Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on the representation of Poincare algebra generators on fields

  1. Aug 20, 2011 #1

    I am working through Maggiore's QFT book and have a small problem that is really bothering me.

    It involves finding the representation of the Poincare algebra generators on fields. I always end up with a minus sign for my representation of a translation on fields compared to Maggiore (and everyone else). Here's what I do that gets the wrong answer:

    For the Poincare transformation [itex] x^\mu \rightarrow x'^\mu = x^\mu + a^\mu [/itex], Maggiore says that any field transforms as [itex] \phi'(x') = \phi(x) [/itex] (Eq. 2.106).

    The representation of a Poincare group element is [itex] \exp(-i a_\mu P^\mu) [/itex] (Eq. 2.96). I assume this is an operator that acts on a field and gives you the transformed (primed) field.

    For the infinitesimal translation by [itex] a^\mu [/itex] and at fixed [itex]x[/itex] (Eq. 2.107):

    [tex] \delta_0\phi \equiv \phi'(x) - \phi(x) = \phi(x-a) -\phi(x) = (-a^\mu \partial_\mu \phi) (x) ,[/tex]

    and, by definition of what a generator is, this must also be equal to

    [tex] \delta_0\phi = \left(\exp(-i a_\mu P^\mu) \phi - \phi \right)(x) = (-i a_\mu P^\mu \phi)(x) .[/tex]

    By equating these two expressions for [itex] \delta_0\phi [/itex] I find that the representation of the generator on fields is

    [tex] P^\mu = -i \partial^\mu .[/tex]

    But this disagrees with the usual result (Eq. 2.110) by a minus sign.

    Can anyone tell me what I'm doing wrong? My definition still obeys all the required commutation relations but it gives the wrong answer for the momentum and energy operators (i.e. an extra minus sign).

    Any thoughts are much appreciated.

  2. jcsd
  3. Aug 20, 2011 #2
    This is flawed. Eq. 2.108 and 2.109 give the correct derivation.
  4. Aug 20, 2011 #3
    Hi Polyrythmic. Can you explain why what I did is flawed?

    I agree that the line you quoted is where my derivation differs from Maggiore's by a minus sign, which leads to the difference in my result for [itex] P^\mu [/itex]. I do not understand Maggiore's Eq. 2.108, which I will copy here:

    [tex] \phi'(x'-a) = \exp(-i(-a_\mu)P^\mu) \phi'(x') = \exp(i a_\mu P^\mu) \phi(x) .[/tex]

    In that equation it looks like the operator [itex]P^\mu[/itex] is operating on the 4-vector [itex] x'[/itex] and not on the field. But it is my understanding that we are trying to find the representation of the Poincare group on the field. I.e. the operator [itex]P^\mu[/itex] should act on the field, transforming one function [itex] \phi [/itex] to another.

    If you could provide a more complete explanation that would be great.

  5. Aug 20, 2011 #4
    Well, let's begin by considering what we want. We want to find an expression for the momentum operator by considering infinitesimal translations for a scalar field that obeys


    The infinitesimal translation is given by

    [tex]x^\mu\rightarrow x'^\mu=x^\mu+\epsilon^\mu.[/tex]

    In order to do that, we begin by defining the infinitesimal change of the fields, as given by Eq. 2.107:


    Next, we want to find an expression for [tex]ϕ'(x'−\epsilon)[/tex] that uses the momentum operator, so that we can relate it to the partial derivative that appears in Eq. 2.107. We can do this by recognizing that the generator of translations is given by the exponential of the momentum operator. This generator acts on the field itself, as given by Eq. 2.108:


    The additional minus sign comes from the fact that we now have a translation into negative epsilon direction, as given in the leftmost part of the last equation.
    We can now expand the exponential into a taylor series and cut it after the first order (since we're dealing with infinitesimal translations). Inserting into 2.107 immediately leads to the result.
  6. Aug 21, 2011 #5
    I am still confused because it looks like the operator [itex] P^\mu [/itex] is transforming the coordinate, not the function, in Polyrhythmic's last equation. Forgive me for making things painfully explicit but perhaps someone can spot the exact source of my confusion if I write everything out as clearly as I can.

    We have a transformation [itex]\exp(-i a_\mu P^\mu) [/itex] that takes a function [itex] \phi [/itex] and turns it into a function [itex] \phi' [/itex], which is related to [itex] \phi [/itex] by
    [tex] \phi'(x+a) = \phi(x), [/tex]
    for any value of [itex]x[/itex]. Physically, the transformation shifts the shape of the function in the direction of [itex]a[/itex]. I.e. if [itex]\phi[/itex] had a spike at [itex]x=0[/itex], [itex]\phi'[/itex] has a spike at [itex]x=a[/itex].

