# Commutation relation of hypercharge and SU(2) generators

• I
• Safinaz
In summary: But the way we understand the electroweak interactions now is that they are described by the symmetry SU(2)xU(1) (including spontaneous symmetry breaking) and that the photon field, i.e., the electromagnetic field, is a specific combination of the fields of the unbroken symmetry. This makes the hypercharge generator Y the generator of U(1) by definition, and gives the relation [Q-T_3,T_i]=0 as a consequence. This is also why when the electric charge operator acts on the vacuum expectation value, it gives zero, because it is a specific combination of the fields of the unbroken symmetry that does not contain the broken field. In summary, the hypercharge commutes with the SU(2)
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 ?

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.

vanhees71
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## ?

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.

vanhees71 and Safinaz
Safinaz said:
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 and vanhees71
samalkhaiat said:
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 ?

Last edited:
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

Safinaz said:
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 $i = 1,2$.

Safinaz said:
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

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 :(

Safinaz said:
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 ..

samalkhaiat said:
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.

Safinaz

## 1. What is the commutation relation between hypercharge and SU(2) generators?

The commutation relation between hypercharge (Y) and SU(2) generators (J) is given by [J,Y] = iΓY, where Γ is the coupling constant.

## 2. What is the significance of the commutation relation between hypercharge and SU(2) generators?

The commutation relation between hypercharge and SU(2) generators is a fundamental property of the electroweak theory, which describes the interactions between elementary particles. It dictates how these quantities behave under transformations and plays a crucial role in calculations and predictions of particle physics.

## 3. How does the commutation relation between hypercharge and SU(2) generators relate to the symmetry of the electroweak theory?

The commutation relation between hypercharge and SU(2) generators is a manifestation of the underlying symmetry of the electroweak theory. It shows that these two quantities are part of a larger symmetry group, SU(2)xU(1), which governs the interactions between particles.

## 4. Can the commutation relation between hypercharge and SU(2) generators be generalized to other symmetry groups?

Yes, the commutation relation between hypercharge and SU(2) generators can be generalized to other symmetry groups, such as SU(N)xU(1) or SU(3)xSU(2)xU(1), for more complex particle interactions. However, the specific form of the commutation relation may differ depending on the group.

## 5. How is the commutation relation between hypercharge and SU(2) generators used in particle physics experiments?

The commutation relation between hypercharge and SU(2) generators is used in calculations and predictions of particle interactions, which can then be tested and verified through experiments. It provides a framework for understanding the behavior of particles and their interactions, and is an essential concept in modern particle physics research.

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