Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac Lagrangian density

  1. Mar 19, 2014 #1
    Hi,

    This is probably a trivial question, but I just wanted to check my understanding.
    Is the following expression for the Dirac Lagrangian correct?

    [itex]\mathcal{L}=\frac{i}{2}\overline{\Psi}\gamma^{\mu}\overleftrightarrow{∂_{μ}}\Psi-\overline{\Psi}m\Psi=\frac{i}{2}\overline{\Psi}\gamma^{\mu}\partial_{μ}\Psi-\frac{i}{2}\overline{\left(\partial_{\mu}\Psi\right)}\gamma^{\mu}\Psi-\overline{\Psi}m\Psi-\overline{\Psi}m\Psi[/itex]

    which can alternatively be expressed as,

    [itex]\cal{L}=\frac{i}{2}\overline{\Psi}\gamma^{\mu}\partial_{\mu}\Psi+\left(\frac{i}{2}\overline{\Psi}\gamma^{\mu}\partial_{\mu}\Psi\right)^{\dagger}-\overline{\Psi}m\Psi[/itex]
     
    Last edited: Mar 19, 2014
  2. jcsd
  3. Mar 20, 2014 #2
    I think you are making a little bit of confusion with the Klein-Gordon (scalar) Lagrangian. The Dirac Lagrangian contains only derivatives on [itex]\Psi[/itex], not on [itex]\bar\Psi[/itex]. It is:
    $$
    \mathcal{L}_{Dirac}=\bar\Psi(i\gamma_\mu \partial^\mu-m)\Psi.
    $$
     
  4. Mar 20, 2014 #3

    Bill_K

    User Avatar
    Science Advisor

    The Dirac Lagrangian does contain both derivatives:
    $$
    \mathcal{L}_{Dirac}=(1/2)\bar\Psi(i\gamma^\mu \partial_\mu - m)\Psi + (1/2)(- i\partial_\mu\bar\Psi\gamma^\mu - m)\Psi
    $$
     
    Last edited: Mar 20, 2014
  5. Mar 20, 2014 #4
    Actually, one usually writes it without the second piece (see for example Peskin, eq. 3.83). The Hermitian conjugate is almost always implicit. Even if you want to include that, then I don't think you should have the 1/2 factor. Still, I could be wrong.
     
  6. Mar 20, 2014 #5
    Isn't the 1/2 needed to ensure that one recovers the correct equations of motion though?
     
  7. Mar 20, 2014 #6

    Bill_K

    User Avatar
    Science Advisor

    It's true, one often sees it written that way. True, but unfortunately incorrect. Peskin & Schroeder, in particular, has many fine features, but their treatment of the introductory topics is rather slapdash. I have to think they were in a hurry to get to things more advanced. It doesn't matter much, if you regard the Lagrangian as any old thing that leads to the equations of motion.

    But if you want to use the Lagrangian to derive other things, such as the energy-momentum tensor, then you'll need to write it in the form I quoted - both the Hermitian conjugate and the factor of 1/2.
     
  8. Mar 20, 2014 #7
    Ah ok, thanks for the information. Much appreciated!
     
  9. Mar 20, 2014 #8
    Maybe I'm wrong but it seems to me that the form you quoted, even including the 1/2, is actually meaningless.
    The Lagrangian is always defined starting from the Action, meaning that you are basically free to integrate it by part and still obtain a perfectly valid one. Let's then assume that your form is correct. Then if you integrate by part the second piece you acquire a derivative term which is equal to the first one but with opposite sign and they cancel. This means that you Lagrangian doesn't have the kinetic term! Where am I wrong? It seems to me that, if you want to express the Lagrangian in your way then you need to include an extra minus sign.

    By the way, I think that you can obtain a reasonable energy-momentum tensor also from the Lagrangian I quoted. See for example, Mandl-Shaw eq. 4.22 and 4.23.
     
  10. Mar 20, 2014 #9

    Bill_K

    User Avatar
    Science Advisor

    Sorry, I left out a minus sign, which is now corrected.
     
  11. Mar 21, 2014 #10

    samalkhaiat

    User Avatar
    Science Advisor

    Both, the usual and the hermitean, Dirac Lagrangians are as good (or as bad) as each other. They lead to the same equations of motion, the same [itex]U(1)[/itex] current, the same energy-momentum vector and the same angular momentum tensor. This is because the difference between them is a total divergence.

    Sam
     
  12. Mar 21, 2014 #11

    Bill_K

    User Avatar
    Science Advisor

    It doesn't bother you that, in the one case, the canonical momentum of [itex]\bar \Psi[/itex] vanishes?

    Adding a four-divergence to the Lagrangian does not lead to the same energy-momentum tensor. For example, the symmetrical Belinfante energy-momentum tensor is obtained from the unsymmetrical version in exactly this way, by adding a four-divergence. It is true that the conserved quantities obtained by integrating either of these tensors remain the same.
     
