Fermion mass terms in Peskin and Schroeder's book

[Geonaut]

TL;DR Summary

I am having some trouble understanding the notation used to describe fermion mass terms in Peskin and Schroeder's book on QFT.

I'm currently looking at how fermion masses are produced via the Higgs mechanism in "An Introduction to Quantum Field Theory" by Peskin and Schroeder. It all makes a lot of sense and I've been fine with it so far, but I ended up getting stuck on something that's driving me nuts. I feel silly asking this since it's just a question about notation, but after hours of searching for a clue my brain is turning to mush. You see, to generate fermion masses in the Standard Model they use the equation

$$ \mathcal{L}_m =-\lambda^{ij}_d \overline Q^i_L \cdot \phi d^j_R - \lambda^{ij}_u \epsilon^{ab}\overline Q^i_{La} \phi^{\dagger}_b u^j_R + h.c.$$

With the following equations given in the book it's clear what's going on here for the most part:

$$ \overline Q^i_L =
\begin{pmatrix}
u^i \\
d^i
\end{pmatrix}_L =

\begin{pmatrix}
\begin{pmatrix}
u \\
d
\end{pmatrix}_L
,&
\begin{pmatrix}
c \\
s
\end{pmatrix}_L
,&
\begin{pmatrix}
t \\
b
\end{pmatrix}_L
\\
\end{pmatrix}
$$

$$u^i_R =
\begin{pmatrix}
u_R ,& c_R ,& t_R
\end{pmatrix}
$$

$$d^i_R =
\begin{pmatrix}
d_R ,& s_R ,& b_R
\end{pmatrix}
$$

The only thing I don't seem to understand is the notation in $$\epsilon^{ab}\overline Q^i_{La} \phi^{\dagger}_b$$. To be specific, what does the "a" in the subscript for $$\overline Q^i_{La}$$ represent? What does the "b" in $$\phi^{\dagger}_b$$ represent? Where does $$\epsilon^{ab}$$ even come from? I would guess that $$\epsilon^{ab} = -\epsilon^{ba}=1$$, but I don't see where this stuff is explained. I would guess that "a" and "b" represent a set of numbers, namely, "a = 1, 2, 3" and "b = 1, 2, 3"... but something isn't adding up. Can someone show me what's going on here? Searching for information as trivial as notation in dense books on QFT is beginning to feel like torture.

 

[Mentz114]

Summary: I am having some trouble understanding the notation used to describe fermion mass terms in Peskin and Schroeder's book on QFT.

I'm currently looking at how fermion masses are produced via the Higgs mechanism in "An Introduction to Quantum Field Theory" by Peskin and Schroeder. It all makes a lot of sense and I've been fine with it so far, but I ended up getting stuck on something that's driving me nuts. I feel silly asking this since it's just a question about notation, but after hours of searching for a clue my brain is turning to mush. You see, to generate fermion masses in the Standard Model they use the equation

$$ \mathcal{L}_m =-\lambda^{ij}_d \overline Q^i_L \cdot \phi d^j_R - \lambda^{ij}_u \epsilon^{ab}\overline Q^i_{La} \phi^{\dagger}_b u^j_R + h.c.$$

With the following equations given in the book it's clear what's going on here for the most part:

$$ \overline Q^i_L =
\begin{pmatrix}
u^i \\
d^i
\end{pmatrix}_L =

\begin{pmatrix}
\begin{pmatrix}
u \\
d
\end{pmatrix}_L
,&
\begin{pmatrix}
c \\
s
\end{pmatrix}_L
,&
\begin{pmatrix}
t \\
b
\end{pmatrix}_L
\\
\end{pmatrix}
$$

$$u^i_R =
\begin{pmatrix}
u_R ,& c_R ,& t_R
\end{pmatrix}
$$

$$d^i_R =
\begin{pmatrix}
d_R ,& s_R ,& b_R
\end{pmatrix}
$$

The only thing I don't seem to understand is the notation in $$\epsilon^{ab}\overline Q^i_{La} \phi^{\dagger}_b$$. To be specific, what does the "a" in the subscript for $$\overline Q^i_{La}$$ represent? What does the "b" in $$\phi^{\dagger}_b$$ represent? Where does $$\epsilon^{ab}$$ even come from? I would guess that $$\epsilon^{ab} = -\epsilon^{ba}=1$$, but I don't see where this stuff is explained. I would guess that "a" and "b" represent a set of numbers, namely, "a = 1, 2, 3" and "b = 1, 2, 3"... but something isn't adding up. Can someone show me what's going on here? Searching for information as trivial as notation in dense books on QFT is beginning to feel like torture.

