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Angular momentum vs the Hamiltonian problem in the Dirac field theory (canonical)

  1. Oct 20, 2004 #1
    I need some suggestions and/or corrections if I understand this correct? My questions are based on the book by Mandl and Shaw.

    Conserved currents are based on Noethers theorem and directly connected to spacetime and field transformations (rotations, translations, phase, ...). One can therefore derive the general expression for the Hamiltonian and angular momentum as two conserved quantities.

    The Hamiltonian density can be written as,
    [tex] \mathcal{H} = \pi_r \times \frac{\mathrm{d}}{\mathrm{d}t}\phi_r - \mathcal{L}, r=1...n,[/tex]
    where
    [tex]n,[/tex]
    is the number of fields.

    In a charge KG theory you can have the fields as either
    [tex]\phi, \phi^\dagger,[/tex]
    or
    [tex]\phi_1 = Re(\phi), \phi_2 = Im(\phi),[/tex]
    but the easiest way is to use the
    [tex]\phi, \phi^\dagger,[/tex]
    for a particle interpretation in a canonical quantization and one starts with
    [tex] \mathcal{H} = \pi \times \frac{\mathrm{d}}{\mathrm{d}t}\phi + \pi^\dagger \times \frac{\mathrm{d}}{\mathrm{d}t}\phi^\dagger - \mathcal{L}.[/tex]

    QUESTION: How does one deal with the expression of the angular momentum? The same question for the Dirac case as follows.

    The Dirac theory imposes the use of the same principle as it is a complex field but we now have
    [tex]\Psi, \bar{\Psi},[/tex]
    and thus

    [tex] \mathcal{H} = \pi \times \frac{\mathrm{d}}{\mathrm{d}t}\Psi + \bar{\pi} \times \frac{\mathrm{d}}{\mathrm{d}t}\bar{\Psi}
    - \mathcal{L}.[/tex]

    QUESTION: However when the book speaks about the angular momentum it does not use the adjoint field at all, only the Dirac field. Is it correct to think that it means that it treats the Dirac field as the complex one directly? Without the need of the adjoint in this sense?

    NOTE: The reason for my question is that the spinpart of the expression on the angular momentum is as well a kind of sum of present fields but also a transformationmatrix is involved. It yields,
    [tex]\mathcal{M}^{0\alpha\beta} = x^\alpha\mathcal{F}^{0\beta} - x^\beta\mathcal{F}^{0\alpha} + \pi_rS^{\alpha\beta}_{rs}\phi_s.[/tex]
    It is stated that one uses the sum as the elements of the Dirac spinor instead of the fields. But, this is not the case for the Hamiltonian nor the momentum where one uses the fields.

    Any suggestions of other resources to read from. I have looked in a lot of books and people speak differently when dealing with these issues. It seems somehow that I miss an important part?

    /C

    EDIT: Did the formulas with LaTex as I discovered it was possible.
     
    Last edited: Oct 20, 2004
  2. jcsd
  3. Oct 22, 2004 #2
    Maybe my question was not clear, but the problem is solved.

    /C
     
  4. Oct 29, 2004 #3

    dextercioby

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    I didn't look in Mandl and Shaw because i trusted someone who said they provide the easiest approach to QFT.

