Talk:Parity (physics)

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[edit] Conservation of parity

I redirected conservation of parity here because it is discussed in the article, but it would be good to increase its visibility. -- Kjkolb 07:35, 25 November 2005 (UTC)

[edit] Fixing the global symmetries

How could it be, then if F=B+L, and if B=L=0 for Majorana neutrinos, then F=1 ?

This is the point. So far, every particle that we've found has obeyed F=B+L. No one knows why, it's just an observation. It is an open question whether Majorana neutrinos exist, but if they do, they would break the F=B+L rule. Which isn't good or bad, it's just a fact, and it means that parity might have eigenvalues other than plus or minus one. JarahE 22:22, 27 April 2006 (UTC)
But P^2 = 1 implies P can only have eigenvalues that square to 1...
Shambolic Entity 02:01, 25 January 2007 (UTC)
OK, then I understand if F=B+L is empirical.Hidaspal 19:22, 29 April 2006 (UTC)
Yeah, not only is it empirical, but there may even be a counter-example if Neutrino's really turn out to have Majorana masses as people seem to suspect now adays. This argument is very old, I think it's Steven Weinberg's from maybe 40 years ago, before there was evidence for Majorana neutrinos.JarahE 15:01, 2 May 2006 (UTC)

Why Q (electric charge) is mentioned as a charge of a global symmetry group?

The Lagrangian of electromagnetism is invariant under the rotatation of the wavefunction of each charge Q particle by e^{ialpha Q} for any alpha. In the particular case in which alpha is constant, this is a global symmetry. You might want to look at the page Noether's theorem which explains this phenomenon in more generality. To my knowledge, all conserved charges in quantum field theory, not just electric charge, arise similarly. JarahE 22:22, 27 April 2006 (UTC)
My point is the following. Electric charge is related to a local gauge symmetry. Of course you can always say, that say alpha(x) = beta(x) + c, but is it possible then assign c again to "another" or "assign again" to the electric charge? It sounds that you can then multiply the number of symmetries arbitrarily. Or do you say here, that for the momnet lets ignore the local symmetries, because the global part is enough for this issue? It seems you are saying this latter case. Hidaspal 19:22, 29 April 2006 (UTC)
Your formula makes it look like the global symmetry is just some Fourier mode of the local symmetry, but physically there's a big difference. In a path integral you fix boundary conditions, which means that global symmetries are fixed and local symmetries are integrated over. In examples this seems to mean that local gauge symmetries are just redundancies of the description, where as global symmetries really change your state, i.e. they change the observables. Gauge symmetries in particular have to leave the boundary conditions fixed, so the zero mode of your alpha(x) is not a gauge symmetry. So anyway it seems to me like for noncompact spaces there's no arbitrariness in separating out the global symmetries (although you can always add a local symmetry to a global one and get another representative of the same global symmetry), the constant rotation corresponds to a real physical transformation. JarahE 15:01, 2 May 2006 (UTC)

Hidaspal 21:13, 27 April 2006 (UTC)

[edit] Simple symmetry relations

"In a quantum theory states in a Hilbert space do not need to transform under representations of the group of rotations, but only under projective representations." - it sounds like a restriction though there is an extension in reality from representations of O(3) to SU(2). Projective to what? There are a lot of missing statements, which makes the whole thing unclear and ununderstandable. Hidaspal 21:30, 27 April 2006 (UTC)

