relations (by means of the complexification)?

    Oh. So are we just using the fact that ##[J^+_i, J^-_j] = 0 \Rightarrow [(J^+)^2, (J^-)^2] = 0## along with the usual ##[J^+_i, (J^+)^2] = [J^-_i, (J^-)^2] = 0## to consider common eigenstates of the form ##|j_+, (j_z)_+, j_-, (j_z)_- \rangle##, fixing ##j_+## and ##j_-## i.e. choosing a pair ##(j_+, j_-)##, and allowing ##(j_z)_+## and ##(j_z)_-## to vary from ##-j_+,...,j_+## and ##-j_-,...,j_-## independently, respectively, akin to how we construct finite dimensional angular momentum representations of ##SU(2)## in regular QM (except now we don't have the interpretation of the representations as angular momentum representations)?

    Sorry, yes you're right. I was confusing the case of ##SO(3,1)## with the case of ##SU(2)## from the non-relativistic QM context. However I'm still confused on the last two points of the OP. Take for example the ##(\frac{1}{2},0)## representation of ##so(3,1) \simeq su(2) \oplus su(2)##. What associated Lie group are we constructing a representation for using the aforementioned representation of ##so(3,1)##? Phrased differently, since ##SO(3,1)## is not isomorphic to ##SU(2) \times SU(2)##, and I presume we want a representation of the universal cover ##SL(2,\mathbb{C})## of ##SO(3,1)## akin to how in non-relativistic QM the ##j = \frac{1}{2}## representation corresponded to the fundamental spinorial representation of ##SU(2)## which is the universal cover of ##SO(3)##, how do we go about constructing a ##(\frac{1}{2},0)## representation of ##SL(2,\mathbb{C})## from said representation of ##so(3,1)##?

    Sorry my knowledge of representation theory is very poor so if you happen to have references on this I would be much obliged. Thanks for the reply!

    Thanks! I wasn't so concerned about ##SO(4)## being the universal cover of ##SU(2)\times SU(2)## but rather the question of what desired Lie group we wish to obtain a representation of by using the ##(j_+,j_-)## representations of ##so(3,1)## since ##SO(3,1)## is not isomorphic to ##SU(2)\times SU(2)## even though ##so(3,1) \simeq su(2) \oplus su(2)##. I'll take a look through the thread you linked.
     
    Last edited: Jan 31, 2014
  6. Jan 31, 2014 #5

    Avodyne

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    Yes.

    Yes.

    That's exactly what we've done. For ##(\frac{1}{2},0)##, we have a representation of ##\vec J_+## as ##\frac{1}{2}\vec\sigma##, where ##\vec\sigma## are the Pauli matrices, and we have a representation of ##\vec J_-## by zero. The angular momentum operator is ##\vec J=\vec J_++\vec J_-##, so this is represented by ##\frac{1}{2}\vec\sigma##. The boost operator is ##\vec K=-i(\vec J_+-\vec J_-)##, so this is represented by ##-\frac{1}{2}i\vec\sigma##. These matrices then exponentiate (with parameters), in exactly the way you described for SU(2), to give representations of elements of SL(2,C).

    Edit: actually I think the group you get is PSL(2,C). But I'm not so good at these names, so that might be wrong.
     
    Last edited: Jan 31, 2014
  7. Jan 31, 2014 #6

    samalkhaiat

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    [itex]SU(2) \times SU(2)[/itex] is the double cover of [itex]SO(4)[/itex]. What you have done is simply shown that
    [tex]SO( 1 , 3 )_{ \mathbb{ C } } \cong SU_{ L } ( 2 ) \times SU_{ R } ( 2 ) ,[/tex]
    where [itex]SO_{ \mathbb{ C } }( 1 , 3 )[/itex] is the representation from a complexified [itex]SO( 1 , 3 )[/itex]. However, on the level of algebras, there is a 1:1 correspondence between [itex]SO_{ \mathbb{ R } }( 1 , 3 )[/itex] and [itex] SO( 1 , 3 )_{ \mathbb{ C } }[/itex]. This follows from the homomorphism
    [tex]SO_{ \mathbb{ R } }( 1 , 3 ) \sim SL( 2 , \mathbb{ C } ) .[/tex]
    But, using [itex]SL( 2 , \mathbb{ C } )[/itex] to classify the representations is more difficult (though elegant) and leads essentially to the same results. This is because of the fact that the Lorentz algebra is the direct sum of two conjugate copies of [itex]sl( 2 , \mathbb{ C } )[/itex]:
    [tex]so( 1 , 3 ) = sl( 2 , \mathbb{ C } ) \oplus sl( 2 , \mathbb{ C } ) .[/tex]
    So, it is easier to label the representations by the pair [itex]( j_{ L } , j_{ R } )[/itex], where
    [tex]\sum_{ i = 1 }^{ 3 } ( J^{ + }_{ i } )^{ 2 } = j_{ L } ( j_{ L } + 1 ) ,[/tex]
    [tex]\sum_{ i = 1 }^{ 3 } ( J^{ - }_{ i } )^{ 2 } = j_{ R } ( j_{ R } + 1 ) .[/tex]
    And, since [itex]J_{ 3 } = J^{ + }_{ 3 } + J^{ - }_{ 3 }[/itex], we can identify the “spin” of the representation as [itex]j_{ L } + j_{ R }[/itex], its dimension being [itex]( 2 j_{L} + 1 ) ( 2 j_{ R } + 1 )[/itex]. All (reducible) representations can be built up (as tensor product) from the two inequivalent fundamental representations
    [tex]\psi_{ \alpha } \sim ( 1/2 , 0 ) \in SU_{ L } ( 2 ) ,[/tex]
    and
    [tex]\bar{ \psi }^{ \dot{ \alpha } } \sim ( 1/2 , 0 )^{ * } = ( 0 , 1/2 ) \in SU_{ R } ( 2 ) .[/tex]
    And the irreducible representations are obtained by subtracting invariant subspaces from the tensor product. The notes below can help you with your learning. Also, I believe, I have posted many posts on the subject in here, but I don’t have the time to dig them up.

    Sam
     

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  8. Jan 31, 2014 #7

    samalkhaiat

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    Last edited: Jan 31, 2014
  9. Jan 31, 2014 #8

    WannabeNewton

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    Awesome, thanks for the help Sam and Avodyne! That really cleared things up for me. I have one or two questions about the ##(\frac{1}{2},\frac{1}{2})## representation, in particular the way we get the vector representation out of the left-handed and right-handed Weyl spinors and the Pauli matrices, but I'll read through the notes and threads Sam posted before moving on. Thanks again!
     
  10. Feb 1, 2014 #9

    dextercioby

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    This is probably the most discussed technical subject in the >10 yr history of PF. WNB, the best book I know of on this topic is 'Sexl, Urbantke - Relativity, Groups, Particles'.
     
  11. Feb 1, 2014 #10

    vanhees71

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    Yes, Sexl+Urbandtke is the best source to learn about the representation theory of the rotation, Lorentz, and Poincare groups. Also Weinberg's presentation in his Quantum Theory of Fields is great. There it's of course done in the context of QFT, while in Sexl+Urbandtke they concentrate on the mathematics of the representations.

    Just to make it clear: SO(1,3) is of course not the same as SO(4) but it's, of course, similar. The point is that you have silently complexified the [itex]\mathrm{su}(2) \oplus \mathrm{su}(2)[/itex] algebra. While the generators [itex]\vec{J}[/itex] and [itex]\vec{K}[/itex] belong to a real Lie algebra the "pseudo spins" are defined by
    [tex]\vec{J}_{\pm} = \frac{1}{2} (\vec{J} \pm \mathrm{i} \vec{K})[/tex]
    and thus are complex.

    The generators [itex]\vec{J}_{\pm}[/itex] of course form a basis of [itex]\mathrm{su}(2) \oplus \mathrm{su}(2)[/itex] and you can write down the (up to equivalence) unitary irreps of this Lie algebra, but the irreps. are unitary only for the real Lie group, but if you want to get back to the original generators of the Lorentz group you have complex linear combinations of [itex]\vec{J}_{\pm}[/itex] and thus work in the complex extension of [itex]\mathrm{su}(2) \oplus \mathrm{su}(2)[/itex]. So you still have irreps but not unitary ones. That all finite-dimensional irreps. of the proper orthochronous Lorentz group are non-unitary comes about, because the Lorentz group is not compact.

    It follows that the Lie algebra of the universal covering group of the proper orthochronous Lorentz group is in fact not [itex]\mathrm{su}(2) \oplus \mathrm{su}(2)[/itex] but [itex]\mathrm{sl}(2,\mathbb{C})[/itex] and the universal covering group of the proper orthochronous Lorentz group is indeed [itex]\mathrm{SL}(2,\mathbb{C})[/itex].
     
  12. Feb 1, 2014 #11

    WannabeNewton

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    Thanks dexter and vanhees! I'll see if I can find the book at my university library. Looking at the table of contents on Amazon, would I be able to skip straight to chapter 6?
     
  13. Feb 1, 2014 #12

    vanhees71

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    If you are lucky, your university has an account for Springer online books. Then you find it here:

    http://rd.springer.com/book/10.1007/978-3-7091-6234-7

    to legally download it. I guess you can directly jump to Cpht. 6, given that you have profound knowledge on relativity and quantum theory.

    You don't need much more than the idea what a Lie group and a Lie algebra and their representations is. You should know the representations of the rotation group SO(3) and its covering group SU(2), which is nothing else than the usual treatment of angular momentum in quantum theory since the angular momentum operator necessarily is a basis of the Lie algebra su(2), because angular momentum generates rotations according to Noether's theorem.
     
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