    That's the tricky part.
    Noether theorem provides with the conserved currents [tex] T^\mu _{\nu} [/tex]-the (a priori unsymmetrized) energy-momentum tensor and [tex] M^\lambda _{\mu\nu} [/tex],the angular momentum tensor (this tensor is antisymmetrical in his last two indices).For a theory which is invariant under the conformal group,u get other currents as well (see P.Ramond "Field theory:A modern primer",Addison-Wesley,1989,p.29 pp.33),The conserved "charges" are:[tex] T^0 _{\nu} [/tex],which is the energy-momentum 4vector [tex] P_{\nu} [/tex],xhose "0-th" component is the energy of the field (pay attention,NOT the hamiltonian),and the "i-th" component is the "i-th" component of the impulse of the field.The definition of a conserved "charge" tells us that it doesn't depend on time.That thing is proven by eveluating the energy-momentum 4vector on the solutions of the LAGRANGE FIELD EQUATIONS.
    The other charge is,of course,[tex] M^0 _{\mu\nu},the "0-th" component of the tensor above which is called "angular momentum tensor" who splits into the orbital angular momentum part (the antisimmetrized product of [tex] x_{\mu} [/tex] and [tex] P_{\nu} [/tex] ) and the spin part.
    There's what we have.I emphsize the fact that Noether's theorem (and hence its consequences) has nothing to do the Hamiltonian,since,at classical level,the Hamiltonian density is basically the Legendre transformation of the Lagrangian density wrt the canonical impulse densities and therefore depends on those impulse densities,on the fields,and the space derivatives of the fields and impulses.As Noether's theorem belongs to the Lagrangian field theory,the zero-th component of the energy momentum 4-vector will contain only fields and time derivatives of the fields,and not impulse densities and their space derivatives.The trick is that when u evaluate this charge on the solutions of the LAGRANGE field equations u get the same result as if evaluating the Hamilton function on solutions the HAMILTON field equations.But to be able to establish such an important equivalence (at classical level) u must come up with a consistent Hamiltonian description of the field.Once u have dunit,then u can canonically quantize such (free) theories.
     
  5. Oct 29, 2004 #4

    dextercioby

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    This is basically correct,but i must add that i resent your notation for the ordinary multiplication on the algebra of complex functions.

    What about angular momentum??
     
  6. Oct 29, 2004 #5

    dextercioby

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    For the formula written above to hold,maybe M&S use the Bjorken-Drell metric,and not the metric used by Weinberg.It's also necessary that canonical momentum densities be defined using the right derivative of the antisymmetrized Lagrangian density wrt to the time derivatives of the fields.I myself use the left derivative.If you use the left derivative,the time derivatives of the fields should be an the left of the canonical momentum densities in the definition of the canonical Hamiltonian,i.e.the Hamiltonian evaluated on the (hyper)surface of the primary (in this case second class) constraints.

    Are u sure???Coz if in the expression of the angular momentum of the Dirac field,the adjoint [tex] \bar{\Psi}_{\alpha} [/tex] does not appear,then this a serious mistake which makes the whole book look like a pile of c**p.

    This part of the formula is good:
    [tex]\mathcal{M}^{0\alpha\beta} = x^\alpha\mathcal{F}^{0\beta} - x^\beta\mathcal{F}^{0\alpha} + [/tex]

    The spin part is wrong.Are u sure that this formula (and the assertion above) are to be found in M&S,or are u making them up??Because if it is so,then,to quote from myself,"the whole book look like a pile of c**p".
    First of all,the canonical momentum densities are part of the Hamiltonian analysis of a field theory,so they have nothing to do with Lagrangian theory and the consequences of its Noether theorem.The correct form for the spin tensor for the Dirac field is (quote from D.Bailin&A.Love:"Introduction to gauge field theory",I.O.P.Publishing,1994,(the reprinted second edition from 1993),p.33):
    "[tex] S_{\rho\sigma} = \int d^3 x \bar{\psi}(x)\gamma^0\frac{1}{2}\sigma_{\rho\sigma}\psi(x) [/tex]",where [tex] \sigma_{\rho\sigma} = \frac{i}{2} [\gamma_{\rho},\gamma_{\sigma}]_{-} [/tex] ,and the spinor indices have been supressed.


    I would advise u to chapters 7 and 8 from Weinberg (first volume,of course).though,as I recall,he discusses second class constraints,but not specifically the Dirac field.What a shame!!!!He should be sued for that!! :tongue2:
     
  7. Nov 2, 2004 #6
    I see that you post a lot of information but it is not at all towards my questions from the beginning. It was difficult to write down my the questions I had on a paper, that is why I prefer discussions in real life. I wanted to delete or remove my first post but the edit was closed after a while. I have written the first post really not in a good way.

    For your information as I posted in the second one I have found the answer and it is explanied exactly in Itzykson and Zuber.

