Question on Lie algebra indices, spinor indices and gamma matrices indices in QCD and QED lagrangians

Question

The question is written in section $2)$

1) Introduction

1.1) QCD

For a non-abelian group, the connection term on the lagrangian will be written as

$$\mathcal{A}_{\mu}=A_{\mu}^{a}T_{a}\tag{1}$$

This notation is far from be explicit and this brings me all sorts of confusion. I came from a classical pseudo-Riemannian background so indices are very important to me, I need to see all the indices to be sure and to be confortable with the math.

Therefore, the standard notation for the QCD lagrangian, with quarks $\psi$ and $\bar{\psi}$ interacting with the gluons $\mathcal{A}_{\mu}$ is:

$$\mathcal{L} =-\frac{1}{4}F^{a}_{\mu\nu}F^{\mu\nu}_{a} + \bar{\psi}[i\gamma^{\mu}D_{\mu}-m]\psi= -\frac{1}{4}F^{a}_{\mu\nu}F^{\mu\nu}_{a}+\bar{\psi}[i\gamma^{\mu}(\partial _{\mu}+ig\mathcal{A}_{\mu})-m]\psi \tag{2}$$

But, $(2)$, just like $(1)$, is not the "explicit" form of the lagrangian. Following the reference $[1]$, the explicit form (with colors!) is given by:

$$\mathcal{L} =-\frac{1}{4}\color{teal}{\sum_{a=1}^{8}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{brown}{\sum_{i=1}^{3}}\color{brown}{\sum_{j=1}^{3}}\color{magenta}{\sum_{(f)=1}^{6}}\bar{\psi}_{\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu}D_{\mu \color{brown}{j}}^{\color{brown}{i}}-m_{\color{magenta}{(f)}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}}_{\color{magenta}{(f)}}$$

$$= -\frac{1}{4}\color{teal}{\sum_{a=1}^{8}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{brown}{\sum_{i=1}^{3}}\color{brown}{\sum_{j=1}^{3}}\color{magenta}{\sum_{(f)=1}^{6}}\bar{\psi}_{\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{i}}_{\color{brown}{j}}\partial_{\mu}+ig\color{teal}{\sum_{a=1}^{8}}A_{\mu}^{\color{teal}{a}}T_{\color{teal}{a}\color{brown}{j}}^{\color{brown}{i}}\Big)-m_{\color{magenta}{(f)}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}}_{\color{magenta}{(f)}} \tag{3}$$

Accordingly to $[1]$, the meaning of the indices are:

$\color{teal}{a = \{1,...,8\}}$ color index in adjoint representation of $SU(3)_{\mathrm{C}}$ (probably here he is actually meaning for the Lie algebra $\mathfrak{su}(3)_{\mathrm{C}}$): indices to take account on how many gluons exist (the famous "Lie algebra indices", maybe "gauge indices"?)

$\color{brown}{i,j = \{1,2,3\} = \{\color{red}{r},\color{green}{g},\color{blue}{b}\}}$ color index in fundamental representation of $SU(3)_{\mathrm{C}}$: indices to take account on how many colors (color charge) a quark can have

$\color{magenta}{(f) = \{1,...,6\} = \{u,d,c,s,t,b\}}$ index to take account on every type of quark that exist

$\mu,\nu = \{0,...,3\}$ our good old spacetime indices.

1.2) QED

Now, using a similar reasoning, I think that explicit indices that appears in QED are:

$\color{teal}{a = \{1\}}$ index in adjoint representation of $U(1)_{\mathrm{em}}$: indices to take account on how many photons exist, i.e., just one photon.

$\color{brown}{i,j = \{1\} = \{\color{brown}{q}\}}$ index in fundamental representation of $U(1)_{\mathrm{em}}$: indices to take account on how many charges a electron can have, i.e., the electric charge.

$\color{magenta}{(f) = \{1\} = \{e\}}$ index to take account on every type of electron that exist, i.e., the electron.

