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Building a Lagrangian out of Weyl spinors

  1. Feb 26, 2012 #1
    I've been watching Sidney Coleman's QFT lectures (http://www.physics.harvard.edu/about/Phys253.html, with notes at http://arxiv.org/pdf/1110.5013.pdf), and I'm now on to the spin 1/2 part of the course. We've gone through all the mechanics of constructing irreducible representations [itex]D^{(s1,s2)}[/itex] of the Lorentz group, and have begun to build Lagrangians out of [itex]D^{(1/2,0)}[/itex] and [itex]D^{(0,1/2)}[/itex], the right- and left-handed Weyl spinors.

    One of the requirements of a Lagrangian density is that it transform like a scalar, so he turns to the task of finding objects that can be built out of spinors that transform like a scalar. He notes that if you have a spinor field [itex]u(x)[/itex], then you can construct a field which transforms like a vector by doing [itex]V^\mu(x) = u^*(x)\sigma^\mu u(x)[/itex], where [itex]\sigma^\mu = (1, \stackrel{\rightarrow}{\sigma})[/itex].

    Seeing that, I would expect us to go on to construct a scalar field by taking [itex]\partial_\mu\cdot V^\mu(x) = \partial_\mu\cdot(u^*(x)\sigma^\mu u(x))[/itex], since taking the four-divergence is normally how one constructs a scalar field out of a vector. However, he instead constructs the object [itex]u^*(x)\sigma^\mu\cdot\partial_\mu u(x)[/itex], and uses that to build a Lagrangian, which when minimized produces the Weyl Equation. I'm having trouble understanding how this works--is that equivalent to putting the derivative outside like I did above, or is it something different? And if it's different, why would one want to do it that way, and how does one show that it transforms like a scalar?
     
  2. jcsd
  3. Feb 26, 2012 #2
    Note that a total divergence is useless as a Lagrangian density term, because we integrate the Lagrangian density to get the Lagrangian, and the integral of a total divergence is just a boundary term, which we always take to vanish. So while the expression you wrote down is a scalar, putting it in the Lagrangian density doesn't give you anything interesting.

    If you can show that [itex]u^*(x)\sigma^\mu u(x)[/itex] transforms like a vector then you ought to be able to follow the same steps to see that [itex]u^*(x)\sigma^\mu\cdot\partial_\mu u(x)[/itex] transforms in the same way except for an extra inverse Lorentz transform matrix from the derivative, which cancels the Lorentz transformation matrix of the vector, so the whole thing is a scalar.
     
  4. Feb 27, 2012 #3
    Ah right, that makes sense. Plus, [itex]\partial_\mu\cdot(u^*(x)\sigma^\mu u(x)) = (\partial_\mu\cdot u^*(x))\sigma^\mu u(x) + u^*(x)\partial_\mu\cdot(\sigma^\mu u(x))[/itex], so my quantity is just the sum of two of these single-derivative quantities, arranged in an unfortunate combination that vanishes in the integral.

    The methods I've seen to show that [itex]u^*(x)\sigma^\mu u(x)[/itex] transforms like a vector involved taking specific examples of [itex]u \rightarrow e^{i\stackrel{\rightarrow}{L_i}}u[/itex] or [itex]u \rightarrow e^{i\stackrel{\rightarrow}{M_i}}u[/itex], and showing that each component in the result did what you would expect. That worked, but it seemed inelegant, and I'm not exactly sure how to generalize that to a [itex]u(x)[/itex] field so that we can take partial derivatives, or how to deal with the interactions between the transformation from the derivatives and the transformation from the spinors.

    Is there some more general way of showing how the transformation works, which can easily handle a case like this? Intuitively it makes sense, since we're dotting together a lower index and an upper index, but that's just coming from the notation--I still don't quite understand how to build the equations that underlie it.
     