I might be completely wrong here - but is a summation implied in

$$\epsilon^{ab}\overline Q^i_{La} \phi^{\dagger}_b$$

The indices a and b are dummies and imply a sum over values of a and b which occur in upper and lower positions. For instance an inner product is written ##\mathbf{u} \cdot \mathbf{v} = u_j v^j## and a linear transformation as ##u^a=M^{ab}v_b##

See https://en.wikipedia.org/wiki/Einstein_notation

 

[Geonaut]

Yes, I agree with that, I think they are using Einstein notation, but what I meant to ask is: if $$\overline Q^i_{L}$$ is a doublet of three left handed quark particle generations then what is $$\overline Q^i_{La}$$? It seems to be a group of doublets, but for what particles? I would also guess that $$\phi^{\dagger}_b$$ represents three complex fields, but for what 3 particles? I think maybe $$\overline Q^i_{La}$$ is all of the possible quark doublets which includes chiral quarks, and so "i = 1, 2" since there is two different chiral fields which differ in the fact that they transform like:

$$\psi_1 \rightarrow (1 + i\alpha^a t^a (\frac{1-\gamma^5}{2}))\psi_1)$$$$\psi_2 \rightarrow (1 + i\alpha^a t^a (\frac{1+\gamma^5}{2}))\psi_2)$$

, but I don't know for sure. In fact, they don't even mention both of these transformations that I just wrote down, I'm just guessing (unless I missed this somewhere, of course).

Moreover, I'd also guess that the three complex scalar fields that $$\phi_b^{\dagger}$$ represents corresponds to three massless scalar particles that exist before spontaneous symmetry breaking, and that after spontaneous symmetry breaking these three particles become 2 goldstone bosons and the 1 higgs particle. I think that's right, but it doesn't seem to all be explicitly stated... as far as I can tell.

Also, I just noticed that the term $$\epsilon^{ab}$$ should be equal to $$\epsilon^{ba}$$ since these fermion mass terms have to be negative... and with that, I think I have answers now to all of the questions that I had about this. The only question left is if I'm misunderstanding any of this.

 

[Dr.AbeNikIanEdL]

On the first look, ##i,j## should be the labels for the generation (running from 1 to 3), and ##a,b## label the components of the ##SU(2)## doublets (running from 1 to 2). Both ##\bar{Q}_L^i## and ##\Phi## are two component objects, i.e.

##\bar{Q}_L^i= \left(\begin{array}{c} u^i_L \\ d^i_L\end{array}\right)##

##\Phi = \left(\begin{array}{c} \Phi_1 \\ \Phi_2\end{array}\right)##

So ##\bar{Q}_{L1}^1 = u_L##, ##\bar{Q}_{L2}^1 = d_L## and so on for the other generations. ##\epsilon^{ab}## should then indeed be ##\epsilon^{ab}=-\epsilon^{ba}##, basically exchanging the first and second component of ##\Phi^\dagger## so that ##u^i## (instead of ##d^i##) is multiplied by the second component of ##\Phi## (which is the one that will conventionally obtain a non-zero vev).

 

[vanhees71]

That's it! Note that the two different types of Yukawa coupling for the quarks besides weak-isospin SU(2) symmetry also take into account hypercharge conservation at each vertex. In the SM there's only one type of Yukawa terms for the leptons, because the neutrino masses are not taken into account and thus there are no right-handed neutrinos.

 

[Geonaut]

@Dr.AbeNikIanEdL, thank you! And sorry, the reply a wrote last night was a little sloppy, I was really tired, but I'm glad that didn't matter. It's funny, I actually considered your answer at one point, I reverse engineered it, but I was confused why they'd use this notation at all and so I ended up way overthinking things. Anyway, that makes a lot of sense because it gives you the right terms after you give the scalar field a VEV. ##\epsilon^{ab}=-\epsilon^{ba}## makes sense now too since ##\phi_1## is imaginary and ##\phi_2## is real, it seems you need this to account for an extra negative sign that appears when you take the complex conjugate of ##\phi##.

 

[vanhees71]

The reason of course is that for two SU(2) spinors ##\phi_1## and ##\phi_2## both ##\phi_1^{\dagger} \phi## and ##\phi_1^{\text{T}} \hat{\epsilon} \phi_2## are scalars. For the first term it's simple. Take an SU(2) matrix ##\hat{U}## and define ##\phi_1'=\hat{U} \phi_1## and ##\phi_2'=\hat{U} \phi_2##. Then
$$\phi_1^{\prime \dagger} u_2'=\phi_1 ^{\dagger} \hat{U}^{\dagger} \hat{U} \phi_2=\phi_1^{\dagger} \phi_2.$$
That's, because as for any unitary matrix you have ##\hat{U}^{\dagger} \hat{U}=\hat{U} \hat{U}^{\dagger}=\hat{1}##.

For the 2nd expression you get
$$\phi_1^{\prime \text{T}} \hat{\eta} \phi_2^{\prime}=\phi^{1 \text{T}} \hat{U}^{\text{T}} \hat{\epsilon} \hat{U} \phi_2,$$
but now
$$(\hat{U}^{\text{T}} \hat{\epsilon} \hat{U})_{ab}=U_{ca} \epsilon_{cd} U_{db} =\epsilon_{ab} \det \hat{U}=\epsilon_{ab}.$$
Thus for any SU(2) matrix (even for any ##\text{SL}(2,\mathbb{C}## matrix in fact) you have
$$\hat{U}^{\text{T}} \hat{\epsilon} \hat{U}=\hat{\epsilon},$$
and thus indeed ##\phi_1^{\text{T}} \hat{\epsilon} \phi_2## is indeed an invariant under ##\text{SL}(2,\mathbb{C})## and thus also under SU(2)-transformations.