This question is about something from page 162 ofPhysics from Symmetry by Schwichtenberg (second edition). The doublet is an electroweak$SU(2)$, say a left-handed neutrino and left-handed electron. Both are Dirac spinors. He multiplies the barred doublet by a right Dirac spinor but without showing the result. It seems he is trying to get an invariant term for the Lagrangian. He says the product is not$SU(2)$ invariant and proceeds then to insert the Higgs field to achieve that.
My question is: What did he mean by a product of the barred doublet and a Dirac spinor? The doublet is really a$1\times8$ and the Dirac a$4\times1$, so not sure what he was trying to do. Thanks for any advice.
Edit - I now think book is figuring doublet as consisting of two left chiral spinors and the right singlet consisting of a right chiral spinor and another right chiral spinor. Thus the product of the barred left doublet and the right singlet gives a valid term for the Lagrangian. Nevertheless, it is rejected because it is not SU(2) invariant.But I am not sure.
- $\begingroup$That's his point! That you cannot combine an SU(2) doublet with a singlet to get an invariant (a singlet)! You must saturate the doublet with another doublet (Higgs field) to get a singlet. The spinor indices are a red herring: you saturate a 4-spinor by a four spinor.... Try a standard text.$\endgroup$Cosmas Zachos– Cosmas Zachos2025-11-14 15:13:43 +00:00CommentedNov 14 at 15:13
- $\begingroup$Thank you. He emphasizes the lack of SU(2) symmetry of in the product and thus it can't be used. But even if the product had the symmetry could it be used in the lagrangian ? That depends on the definition of the product and also if another term is inserted someplace or not. Could be I reading too much into this.$\endgroup$D. Apple– D. Apple2025-11-14 20:48:31 +00:00CommentedNov 14 at 20:48
- $\begingroup$The theory is predicated on the SU(2) symmetry, so an invariant termcould be included in the lagrangian, unlike the ones first contemplated, leading thus to the invariant Yukawa terms.$\endgroup$Cosmas Zachos– Cosmas Zachos2025-11-14 22:21:22 +00:00CommentedNov 14 at 22:21
2 Answers2
The product is strictly in the four-dimensional spinor space. You have a left-handed lepton spinor$l_{L}$, which has two indices, a Dirac index and a flavor index; and a right-handed electron spinor$e_{R}$ with just one index, its Dirac index. Using Greek letters for the Dirac indices and Carolingian minuscule for the flavor index, the product he forms is$$\left(\bar{l}\!_{L}\right)_{a}e_{R}\equiv\sum_{\alpha,\beta=1}^{4}\left(l_{L}^{*\alpha}\right)_{a}\left(\gamma^{0}\right)^{\alpha\beta}e_{R}^{\beta}.$$Note that the flavor index$a$ is not contracted with anything, which means that the resulting product still has a flavor index, making the product an$SU(2)$ singlet.
In order to get a gauge-invariant expression, you need to include another field that also has an uncontracted$SU(2)$ index that may be contracted with the$a$ of the product. The simplest way to do this is to insert a scalar field that is also in the fundamental$SU(2)$ representation$h_{a}$. This is the Higgs field, and the coupling has the Yukawa(-like) form$$Y_{e}\sum_{a=1}^{2}\left(\bar{l}\!_{L}\right)_{a}e_{R}\,h_{a},$$with$Y_{e}$ being the coupling constant. Now all indices are contracted, so a term like this in the Lagrange density is Lorentz invariant (because there are no free Dirac indices) and$SU(2)$ gauge invariant (because there are no uncontracted flavor indices). When the scalar field$h_{a}$ gets a vacuum expectation value$\langle h_{1}\rangle=v$, the product$m_{e}vY_{e}$ becomes an electron mass term.
- 1$\begingroup$Thank you. The flavor index an integer ranges from one to two ?$\endgroup$D. Apple– D. Apple2025-11-14 20:20:38 +00:00CommentedNov 14 at 20:20
- $\begingroup$The book was emphasizing the lack of SU(2) symmetry in the product. But regardless of that issue there is the overarching problem what is the lagrangian term he intended to use. A singlet is not a lagrangian term. I thought he was intending the product to be a lagrangian term. Your definition of the product is not a lagrangian term and thus can't be used for that reason alone.$\endgroup$D. Apple– D. Apple2025-11-14 20:37:57 +00:00CommentedNov 14 at 20:37
- $\begingroup$@D.Apple I'm not sure what you mean. As to the first question, there are different conventions for indexing flavor. I used 1 and 2, but you could equally use isospin $\pm1/2$. As to the second comment/question, I just don't understand what you mean. A term in the Lagrange density should be invariant under all symmetry transformations, which what I described is. So I don't see what you mean by saying it is "not a lagrangian term."$\endgroup$2025-11-14 20:46:47 +00:00CommentedNov 14 at 20:46
- $\begingroup$The integral over space and time of the lagrangian gives you a number. So a barred dirac spinor times a dirac spinor will fulfill that requirement for the lagrangian. A single dirac spinor will not work. I was thinking the book had the barred doublet times a singlet could be put in the lagrangian, it depends how the product is defined. P.S. physic books needs to clarify notation.$\endgroup$D. Apple– D. Apple2025-11-14 23:31:03 +00:00CommentedNov 14 at 23:31
Based on reading page 166 I think the intent is this. Define the right singlet with 4 components, each a function of space and time. The doublet consists of 4 components each a function of space and time. So the barred doublet time the right singlet is a valid Lagrangian term. We reject it since it is not SU(2) invariant.The doublet can be thought of made of two components. Each a left chiral spinor. The singlet is also made of 2 chiral spinors each a right chiral spinor. But I am not sure !
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