  5. Feb 27, 2012 #4
    I think a good way to understand this is to realize that [itex]\sigma^\mu_{\alpha \dot{\alpha}}[/itex] is a Lorentz invariant quantity, like for example [itex]\eta_{\mu \nu}[/itex]. [itex]\sigma[/itex] carries three indices: a vector index, an SU(2)_L index and an SU(2)_R index. [itex]\eta_{\mu\nu}[/itex] has the property that if you hit it with two Lorentz transformation matrices, one for each index, it is unchanged. This property ensures that if you use [itex]\eta[/itex] to contract two contravariant vectors, you get a scalar. Similarly you can show that if you hit [itex]\sigma[/itex] with three Lorentz transformation matrices, one for each index, you get the same thing back unchanged (but note that each matrix will be from a different representation of the Lorentz group).

    Once you have convinced yourself that [itex]\sigma[/itex] is a Lorentz invariant, it is easy to show that contracting it with other Lorentz covariant quantities gives results that transform as expected under the Lorentz group. For instance if you contract both of its spinor indices with Weyl spinors in an expression like [itex]u^*(x)\sigma^\mu u(x)[/itex], the only index remaining is a vector index, so the expression transforms like a vector. If you convince yourself that [itex]\partial_\mu u_\alpha(x)[/itex] (alpha = spinor index) transforms as its indices suggest, with one vector transformation matrix and one SU(2)_L transformation matrix, then you know that [itex]u^*(x)\sigma^\mu\cdot\partial_\mu u(x)[/itex] transforms like a scalar, because all the indices are contracted.

    I like Srednicki's QFT book's discussion of this stuff.
     
  6. Feb 27, 2012 #5
    Perfect, that's exactly what I was looking for. The equations in my text weren't listing spinor indices on the same footing as vector indices like you did, so it was difficult to see the symmetry between the Lorentz transformation properties of the different indices. I still need to actually sit down and work out the transformation properties of [itex]\sigma[/itex], but this at least gives me the perspective on what I'm trying to accomplish--it's clear to me how Lorentz invariance of [itex]\sigma[/itex] would imply Lorentz invariance of the entire quantity. Thanks!

    Incidentally, I really like Coleman's description of Weyl spinors as "the square roots of vectors". Seeing the indices listed out this way helps reinforce that idea.
     
  7. Feb 28, 2012 #6
    Ok, so I've worked out the transformation properties of [itex]\sigma^\mu[/itex] to my satisfaction. There was one thing I came across in Srednicki that jumped out at me, though. He states that quantities like [itex]g_{\mu\nu}[/itex], [itex]\sigma^\mu_{\alpha\dot{\alpha}}[/itex], etc. which are invariant under Lorentz transformations will exist in any representation whose direct sum decomposition contains a scalar representation. This makes perfect sense--scalar representations are necessarily the identity element, so that basis element will always remain unchanged by any transformation. This is also the reason [itex]\epsilon_{abc}[/itex] is an invariant, and the fact that it's a scalar representation in the antisymmetric part of the decomposition is the reason that it's antisymmetric in its indices.

    This leads me to another question--how does one go about finding these elements for an arbitrary representation? If the representation in question is the product of two spin representations, then this is equivalent to finding the singlet state, so I believe the matrix elements of the invariant element are just the Clebsch-Gordan coefficients for that element, correct? Is it possible to generalize that process to representations of other groups like, say, SO(3,1)? If so, then one could construct the [itex]\sigma[/itex] matrices explicitly, by decomposing [itex]D^{(0,^1/_2)*}\otimes D^{(^1/_2,0)*}\otimes D^{(^1/_2,^1/_2)}[/itex] and finding the scalar element.

    From what I've read so far about the [itex]\gamma^\mu[/itex] matricies, it seems like they're the same thing for the space of Dirac spinors, which I guess means that they're the scalar component of something like [itex](D^{(0,^1/_2)*}\oplus D^{(^1/_2,0)*})\otimes (D^{(0,^1/_2)*}\oplus D^{(^1/_2,0)*}) \otimes D^{(^1/_2,^1/_2)}[/itex], right? Is there some procedure for finding these things that applies to the general case?
     
