Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In QFT one may refer to a particle as a representation of the Lorentz group (LG). More accurately - every particle is a quantum of some field $\phi(x)$ that belongs to some representation of the LG. I wonder if that is really neccesary? Arguments are usually like following:

Let $L$ be some Lorentz tranformation and $\Lambda(L)$ is the corresponding transformation of the field $\phi(x)$ , i.e. under the transformation $x\to x'=Lx$ field $\phi(x)$ goes to $\phi'(x')=\Lambda(L)\phi(x)$. Then under two succsessive transformations $L_2,L_1$ field $\phi(x)$ goes to $\Lambda(L_2)\Lambda(L_1)\phi(x)$. But one can also find himself in the same reference frame via transformation $L_2L_1$ and therefore observe the field $\Lambda(L_2L_1)\phi(x)$. Now demanding $\phi'(x')$ to be the same thing in both cases (since we arrived in the same reference frame) one concludes that $$\Lambda(L_2)\Lambda(L_1)=\Lambda(L_2L_1)\,\,\,\,\, (1)$$. In mathematical language one says that tranformations of the field $\phi(x)$ constitute the representation of the LG.
Well i do not see this to be neccesary. At least not within the "proof" that was been given. I've intentionally used the word "observe" above. Usually in physics not all the information that is encoded in the field value is virtually observable. For instance let $\phi(x)$ be a complex scalar field. Then an overall phase of the $\phi(x)$ is not observable. So i can assume that two ways of going to new reference frame may differ by the "gauge" tranformartion since it does not affect observables. In the case of the complex scalar field one can try to weaken (1) to something like $$\Lambda(L_2)\Lambda(L_1)=\Lambda(L_2L_1)e^{i\alpha(L_1,L_2)}$$ where $\alpha(L_1,L_2)$ is the real number depending on Lorentz tranformations $L_1,L_2$. My questions are:

  1. is it really possible for the physical fields to behave as was described?

  2. if so, does their behavior differ (at classical or quantum level) from the "ususal" fields behavior?

share|improve this question
2  
For good discussion, see Weinberg, The Quantum Theory of Fields, Volume 1, Chapter 2 appendix A and B. –  Steve B Aug 3 '12 at 12:26
    
A particle correspond to a irreducible representation of the Poincaré group and not the Lorentz group. –  Marcel Aug 5 '12 at 10:41
add comment

2 Answers 2

The possibility

$$ \tag{PR} \Lambda(L_2)\Lambda(L_1)~=~\Lambda(L_2L_1)e^{i\alpha(L_1,L_2)},$$

which OP describes, is important in physics. In particular in quantum mechanics, cf. the notion of rays. Eq. (PR) is known as a projective representation.

See e.g. the text around eq. (2.23) in these pdf lecture notes by 't Hooft. Strictly speaking 't Hooft is in that section just considering the Lie group $SO(3)$ of rotations, to be as simple as possible. However, the concept of a projective representation generalizes to an arbitrary group, e.g. finite groups or Lie groups.

The phase factor in eq. (PR) is a so-called two-cocycle. See also this Phys.SE question. If the two-cocycle is not a two-coboundary, i.e. non-trivial, then it is impossible to get rid of the phase factor in eq. (PR) via phase redefinitions alone.

share|improve this answer
    
As far as i can see in these notes 't Hooft propose as an example of such a projective representation the spinor representation of $SO(3)$. Actually my question arose from the thinking about fermions and their "weird" property to change a sign under the $2\pi$ rotation. And i came to a thought that it is basically just a convention. If my reasoning is correct then one can (at least at classical level) choose the function $\alpha(L)$ in such a way that the fermion fields will not change the sign under a complete $2\pi$ rotation. So i'm wondering if that's true. –  Weather Report Aug 2 '12 at 21:15
    
I updated the answer. –  Qmechanic Aug 12 '12 at 12:27
add comment

Elementary particles are represented by certain irreducible unitary projective representations of the Poincare group generated by the Lorentz group and the translations.

On representations with non-integral spin, there is indeed a nontrivial phase factor $\pm 1$, depending on whether in the group, the path from 1 to $L_2$ to $L_2L_1$ to $1$ is ($+$) or is not ($-$) contractible to a point. Upon restriction to the rotation subgroup, this becomes the familiar phase factor for the spinor representation of $SO(3)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.