    I also write this relation as
    [tex] \phi' = \exp(-i a_\mu P^\mu) \phi, [/tex]
    i.e. the object [itex][\exp(-i a_\mu P^\mu) \phi][/itex] is a function.
    Plugging in any value of [itex]x[/itex],
    [tex] \phi'(x+a) = [\exp(-i a_\mu P^\mu) \phi](x+a) = \phi(x). [/tex]
    This last equality is a self-contained definition for [itex]P^\mu[/itex]: if we expand [itex][\exp(-i a_\mu P^\mu) \phi](x+a)[/itex] to first order in [itex]a[/itex] we get
    \phi(x) &=&[(1 -i a_\mu P^\mu) \phi](x+a) \\
    &=& [\phi - i a_\mu P^\mu \phi](x+a) \\
    &=& \phi(x+a) -(i a_\mu P^\mu \phi)(x+a) \\
    &=& \phi(x) +a^\mu \partial_\mu\phi(x) -(i a_\mu P^\mu \phi)(x).

    Matching terms proportional to [itex]a^\mu[/itex] gives
    P^\mu = -i \partial^\mu,
    the wrong answer...

    I also tried reproducing Polyrhythmic's last equation: I think we would agree that
    [tex]\phi'(x'-\epsilon) = \phi'(x) = \phi(x-\epsilon) .[/tex]
    If this is to equal [itex] \exp(-i (-\epsilon_\mu) P^\mu) \phi(x)[/itex], then it seems that the operator [itex]P^\mu[/itex] needs to act on the argument [itex]x[/itex]. I.e. it appears you are defining [itex]P^\mu[/itex] to do the following:
    [tex] [\exp(-i (-\epsilon_\mu) P^\mu) \phi](x) = \phi(x-\epsilon).[/tex]
    This equation should hold for any value of [itex]x[/itex] and [itex]\epsilon[/itex]. So try plugging in [itex]x \rightarrow x+a[/itex] and [itex] \epsilon \rightarrow -a[/itex]:
    [tex] [\exp(-i a_\mu P^\mu) \phi](x+a) = \phi(x+a+a).[/tex]
    But this is different from the third equation in this post.

    What is going wrong? Hopefully, I've thrown enough of my thoughts out there for someone to be able to identify a misconception I have. I'm betting it has to do with definitions. Let me know if you have any thoughts on the matter.
  7. Aug 21, 2011 #6
    But [itex]x \rightarrow x+a[/itex] is not compatible with [itex] \epsilon \rightarrow -a[/itex]. For [itex]x \rightarrow x+a[/itex], you have to set [itex] \epsilon \rightarrow +a[/itex], then everything is consistent.
  8. Aug 21, 2011 #7
    Not sure what you mean by "compatible" and "consistent". If you are saying that the equality with the substitutions you mention is consistent with the third equation of my post I'm pretty sure that's not true. The reason I suggested the substitutions I did was so that the left hand sides would both be [itex][\exp(-i a_\mu P^\mu)\phi](x+a)[/itex]. But then the right hand sides are different ([itex]\phi(x)[/itex] vs. [itex]\phi(x+2a)[/itex]).

    But forget about that for a moment. Just take a look at my last post up to the point I "find" [itex]P^\mu = -i \partial^\mu[/itex]. Can anyone spot the problem with that derivation?
  9. Aug 21, 2011 #8
    Sorry, I got something wrong, disregard my last post. However:

    I don't see how this contradicts anything I've written so far, it's correct. The operator on the left hand side of this equation shifts the field, which is originally at x+a, to its value at x+2a. This is correct.
  10. Aug 21, 2011 #9
    This equation is false. The middle part is neither equal to the right nor the left. A negative exponent shifts the field in the positive direction.
  11. Aug 21, 2011 #10


    User Avatar
    Science Advisor

    Everything is correct. The algebra of the translation group does not allows you to distinguish between [itex]P^{\mu}[/itex] and [itex]-P^{\mu}[/itex], so the sign of [itex](a.P)[/itex] in the unitary operator
    [tex]U(a)=\exp (ia_{\mu}P^{\mu}),[/tex]
    is a matter of convention.

