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QFT question about algebra of charges

  1. Feb 15, 2009 #1

    nrqed

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    If a Lagrangian has some symmetry leaving the action invariant, we can introduce the corrsponding currents and charges. Now, the transformation can also be implemented as a unitary transformation of the fields in the form (let's consider a scalar field)

    [tex] \phi \rightarrow \phi' = U \phi U^\dagger [/tex]

    Now a couple of stupid questions.

    What happens if we consider a correlation function of, say, [itex] \phi \phi^\dagger [/itex]? It seems like we won't have that [itex] \langle \phi' \phi^{\dagger '} \rangle = \langle \phi \phi^\dagger \rangle [/itex].

    Also, why do we have to impose that the commutator of two transformations give a transformation? what would go wrong if it didn't?
    To take a specific example, susy transformations leave a lagrangian invariant (modulo total derivatives) even before introducing auxiliary fields. And yet, we must introduce auxiliary fields to get the algebra to close. I am trying to understand why this is necessary.


    Thanks in advance
     
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  3. Feb 15, 2009 #2

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    The expectation value (ie, in the vacuum) of A and UAUt are the same if the vacuum is invariant under U, as is usually the case (since otherwise applying U would give you a new vacuum). Since:

    [tex] \phi' \phi^\dagger' = (U \phi U^\dagger) (U \phi U^\dagger)^\dagger [/tex]

    [tex]= (U \phi U^\dagger) (U \phi^\dagger U^\dagger) = (U \phi \phi^\dagger U^\dagger) [/tex]

    you see that [itex]<\phi \phi^\dagger> = <\phi' \phi^\dagger'>[/itex].

    As for your second question, the commutator of two transformations (ie, transformations of the field configuration space, not the Hilbert space) will always give you a transformation that leaves the action invariant. The same is true of the corresponding operators on the Hilbert space. However, in mapping between these two kinds of transformations, all that matters is the transformations of the on-shell fields. So while the commutator of, say, two supersymmetries might give you a translation on Hilbert space, this only requires that the corresponding thing is true for the on-shell fields, ie, up to terms proportional to the equations of motion.

    It is sometimes possible to add auxiliary fields, which vanish on shell, so that the transformation is what we would expect everywhere in field configuration space. This is useful for certain purposes, but not really necessary, and in fact there are some theories that can't be closed off shell with any finite number of auxiliary fields.
     
  4. Feb 15, 2009 #3

    nrqed

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    I am slapping myself hard! I should have thought about it for more than a nanosecond before asking. Thanks!
    Makes sense.

    Interesting. But I am still perplex a bit. Indeed if we consider only free fields, the equation of motion of the auxiliary field F just says that it must be zero. But once we introduce interactions, the eom is no longer F=0 the introduction of the auxiliary fields seems absolutely necessary in SUSY. For example, it is not possible to write down mass terms or interactions if F is not introduced. However, after the introduction of F, and after getting rid of it using its eom, we now have mass terms in addition to
    the interactions. So the introduction of the auxiliary field was absolutely key to generate interactions and masses.

    I guess I am trying to understand the reasoning here. On one hand there is the need to introduce interactions (and masses). On the other hand there is the issue of the algebra not closing off-shell. How are the two related (if at all)? Why is it important to have closure of the algebra off-shell? Is it just a happy byproduct that closing the algebra off-shell allows the introduction of interacions respecting supersymmetry (and masses, after solving for the auxiliary field) ?


    Thanks for your input!

    Patrick
     
  5. Feb 15, 2009 #4

    strangerep

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    Maybe I'm missing something deeper in your question, but... if an algebra doesn't
    close, then you don't have an algebra at all.

    The on-shell fields are just eigenstates of (usually several) Casimir operators in the
    enveloping algebra. (The equations of motion really do nothing more than express
    this constraint.) But to define the Casimirs in the first place, you need a bona-fide
    algebra.

    HTH (and sorry if I've misunderstood what's bugging you).
     
