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Independent fields in euclidean space

  1. Sep 10, 2012 #1
    If the Lagrangian is Hermitian, then fields and their complex conjugates are not independent. That is, you can solve the EOM for one field, and if you take the complex conjugate of that field, then that's how the complex conjugate field evolves: you don't have to solve the Euler-Lagrange equations for the complex conjugate field.

    If the Lagrangian is not Hermitian, does it even still make sense to say that fields are complex conjugate to each other, because then it'll no longer be true that one field is the complex conjugate of the other at a later time?

    I ask this because it seems that in Euclidean space, the Dirac Lagrangian is not Hermitian, yet the fields are denoted by the same symbol ψ, except the other field has a dagger. That doesn't seem to make sense, since if they are independent, they shouldn't be the complex conjugate of each other. Should there really be two independent symbols?
     
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  3. Sep 10, 2012 #2

    strangerep

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    Ummm, what makes you think it's not Hermitian? Remember that one can do an integration by parts and discard surface terms at infinity... :-)

    (Maybe you'd better write down the Lagrangian here, to make sure we're on the same page...)
     
  4. Sep 11, 2012 #3
    [tex]\mathcal L= \psi^\dagger i \gamma^0 \gamma ^\mu \partial_\mu \psi=
    \psi^\dagger D \psi [/tex]
    The operator D is Hermitian in Minkowski space.

    In Euclidean space, the Lagrangian is slightly different:

    [tex]\mathcal L=\psi^\dagger \gamma^0 D \psi [/tex]
    where D is

    [tex]-\gamma_i \partial_i [/tex]
    where i=1,2,3,4 with γ4=iγ0.

    D is anti-Hermitian in Euclidean space, so

    [tex]\mathcal L^\dagger= \psi^\dagger D^\dagger \gamma^{\dagger 0} \psi=
    -\psi^\dagger D \gamma^{0} \psi [/tex]

    If D and γ0 anti-commuted, then it would be Hermitian. But γ4 and γ0 commutes, so it seems that it's not true that the Lagrangian is Hermitian.
     
  5. Sep 11, 2012 #4

    Ben Niehoff

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    The gamma matrices have different properties in Euclidean space. Make sure you check carefully what's Hermitian and what's anti-Hermitian.
     
  6. Sep 12, 2012 #5

    samalkhaiat

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    Yes, you presented the poorman version of the problem. In my opinion, complex conjugation of spinors in Euclidean space is the most complicated and (mathematically) ill-defined concept in theoretical physics. However, one has to lean about it if one needs to work in Euclidean space.
    There have been to different approaches to the subject. The first is that of Osterwalder and Schrader. The degrees of freedom in the O-S method are “doubled”: Dirac spinor [itex]\psi[/itex] and its conjugate [itex]\bar{\chi}[/itex] are taken to be independent and Hermiticity of the action is abandoned. Of course there is nothing wrong with that. Indeed on physical grounds, requiring the Euclidean action to be Hermitean is not necessary, because Hermiticity is essentially needed for unitarity and unitarity only make sense in a theory with real time.
    The second approach is due to Schwinger and Zumino. Here the spinor degrees of freedom are “undoubled” and the Euclidean action is Hermitean! The puzzling difference between the two approaches can be understood by defining a new Wick rotation which acts as an analytic continuation [itex]x^{0}\rightarrow i \tau[/itex] and a simultaneous [itex]O(4)[/itex] rotation on spinor indices.
    Of course in the path integral context, the distinction between integrating over [itex]( \psi , \psi^{\dagger})[/itex] versus [itex]( \psi , \chi^{\dagger})[/itex] is only artificial due to the Grassmannian nature of both sets.

    As a good reference (written by good physicists), see hep-th/9608174v1. I suggest that you have a look at it to see that the problem is not as naïve as you presented it.

    Sam
     
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