    I agree that I was "sloppy" when I talked about the Hamiltonian and conserved quantites connected to the Poincare group. But why do you speak of the conformal group which is not even requested nor used at this moment.

    The spinpart is not wrong, you have written the same physical expression as I have. I only took the relevant expressions from M&S and they are exactly the same as the one you took from "D.Bailin&A.Love".

    However this quote from you I do not understand?

     
    Last edited: Nov 2, 2004
  8. Nov 3, 2004 #7

    dextercioby

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    In this formula [tex] \pi_rS^{\alpha\beta}_{rs}\phi_s [/tex] I took [tex]\pi_r [/tex] for some some canonical momentum densities and that's why I said it was wrong.


    What's not be understood? I was looking for something challenging for my dying neurons,not for a "walk in the park".
     
  9. Nov 3, 2004 #8
    [itex]\pi_r[/itex] is the momentum density conjugate to the corresponding field in question. Nothing wrong there in my opinion and I show that as follows for the spinpart of the Dirac spinor equations:

    We have the complex Dirac field and therefore we are able to use the Dirac field and its adjoint as two different fields. Compare with approach of the complex KG-field but note the difference that we have the adjoint one. It is due to ...... (I do not give this part at the moment). I use the Lagrangian density, with suppressing the components,

    [itex]\mathcal{L} = \bar{\Psi}[i\gamma^\mu\partial_\mu + m]\Psi[/itex]

    The two fields

    [itex]\Psi, \bar{\Psi}[/itex]

    and the momentum densities conjugate to the fields are

    [itex]\pi=\frac{\partial\mathcal{L}}{\partial\Psi_{,0}}=i\Psi^\dagger, \bar{\pi}=0[/itex]

    Thus the spinpart of [itex]\mathcal{M}^{0\alpha\beta}[/itex] which only remains is in fact just the piece I have written

    [itex]\pi S^{\alpha\beta}\Psi \rightarrow i\Psi^\dagger S^{\alpha\beta}\Psi[/itex]

    The [itex]S^{\alpha\beta}[/itex] part is the that is connected to the infinitesimal transformation of the field considered. It is an interesting calculation to actually show that for the Dirac field it can be expressed as

    [itex]S^{\alpha\beta} = -\frac{i}{2}\sigma^{\alpha\beta} = -\frac{i}{2}[\gamma^\alpha,\gamma^\beta][/itex]

    where of course the constants and sign may depend on how you state your starting expression. Therfore I can write the spinpart as

    [itex]\Psi^\dagger\frac{1}{2}[\gamma^\alpha,\gamma^\beta]\Psi[/itex]

    or as given by the one you stated using [itex]\gamma^0\gamma^0=1[/itex]

    [itex]\bar{\Psi}\gamma^0\frac{1}{2}[\gamma^\alpha,\gamma^\beta]\Psi[/itex]

    To understand the integral we have that [itex]\mathcal{M}^{0\alpha\beta}[/itex] is the density and thus if we space integrate we get the relevant expression. Note as well that the only relevant parts for the general angular momentum of all combinations of [itex]\mathcal{M}^{0\alpha\beta}[/itex] in the indices are (23,31,12).

    I therefore claim that the expression I gave is not wrong but I agree that it can be confusing to use different conventions of different textbooks in order to understand the expressions in depth. I see also that you always want to make a clear different use of the Hamiltonian and Lagrangian language in a sense? I can understand that but I do not see any wrong in mixing it. As in this case I can from Noethers theorem find a conserved quantity, the energy, which is the same as the Hamiltonian expression. Nothing wrong to only speak of the Hamiltonian (density). As long as I state that I work with the canonical field quantization.

    Again, I totally agree that my first post was written in a really bad way.
     
  10. Nov 3, 2004 #9

    dextercioby

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    To me,spin part of the angular momentum tensor has nothing to do with the canonical momentum densities.

    Should have been a minus at the mass term (...[tex] -m\bar{\Psi} \Psi [/tex].
     
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