Why Use a Bi-Doublet Scalar Field (2,2) Under SU(2)L x SU(2)R?

[btphysics]

Hello,

why one can use a bi-doublet scalar field (2,2) under SU(2)L x SU(2)R ? In terms of group theory, we should have only triplets (3,1) or (1,3) since 2 x 2=3+1 ? But in left right symmetric models, indeed yukawa coupling are formed with bi-doublet scalars.

Best regards

 

[kurros]

I don't really know about such models, but in either case individually one can have SU(2) doublet scalar fields, e.g. the Standard Model Higgs doublet, so why couldn't you have scalars which are doublets under both groups?

 

[btphysics]

The only problem is that in terms of group theory representation, for me, that does not makes sense.

 

[samalkhaiat]

Hello,

why one can use a bi-doublet scalar field (2,2) under SU(2)L x SU(2)R ? In terms of group theory, we should have only triplets (3,1) or (1,3) since 2 x 2=3+1 ? But in left right symmetric models, indeed yukawa coupling are formed with bi-doublet scalars.

Best regards

This is very much SM question. So, this thread should be moved to particle physics sub-forums.

As for the answer, recall that in the interacting theory of massless nucleon and mesons, we introduce the following $8 \times 8$ matrix of meson fields
$$\Phi ( x ) = I_{ 8 \times 8 } \ \sigma ( x ) + i \gamma_{ 5 } \tau_{ i } \pi_{ i } ( x ) ,$$
where $\sigma ( x )$ is an iso-scalar in the $[1]$ representation of $SU(2)$ and $\pi_{ i } ( x )$ is an iso-vector in the $[3]$ representation of $SU(2)$. We couple this to the nucleon field $N ( x ) \in \ [2]$ by (Yukawa)
$$\mathcal{ L }_{ \mbox{int} } = \bar{ N } ( x ) \Phi ( x ) N ( x ) .$$
a) The requirement that $\mathcal{ L }_{ \mbox{int} }$ be invariant under the (vector) $SU(2)$ transformation $U$, implies that
$$N ( x ) \rightarrow U N ( x ) , \ \ \ \Phi ( x ) \rightarrow U \Phi ( x ) U^{ \dagger } .$$
b) Since the nucleon in the model is massless, we also demand that $\mathcal{ L }_{ \mbox{int} }$ be invariant under the axial iso-spin transformation $U_{ 5 } = \exp ( i \gamma_{ 5 } \alpha_{ i } \tau_{ i } / 2 )$. This implies
$$N \rightarrow U_{ 5 } N , \ \ \ \Phi \rightarrow U^{ \dagger }_{ 5 } \Phi U^{ \dagger }_{ 5 } .$$
With a bit of algebra we can combine the transformations in (a) and (b) to form the invariance group $SU_{ L } (2) \times SU_{ R }(2)$ of $\mathcal{ L }_{ \mbox{int} }$ as follows
$$N_{ R } \rightarrow R N_{ R } ,$$
in the $(1 , 2)$ representation of $SU_{ L } (2) \times SU_{ R }(2)$,
$$N_{ L } \rightarrow L N_{ L },$$
in the $(2 , 1)$ representation, and
$$\Phi \rightarrow L \Phi R^{ \dagger },$$
in the $(2 , 2)$ representation of $SU_{ L } (2) \times SU_{ R }(2)$.

So, in short, it is the massless (chiral) fermions that require mesons from $(2 , 2 )$ representation.

See:
https://www.physicsforums.com/showpost.php?p=3819325&postcount=6

See also pages 115-121 in the textbook by Ta-Pei Cheng & Ling-Fong Li:
“Gauge Theory of Elementary Particle Physics, Problems and Solutions” , Oxford University Press, 2000.

Sam