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Conservation of angular momentum in field theory: imposed, or emergent?

  1. Aug 23, 2012 #1

    bcrowell

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    We would like both 4-momentum and angular momentum to be conserved in a field theory such as QED.

    My understanding is that in the case of 4-momentum, there are two different descriptions of the same theory. (1) We can impose conservation of 4-momentum p at each vertex, which requires that virtual particles be off-shell. (An electron can't emit an on-shell photon while conserving p.) This automatically guarantees that you can't compute a nonzero amplitude for any process that violates conservation of p. (2) There are also descriptions in which p fluctuates according to the uncertainty principle, but by the time you calculate amplitudes, it emerges that the amplitude for any p-nonconserving process cancels out.

    I'm confused about how this works when it comes to conservation of angular momentum. The analog to approach #1 above would be that we impose conservation of spin at each vertex. Just as we were forced to allow particles to be off shell, I guess we're forced to allow them to have the "wrong" spin. For example, the static repulsion between two electrons arises from interference between the no-exchange diagram and the one-particle-exchange diagram. If the internal state of the emitting or absorbing particle changes, then you can't have this interference. Therefore the virtual photon has to have spin 0.

    Is this analogy valid? I can see how there's at least some logical link between being off-shell and having "wrong" spin properties, since off-shell photons don't propagate at c, and therefore Lorentz invariance forbids you from constraining them to have transverse polarization. But that's about a component of the spin vector, which isn't a built-in property of the particle, whereas the analogy with #1 requires the photon to have a spin vector with the wrong *magnitude*.

    If virtual photons can have spins 0 and 1, is there any way to rule out other spins? Why not spin 2, or spin 1/2? Do we rule them out based on knowledge of the properties of the classical near fields, e.g., knowing that the static field of a point charge has zero angular momentum density everywhere?

    Is there an analog of #2, where we would not impose conservation of spin at each vertex, but amplitudes for nonconservation of angular momentum would end up canceling out?

    Is it valid to talk about conservation of spin at a vertex, or should it be spin plus orbital angular momentum? Classically, interactions at a point automatically conserve angular momentum if they conserve momentum. However, a vertex in a Feynman diagram doesn't necessarily represent anything localized, and if we're taking approach #2 for momentum, then there's also no guarantee that momentum is conserved.

    An example that's confusing the heck out of me is on p. 24 of the Feynman Lectures on Gravitation (not the same book as the undergrad Feynman Lectures). He's discussing crazy ideas for describing gravity using ordinary QFT, and here he's specifically talking about describing it using the exchange of virtual neutrinos. (I know, it's crazy.) He gives the argument I gave above about how you can't have static attraction if the emitting or absorbing body changes its internal state, and says that this rules out a description of gravity using the exchange of a spin-1/2 particle. (He then goes on and tries to salvage the model by making it a three-body interaction.) The argument he *doesn't* invoke, which would seem much simpler to me, is that we expect force carriers to be bosons, not fermions, because otherwise we'd be violating the expectation that at any three-leg vertex, we should have an even number of fermions -- otherwise the [itex]\Delta S[/itex] at the vertex would have to be a half-integer, and therefore nonzero. So if he's taking the analog of approach #1 by imposing spin conservation at each vertex, then it automatically fails. But if he's taking the analog of approach #2 by allowing nonconservation of spin at a vertex, then his argument about a change of internal state seems invalid, because he could just decree it to stay the same.

    The trouble I'm having figuring out this kind of thing from field theory books is that the spins are all sort of hidden behind a bunch of gamma matrices, and I've long since lost whatever fluency I had 20 years ago in interpreting those. There also seems to be a tendency to explain this kind of thing by writing down Lagrangians and reading off their properties, but I'm not fluent in that either at this point.
     
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  3. Aug 23, 2012 #2

    Bill_K

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    Where QFT starts, after all, is a description in terms of fields: scalars, spinors, four-vectors. We assemble these to form an interaction, which must be a Lorentz scalar. And being Lorentz invariant includes being rotationally invariant, therefore the interaction at each vertex conserves angular momentum.