$\mu,\nu = \{0,...,3\}$ our good old spacetime indices.

then, we actually recover the standard notation for Dirac lagrangian of QED:

$$\mathcal{L}= -\frac{1}{4}\color{teal}{\sum_{a=1}^{1}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{brown}{\sum_{i=1}^{1}}\color{brown}{\sum_{j=1}^{1}}\color{magenta}{\sum_{(f)=1}^{1}}\bar{\psi}_{\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{i}}_{\color{brown}{j}}\partial_{\mu}+ig\color{teal}{\sum_{a=1}^{1}}A_{\mu}^{\color{teal}{a}}T_{\color{teal}{a}\color{brown}{j}}^{\color{brown}{i}}\Big)-m_{\color{magenta}{(f)}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}}_{\color{magenta}{(f)}} = $$

$$= -\frac{1}{4}F^{\color{teal}{1}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{1}} + \bar{\psi}_{\color{brown}{1}\color{magenta}{(1)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{1}}_{\color{brown}{1}}\partial_{\mu}+igA_{\mu}^{\color{teal}{1}}T_{\color{teal}{1}\color{brown}{1}}^{\color{brown}{1}}\Big)-m_{\color{magenta}{(1)}}\delta^{\color{brown}{1}}_{\color{brown}{1}}]\psi^{\color{brown}{1}}_{\color{magenta}{(1)}} \implies$$

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}[i\gamma^{\mu}(\partial _{\mu}+iqA_{\mu})-m_{q}]\psi \tag{4}$$

2) My question

Now, I have two sorts of confusion here. One concerning the "explicit" form of QED lagrangian, and the other concerning the gamma matrices $\gamma^{\mu}$. This seems to be two questions, but they are, in my opinion quite related as one. I'll explain.

The gamma matrices are formalized into the framework of clifford algebras, there the one-index object $\gamma_{\mu}$ or $\gamma^{\mu}$ are just basis vectors. But in the context of field theory, the writting of gamma matrices as $\gamma^{\mu}$ are just a (historical?) notation. These symbols are matrice and matrices have two indices, period. Therefore, they are the matrices given by: $\gamma^{\mu\color{purple}{A}}_{\color{purple}{B}}$ where $\color{purple}{A,B = \{1,2\}}$ in Pauli-Dirac representation and $\color{purple}{A,B = \{1,2,3,4\}}$ in Dirac representation.

Well, in both $(3)$ and $(4)$, I'm not sure what are the spinor indices. If we write the QED lagrangian with $\gamma^{\mu\color{purple}{A}}_{\color{purple}{B}}$, then other sorts of indices appear,

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}_{\color{purple}{A}}[i\gamma^{\mu \color{purple}{A}}_{\color{purple}{B}}(\partial _{\mu}+iqA_{\mu})-m_{q}\delta^{\color{purple}{A}}_{\color{purple}{B}}]\psi^{\color{purple}{B}} \tag{5}$$

It seems to me that the indices $A,B$ are in fact the indices that takes account for spin the up and down. Since in the Pauli-Dirac representation $\color{purple}{A,B = \{1,2\} = \{\uparrow,\downarrow \}}$ and in Dirac representation $\color{purple}{A,B = \{1,2,3,4\} = \{\uparrow_{E_{+}},\downarrow_{E_{+}},\uparrow_{E_{-}},\downarrow_{E_{-}}\}}$, where in the first case are just spin up and down states, and the later case is the spin up and down for positive and negative energy solutions.

Therefore my question is:

The spinor indices are the $A,B$?

3) Side notes

1. If the question is answered to be true, then the spinor describing the relativistic electron is: $\psi^{\color{purple}{B}}$ and for a quark is a object with many indices: $\psi^{\color{brown}{j}\color{purple}{B}}_{\color{magenta}{(f)}}$

2. The lagrangian of QCD would be, in my opinion: $$\mathcal{L} = -\frac{1}{4}\color{teal}{\sum_{a=1}^{8}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{purple}{\sum_{i=1}^{4}}\color{brown}{\sum_{i=1}^{3}}\color{brown}{\sum_{j=1}^{3}}\color{magenta}{\sum_{(f)=1}^{6}}\bar{\psi}_{\color{purple}{A}\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu \color{purple}{A}}_{\color{purple}{B}}\Big(\delta^{\color{brown}{i}}_{\color{brown}{j}}\partial_{\mu}+ig\color{teal}{\sum_{a=1}^{8}}A_{\mu}^{\color{teal}{a}}T_{\color{teal}{a}\color{brown}{j}}^{\color{brown}{i}}\Big)-m_{\color{magenta}{(f)}}\delta^{\color{purple}{A}}_{\color{purple}{B}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}\color{purple}{B}}_{\color{magenta}{(f)}}$$