Your appraisal sounds a bit extreme, but anyway, I also think that the term projective representation is misleading because it sounds like a kind of representation. This leads to your observation that the phrase sounds like a restriction. Instead the opposite is true, a representation is a kind of projective representation. Anyway, I've tried to explain this a bit better. In particular, I've added the fact that the projection is with respect to the overall U(1) phase of a state in the Hilbert space, as rotations of a Hilbert space vector by an overall phase do not change the corresponding physical state. Do you understand it now?JarahE 22:35, 27 April 2006 (UTC)
Sorry for being extreme.:-) No, I do not understand, but the basic problem, that I have only learned another approach, which is probably identical, but does not mention projective representation at all. My book "HF Jones, 1990" does not do that. The picture in my mind is that we start from SO(3) which is a Lie-group, then realize, that it has the same Lie-algebra than SU(2) and it allows reprezentations of SU(2) which are not reprezentations of SO(3), but for us the Lie-algebra is important (the same handwaving as "use projective instead of real representation") not the group, and if the Lie-algebra allows, then spin-half will occur, etc.
So I understand that only means extension. But I do not know yet, whether my picture is the same as the "projective representation picture". I think you might be more specific, and less abstract, then more of us will understand. Mention specific groups, the world is specific anyway, we have O(3) rotations and SU(2)-spins :-) Or the best, if you do it twice, once beeing abstract, and mathematically more correct, and once more specific and more understandable for many. Sorry again for beeing extreme. Hidaspal 20:46, 29 April 2006 (UTC)
The two specific groups, in the case of massive particles are SO(N) and Spin(N), which in 3+1 dimensions means that the little group of the rotation group is SO(3) and the projective representations of SO(3) are the ordinary representations of its universal cover SU(2). It so happens that SO(3) and SU(2) have the same algebra, which many books use as a pretext for why fermions transform under a rep of SU(2) that is not a rep of SO(3). But this argument is too fast, because in classical physics for example states transform as reps of SO(3), ie there are no fermions. The fact that quantum states only need to be reps of SU(2) and not of SO(3) is really a quantum effect, its a result of the choice of phase in a vector in the Hilbert space.
The standard argument that you cite fails in general, as not all projective reps are reps of groups that have the same algebra, more generally they are reps of groups that have an algebra that has one extra generator (called a central extension). This is occurs for example in the Wess-Zumino-Witten model, the classical symmetry is a loop group but the states transform under projective reps of the loop group, which are affine Lie algebras. The argument that you cite always works when there is only a discrete set of possible phases to choose (like the plus or minus one in the choice of fermions and bosons).
So anyway, for the page, your suggestion is to mention the specific groups SO(3) and SU(2) or SO(N) and Spin(N) as examples? JarahE 15:09, 2 May 2006 (UTC)
Yes Hidaspal 23:14, 6 May 2006 (UTC)

[edit] Contradiction in intro

The first line states "...a parity transformation is the simultaneous flip in the sign of all spatial coordinates." Almost directly below this it is stated that "In a two-dimensional plane, parity is not the same as a rotation by 180 degrees." These two statements blatantly contradict each other. Shambolic Entity 02:01, 25 January 2007 (UTC)

The first statement is about the three-dimensional world in which we live and is much more important than the second statement which is about a two-dimensional imaginary world. Since the context is different, they do not contradict each other. JRSpriggs 09:55, 25 January 2007 (UTC)
There is one definition that fits both: "Parity is a flip in the sign of one spatial coordinate." This works for 3D as well as 2D cases as well as in arbitrary dimensions. The definition given in the first sentence only works in odd number of dimensions (You can think of x, y, and z flips as 3 parity transformations that form again a parity transformation. In two dimensions, flipping x and y brings you back to the original parity). --149.217.1.6 (talk) 16:28, 29 April 2008 (UTC)
I just changed the introductory sentence. I don't understand why the three-dimensional definition is taken with such importance, and later reflection is introduced as being something independent from parity that has to be combined with rotation to form parity...?!? Reflection CHANGES the parity. Parity says something about the handedness of objects in space-time. If you transform an object through a reflection, in general you can not bring it back into the original position just by a rotation, so reflection IS a parity transformation. On top of that, reflection as a definition for parity works in all space dimension. The definition of flipping the signs of ALL coordinates is IMHO a simplified definition that only works in odd number of dimensions. Really you perform 3 parity transformations in a row, which is why you end up with something that you could have obtained by a single (reflection) parity transformation. I think the paragraph about reflection and so on needs to be re-written. Do people agree? --149.217.1.6 (talk) 16:46, 29 April 2008 (UTC)

[edit] Intrinsic parity

This article could really use a definition of instrinsic parity. I would do it myself but, since that is the topic I came here to learn about, I don't know the definition myself. Tpellman (talk) 17:22, 29 April 2008 (UTC)