Parity of a Particle (and Parity of the Higgs in particular)


by the_pulp
Tags: higgs, parity, particle
the_pulp
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#1
Nov14-12, 06:19 AM
P: 132
Im quite confused with the Parity idea. I never payed much attention to it. My thought always was that Parity is related to a discrete symetry of space and it is something that can be measured and can give 1 or -1 (like the x-momentum of a particle that can give "2" "1,9" "1,8" and so on).

Nevertheless reading about the Higgs particle I saw that the Higgs Parity is "+1". My concern is that it is written as if the Parity is not something that can give "+1" or "-1" depending on randomness (the operator-eigenvalue idea) but it is just a parameter (like mass or electric charge that it is not subject to randomness).

I think that it is related to the idea that weak interactions apply to left handed particles only but Im not sure and I would like if anyone can give me an explanation of it.

I have a more or less general knowledge about SM and the math behind it. Ive never actually done the math but I saw it in Peskin and such and I think I sort of followed it (I dont know, perhaps a 50%-80% of it).

Thanks for every Math or non Math explanations you can give me. Both will be very helpful.
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ofirg
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#2
Nov14-12, 07:19 AM
P: 82
parity is a discrete symmetry. it is defined as taking x[itex]\rightarrow[/itex]-x,y[itex]\rightarrow[/itex]-y,z[itex]\rightarrow[/itex]-z

However the result of applying parity is not random; it depends on its transformation properties under parity. It is a property of the object.

for example the position vector [itex]\vec{r}\rightarrow-\vec{r}[/itex] is a vector
the momentum vector also. (parity -1)
Angular momentum [itex]\vec{L}=\vec{r}\times\vec{p}[/itex] is a pseudo-vector [itex]\vec{L}\rightarrow\vec{L}[/itex] (parity +1 doesn't change sign under parity)

A scalar particle like the higgs can be either a scalar which has a parity value of +1 or a pseudo-scalar which has a value of -1 (other theories, such as susy, predict also pseudo-scalar higgs particles)

The difference from momentum is that momemtun can be changed by a lorentz transformation and thus not considered to be a property of a particle, While parity can't be.

This has nothing to do with the parity violation due to the weak interaction.

In particles physics one usually talks about the combined discrete symmetry CP, since P is not such as good symmetry (because of the weak inetraction) and CP is better( though not exact also)
the_pulp
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#3
Nov14-12, 09:01 AM
P: 132
Ok, thanks, I almost followed you. Nevertheless, in the first part you talk about operators that change or does not change after Parity (ie commutes with parity I imagine). However in the second part you say that "the particle" does not change with parity. What is that does not change with parity (ie commutes with parity I imagine again)? the [itex]\phi[/itex] operator perhaps (wether it refers to the scalar or vector version of the particle)?

Thanks

ofirg
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#4
Nov14-12, 09:30 AM
P: 82

Parity of a Particle (and Parity of the Higgs in particular)


First,

You need to seperate between the operator and the vector space it acts on.

In general any vector (in any vector space) which is an eigenstate of parity has eigenvalue 1 or -1 (position, momentum, scalar partice, pseudo-scalar particle)

comutation is defined between two operrators, not an operator and and a vector.


Second,

If by saying that "the particle" does not change with parity you mean this

The difference from momentum is that momemtun can be changed by a lorentz transformation and thus not considered to be a property of a particle, While parity can't be
I mean that a lorentz transformation will never change a scalar to a pseudo-scalar. Unlike momentum, which has a different value for every reference frame, a scalar (P=+1) will be a scalar in all reference frame.


-This is not true for a reperesenation of lorentz. In general, lorentz transformation don't commute with parity
the_pulp
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#5
Nov14-12, 09:45 AM
P: 132
Ok, let me think about it and perhaps I'll ask a bit more later. Any other thought is very welcomed.

Thanks
Bill_K
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#6
Nov14-12, 09:53 AM
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P: 3,853
Parity tells us whether a quantity changes sign under a space reflection. It is used in three senses:

1) The parity of a variable, such as r, p, L, as ofirg illustrated.

2) The parity of a wavefunction. The wavefunction for an atomic state can have different parity. States with an orbital angular momentum that is even, L = 0, 2, 4... have parity +1. States with odd orbital angular momentum have parity -1.

3) The parity of a particle. This is also called the particle's intrinsic parity, and is the same regardless of what state the particle is in. The Higgs particle in the Standard Model has intrinsic parity +1. The data analysis at the LHC is trying to confirm that the particle they see at 125 GeV does have spin 0 and parity +1.
JTP
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#7
Nov21-12, 03:21 PM
P: 4
The Higgs particle in the Standard Model has intrinsic parity +1.
This may sound like a silly, trivial question... but why? How does one see that the parity of the SM Higgs is +1?

