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The Poincare algebra

  1. May 8, 2009 #1

    nrqed

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    I am facing a dilemma which leaves me quite puzzled. I hope someone can straighten this out.

    The short version is : when we use representations of the charges as differential operators to calculate their commutators, we always get -1 times the correct result. So, doe sthat mean that we cannot use the diferential operators to calculate the algebra??


    The long version is:

    The commutator of [itex] P_\mu [/itex] with a scalar field is

    [tex] [P_\mu, \phi] = -i \partial_\mu \phi [/tex]

    whereas the differential operator form is

    [tex] P_\mu = i \partial_\mu [/tex]

    This agrees with Baer and Tata (Weak Scale Supersymmetry) and with Aitchison (Intro to Supersymmetry), as well as with Greiner (Field Quantization). I have seen results with different signs in other books, though)

    The Lorentz generators have commutation relation with a scalar field equal to

    [tex] [M_{\mu \nu} , \phi ] = i (x_\nu \partial_\mu - x_\mu \partial_\mu) \phi [/tex]

    this agrees with Greiner but disagrees with Srednicki.

    Whereas the differential form representation is

    [tex] M_{\mu \nu} = i ( x_\mu \partial_\nu - x_\nu \partial_\mu) [/tex]



    If I compute the Lorentz algebra using the operators, I get commutators with the signs opposite to what I get when I use the differential operator form. The commutators based on the operators agree with my references. So, is there some reason preventing us to use the differential form to compute the algebra??
     
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  3. May 9, 2009 #2
    I will say right off the bat that my knowledge of this is shaky, but I don't think what you're doing is valid. You have to use operators and not differential operators for the left hand side. The right hand side contains differential operators, but not the left. So for the momentum and angular momentum on the left hand-side, you need the operators which are calculated classicaly via Noether's theorem and then 2nd-quantized.

    One reasoning that might prevent you from using the differential form to compute the algebra is how can you have a derivative in the commutator? When it is on the right, what does it act on? When it's on the left it can act on the operator, but on the right it acts on nothing.
     
  4. May 9, 2009 #3

    Haelfix

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    Just glancing at this without the benefit of a book, some of those equations look right, and some don't/appear upside down; there looks like there is a signature/conventions/operator ordering issue going on. I'd stick with the formalism of one book and ask the question again.

    In general factors of -1, 2pi, 4pi, and i are notoriously irratating to disentangle, b/c the error or confusion can arise in about 100 different ways and are sometimes even present within a textbook to begin with. I remember many hours spent trying to get things to match during homework sessions, time that ultimately didn't really help much b/c i've promptly forgotten what the issue was.
     
  5. May 9, 2009 #4

    nrqed

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    Thanks for your reply.

    What I do is to simply assume there is a test function on the right, the same way we do it in quantum mechanics when computing, say, the commutator of X and P. Actually, I just noticed that Baer and Tata write

    it is straightforward to work out the commutation relations for the generators using their representation as differential operators. On finds..."

    and their result is consistent with the use of

    [tex] P_\mu = i \partial_\mu, ~~~~~~M_{\rho \sigma} = i (x_\rho \partial_\sigma - x_\sigma \partial_\rho ) [/tex]

    So it seems as if there is no problem using differential operators. My problem is that I always get the opposite sign with the operator approach.
     
  6. May 9, 2009 #5

    nrqed

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    I wish I could do that. Unfortunately, I haven't found a single book which provides explicit expressions for all the following:

    1) The commutator of [itex] M_{\mu \nu} [/itex] with a scalar field (or any other field)

    2) The commutator of [itex] P_\lambda [/itex] with a scalar field

    3) The differential operator representations of both operators

    4) The transformation of a scalar field used, in terms of the unitary operator U (is it [itex] U^\dagger \phi U [/itex] or [itex] U \phi U^\dagger [/itex] ? )

    5) the definition of U used (is it [itex] exp(-i a^\mu P_\mu + i \omega^{\mu \nu} M_{\mu \nu}/2)[/itex] ? )

    6) the coordinate transformations used. It is [itex] x^\mu + a^\mu + \omega^\mu_\nu x^\nu [/itex] ?


    Books always provide only partial information!
     
    Last edited: May 9, 2009
  7. May 9, 2009 #6

    nrqed

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    Actually, my problem is deeper. I have done the calculation with the signs of all the terms involving [itex] M_{\mu \nu} [/itex] arbitrary and I find that I cannot find any choice of signs that make the calculation using operators agree with the calculation using differential operators! This is what puzzles me. I will show the details a bit later, after I will have checked everything a third time.
     
  8. May 9, 2009 #7
    Differential operators operate on arguments, while operators operate on operators. So if you have an unprimed observer with a function f, then a primed observer will observe a function f':

    x'=x+a ; f'(x')=f(x'-a)

    So the 'a' has a minus sign in the argument of the function which is opposite the coordinate transformation.

    So maybe this is it?

    Even the Lorentz algebra can vary from sign to sign, depending on if you want to define e^(iJ theta) or e^(-iJ theta) as your group elements (which amounts to a definition of the positive direction for theta).

    U^(-1) operator U, makes better physical sense. Kaku's book is the only one I've seen that does U operator U^(-1), which has the nice feature that it changes the argument of an operator f' mentioned earlier to f'(x'+a), matching the coordinate transformation, which makes the equations look nicer.
     