  6. Feb 15, 2009 #5

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    Auxilliary fields, are useful when it comes to writing supersymmetric actions, especially when we group them with the dynamical fields into superfields, since these superfields have simple transformation properties under supersymmetry. It's analagous to grouping fields of the same spin into objects with indices, which then only need to be contracted appropriately to get rotationally invariant quantities.

    Of course, if we're able to use superfields, the algebra automatically closes off-shell, basically by construction. But this is only a convenience - if we could have guessed the final lagrangian we get after integrating out the auxilliary fields, we could just check its invariance without ever having closed anything off-shell. And in many cases, especially in higher dimensions, this is the only way to proceed.
     
  7. Feb 15, 2009 #6

    nrqed

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    Thanks for the feedback.

    I know that superspace is the way to go if one wants to be efficient but I first want to uderstand the brute force approach. So you are saying that we could in fact built the Wess-Zumino Lagrangian (in the form obtained after the auxiliary field has been replaced by the solution of its eom) directly, simply by being clever and putting together interactions and masses that are invariant under susy (up to total derivatives).

    So, is there any reason to introduce an auxiliary field in the first place? The route usually followed in SUSY presentations is

    a) Notice that the SUSY algebra does not close off-shell

    b) Introduce an auxiliary field to take care of the problem

    c) now build a susy invariant theory that includes F

    d) Get rid of F

    I guess I am trying to understand the logic in this approach. Is there a gain (either conceptual or simply coputational) in going through this route instead of simply working directly with the initial fields to build a SUSY invariant action?

    Thanks
     
  8. Feb 15, 2009 #7

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    Let's say you try to build an action out of fields that only represent the susy algebra on-shell. Then there are two things you need to pick: the action and the susy transformations themselves. The latter are only expected to satisfy the algebra on-shell, so there will be eom terms in the commutators. But now as you adjust the action to try to make it invariant, this changes the eom's, and so also requires we change the susy transformations, which changes the variation of the action in some complicated way. It's all horribly tangled together. If the representation is off-shell, things are considerably simpler, and if we write things in terms of superfields, it's almost trivial.
     
  9. Feb 15, 2009 #8

    nrqed

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    Thanks for the reply. My question is not very clear but I don't understand the situation well enough to formulate a clear question in the first place. I am hoping that by discussing it and getting some feedback, it will become clear enough for me to either see the answer to my question or to be able to formulate a more clear question.

    I guess I don't see the connection between the algebra of charges (whether it closes off-shell or not or even whether there is an algebra in the first place) and the invariance of the action.

    Let's start with a theory of a massless Weyl spinor and a massless complex scalar field. We can write down transformations of the fields that leave the action invariant without ever using the equations of motion. However, the algebra does not close off-shell. I guess I am wondering why this matters at all! But references on SUSY make it sound as though there is a big problem and we need to modify the theory to make the algebra of charges close off-shell. I guess my reaction is "who cares since the transformations leave the action invariant!". But I know that I am missing something!
     
  10. Feb 15, 2009 #9

    nrqed

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    Ok, but this is the crux of the question. If we consider, to keep things extremely simple, a theory of a a free and massless weyl fermion and complex scalar field. W ecan write down susy transformations that leave the action invariant *without* using the eoms. The algebra of the charges does not close off-shell but why does that matter since the action is invariant anyway. Susy references make it sound as if there is a problem and we must make the algebra close off-shell...but the action is invariant anyway so I wonder why the algebra matters. I know that I am missing something obvious but I can't see what. I am not sure if the point they are making is a trick to make it easier to go to the next step: which is to add interactions, or there is something more fundamental.
     
  11. Feb 16, 2009 #10

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    Sure, in principle, if we could find any fermionic transformation that left the action invariant, we'd be set. But we http://en.wikipedia.org/wiki/Haag-Lopuszanski-Sohnius_theorem" [Broken]any fermionic symmetry must anticommute to a translation (again, up to eom's if we're not closing off-shell), so we might as well use this to guide us in finding such transformations. In particular, when we change the action we know we'll have to change the susy transformations as I described to have any chance of success, because if they don't anticommute to translation + eom, they have no chance of being symmetries.

