Gauge invariance of electroweak Lagrangian

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  • #1
lalo_u
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I was trying to prove all those little things you spend long as the local invariance in the free Lagrangian of electroweak interaction.

Taking into account the appropriate SU(2) transformations (without covariant derivatives), came to the following expression

[tex]\mathcal{L}_{\text{ferm.}} = i\bar{\Psi}^Lexp\left [-ig\tau_j\frac{\omega_j(x)}{2}\right] \gamma^\mu \left [\frac{ig\tau_j}{2}\partial_\mu \omega_j(x)exp\left [ig\tau_j\frac{\omega_j(x)}{2} \right]\Psi^L+...\right] [/tex]

Where: [itex]\Psi^L[\itex]: left leptonic doublet, [itex]\tau_j[\itex]: Pauli matrices

Well, i'm stuck with this: the right exponential expression can commute with the real [itex]\partial_\mu\omega_j(x)[\itex] and tha matrix [itex]\tau_j[\itex] but it can't commute with [itex]_gamma^\mu[\itex] because the superindex [itex]\mu[\itex] doesn't need the same as j. T This is necessary to cancel the exponential terms.

What's missing?
 

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  • #2
ChrisVer
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There's no problem with [itex]\gamma^\mu[/itex]. The last is seeing a different vector space (it acts on the 4-spinors of Dirac dields [itex]\psi[/itex] with its spinorial indices [itex] \gamma^\mu_{ab}[/itex] and on spacetime (the derivative) with its [itex]\mu[/itex]-index) whereas the t generators act on the SU(2) doublets...
 
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  • #3
lalo_u
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Yes. I understand what you're saying about the different vector spaces ChrisVer. But if you expand the SU(2) doublet we have the same operation but now with singlets, isn't it?
 
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  • #4
samalkhaiat
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I was trying to prove all those little things you spend long as the local invariance in the free Lagrangian of electroweak interaction.

Taking into account the appropriate SU(2) transformations (without covariant derivatives), came to the following expression

[tex]\mathcal{L}_{\text{ferm.}} = i\bar{\Psi}^Lexp\left [-ig\tau_j\frac{\omega_j(x)}{2}\right] \gamma^\mu \left [\frac{ig\tau_j}{2}\partial_\mu \omega_j(x)exp\left [ig\tau_j\frac{\omega_j(x)}{2} \right]\Psi^L+...\right] [/tex]

Where: [itex]\Psi^L[\itex]: left leptonic doublet, [itex]\tau_j[\itex]: Pauli matrices

Well, i'm stuck with this: the right exponential expression can commute with the real [itex]\partial_\mu\omega_j(x)[\itex] and tha matrix [itex]\tau_j[\itex] but it can't commute with [itex]_gamma^\mu[\itex] because the superindex [itex]\mu[\itex] doesn't need the same as j. T This is necessary to cancel the exponential terms.