  12. Aug 21, 2011 #11
    But once you have chosen a certain convention you have to stick with it, this is where the mistake in the OP's calculation lies, why he arrives at the "wrong" result.
  13. Aug 21, 2011 #12


    User Avatar
    Science Advisor

    Yes but, there is no "wrong" result. What he needs to know is “the action on fields is inverse to that on points” :
    [tex]\left( T . \phi \right) (x) = \phi ( T^{-1}. x) \ \ (1)[/tex]
    The meaningful quantity here is the value [itex]\phi (x)[/itex] that the field takes at a point. This should be invariant, i.e., the transformed field at the transformed point [itex](T . \phi ) ( \bar{x} ) \equiv ( T . \phi ) ( T . x )[/itex] should be the same as the original field at the original point [itex]\phi (x)[/itex]. Thus
    [tex](T . \phi ) ( T . x) = \phi (x)[/tex]
    Since [itex] x = T^{-1}\bar{x}[/itex], then
    [tex](T . \phi ) (\bar{x}) = \phi ( T^{-1}. \bar{x})[/tex]
    Since this is true for any point, we can rename the point by [itex]\bar{x} = x[/itex] and arrive at eq(1).
  14. Aug 21, 2011 #13
    I agree, that's true.
  15. Aug 22, 2011 #14
    Some basic questions:

    When you write [itex](T.\phi)(x)[/itex] is this the same as [itex][\exp(-i a_\mu P^\mu)\phi](x)[/itex]? And if so, would [itex]T^{-1}.x = x-a[/itex] or [itex]x+a[/itex]?

    If you can write [itex]T.\phi[/itex] in terms of [itex]P^\mu[/itex], is that enough to get the form of [itex]P^\mu[/itex] using samalkhaiat's Eq.1? If so, can someone write it out explicitly?

    If [itex]P^\mu = +i\partial^\mu[/itex] then it seems
    [tex] [\exp(-i a_\mu P^\mu)\phi](x) = \phi(x+a). [/tex]
    Is this saying that the shape of the transformed field is shifted in the direction [itex]-a[/itex] compared to the original field? This seems to be the opposite of the convention where [itex]\phi'(x') = \phi(x)[/itex] and [itex]x \rightarrow x'=x+a[/itex]. There, the shape of [itex]\phi'[/itex] is shifted in the direction of [itex]+a[/itex].
  16. Aug 23, 2011 #15


    User Avatar
    Science Advisor

    As I told you before, your calculation was correct, and you are needlessly confusing yourself. Ok, forget about the text you are using and follow the simple stuff below;
    Consider the following infinitesimal coordinates transformation
    [tex]x \rightarrow g.x = x + a,[/tex]
    and its inverse
    [tex]x \rightarrow g^{-1}. x = x - a.[/tex]
    According to Wigner, scalar field (which plays the role of “wave function” in here) transforms by (linear) unitary representation U of g. That is
    [tex]\bar{\phi}(\bar{x}) = \left(U(g) . \phi \right) ( \bar{x}) = \phi (x),[/tex]
    or, which is the same thing
    [tex]\bar{\phi}(x) = \left(U(g) . \phi \right) (x) = \phi (g^{-1}. x). \ \ (1)[/tex]
    Now, introduce some hermitian operator P and write
    [tex]U(g) = \exp ( - i a . P ) \approx 1 - i a . P \ \ (2)[/tex]
    (Nothing prevent you from choosing U with opposite sign for [itex](a . P)[/itex])
    With this choice of [itex]U(g)[/itex], the infinitesimal form of (1) leads to
    [tex]\delta \phi (x) = - i a . P \phi(x) = - a . \partial \phi (x).[/tex]
    Thus we are led to the differential realization
    [tex]i P = \partial . \ \ (3)[/tex]
    The question “does this have the wrong sign?” is meaningless because both [itex](P)[/itex] and [itex](-P)[/itex] satisfy the same algebra. However, your metric signature determined which sign of P is appropriate for energy-momentum operator. So, if your metric is [itex](-1, +1, +1, +1)[/itex], you should use (3) as your energy-momentum operator. And, if your metric has the signature [itex](+1, -1, -1, -1)[/itex], then [itex](-P)[/itex] is the energy-momentum operator.
    Now, if (for some reason) you want to express the coordinates transformation in terms of the operator P as given in (3), do the following, since [itex]\partial_{c}x^{a} = \delta^{a}_{c}[/itex], then
    [tex]g . x^{a} = x^{a} + a^{c}\partial_{c}x^{a} = (1 + i a . P) x^{a},[/tex]
    [tex]g^{-1}. x = (1 - i a . P) x.[/tex]

    Last edited: Aug 23, 2011
  17. Aug 23, 2011 #16
    Thanks. That makes a lot of sense. So just define P with a + or - so as to give the usual operator from regular quantum mechanics (P^i = -i d/dx^i, for i=1,2,3). The unitary operator that shifts the field for that choice of P is [itex]exp(+i a_\mu P^\mu)[/itex] (using the +,-,-,- metric).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook