Poincare group representations

  • Thread starter Ravi Mohan
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  • #1
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Main Question or Discussion Point

My question concerns both quantum theory and relativity. But since I came up with this while studying QFT from Weinberg, I post my question in this sub-forum.

As I gather, we first work out the representation of Poincare group (say ##\mathscr{P}##) in ##\mathbb{R}^4## by demanding the Minkowski norm of 4 vector be invariant under homogeneous Lorentz transformations. Since ##\mathscr{P}## is a continuous symmetry group, we expand the elements of the group in terms of continuous parameters and discover that the Minkowski metric imposes constraints on the elements and we end up with 10 independent parameters. So far so good.

Now we work out the unitary representation of ##\mathscr{P}## in the Hilbert space. Here we use the same parameters we obtained in ##\mathbb{R}^4## representation (and the corresponding algebra) to obtain the generators of ##\mathscr{P}## in Hilbert space. And finally by using the continuity constraint, we end up with the commutation relations of the ##\hat{J}^{\mu\nu}## and ##\hat{P}^{\mu}## (and obtain the Poincare algebra).

My question: Is there some way to work out the unitary representation of the Poincare group without working out the ##\mathbb{R}^4## representation first? If the answer is no, then it seems to me that we are giving a "preference" to the ##\mathbb{R}^4## representation.
 

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  • #2
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I think I should reformulate my question. Can we obtain a symmetry group from Hilbert space such that its representation in ##\mathbb{R}^4## is like or similar to Lorentz matrices + translations?
 
  • #3
naima
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Once you know the structure constants of a group you can search operators on a Hilbert space which have these constants in their algebras. You exponentiate them with parameters and you get a representatin of the group.
 
  • #4
dextercioby
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To get to know the structure constants of a Lie group, you would still have to parameterize it.
 