  8. Mar 1, 2012 #7
    Yes, It's correct.

    I'm not sure of the following, so I'd be glad if someone more competent then me could correct me if I'm wrong.
    This is what I think I have understood:

    In general if you have a term

    [itex]L^{a_1\cdots a_n}= \phi_1^{a_1}\cdots \phi_n^{a_n}[/itex],

    where each field transforms according to a irreducible representation [itex]D^{A_1}(G)[/itex] of the symmetry group [itex]G[/itex]

    [itex]\phi_i^{a_i} \rightarrow D^{A_i}(g)^{a_i}_{\phantom{a_1}b_i} \;\phi_i^{b_i} \quad,\quad g\in G[/itex],

    then [itex]L[/itex] transforms with the direct product representation of the [itex]n[/itex] representations

    [itex]L^{a_1\cdots a_n} \rightarrow D^{A_1}(g)^{a_1}_{\phantom{a_1}b_1}\; \cdots \;D^{A_n}(g)^{a_n}_{\phantom{a_1}b_n}\;L^{b_1 {\cdots} b_n}[/itex].

    Even if the representations [itex]D^{A_i}(G)[/itex] are irreducible, in general the direct product representation it isn't.
    It can be though decomposed into the direct sum of irreducible representations.

    If you want [itex]L[/itex] to transform according to a single irreducible representation of [itex]G[/itex] you have to project [itex]L[/itex] onto the corresponding irreducible subspace.

    This is equivalent to what we are used to do with direct product representation of irreducible representations of [itex]SU(2)[/itex] when we sum two angular momentum to obtain the total spin.

    [itex]|L_1 L_2 J J_z > = \; \sum_{m_1 ,m_2} <m_1 m_2(L_1 L_2) J J_z> \,|L_1 L_2 m_1 m_2>[/itex]

    Another example could be the one you gave with QED interaction lagrangian

    [itex]\overline{\psi} _\alpha A^\mu \psi_\beta \rightarrow {\large(}D^{(0,1/2)}(\Lambda)\oplus D^{(1/2,0)}(\Lambda){\large)}_{\beta,\rho}\;{\large(}D^{(1/2,0)}(\Lambda)\oplus D^{(0,1/2)}(\Lambda){\large)}_{\alpha,\sigma}\;D^{(1/2,1/2)}(\Lambda)^\mu_{\phantom{\mu}\nu}\; \overline{\psi} _\rho A^\nu \psi_\sigma [/itex]

    this term contains "spins" from zero to two, so to project it on the scalar subspace you need to multiply it by [itex]\gamma^\mu_{\alpha\beta}[/itex] so that

    [itex]\gamma_{\mu;{\alpha,\beta}} \;{\large(}D^{(0,1/2)}(\Lambda)\oplus D^{(1/2,0)}(\Lambda){\large)}_{\beta,\rho}\;{\large(}D^{(1/2,0)}(\Lambda)\oplus D^{(0,1/2)}(\Lambda){\large)}_{\alpha,\sigma}\;D^{(1/2,1/2)}(\Lambda)^\mu_{\phantom{\mu}\nu} = \gamma_{\nu; \rho,\sigma}[/itex]

    i.e. [itex]\overline{\psi} \gamma^\mu \psi A_\mu [/itex]is a scalar.

    Constructing the projection operator over a given irreducible space it's similar to find Clebsch-Gordan coefficent for [itex]SU(2)[/itex]. If you are interested in details of this topic, I think you could find someting on J.F. Cornwell, Group Theory in Physics.

    I hope I didn't completly misunderstood this topic, I'm studing something similar for an exam xD

    Ilm
     
    Last edited: Mar 1, 2012
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