On a sidenote, the SM Higgs charge conjugation number is also said to be +1. Again, how does one extract this from the theory?

Thank you in advance for your attention.
ofirg
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#8
Nov21-12, 04:45 PM
P: 82
How does one see that the parity of the SM Higgs is +1?
Basically, One looks at the theory and "reads out" the value from the requirement the parity is a symmetry(and thus conserved). Since in the standard model parity isn't a good symmetry, one usually talks about CP (the product of parity and charge conjugation). The higgs has to be assigned a value of CP=1 (scalar) for CP to be a good symmetry (although still not exact)

It is important to note that the existence of the symmetry doesn't depend if we actually do the assignment in our heads. It just tells us what the meaningful value is.


What is your background in the field? Are you familiar with the lagrangian formulation of the standard model ( or QFT in general?). This can be deduced from the terms in the SM lagrangian.
the_pulp
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#9
Nov22-12, 06:30 PM
P: 132
I'm pretty familiar with the lagrangian formulation of qft And of sm in particular, but i cant see how i can deduce that CP=1, so anything you can say would be helpful.
ofirg
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#10
Nov22-12, 07:32 PM
P: 82
Two examples:

in the SM (after the higgs gets a vaccum expectation value) higgs potential there are cubic terms in the higgs field. Obviously this isn't possible for CP=-1, since the term would have a total [itex]cp^{3}=-1[/itex].

The higgs yukawa coupling to fermions (after diagonalization, where the CP violating phase appears in the CKM matrix) doesn't include [itex]\gamma_{5}[/itex] matrix in the interaction term which makes the fermion bilinear odd under parity. If the higgs where a pseudoscalar the
[itex]\gamma_{5}[/itex] factor would be present in the interaction.

Hope that helps.
JTP
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#11
Nov22-12, 09:35 PM
P: 4
Quote Quote by ofirg View Post
Basically, One looks at the theory and "reads out" the value from the requirement the parity is a symmetry(and thus conserved). Since in the standard model parity isn't a good symmetry, one usually talks about CP (the product of parity and charge conjugation). The higgs has to be assigned a value of CP=1 (scalar) for CP to be a good symmetry (although still not exact)
So, if I understood it correctly, one imposes CP invariance in the Standard Model lagrangian. The cubic term ([itex]H^3[/itex] after SSB) is a clear way of noticing that the desired CP invariance is only attained if the Higgs field itself is also CP invariant.
Quote Quote by ofirg View Post
What is your background in the field? Are you familiar with the lagrangian formulation of the standard model ( or QFT in general?).
As the_pulp, I am also familiar with the lagrangian formalism in QFT.
Quote Quote by ofirg View Post
The higgs yukawa coupling to fermions (after diagonalization, where the CP violating phase appears in the CKM matrix) doesn't include [itex]\gamma_{5}[/itex] matrix in the interaction term which makes the fermion bilinear odd under parity. If the higgs where a pseudoscalar the
[itex]\gamma_{5}[/itex] factor would be present in the interaction.
This was actually my first reasoning, which only made sense in light of the CP invariance requirement (of which I wasn't sure).


Also, if one has established CP=+1, then, if the particle is additionally an eigenstate of both C and P, one should have either C=P=-1 or C=P=+1 (the latter corresponds in the literature to the assigment for the SM Higgs).

Firstly, how can one be sure that a particle, such as the Higgs, is an eigenstate of C or P?
Finally, if it is shown to have a well defined value of P and C, I would suppose that, in order to confirm the assignment C=P=+1, one would again look at the lagrangian. But since, as you stated, parity isn't a good symmetry in the SM, we cannot (as we could before) rule out P=-1 by looking at the cubic term or the yukawa coupings, for example.


Thank you for your previous answers, they were indeed very helpful.
ofirg
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#12
Nov23-12, 06:28 AM
P: 82
So, if I understood it correctly, one imposes CP invariance in the Standard Model lagrangian
No. CP is an approximate accidental symmetry of the Standard model (SM) lagrangian.

After requiring the symmetries which are part of the SM definition (gauge symmetries and lorentz) one can look at the lagrangian and identify other symmetries which were not imposed.

When one does that you get that there is an approximate CP symmetry with the assignment CP=1 for the higgs. The symmetry exists with this useful assignment because of the form of the lagrangian, and not because of how the particle was defined. It is, however, useful to identify it with the particle.

The same is true for any accidental symmetry. You don't define the electron,muon,tau to have lepton number=1 before writing the lagranagian, later you see that the symmetry exists with those assignments.

