Commutation relation of hypercharge and SU(2) generators

[Safinaz]

Hi all,

I read in Cheng and Li's book "Gauge theory of elementary particle physics" Ch 11, specifically : Eq. (11.46) that the hypercharge commutes with the SU(2) generators, i.e.,

$[Q-T_3,T_i]=0$, I'd like to understand what that mean and how this could be proved ?

 

[MathematicalPhysicist]

Didn't I ask a similar question a few years ago from Srednicki's book?
Here's a link to that thread:
https://www.physicsforums.com/threads/another-question-from-srednickis-qft-book.761915/page-3

from post 60 onwards, seems relevant to your question, though not exactly the same.

 

[Orodruin]

The SM gauge group contains the direct product between the hypercharge U(1) and the SU(2). By definition, the generators of the groups in the direct product commute. The U(1)xSU(2) symmetry is spontaneously broken to the U(1) symmetry of electromagnetism, whose generator is therefore a linear combination of the U(1) and SU(2) generators.

 

[Safinaz]

Please can you look at the definition of the generator Q and the isospin $T_3$ at Cheng & Li's book and let me know how it led to the commutation between the hyperchagre and $T_i$ ?

 

[Orodruin]

I don't have Cheng and Li, you would have to quote the definition.

Regardless, Q should be defined in such a way that the proper combination of Q and T3 is the generator of the hypercharge U(1), which by definition commutes with the SU(2) generators.

 

[samalkhaiat]

Hi all,

I read in Cheng and Li's book "Gauge theory of elementary particle physics" Ch 11, specifically : Eq. (11.46) that the hypercharge commutes with the SU(2) generators, i.e.,

$[Q-T_3,T_i]=0$, I'd like to understand what that mean and how this could be proved ?

Denote the lepton doublets, quark doublets, by \psi(x), then you have (repeated indices are summed over) Q - T^{3} \sim \int d^{3}y \ \psi^{\dagger}_{c}(y) \psi_{c}(y) , \ \ c = 1,2

You also have T^{i} = \frac{1}{2} \int d^{3}x \ \psi^{\dagger}_{a}(x) (\tau^{i})_{ab} \psi_{b}(x) , \ \ i = 1,2,3 . So, you need to evaluate the following commutator [Q - T^{3} , T^{i}] \sim (\tau^{i})_{ab} \int d^{3}x \ d^{3}y \ [\psi^{\dagger}_{c}(y)\psi_{c}(y) , \psi^{\dagger}_{a}(x)\psi_{b}(x)] . To do that, expand the RHS using the identity [AB , CD] = CA\{ B , D \} – C \{ A , D \} B + A \{ B , C \}D - \{ A , C \} BD , then use the equal-time anti-commutation relations \{\psi^{\dagger}_{a}(x) , \psi_{b}(y) \} = \delta_{ab} \delta^{3}(x-y) , \{ \psi_{a}(x) , \psi_{b}(y) \} = \{ \psi^{\dagger}_{a}(x) , \psi^{\dagger}_{b}(y) \} = 0 . If you don’t make a mistake, you get [Q - T^{3} , T^{i}] = 0. This tells you that the model has an additional U(1) symmetry, generated by (Q - T^{3}). Thus, you conclude, SU(2) \times U(1) is the symmetry group of the model.

 

[Safinaz]

you get [Q−T3,Ti]=0[Q - T^{3} , T^{i}] = 0. This tells you that the model has an additional U(1)U(1) symmetry, generated by (Q−T3)(Q - T^{3}). Thus, you conclude, SU(2)×U(1)SU(2) \times U(1) is the symmetry group of the model.

Hi thanks for the perfect answer. But Q and $T_3$ also commute with $T_i$ as:

$Q = \int (- e^\dagger e + \frac{2}{3} u^\dagger u - \frac{1}{3} d^\dagger d ) d^3 x ,$ and

$T_3 = \frac{1}{2} \int (\nu^\dagger_L \nu_L - e^\dagger_L e_L + u^\dagger_L u_L - d^\dagger_L d_L ) d^3 x$

So why did they added to each other to get Y ?

 

[Safinaz]

Also here have you an idea how when the electric charge operator acting on the vacuum expectation value $\phi_0 = <0|\phi|0> = (0~~~~~~ v)^T$, it gives zero ,i.e., $Q <\phi>_0 = 0$ , while the isospin operator when acting on the VEV it doesn't vanish ? so that the electric charge still conserved while the hypercharge has broken

 

[samalkhaiat]

Hi thanks for the perfect answer. But Q and $T_3$ also commute with $T_i$

No, they don’t. Don’t make such statement before doing the actual calculation first: [Q , T_{i}] = [T_{3} , T_{i}] = i \delta_{1i}T_{2} - i \delta_{2i}T_{1} . Clearly, this does not vanish for i = 1,2.

 

[samalkhaiat]

Also here have you an idea how when the electric charge operator acting on the vacuum expectation value $\phi_0 = <0|\phi|0> = (0~~~~~~ v)^T$, it gives zero ,i.e., $Q <\phi>_0 = 0$ , while the isospin operator when acting on the VEV it doesn't vanish ? so that the electric charge still conserved while the hypercharge has broken

See the following thread
https://www.physicsforums.com/threads/why-su-2-times-u-1-for-the-sm.846099/#post-5320206

 

[Safinaz]

Hi,

Thanks again for replying. I'm refereed to the nominated thread , actually I still don't understand why should Y commutes with $T_i$ to become one of $SU(2)\times U(1)$ generators , sorry :(

 

[samalkhaiat]

Hi,

Thanks again for replying. I'm refereed to the nominated thread , actually I still don't understand why should Y commutes with $T_i$ to become one of $SU(2)\times U(1)$ generators , sorry :(

I give you the group G_{1} \times G_{2} and ask you: Do the generators of G_{1} commute with the generators of G_{2}?

 

[Safinaz]

Yes, actually they should commute ..

 

[Orodruin]

I give you the group G_{1} \times G_{2} and ask you: Do the generators of G_{1} commute with the generators of G_{2}?

Reading through some parts of this thread again, I think that Safinaz's problem might be the opposite inference, i.e., not taking a given gauge symmetry and inferring the relations among the generators, but starting from the low energy broken theory and inferring what types of gauge symmetry operators there are and what their commutation properties are, i.e., finding out what the gauge symmetries of the unbroken theory are. The only way of doing that is by experiment and finding out what gives a good description of observations. If you are given U(1)xSU(2), the inference that the generator of U(1) commutes with the SU(2) generators follows directly from definition. If you are not given the gauge group and need to infer it, it is a different story. For example, from what I understand, it was originally proposed that a single SU(2) was broken to U(1) before the discovery of neutral currents.