    You've been hinting at some underlying connection between interactions and off-shell closure, but I'm not sure there is one, since, as I've mentioned, there are perfectly good interacting susy theories that can't be closed off-shell.
     
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  12. Feb 16, 2009 #11

    nrqed

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    Thanks again for your feedback, it is very much appreciated.

    So you are saying that if I simply build a Lagrangian that I want to be invariant under some transformation (and by this I mean up to total derivatives but never using the eoms), I never need to introduce auxiliary field.

    Let me tell you where I am coming from: I am looking at obtaining the Wess-Zumino Lagrangian starting simply from a Weyl spinor and a complex scalar field. If I start with only two kinetic terms, it is easy to write a SUSY transformation that leave the lagrangian invariant (in the sense described above). So I can easily make this (trivial) theory susy invariant. I am not working with charges at all, just plain and simple transformations of the fields.

    Now I want to add interactions. Again, I don't want to work with charges at all, just writing transformations of the fields and putting together interactions that will be invariant (again, up to total derivatives but not using the eoms).

    I have figured it out if the auxiliary field is first added. That is, I have worked out the SUSY transformations of the spinor, scalar field and auxiliary field and the interaction terms that will yield the WZ Lagrangian. I started with the auxiliary field included because every reference makes it sound as if it is a requirement to include it.

    So, I guess what you are saying is that I could simply work with the spinor and the scalar field and put together interactions that are susy invariant (again, up to total derivatives but not including terms proportional to the eoms). Do I understand correctly?

    Thanks for your help (and patience!)

    Patrick
     
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  13. Feb 16, 2009 #12

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    Right, I'm saying that's possible. And, yes, the variation must be zero everywhere, not just on-shell, in order for Noether's theorem to work.

    But I'm also saying that it's easier to work with an auxilliary field when you can. The reason is that, say you're using a scalar [itex]\phi[/itex], spinor [itex]\psi[/itex], and auxilliary scalar [itex]F[/itex]. Then you can assume without loss of generality that your transformations take the form (schematically):

    [tex] \delta \phi = \psi, \bar{\delta} \phi = 0 [/tex]

    [tex] \delta \psi = F, \bar{\delta} \psi = \partial \phi [/tex]

    [tex] \delta F = 0, \bar{\delta} F= \partial \psi [/tex]

    Essentially, the fields are defined so this is true, and so that [itex]\{\delta,\bar{\delta}\}=\partial[/itex], as required by HLS. Now all you need to do is find an action that's invariant under these transformations.

    If you decided instead to go straight for the action where F has been integrated out, you have to tune not only your action, but also your transformations. This is because your transformations are going to be what I wrote above, except with the on-shell value of F plugged in. And this on-shell value will change as you change the action. This is why you see people going back to the auxilliary field all the time.
     
  14. Feb 17, 2009 #13

    nrqed

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    Thank you. I think see what you are saying now.

    So if I go to the action with the F integrated out and write down the transformations of psi and phi (with F replaced by its "on-shell" value), the Lagrangian will be invariant (up to total derivatives but without the need to use the eoms) . So solving for F and applying a SUSY transformation to the Lagrangian are two operations which commute.

    There is just one part of the logic that I am still struggling with (and I am litterally losing sleep over this :yuck: )

    You said that it does not really matter if the algebra closes off-shell or not (it did surprise me a bit because I thought that in order to be implemented quantum mechanically, the algebra had to close on off-shell states).

    So, as you said, the introduction of the auxiliary field is made only in order to facilitate figuring out the susy transformations and the Lagrangian when interactions are included.

    Let me try to explain what still confuses me (the answer is probably very simple).

    We can write down the *free* Lagrangian for psi, phi and F, and we can choose the transformations of the fields so that the algebra closes of-shell and the action is invariant. Things are fine so far.

    Now we add interactions. If we add only worked with psi and chi (no F introduced), as you said the transformations would be messy because we have to worry about including terms proportional to the eoms, which themselves depend on the terms we are adding...so a big mess. Ok, that makes sense.

    Consider instead the Free Lagrangian with the auxiliary field introduced. The algebra closes off-shell. Now we add interactions. My main question is : why is it now that we don't have to worry about the transformations of the fields getting more complicated due to the introduction of interactions? Why is it that the fact that the algebra was closing off-shell when there were no interactions guarantees that once the interactions are added, we don't have to worry about modifying the transformation laws of the fields?

    This is what still confuses me. There seems to be something special about the fact that the algebra was closing off-shell when the theory was free, and the addition of interactions. This is the key point that I would love to understand (which would allow me to get some sleep)

    Thanks again !

    Patrick
     
  15. Feb 17, 2009 #14

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    By the way, I should apologize if my explanations aren't clear. I wouldn't consider myself an expert, although I have been working with supersymmetry for a while now, and some of this stuff has been floating around my head but I haven't tried to put it into words until now. So thanks for making me think about it more carefully.

    Remember that the equations of motion hold as operator equations in the quantum theory, so those extra terms shouldn't contribute.

    Like I said, the fields are defined so that they satisfy those simple transformation laws. See chapter 26 of Weinberg, where he goes through this procedure. Basically, set [itex]\psi=\delta \phi [/itex] and [itex]F=\delta \psi[/itex]. That this process ends here is a non-trivial property of the supersymmetry algebra in 4D, and doesn't hold in higher dimensions, where we get an infinite series of auxilliary fields. The remaining transformations are determined by the fact that the supersymmetries anticommute to a translation.

    Why are we allowed to assume these algebraic properties hold before we even know what our transformations are? Well, this is the content of the theorem of Haag et al I mentioned. Basically, we know the anticommutator has to be some bosonic symmetry of the theory, and these are restricted by the Coleman-Mandula theorem to be translations, rotations, and internal symmetries. Since the supersymmetries have spin 1/2, their anticommutator has to have spin 1, and momentum is the only possibility (this is clearly only a very rough sketch of the argument). This argument works in an arbitrary interacting theory, and so we can always assume the transformations take the simple form I wrote down, at least when written in terms of off-shell fields.
     
  16. Feb 17, 2009 #15

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    Sorry, I still don't think I've answered your question. In fact, what I said isn't entirely true. The Coleman-Mandula and HLS theorems are actually about the quantum symmetries, so we should only expect the algebras they identify to be satisfied by the field transformations up to eom terms, which is exactly what you were worried about. I don't really have a good answer for this, except to say that if the theory we're looking for has an off-shell formulation (as WZ does, but as some theories do not), then this method, with fixed transformations and varying action, will work and will find this formulation, and will be the best way to find the action for the reasons I've explained. I'll keep thinking about it, but that might be the best you can say.
     
  17. Feb 17, 2009 #16

    nrqed

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    No apologies necessary. You don't know how much I appreciate your feedback!!
    I would have to travel over 100 miles to find someone to talk about SUSY with, and probably much farther to find anyone who could offer to me the level of hindsight you have. I am really grateful. I was not sure if my question was trial or if it involved something subtle. Either way, I wanted to find out. You have given me a new (and deeper) perspective of the situation and I appreciate that very much. I will keep thinking about it too.

    It is just a bit annoying that authors (at least the ones I have read) introduce the auxilary field to close the algebra off-shell without explaining the reasoning behind this and why it simplifies finding a susy invariant action that included interactions. Unfortunately I don't have Weinberg's book on SUSY, maybe he does discuss this.


    I am the one who needs to thank *you* for taking the time to discuss all this! You obviously have a much deeper understanding than I have so I am the one reaping the benefits of the exchange! But if you find this useful a bit, then I am glad.




    Ok, then I guess the crux of the matter is why it is sufficient for the transformations to have this simple form if written in terms of off-shell fields and not if written of on-shell fields.

    Thanks for all your input!

    Regards,

    Patrick
     
  18. Feb 21, 2009 #17

    nrqed

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    I have found an old Physics Reports by Sohnius. In one section, he makes some comments about the on-shell vs off-shell argument.

