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Factor of 'i' and antisymmetrization in Dirac Lagrangian

  1. Feb 13, 2009 #1
    Hello everyone, I'm not sure if these questions are really trivial or of they're a little subtle... but here goes.

    1. In Ramond's text (Field Theory: A Modern Primer), he explains that the Lagrangian for fermions should have the derivative operator antisymmetrized in order for the kinetic term to be real: [itex]\psi^\dag \sigma^\mu \overleftrightarrow\partial_\mu \psi[/itex] .This is equation (1.7.3), with a discussion above equation (1.4.40). I don't quite understand why the antisymmetrization should do this since the derivative really only acts on a factor of [itex]e^{ip\cdot x}[/itex] in the Fourier transform of the spinors?

    [I understand that at the end of the day one can always integrate by parts to get the "only-right-acting" derivative... but this becomes more subtle in cases where one dimension is compact or when the space has curvature since then integration by parts picks up surface terms and/or derivatives of [itex]\sqrt{g}[/itex]]

    2. This, I'm almost sure, is a stupid question: I'm confused about the factors of i in Ramond's fermion Lagrangian. He seems to use the same conventions as particle physicists, e.g. (+---) metric and chiral basis of gamma matrices, but his kinetic term appears to be missing a factor of i and the mass term appears to have gained a factor of i. Doesn't this mean that the Lagrangian is no longer real? Where is this factor of i coming from?

    Thanks,
    Joe
     
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  3. Feb 14, 2009 #2
    I think I can talk about your 2nd question. I asked this before on here but got no reply so I'm still a bit confused. The important thing about the gamma matrices is the relation:
    [tex][\gamma_{\mu}, \gamma_{\nu}]=2g_{\mu \nu}[/tex]. Now suppose I wanted to switch the signature of the metric, but keep this relationship (so if I wanted to switch from -+++ to +---, I would do so, but I would still want [tex]\gamma_{0}\gamma_{0}=-1[/tex] and not +1. Then this can be done by inserting the imaginary "i" wherever you see a gamma matrix: [tex][i\gamma_{\mu}, i\gamma_{\nu}]=-2g_{\mu \nu}[/tex]. . The mass term has the zeroth gamma matrix so one i is inserted there, and the kinetic term has two gamma matrix so there should be an overall minus sign. However, this minus sign might cancel from having to change this sign anyway as you would in (p^2=-m^2 to p^2=m^2).

    So basically what I'm saying is that it's really confusing and I have no idea and I hope someone can clear it up :-).

    1. I thought all the operators had to be antisymmeterized or otherwise the expression would be zero because of the anticommutivaty of Grassman variables?
     
  4. Feb 14, 2009 #3

    StatusX

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    Actually integration by parts works pretty generally. For example, you can compute that:

    [tex] \sqrt{g} f (\nabla_\mu v^\mu) = - \sqrt{g} (\partial_\mu f) v^\mu + d(...) [/tex]

    where [itex]d(...)[/itex] represents a total derivative which turns into an integral over the boundary of the space. If the space has a boundary, we usually pick boundary conditions so that such terms can be ignored (eg, all fields go to zero at the boundary), but sometimes you need to be more careful.

    As for the factor of i, I think it's what Redx mentioned, that he's using an unconventional signature. For example, Srednicki uses (-1,1,1,1), and his spinor kinetic terms have the i. Note that if a matrix squares to -1, it has imaginary eigenvalues, and so cannot possibly be hermitian, so this convention choice really has a bearing on things like reality properties.
     
  5. Feb 14, 2009 #4
    Hi there StatusX 00 is the [tex]\nabla_\mu[/tex] a covariant derivative or a partial derivative? Should the [tex]\partial_\mu[/tex] also be a covariant derivative?

    Thanks,
    Joe
     
  6. Feb 14, 2009 #5

    StatusX

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    Technically they're both covariant derivatives, but the covariant derivative of a scalar is just an ordinary derivative.
     
  7. Feb 14, 2009 #6
    You want even more confusion? Take a look at Weinberg's Dirac equation, eqn. 5.5.43, page 225, volume I. There is no 'i' in the kinetic term or the mass term, and Weinberg uses (-1,1,1,1).

    I've also seen the form of the Dirac equation that the original asker asked about, elsewhere. So two examples does not necessarily make something conventional, but you have to wonder...where did those authors learn about their funny form of the Dirac equation, and where did those people learn it from...ha.
     
  8. Feb 14, 2009 #7

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    Yea, Srednicki uses a bizarre convention where [itex]\{ \gamma_\mu, \gamma_\nu\}=-2\eta_{\mu \nu}[/itex]. Meanwhile, Ramond uses a convention where [itex]\beta[/itex], the matrix used to define [itex]\bar{\Psi}[/itex], is imaginary, where as most people (including Weinberg and Srednicki) use a real [itex]\beta[/itex].

    I would say Weinberg's conventions are the ones most commonly encountered. An unfortunately large amount of time is spent sorting out all these sign conventions, especially when it comes to spinors.
     
    Last edited: Feb 14, 2009
  9. Feb 15, 2009 #8
    I think Srednicki uses that bizarre convention so that his Dirac equations looks like the Dirac equation that (+---) people use, even though Srednicki is using (-+++). But this comes at the expense of some atypical gamma matrix identities. This stuff seems highly non-trivial to sort out, but I do wish that it was touched upon a little in books, about the different conventions and stuff.
     
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