  • #5
samalkhaiat
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I think I should reformulate my question. Can we obtain a symmetry group from Hilbert space such that its representation in ##\mathbb{R}^4## is like or similar to Lorentz matrices + translations?
Yes, the notion of space-time, fields and their transformations can be derived using the elegant method of induced representations of the Poincare’ group [itex]\mathcal{P}(1,3)[/itex]. Indeed, Minkowski space-time, [itex]M^{(1,3)}[/itex], can be identified with the coset space [itex]\mathcal{P}(1,3) / SO(1,3)[/itex], i.e., the space of orbits that the Lorentz group [itex]SO(1,3)[/itex] sweeps out in the Poincare’ group [itex]\mathcal{P}(1,3)[/itex]. Before we do this, let us choose ten arbitrary real numbers [itex]a^{\mu}[/itex] and [itex]\omega^{\mu\nu} = - \omega^{\nu\mu}[/itex] and rewrite the Poincare’ algebra in the form [tex][ia^{\mu}P_{\mu}, P_{\nu}] = 0 ,[/tex] [tex][\frac{i}{2}\omega^{\mu\nu}J_{\mu\nu} , P_{\rho}] = \omega^{\sigma}{}_{\rho} \ P_{\sigma} ,[/tex] [tex][\frac{i}{2}\omega^{\mu\nu}J_{\mu\nu} , J_{\rho\sigma}] = \omega^{\mu}{}_{\rho} \ J_{\mu\sigma} - \omega^{\mu}{}_{\sigma} \ J_{\mu\rho} .[/tex]
Okay, now we choose an origin in the coset space, and coordinatize its neighbourhood by exponentiating the coordinates in the tangent space at that point, i.e., represent a point in the coset space by [tex]h(x) = e^{i x^{\mu} P_{\mu}} \mathbb{I} .[/tex] The group action is then given by left multiplication [tex]h(gx) = h(\bar{x}) = g h(x) \ \mbox{mod} \ SO(1,3) .[/tex] So, for [itex]g = \exp (i a^{\mu} P_{\mu})[/itex], we get [tex]h(\bar{x}) = e^{ i a^{\mu}P_{\mu}} \ e^{ i x^{\nu} P_{\nu}} \mathbb{I} = e^{i (x^{\mu} + a^{\mu}) P_{\mu}} \mathbb{I} .[/tex] Thus [tex]g(a) x^{\mu} = \bar{x}^{\mu} = x^{\mu} + a^{\mu} .[/tex]
For [itex]g = \exp(\frac{i}{2} \omega^{\mu\nu} J_{\mu\nu})[/itex], we get
[tex]h(\bar{x}) = e^{\frac{i}{2} \omega^{\mu\nu} J_{\mu\nu}} \ e^{i x^{\rho}P_{\rho}} \mathbb{I} \ \ \mbox{mod} \ SO(1,3) ,[/tex] or [tex]h(\bar{x}) = e^{\frac{i}{2} \omega^{\mu\nu} J_{\mu\nu}} \ e^{i x^{\rho}P_{\rho}} \ e^{- \frac{i}{2} \omega^{\mu\nu} J_{\mu\nu}} \ \mathbb{I} \ \ \mbox{mod} \ SO(1,3) .[/tex] Now, on the RHS, if we use the identity [tex]e^{A}e^{B}e^{- A} = e^{(e^{A}Be^{-A})} ,[/tex] we find [tex]h(\bar{x}) = \exp \left( e^{\frac{i}{2} \omega \cdot J} ( i x^{\rho} P_{\rho}) e^{- \frac{i}{2} \omega \cdot J} \right) \mathbb{I} \ \mbox{mod} \ SO(1,3) .[/tex] Expanding the two Lorentz group elements, we obtain [tex]h(\bar{x}) = \exp \left( i x^{\sigma}P_{\sigma} + i x^{\rho} ( [\frac{i}{2} \omega^{\mu\nu} J_{\mu\nu}, P_{\rho}] ) + \mathcal{O}(\omega^{2}) \right) \ \mathbb{I} .[/tex] Now, we can use the Poincare’ algebra to get [tex]h(\bar{x}) = e^{i (x^{\sigma} + \omega^{\sigma}{}_{\rho} x^{\rho}) P_{\sigma} + \mathcal{O}(\omega^{2}) } \ \mathbb{I} .[/tex] Form this, we obtain the Lorentz transformations [tex]\bar{x}^{\sigma} = g(\omega) x^{\sigma} = x^{\sigma} + \omega^{\sigma}{}_{\rho} \ x^{\rho} + \mathcal{O}(\omega^{2}) .[/tex]
The fields [itex]\varphi_{a}[/itex] can then be defined as functions from the x’s (i.e., Minkowski space) into a vector space (the field space) on which the Lorentz group acts by matrix multiplication: [itex]D_{a}^{c}(\omega) \varphi_{c}(x)[/itex]. For more detail on this and other related topics, see
www.physicsforums.com/showthread.php?t=172461
 
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  • #6
naima
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As these posts are closed (2007) I cannot quote them to get the Latex source. Is there another way to get them?
 
  • #7
samalkhaiat
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As these posts are closed (2007) I cannot quote them to get the Latex source. Is there another way to get them?
I’m afraid that thread is the only source for those notes. I made them as I went right there and then. This is why there are many typos. Unfortunately, the thread was closed before I had the chance to correct those typos. I also had many people asking me why the thread was closed! And others who wanted the notes organised in PDF file, but the fact is I don’t even have a copy of those notes.
 
  • #8
naima
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Maybe some one in the staff could send you a copy of your original posts. I think that you are still the owner of these notes.
 
  • #9
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As these posts are closed (2007) I cannot quote them to get the Latex source. Is there another way to get them?
In my Firefox, if I right-click on a formula, there's the option to show it as a LaTeX command.
 
  • #10
naima
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Yes and if you right-click somewhere else, you can get the source code of the page. And the formulas are well hidden there!
I think that this forum could provide something easier.
.
 
  • #11
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