  9. May 9, 2009 #8

    Hans de Vries

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    [tex]
    P_\mu = i \partial_\mu, ~~~~~~M_{\rho \sigma} = i (x_\rho \partial_\sigma - x_\sigma \partial_\rho )
    [/tex]

    These generators should actually be real valued from a mathematical point of view.
    They are somewhat artificially multiplied with -i in order to get the i from quantum
    mechanics in the commutation rules.

    The -i compensates the i at the left hand side in the operator. The arguments in the
    exponents of the operators are real valued.

    [tex]\tilde{H} ~~=~~ e^{iP^{\mu}t} ~~~~~~.[/tex] translate
    [tex]\tilde{H} ~~=~~ e^{iM^{\mu\nu}t} ~~~~~.[/tex] rotate, boost

    The differential operators have a general significance beyond quantum mechanics,
    you can use them to rotate or translate a picture for instance. see the attachment.

    The left image is a 1d function which undergoes an infinitesimal translation while
    the other two are 2d images which are rotated and boosted respectively.


    Regards, Hans
     

    Attached Files:

    Last edited: May 9, 2009
  10. May 9, 2009 #9

    nrqed

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    Thanks Hans.



    This defines your sign convention for the generators as differential operators.

    Can you define your sign convention for the transformation of a scalar field with the generators being quantum field operators? More specifically,

    a) do you use [itex] \phi' = U^\dagger \phi U [/itex] or the other convention?

    b) what is your U in terms of the generators P and M ?

    c) What do you take to be the variation induced in a scalar field ? which sign in [itex] \phi' = \phi(x^\mu \pm \omega^\mu_\nu x^\nu) [/itex] ?


    I have no problem making the differential operator form work, or the quantum field operator form work, I just have problem reconciling the two.

    Thank you again,

    Pat
     
    Last edited: May 9, 2009
  11. May 9, 2009 #10

    nrqed

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    Thanks for the suggestion.


    I am taking this into account. With the translation generator, everything works ok, It is just when I work with the Lorentz generators that I run into a problem.



    I am following Baer and Tata's convention, which is

    [tex] \phi' = \phi(x+a) = e^{i a^\mu P_\mu} \phi e^{-i a^\mu P_\mu} [/tex]


    This leads to [itex] [P_\mu, \phi] = -i \pa_\mu \phi [/itex] and I am ok with this.

    Unfortunately, they are not as explicit for [tex] M_{\mu \nu} [/tex].

    My problem is: what sign to take in


    [tex] \phi' = e^{\pm i \omega^{\mu \nu} M_{\mu \nu}/2 } \phi e^{\mp i \omega^{\mu \nu} M_{\mu \nu}/2 } [/tex] ?

    And what sign to take in

    [tex] \phi' = \phi( x^\mu \pm \omega^\mu_\nu x^\nu) [/tex] ?


    I know that there are some arbitrariness, but I am having a hard time finding a combination that will yield the same commutation relation as with the operators as differential operators.
     
  12. May 9, 2009 #11

    nrqed

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    I finally got it to work. Thanks for all your input.

    I redid the calculation leaving the sign in front of [itex] M_{\mu \nu} [/itex] arbitrary in its definition as a quantum field operator, leaving the sign of the differential operator representation also arbitrary, and leaving in addition, the sign of the transformation of the coordinate in
    [itex] x^\mu \pm \omega^\mu_\nu x^\nu [/itex].


    It turns out that these three signs are related, once one is chosen, the other two are fixed. My mistake had been to fix the sign of one of the operator and to play with the sign of the differential form representation only while leaving the transformation of the coordinate as [itex] x^\mu + \omega^\mu_\nu x^\nu [/itex].
    Unfortunately, I happened to have picked the wrong relative sign between the coordinate transfo and the operator ! So no matter what I would do with the sign od the representation as a differential operator, there was an inconsistency.


    Thanks to all for you input!! It helped.
     
  13. May 10, 2009 #12

    nrqed

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    Hans, I am curious.

    what sign do you use in the transformation of the coordinates for the Lorentz part, i.e. what signs do you use here

    [tex] x' = x^\mu \pm a^\mu \pm \omega_\nu^\mu x^\nu [/tex]
    ?
     
  14. May 10, 2009 #13

    Hans de Vries

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    I'm not sure if one can unambiguously relate the two because the
    differential operators do not act on the coordinates but on the field
    itself. So it seems a question of interpretation.

    If the differential operators are considered to perform a Lorentz
    transform from one reference frame to another then the signs should
    be negative.

    If they are considered to rotate and boost a field in a fixed reference
    frame then the signs should be positive.


    Regards, Hans
     
  15. May 20, 2009 #14
    I just read the appendix on group theory in Zee's book on QFT, and it seems on the left hand side of your equations you have the traceless Hermitian tensor representation (called the adjoint representation by Zee) [tex]\phi^i_j [/tex], and on the right hand side you have the fundamental vector representation [tex]\phi^i[/tex]. It's eqn. (20) of Zee if you have the book.

    It's interesting though that this is for SU(N), whereas with the Lorentz algebra it's SO(N).
     
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