    Here are some excerpts:

    Refering to the fact that algebra of the charges does not close off-shell if no auxiliary field is introduced, he says

    But then, in the next sentence he says


    As for the question why the introduction of auxiliary fields allow a simplification of the SUSY transformations, he does not explain it but mentions briefly a few things, in passing. But there is no clear explanation.

    So it is a more subtle issue than I thought it was, initially.
     
  19. Feb 21, 2009 #18

    Haelfix

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    This is a very subtle technical question that isn't entirely understood afaik, or at least, I don't understand it well. Consider the case in dimensions >6 where no set of auxillary fields are known whereby they can completely close the full algebra offshell.

    The commutator of two supersymmetries will then in general contain not just a gauge transformation + a spacetime translation, but extra junk that will only vanish if you apply the equations of motion.

    The problem then is evident. The equations of motion will in principle recieve quantum corrections, and you will have to find a self consistent way to quantize the mess much less renormalize it. And in fact its usually the case that you can't renormalize it by dimensional arguments.

    A lot of this is related to the gauge fixing condition. Depending on the choice you can generates something with either lack of off shell closure (breaking the supersymmetry) or sometimes it preserves the algebra but then is nonlocal. C'est la vie!

    But why are auxillary fields useful? Well I think Status answered correctly but to summarize they

    1) They preserve *manifest* supersymmetry so you can quantize the theory in a pleasant way

    2) The help you construct the action. Moreover, without the auxillary fields, the susy symmetry transformations are realized nonlinearly, ward identities become difficult to use, and so on and so forth.
     
  20. Feb 22, 2009 #19

    nrqed

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    Hi Haelfix, thanks for your feedback.

    I guess I am trying to understand the last point you make in the simplest context possible: the Wess-Zumino model.
    I can start with the free model (so just a Weyl spinor and a complex scalar field, no F yet). I can write down linear transformations that leave the action invariant. Without working out the commutator of two transformations on the spinor field, there is nothing to "warn" us of any potential problem. The transformations are linear even though the algebra does not close off-shell.

    Now we add interactions. As you said, the fact that that we introduce auxiliary fields allow us to keep the SUSY transformations linear. This is the point I am trying to understand: why is it that having the algebra close off-shell allows the transformations to remain linear? Why is it that when the algebra closes off-shell, modifying the Lagrangian does not introduce any change in the susy transformations of the fields? I would love to understand this. I thought that there was maybe a simple and obvious proof, but I would also be happy to see a complicated derivation:smile:
     
  21. Feb 24, 2009 #20

    samalkhaiat

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    Your on-shell transformatioms are linear because you were using trivial model. if you add the usual potential [itex]W(\phi)[/itex], the on-shell action will be invariant under the non-linear (on-shell) transformations

    [tex]T(\eta)\phi = \eta \psi[/tex]
    [tex]T(\eta)\psi = i\sigma^{\mu}\eta^{\dagger}\partial_{\mu}\phi - \eta \frac{\partial W}{\partial \phi^{*}}[/tex]

    on the on-shell representations [itex](\phi ,\psi )[/itex]. The algebra of these transformations closes only modulo the dynamical equation of the spinor field. If you introduce proper number of auxiliary fields transforming as a multiple of the dynamical equation, you end up with an off-shell theory invariant under the linear transformations

    [tex]T(\eta)\phi = \eta \psi[/tex]
    [tex]T(\eta)\psi = i\sigma^\mu}\eta^{\dagger}\partial_{\mu}\phi +\eta F[/tex]
    [tex]T(\eta)F = \eta^{\dagger}E(\psi)[/tex]

    on the off-shell multiplet [itex](\phi ,\psi , F)[/itex]. the algebra of these linear transformations closes without any use of any dynamical equation. All off-shell models(including the free W-Z that you used) are invariant under these transformations.
    Introducing the F does modify the SUSY transformations for the spinor field.

    regards

    sam
     
    Last edited: Feb 24, 2009
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