    It's the spin of a particle that's an emergent concept. A photon, for example, is an excitation of a vector field. The fact that free photons have spin one is a derived property, not holding true for virtual photons.
     
  4. Aug 23, 2012 #3

    bcrowell

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    Thanks for the reply, Bill_K!

    Hmm...Lorentz invariance also includes translational invariance, which implies conservation of 4-momentum p. And yet people do sometimes use the description I referred to as #2 in the OP, where p is not conserved.

    Then I guess part of my question is whether there is any simple rule to predict what spins can emerge from a given field, and whether that rule has any simple explanation. My guess is that the rules (for bosons) are the following:
    -A. Look at the field's representation as a tensor, count indices, and call this the free spin jf.
    -B. The spin of virtual particles is allowed to be 0, 1, ... jf.
    Assuming I'm right about A and B, my question would be how B emerges, so that we know, e.g., that virtual gravitons will never have spin 3 or 1/2.

    I think I understand what was confusing me about the Feynman example. He has an object emitting a neutrino, and the legs of the vertex representing this object are drawn in the same style before and after emission. But I think this is misleading, because the object is not just changing its spin, it's also changing from a fermion into a boson or vice versa. This is what happens in ordinary processes like electron capture, where, for example, a fermionic atom can emit a neutrino and become a boson.
     
  5. Aug 23, 2012 #4

    tom.stoer

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    Momentum and angular momentum conservation in QFT follow from Poincare invariance of the (renormalized) operator algebra. So what one has to do is to construct Pα = (H, Pi), Li and Ki as quantum operators and to show that [H, Pi] = [H,Li] = 0
     
  6. Aug 23, 2012 #5

    bcrowell

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    Thanks for the reply, Tom. However, I'm not asking how to prove conservation, I'm asking questions about how conservation is or isn't imposed in the construction of Feynman diagrams.
     
  7. Aug 23, 2012 #6

    tom.stoer

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    A sum over Feynman diagrams corresponds to the Dyson series which is nothing else but a reformulation of the time evolution operator, i.e. exp(-iHt). In addition Feynman diagrams are related to (Lagrangian) path integrals for which a fully covariant expression is known. So at which step do you think that covariance has to be proven explicitly?

    Don't get me wrong; I think you are perfectly aware these facts, but I want to understand in detail where you think an explicit proof is required.
     
  8. Aug 23, 2012 #7

    bcrowell

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    I don't think an explicit proof is required. I'm asking specific questions about Feynman diagrams.
     
  9. Aug 23, 2012 #8

    strangerep

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    Maybe I misinterpret your questions, but my understanding is that both 4-momentum and angular momentum are indeed conserved at each vertex. This is part of the reason why the internal photon line in a QED tree diagram must be off-shell.

    That such conservation occurs at each vertex is a consequence of the form of the interaction term in the Hamiltonian. Since the fields must carry a representation of the Poincare group, it turns out that this term needs to be a Lorentz scalar (if we use the instant form of dynamics, as explained in Weinberg vol 1).

    Each type of elementary field corresponds to a unitary irreducible representation of the Poincare group. Each such rep already has a value for spin magnitude, so the spin value does not need to "emerge". But I'm not sure how much of this you already know...
     
  10. Aug 24, 2012 #9

    tom.stoer

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    The conservation laws in Feynman diagrams follow by construction. The starting point (e.g. the path integral) and the derivation respect Poincare invariance, so do Feynman diagrams (order by order) and at each vertex.
     
    Last edited: Aug 24, 2012
  11. Aug 24, 2012 #10

    Bill_K

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    No, no, strangerep, you have it backwards. A field is not a representation of the Poincare group. And it is not always irreducible. A field provides, at each point, a reducible representation of the Lorentz group. A field may describe one or more particles.

    A particle is a representation of the Poincare group, and the transition from a Lagrangian description in terms of fields to a description in terms of particles is where the spin value "emerges".
     
  12. Aug 24, 2012 #11

    tom.stoer

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    Which field is a reducible rep.?

    In QFT both field operators (fields in the Lagrangian) and states form irred. reps. For the fields you get it for free, for the field operators you have to prove that the operator algebra is anomaly-free. For the states it's awfully complicated b/c in principle you first have to solve the theory!!