What's missing?
In order to carry out the transformations correctly, it is essential to understand the following fact: The Dirac matrices commute with all the [itex]SU(2)[/itex] matrices, i.e. for all [itex]( u , u^{ \dagger } ) \in SU(2)[/itex] , [tex][ \gamma^{ \mu } , u ] = [ \gamma^{ \mu } , u^{ \dagger } ] = 0 . \ \ \ \ (1)[/tex] This is because they operate on two different spaces: [itex]u ( x )[/itex] and [itex]u^{ \dagger } ( x )[/itex] operate on the doublet, while [itex]\gamma^{ \mu }[/itex] operates on the Dirac’s spinors within the doublet. To prove the invariance of the total Lagrangian under local [itex]SU(2)[/itex] transformation, equation (1) will be used in almost all the steps. Let’s do that from scratch. Under local [itex]SU(2)[/itex], the doublet transforms as [tex]\psi_{ u } ( x ) = u ( x ) \psi ( x ) .[/tex] Taking the dagger gives [tex]\psi^{ \dagger }_{ u } = \psi^{ \dagger } u^{ \dagger } .[/tex] Multiplying both sides from the right by [itex]\gamma^{ 0 }[/itex] and using [itex][ \gamma^{ 0 } , u^{ \dagger } ] = 0[/itex] , we get [tex]\bar{ \psi }_{ u } ( x ) = \bar{ \psi } ( x ) \ u^{ \dagger } ( x ) .[/tex] Now, let us determine how the free Lagrangian [itex]\mathcal{ L } = i \bar{ \psi } \gamma^{ \mu } \partial_{ \mu } \psi[/itex] , transforms under the local [itex]SU(2)[/itex] group. First note that using (1) allows us to write [tex]( \gamma^{ \mu } \ \partial_{ \mu } \psi )^{ u } = u ( x ) \ \gamma^{ \mu } \partial_{ \mu } \psi + \gamma^{ \mu } \ ( \partial_{ \mu } u ) \psi .[/tex] So, by using [itex][ \gamma^{ \mu } , u^{ \dagger } ( x ) ] = 0[/itex] this time, we can write [tex]\mathcal{ L }^{ u ( x ) } = i \bar{ \psi } \ \gamma^{ \mu } \ \partial_{ \mu } \psi + i \bar{ \psi } \ \gamma^{ \mu } \ ( u^{ \dagger } \partial_{ \mu } u ) \ \psi ,[/tex] which means that the free Lagrangian is not invariant under the local [itex]SU(2)[/itex] transformations: [tex]\mathcal{ L }^{ u ( x ) } = \mathcal{ L } + i \bar{ \psi } \ \gamma^{ \mu } \ ( u^{ \dagger } \partial_{ \mu } u ) \ \psi . \ \ \ (2)[/tex] Now consider interaction Lagrangian [tex]\mathcal{ L }_{\mbox{int}} = i \ g \ \bar{ \psi } \ \gamma^{ \mu } \ A_{ \mu } \ \psi ,[/tex] which describes the coupling between the fermions and some “vector” field [itex]A_{ \mu }[/itex]. This transforms into [tex]\mathcal{ L }^{ u ( x ) }_{\mbox{int}} = ( i g \ \bar{ \psi } \ \gamma^{ \mu } \ A_{ \mu } \ \psi )^{ u ( x ) } = i g \ \bar{ \psi }_{ u } \ \gamma^{ \mu } \ A^{ u ( x ) }_{ \mu } \ \psi_{ u }.[/tex] Thus, using (1) and other transformations for [itex]\psi[/itex] and [itex]\bar{ \psi }[/itex] we find [tex]\mathcal{ L }^{ u ( x ) }_{\mbox{int}} = i g \ \bar{ \psi } \ \gamma^{ \mu } \ ( u^{ \dagger } \ A^{ u ( x ) }_{ \mu } \ u ) \ \psi .[/tex] If, under the local group [itex]SU(2)[/itex], the “vector” field [itex]A_{ \mu }[/itex] transforms in the adjoint representation, i.e. if [tex]A^{ u ( x ) }_{ \mu } = u \ A_{ \mu } \ u^{ \dagger } - \frac{ 1 }{ g } \ ( \partial_{ \mu } u ) \ u^{ \dagger } ,[/tex] then
[tex]\mathcal{ L }^{ u ( x ) }_{ \mbox{int}} = \mathcal{ L }_{\mbox{int}} - i \bar{ \psi } \ \gamma^{ \mu } \ ( u^{ \dagger } \ \partial_{ \mu } u ) \ \psi . \ \ \ (3)[/tex]
Adding (2) to (3) gives us [tex]\mathcal{ L }^{ u ( x ) } + \mathcal{ L }^{ u ( x ) }_{\mbox{int}} = \mathcal{ L } + \mathcal{ L }_{\mbox{int}} .[/tex] This means that the total Lagrangian [tex]\mathcal{ L }_{\mbox{tot}} \equiv \mathcal{ L } + \mathcal{ L }_{\mbox{int}} = i \bar{ \psi } \gamma^{ \mu } D_{ \mu } \psi ,[/tex] is invariant under the local [itex]SU(2)[/itex] transformations. Notice that (1) has been used in almost all the steps.
 
  • #5
ChrisVer
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Yes. I understand what you're saying about the different vector spaces ChrisVer. But if you expand the SU(2) doublet we have the same operation but now with singlets, isn't it?
Expand the SU(2) doublet? what do you mean?
I am just trying to tell you that an SU(2) doublet is a vector with two Weyl spinor components: [itex] \Psi_L = \begin{pmatrix} \psi_{L1} \\ \psi_{L2} \end{pmatrix} [/itex].
On this space act the generators of SU(2).