The same is true for C=1 and P=1 ( for the higgs) but these symmetries are useful only when the weak interaction is not involved.
JTP
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#13
Nov25-12, 07:14 AM
P: 4
Quote Quote by ofirg View Post
No. CP is an approximate accidental symmetry of the Standard model (SM) lagrangian.
Oh, ok, my mistake.
Quote Quote by ofirg View Post
The same is true for any accidental symmetry. You don't define the electron,muon,tau to have lepton number=1 before writing the lagranagian, later you see that the symmetry exists with those assignments.
But still, isn't the accidental lepton number symmetry a little different? Consider the charged current term in the lagrangian: [itex]\mathcal{L}_e^{\textrm{CC}} \propto \overline{\nu_e}\,\gamma^\mu(1-\gamma_5)\,e\,W_\mu[/itex].

Suppose we assign lepton number to the fields like so: [itex]\quad \overline{\nu_e}\rightarrow -1 \qquad e \rightarrow 1 \qquad W^\mu \rightarrow 0[/itex]
Then, looking at the term [itex]\mathcal{L}_e^{\textrm{CC}}[/itex], considering lepton number additive, we get [itex]-1+1+0=0[/itex] and this means the interaction associated with this term conserves lepton number. Lepton number is assigned to the fields and not to the whole term.

Consider the Higgs interaction term with the electron, if the Higgs were a pseudoscalar: [itex]\mathcal{L}_e^{\textrm{H}} \propto \overline{e}\,\gamma_5\,e\,H[/itex].
Here, again, one assigns lepton number to the fields: [itex]\quad \overline{e}\rightarrow -1 \qquad e \rightarrow 1 \qquad H \rightarrow 0[/itex]
The interaction again conserves lepton number. Lepton number is assigned to the fields and not to the bilinear [itex]\overline{e}\,\gamma_5\,e[/itex] as a whole.

Now, regarding CP, it would be different in the sense that [itex]\overline{e}\,\gamma_5\,e[/itex] (a bilinear, not a field) already has a well defined way of transforming under CP:
[itex]\overline{e}\,\gamma_5\,e\, \overset{CP}{\rightarrow}\, -\overline{e}\,\gamma_5\,e\ [/itex], because [itex]\det(CP)=-1[/itex].

Assigning a CP number to the field [itex]H[/itex] would then translate our desire to either have a (almost) CP invariant lagrangian or not. So I can't see here how the symmetry exists independently of the assignment, since [itex]\overline{e}\,\gamma_5\,e[/itex] already has a fixed assignment of CP-number ([itex]-1[/itex]).

I hope the mistake in my reasoning is easy to spot.
ofirg
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#14
Nov25-12, 08:18 AM
P: 82
But still, isn't the accidental lepton number symmetry a little different?
The difference is that parity is a spacetime symmetry and not an internal symmetry. Therefore spacetime objects transfrom under it in a known way, while they are invariant under lepton number symmetry.

Lepton number is assigned to the fields and not to the bilinear [itex]e^{-}\gamma_{5}e[/itex] as a whole
It is entirely possible to tranform a single spinor field under parity. When one does that, one sees that the tranformation includes a factor which is assigned to the fermion (see peskin page 65)
The difference from lepton number is that a single spinor isn't an eigenstate of parity or CP (only the bilinear is) so that the tranformation mixes different component and isn't simply a multiplicative factor as in lepton number.
The other reason why one mainly cares about the transformation of fermion bilinears is that fermions always appear in the lagrangian ( and in observables) in pairs.

it would be different in the sense that [itex]e^{-}\gamma_{5}e[/itex] (a bilinear, not a field) already has a well defined way of transforming under CP:
It just happens that the bilinear tranformation doesn't depend on the value assigned mentioned above.

Assigning a CP number to the field H would then translate our desire to either have a (almost) CP invariant lagrangian or not
The main thing to understand here is that the lagrangian determines all the dynamics, including conservation laws. This dynamics of course doesn't depend on any arbitrary assignment we make. It is simply useful to assign numbers such that the lagrnagian is invariant ( if possible) and thus these numbers are conserved.

for example, I could define a transformation and call it CP under which [itex]H\rightarrow-H[/itex] but then it woudn't be conserved. The conservation of CP with CP=1 for H would still of course happen.

Any transformation I can define under which my lagrangain is invariant, would be useful. However if I were to identify it with parity it needs satisfy the basic group mutiplication of parity. for example [itex]P^{2}=1[/itex]. Both assignments for the higgs would satisfy this.
JTP
JTP is offline
#15
Dec10-12, 03:37 AM
P: 4
I think I got it. I'll go and read Peskin & Schröder to see if I missed anything.
Thank you very much.


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