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Aspects of BRST Quantization

  1. Aug 3, 2006 #1

    selfAdjoint

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    In J.W. van Holten's essay Aspects of BRST Quantization ,hep-th/0201124, he begins (Section 1.1) with the free relativistic particle. In order to derive the dynamics from the action principle he introduces an auxiliary variable:

    a bit later he notes
    I thought I was really familiar with the relativistic free particle, but I've never encountered this einbein before. I can't make out from the rest of his discussion just why it's there. It's a constraint on the action to be sure, and that is very apposite to his future development of BRST symmetry. But why have it at all?

    Anybody know?
     
    Last edited: Aug 3, 2006
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  3. Aug 3, 2006 #2

    pervect

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    I never heard of them before, but Google find:

    http://arxiv.org/PS_cache/hep-ph/pdf/9708/9708319.pdf

    [add]
    This confused me as written, so I wound up re-writing it as

    [tex]L = -m \sqrt{1 - \dot{x}^2}[/tex]

    and

    [tex]L = -\mu (1 - \dot{x}^2) / 2 - m^2 / 2 \mu[/tex]
     
    Last edited: Aug 4, 2006
  4. Aug 4, 2006 #3

    selfAdjoint

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    Thanks pervect, I should have thought of googling on einbein.

    So it's just a convenience! I wondered if it was some new necessity (perhaps to insure reparametrization invariance) that I had somehow missed.
     
  5. Aug 4, 2006 #4

    pervect

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    I'm glad you mentioned it, actually - it looks interesting, and I probably wouldn't have stumbled across it otherwise.

    Apparently that's all it is, a convenience. It looks like e = m/[itex]\mu[/itex] and the rest of the differences can be ascribed to geometric units (i.e c^2=1).

    Anyone else heard of this before?
     
  6. Aug 4, 2006 #5

    pervect

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    I just realized that not long ago, I saw very similar formula.

    If you look at extremizing

    [tex]\int \sqrt(I) dt[/tex]

    you get the Euler-Lagrange equations

    [tex]\frac{d}{dt}\left(\frac{\partial \sqrt{I}}{\partial \dot{x}}\right)= \frac{d}{d t} \left[ \frac{1}{2 \sqrt{I}} \left( \frac{\partial I}{\partial \dot{x}} \right) \right] = \frac{1}{2 \sqrt{I}} \frac{\partial I}{\partial x}[/tex]

    (see for example the thread: )

    https://www.physicsforums.com/showpost.php?p=1048523&postcount=23

    Now, if I were constant, you could simplify this significantly: i.e you could multiply under the derivative by the constant [itex]2 \sqrt{I}[/itex] and get:

    [tex]\frac{d}{d t} \left[ \left( \frac{\partial I}{\partial \dot{x}} \right) \right] = \frac{\partial I}{\partial x}[/tex]


    These are the same equations one gets when extremizing I.

    By parameterizing all the variables in terms of proper time, one can essentially make I constant. This is why one can derive the geodesic equations by minizming I rather than sqrt(I), where for force-free geodesic motion

    [tex]I = g_{uv} \dot{x^u} \dot{x^v}[/tex]

    Those flyspecs over the x^i above are "dots" :-)

    Basically, the einbein re-creates this happy state of affairs, without formally introducing proper time. (But it does introduce another free parameter, so the difference isn't that large).

    If you think of 'e' in your expression as

    [tex] \left( \frac{d \lambda}{d \tau} \right)^2 [/tex]

    the relationship becomes a little more clear, I think.
     
    Last edited: Aug 4, 2006
  7. Aug 22, 2006 #6
    This question has already been answered by others, but here's my tuppence. This is pretty much the same technique used in introductory string theory courses to motivate the existence of the Brink-Di Vecchia-Howe action for a bosonic string. The most obvious action for such a string is the Nambu-Goto action, which is proportional to the area of the world-sheet. However, because of this proportionality, you essentially have a square-root of a metric determinant in the action, and this is difficult to quantise. Instead of using the Nambu-Goto action, an easier way is to use the BDH action, where you introduce an independent world-sheet metric. This whole procedure is usually motivated by first showing that the standard action for a free particle (which has a square-root because the action is proportional to the length of the particle's world line) can be put in an easier-to-handle form with the introduction of an einbein field [tex]e(\lambda)[/tex].

    There are good (if brief) discussions of this in Chapter 2 of Green, Schwarz, and Witten, as well as at the start of D'Eath's book on SUSY Quantum Cosmology. I seem to recall a discussion in Henneaux & Teitelboim's book too.
     
  8. Aug 22, 2006 #7

    selfAdjoint

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    The thing about van Holten's text (which I am still reading) is that he just sort of assumes the reader is hip to all this. That's justifiable, I suppose, but this is recommended as the best INTRODUCTION to BRST theory. I still don't think there's anything too daunting about BRST itself, but it's always encountered buried under a load of (for this purpose) extaneous physics.

    I wish one of the n-category wizards would post on how BRST symmetry comes up with just a bundle (or n-bundle if necessary) and how the Grassmann variables come in (other than their use in integrating fermionic integrals, which is an example of what I mean by extraneous physics). If BRST is a mapping on cohomology, then is should be possible to define it by itself and only then apply it to the quiddities of field theory!
     
  9. Aug 22, 2006 #8
    For what it's worth, my own opinion is that if one wishes to seriously study BRST symmetry and quantisation, then a necessary prerequisite is a solid grounding in singular Lagrangians and their relation to constrained Hamiltonian systems. Dirac opened up a huge can of worms when he wrote his little book "Lectures on quantum mechanics," and it can be argued that many of the objections people have against BRST and rigourous analysis of gauge theories in general are due to the fact that we still don't really understand what's going on with the geometry of constrained gauge theories.

    (As an aside, I guess the best example is the Hamiltonian constraint in GR. What the hell is it? Why is it so important? What are we missing about the relationship between the Hamiltonian constraint and the well-known "problem of time" in the canonical quantisation of GR? A lot of people, including, sadly, most of the LQG crowd, will tell you that they understand it but the truth of the matter is that nobody is even close.)

    Anyhow, it might be worth trying to track down Henneaux & Teitelboim's "Quantization of Gauge Systems." It's probably the best introduction to BRST methods that I know of.
     
  10. Aug 22, 2006 #9

    Physics Monkey

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    Hi selfAdjoint,

    As people have already said, the einbein is a way to make your action simple (as well as to encode the constraint). When doing the path integral, an action with a square root is a real nightmare! Anyways, not to rehash what others have already mentioned, I just wanted to point out another virtue of the einbein: it enables you to treat massless particles. Also, like the vierbein is the "square root" of the metric in four dimensional spacetime, the einbein is the square root of the metric on the particle's world line. Just some interesting tidbits!
     
    Last edited: Aug 22, 2006
  11. Aug 23, 2006 #10

    selfAdjoint

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    Thanks for those insights, Monkey. It's "tidbits" like that that I need to build the physics into my brain box. Just bare definitions and "do the excercises" never works for me.
     
  12. Apr 30, 2010 #11
    Re: Einbein

    There is a easy way to see what is the einbein for a free particle. For example define the following Lagrangian:

    [tex]L =-( \mu \dot{x}^2 / 2 + m^2 / 2 \mu)[/tex]

    Where [tex]\mu[/tex] is an arbitrary dynamical variable. If you want to know how the einbein transform under an infinitesimal re-parametrization, you need to impose the condition that the action is invariant under this transformation. Than mean that the change in the Lagrangian should be equal to a total derivative. Then you will find the rule to transform the einbein and after that you can use the method of variation of parameter to find the einbein explicitly. In the case of a free particle with mass you will find that the einbein is the total energy of the particle in some reference frame. Also in the case of a massless particle you can't use the lagrangian:

    [tex]L= -m \sqrt{ \dot{x}^2}[/tex]

    But you can use the Lagrangian presented before and you will obtain the equation of motion:

    [tex]\dot{x}^2=0[/tex]

    As expected.

    Note that the Lagrangian defined at the beginning is not invariant in general with respect to re-parametrization, the reason is that a difference of the other Lagrangian this one is not homogeneous of first order with respect the generalized four velocity. The einbein is an auxiliary variable that make possible to have a Lagrangian which is not homogeneous of first order in the velocity, but is invariant with respect to re-parametrization. Then is easy to show that this Lagrangian produce the equation of motion of a free particle, then both Lagrangian are equivalent. And that's the story.
     
    Last edited: May 1, 2010
  13. Dec 7, 2010 #12

    haushofer

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    Re: Einbein

    I'm sorry to kick an old topic, but I have a question about this: how would you incorporate a potential V(x) with this analysis? The natural thing to do is to couple this potential also to the einbein, and to subtract the term eV(x) from the free part, but I can't see how that works. Or is there some text in which people do this?
     
  14. Dec 7, 2010 #13
    Re: Einbein

    Well there is not need to couple the einbein to the potential. The einbein is used for the T term of the Lagrangian or that's my experience. You can find a discussion of the embein in the book Introduction to classical theory of field and particles by kosaykov. In the section re parametrization you can find the Lagrangian of a particle in a external EM Field and the potential term has not einbein. Now dealing with fields there is always parameters that plays role similar to the einbein, but is not the einbein.
     
  15. Dec 7, 2010 #14

    haushofer

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    Re: Einbein

    Indeed, for the EM-coupling I see that things work out OK; the interaction term you add is simply (putting the coupling to 1)

    [itex]
    \int A_{\mu }\frac{dx^{\mu}}{d\tau} d\tau = \int A_{\mu}dx^{\mu} = \int A
    [/itex]

    which is manifestly tau-reparametrization invariant. But then the analogy with "wordline coupling to 1D gravity" doesn't hold anymore, I would say. Also, one could wonder what one can say about general potentials; the solution would then be that indeed one just subtracts a potential V of the lagrangian and demands that the potential term is reparam.invariant without einbeins.
     
  16. Mar 14, 2012 #15
    Re: Einbein

    dear All! in order to understand it,one should go for tetrad formalism which is of very much use in understanding the spinning particles dynamics. for this one needs to know what these tetrads are and how these are related to the usual spacetime manifolds? these einbein fields appear as transformation matrices between the tetrad space and spacetime manifold, of which this tetrad space is defined. then one will be able to know the real role of einbein and its transformation under reparametrizaton. this type of action which is mentioned earlier in conversations ,is known as Polyacov action and comes with step by step calculations.
    for the better details i attach one good paper to have proper understanding of this.
    if U have any further query , please dont hesitate to ask me.
    sincerely
    dheeraj shukla
    BHU, India
     

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