Now each component of this doublet is a Weyl spinor:
[itex] \psi^{a}_{L1} = P_L \psi^a_1 = P_L \begin{pmatrix} \psi_1^1 \\ \psi^1_2 \\ \psi^1_3 \\ \psi^1_4 \end{pmatrix}[/itex] where [itex]P_L[/itex] the projection operator, and on that thing acts the :
[itex] \gamma^\mu_{ab} \partial_\mu = ( \gamma^0 \partial_0 - \vec{\gamma} \cdot \vec{\partial} )_{ab} [/itex] with the ab indices...
 
  • #6
lalo_u
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I mean for example:

[tex]\left(\bar{\psi}^L_{\nu_l},\bar{\psi}^L_l\right)
\begin{pmatrix} 0 & \sigma_k \\ -\sigma_k & 0 \end{pmatrix}\partial_{\mu}
\begin{pmatrix} \psi^L_{\nu_l} \\ \psi^L_l \end{pmatrix}
[/tex]

Are we forbidden to expand this?

I'm pretty sure you're right like all the QFT books. I'm just trying to figure it out without group theory, only lnear algebra.
 
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  • #7
ChrisVer
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Again I am not able to understand what you mean by "expand" ...And I don't understand what is that matrix you wrote between your doublets... It looks like the [itex] \gamma^k[/itex] with k spatial indices... and then a partial derivative with mu spacetime indices....

You would have (without the SU(2) generators inbetween), and denoting the doublet as [itex]L[/itex]:

[itex] \bar{L} \gamma^\mu \partial_\mu L = \bar{\psi}_{\nu_l}^L \gamma^\mu \partial_\mu \psi_{\nu_l}^L + \bar{\psi}_{l}^L \gamma^\mu \partial_\mu \psi_{l}^L [/itex]

If I put an SU(2) generator inbetween the doublets, I will somehow mix them....but the [itex]\gamma^\mu \partial_\mu[/itex] will keep following each term...you could as well write:

[itex]\gamma^\mu \partial_\mu \begin{pmatrix} \nu_l \\ \psi_l \end{pmatrix}_L = \begin{pmatrix} \gamma^\mu \partial_\mu \nu_{l}^L \\ \gamma^\mu \partial_\mu \psi_l^L \end{pmatrix} =\begin{pmatrix} \frac{1}{2} \gamma^\mu (1 - \gamma^5) \partial_\mu \nu_{l} \\ \frac{1}{2} \gamma^\mu (1- \gamma^5) \partial_\mu \psi_l \end{pmatrix} [/itex]

Linear algebra can't help... the gammas are 4x4 matrices acting on the 4 spinors (the [itex]\nu_l[/itex] and [itex]\psi_l[/itex] in my last equation, after they get projected) and the SU(2) generators acting on doublets are 2x2....
 
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  • #8
lalo_u
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Sorry for my english, I realize i don't let me understand when I use the word expand. I used it in the sense of "calculate something" like the command in "Mathematica".

But you've got the point of my question ChrisVer, when you wrote

"Linear algebra can't help... the gammas are 4x4 matrices acting on the 4 spinors (the νl and ψl in my last equation, after they get projected) and the SU(2) generators acting on doublets are 2x2...."

I know the gamma matrices commute with all the SU(2) matrices. It's a fact in all QFT. However it's a fact i've never questioned.

Then, despite the multiplication between square matrices of different order is not defined in linear algebra (and I don't speak from direct product), they can be written in the Lagrangian and say simply commute. Is that right?

Or, there's a theorem in order to demonstrate this?
 
  • #9
ChrisVer
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The main reason they commute is because they are "seeing"/acting on different vector spaces, as I already mentioned in my very first post.

Why do spin operators in QM (not QFT) commute with eg. position operator? Or when you have two particles of spin [itex]S_1, S_2[/itex] why does [itex][\hat{S}_1 ,\hat{S]_2]=0 [/itex] hold?

The main reason is that they act on different vector spaces. That's the reason it is a fact. Even if you had the same dimensionality, they would still not commute.

So the SU(2) transformations ([itex]t^a[/itex]) act on the doublets. The doublets contain spinors, which are acted on by the [itex]\gamma^\mu \partial_\mu[/itex] 4x4 matrix.
 

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