    Last edited: Mar 21, 2014
  13. Mar 21, 2014 #12

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, it's kind of weird that [itex]\bar{\Psi}[/itex] should be treated as something independent of [itex]\Psi[/itex]. There are really 8 independent components of [itex]\Psi[/itex]: The real and imaginary parts of each row of the spinor. So if you varied them, you would get 8 corresponding momenta, which presumably could be reassembled into a single, complex, spinor-valued momentum. Having separate momenta for [itex]\Psi[/itex] and [itex]\bar{\Psi}[/itex] seems redundant.

    Maybe there's a way to make sense of it by starting out with [itex]\Psi[/itex] and [itex]\bar{\Psi}[/itex] as independent fields, and then having a constraint that forces them to be related?
     
  14. Mar 21, 2014 #13

    samalkhaiat

    User Avatar
    Science Advisor

    No, not in this case because the enlarged phase-space associated with the Hermitean Lagrangian does not lead to any new information.

    In QFT all we care about is [itex]J^{ \mu } , \ \ P^{ \mu } , \ \ M^{ \mu \nu }[/itex] and the anticommutation relations
    [tex]\{ \psi_{ i } ( x ) , \psi^{ \dagger }_{ j } ( y ) \} = \delta_{ i j } \delta^{ 3 } ( x - y ) ,[/tex]
    [tex]\{ \psi_{ i } ( x ) , \psi_{ j } ( y ) \} = \{ \psi^{ \dagger }_{ i } ( x ) , \psi^{ \dagger }_{ j } ( y ) \} = 0 .[/tex]

    You can check that this structure is identical for both Lagrangians.
     
  15. Mar 21, 2014 #14

    samalkhaiat

    User Avatar
    Science Advisor

    If you want to do it systematically, then yes you can treat it as constrait system and quantize it using the so-called Dirac brackets. But, in this case, you will not gain any new structure.
     
  16. Mar 21, 2014 #15

    Bill_K

    User Avatar
    Science Advisor

    :confused: It's always done that way, regardless of which Lagrangian you use. The Lagrangian is varied treating ##\Psi## and ##\bar \Psi## as independent variables. I've never seen a treatment that did not do this. What is really weird IMHO, is treating ##\Psi## and ##\bar \Psi## differently!

    Well, use whichever Lagrangian you want, but the issue is often badly mishandled. Here, for example, is what Peskin and Schroeder do (p43) First they "derive" the Lagrangian from the Dirac Equation:

    What they meant to say (I hope) is that the Euler-Lagrange equation obtained by varying ##\Psi## is the conjugate Dirac Equation for ##\bar \Psi##, and the Euler-Lagrange equation obtained by varying ##\bar \Psi## is the Dirac Equation for ##\Psi##. Anyway, what they had to do without mentioning it is to integrate by parts (equiv. add a divergence). P & S derive the Dirac Equation and its conjugate from two different Lagrangians. Unfortunately, this is typical. :frown:

    Hm, by "In QFT" I think you mean "excluding gravity", because in that case the energy-momentum tensor really does matter. IMHO again, the energy-momentum tensor exists in physics as a unique local quantity, not just its integral, and it behooves us to derive the correct expression for it, even if we're only doing QFT. Similarly, one should not be satisfied with an energy-momentum tensor that is not symmetric, even though it gives the same integrated quantity.
     
    Last edited: Mar 21, 2014
  17. Mar 21, 2014 #16

    ChrisVer

    User Avatar
    Gold Member

    for treating psi and psibar differently, you can just think of waves instead... an incoming and an outgoing wave are orthogonal to each other/independent...
     
  18. Mar 21, 2014 #17

    samalkhaiat

    User Avatar
    Science Advisor



    Are you under the impression that the Hermitean Dirac Lagrangian gives you some symmetric energy-momentum tensor? The symmetrization procedure of the energy-momentum tensor does not depend on the Hermiticity of the Lagrangian. Both Lagrangians, the usual and the Hermitean, lead to non-symmetric energy-momentum tensor:
    [tex]
    \mathcal{ L } = \bar{ \psi } i \gamma^{ \mu } \partial_{ \mu } \psi , \ \Rightarrow \ T^{ \mu \nu } = \bar{ \psi } i \gamma^{ \mu } \partial^{ \nu } \psi .
    [/tex]
    [tex] \mathcal{ L } = \frac{ i }{ 2 } \bar{ \psi } \gamma^{ \mu } \overleftrightarrow{ \partial_{ \mu } } \psi , \ \Rightarrow \ T^{ \mu \nu } = \bar{ \psi } \gamma^{ \mu } \overleftrightarrow{ \partial^{ \nu } } \psi .
    [/tex]

    Both these tensors can be symmetrized through the usual Belinfante procedure (which depends only on the Poincare’ invariance of the theory and nothing else). After symmetrization you can, if you wish, couple either one to gravity but, again, you get equivalent results.

    Sam
     
  19. Mar 21, 2014 #18

    Bill_K

    User Avatar
    Science Advisor

    No, of course not.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dirac Lagrangian density
  1. Dirac Lagrangian (Replies: 11)

Loading...