    The problem is the following: yes, you have an irred. spinor rep., e.g. a quark; you may use it in perturbation theory; but unfortunately this irred. rep. of the Poincare group is unphysical in terms of SU(3) b/c it's not a color singulet and belongs therefore to the unphysical sector of the Hilbert space. So strictly speaking the fundamental fields do not help so much and you would have to consider the physical states (protons, neutrons, ...) but of course they are not known analytically (only the reps are known)
     
  13. Aug 24, 2012 #12

    Haelfix

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    There are some mathematical subleties about angular momentum and in particular orbital angular momentum in relativistic field theory and one of the reasons this is often not treated in introductory texts (the other reason is that typically in most experiments, we do not know the spins of the incoming waves, and so bypass all of that by using Casimirs trick (average over initial spins, sum over final).

    The first observation is that it is total angular momentum that must be conserved by Lorentz invariance, and not necessarily spin or orbital angular momentum seperatedly. This is true even in classical field theory.

    One of the problems is that, in relativistic quantum mechanics, you usually cannot seperate the spin and orbital angular momentum parts of the conserved Noether current (arising from rotational invariance) into seperate terms that are conserved. However, if you know that you prepare an incoming particle in say a plane wave state, even though it will in general contain a mixture of all orbital angular momentum eigenvalues, the projection along the direction of linear momentum is zero. Therefore if you are given the polarization states of the incoming particles, you can sometimes forbid a particular decay from happening by simply invoking angular momentum conservation.

    However you aren't usually given this information in a typical problem, so quantum field theorists build their theories with all the covariance built into all the objects from the getgo, so for instance the generalized Ward identities (which replace Noethers laws) are fully covariant by construction and so interactions automatically respect all momentum conservations at each vertex.
     
  14. Aug 24, 2012 #13

    Bill_K

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    A Dirac spinor for one. ψ ~ (½, 0) ⊕ (0, ½)
     
  15. Aug 24, 2012 #14

    tom.stoer

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    OK, you are right, trivial but correct example ;-)
     
  16. Aug 24, 2012 #15

    bcrowell

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    Are the rules A and B that I gave in #3 correct, and if so, is there some simple way to see why they arise?
     
  17. Aug 24, 2012 #16

    Bill_K

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    Not as good an example, but also formalisms for higher spin objects often use reducible representations. The Rarita-Schwinger formalism for spin 3/2 uses a field ψμ which is a hybrid thing - a Dirac spinor with a 4-vector index attached. A constraint is then used to project out the undesired part.
    [Question: Spin 3/2 particles are usually said not to exist. Is the gravitino a legitimate example of a spin 3/2 particle?]
    The Lorentz group is the direct product of two copies of the three-dimensional rotation group. And a representation of the Lorentz group is therefore characterized by specifying two half-integers, j1, j2. The spin content is obtained by reducing the direct product. For example a vector is (½,½) so its spin content is ½ ⊗ ½ = 1 ⊕ 0.
     
  18. Aug 24, 2012 #17

    bcrowell

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    Thanks. I think I partially get this. I looked at the examples at http://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group . So if you had a field that was (1,0), it wouldn't be a vector, its spin content would be simply spin 1, and it would only give rise to scattering, not static attraction or repulsion?

    How about gravity? It's a symmetric tensor, but I don't know what two half-integers that tells me to use. I know it can't be (2,0), since then it would have no spin-0 content and we couldn't get static attraction. So would it be (1,1), or (3/2,1/2)?
     
  19. Aug 24, 2012 #18

    strangerep

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    I think you didn't read my post carefully. I used words like "type", "elementary", "corresponds to", but you seem to have changed these into something else in your mind. And of course I meant "quantum field", since this thread is in the context of Feynman diagrams.

    But never mind.
     
  20. Aug 25, 2012 #19

    dextercioby

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    Linearized gravity (the Pauli-Fierz field) is described by a (1,1) representation of SL(2,C) which is shown to be containing spin 2 and spin 0 (the trace of [itex] h_{\mu\nu} [/itex]) when reduced with respect to SU(2).
     
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