3. If the question is answered to be true, then the spinor trasformations describing the relativistic electron are: $\psi'^{\color{purple}{B}} = S[\Lambda]^{B}_{C}\psi^{\color{purple}{C}}$, which is ok to say that "a electron is a spinor that transforms with SU(2)". But and for a quark: $\psi'^{\color{brown}{j}\color{purple}{B}}_{\color{magenta}{(f)}}=S[\Lambda]^{B}_{C}\psi^{\color{brown}{j}\color{purple}{C}}_{\color{magenta}{(f)}}$ it seems not ok saying that they transforms via $SU(2)$, I mean, they shouldn't transform with a $SU(3)$ matrix (the matrix associated with the representation of the Lie algebra)?

4. I think the summation notation in $(4)$, would appear to be wrong in a first glance: $$\mathcal{L} = -\frac{1}{4}F^{\color{teal}{1}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{1}} + \bar{\psi}_{\color{brown}{1}\color{magenta}{(1)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{1}}_{\color{brown}{1}}\partial_{\mu}+igA_{\mu}^{\color{teal}{1}}T_{\color{teal}{1}\color{brown}{1}}^{\color{brown}{1}}\Big)-m_{\color{magenta}{(1)}}\delta^{\color{brown}{1}}_{\color{brown}{1}}]\psi^{\color{brown}{1}}_{\color{magenta}{(1)}} $$

But, if we consider the generators of lie algebra to be $T_{a} = 1$; since, $g_{U(1)_{\mathrm{em}}} = e^{i\theta} = e^{i\theta 1} \implies \theta^{a} = \theta, T_{a}= 1$. Moreover, $(T_{a})^{i}_{j} = (T_{1})^{i}_{j} = (1)^{i}_{j} = \delta^{i}_{j}$. Finally, considering the case that $i,j = \{1\}$, therefore, in fact, seems to be $(T_{1})^{1}_{1} = \delta^{1}_{1} = 1$. So, the identity matrix multipling the Dirac equation is from $A,B$ indices not from $i,j$: $(i\gamma^{\mu}\partial_{\mu}- m\mathbb{I}_{4\times4})\psi$ is actually, $(i\gamma^{\mu \color{purple}{A}}_{\color{purple}{B}} \partial_{\mu}- m\delta^{\color{purple}{A}}_{\color{purple}{B}})\psi^{\color{purple}{B}}$.

5. Also, the Pauli equation will be written as: $$\hat{H}\psi \equiv \hat{H}\psi^{B} \equiv H^{A}_{B}\psi^{B}. \tag{6}$$

Then,

$$H^{A}_{B}\psi^{B} = \frac{1}{2m}\Bigg[ \sigma^{jA}_{B}\Big(p_{j}-eA_{j}\Big)\Bigg]^{2}\psi^{B}+q\phi\delta^{A}_{B}\psi^{B}. \tag{7}$$

Now, the $\sigma^{jA}_{B}$ are the Pauli matrices and $\delta^{A}_{B} = \mathbb{I}_{2\times2} $ is the "$SU(2)$ identity matrix". The $j ={1,2,3}$ and $A,B={1,2}$, are, respectively, the spatial index and the spinor indices.

6. Note that in the relativistic case (Dirac equation), the identity matrix and the gamma matrices are $4 \times 4$. This fact is a consequence of the Lie group structure of $SU(2) \times SU(2)$, which is the underlying structure of Weyl spinors $[2]$. A Pauli spinor transforms with a $SU(2)$ matrix, and therefore is a "two-colunm complex vector" $[3]$. Moreover, we have the spin up and down for Pauli spinors, which descrive only matter: $A,B= \{1,2\} \equiv \{\uparrow_{\mathrm{matter}}, \downarrow_{\mathrm{matter}} \} $; in the Dirac equation we thus have Weyl spinors, which describe both matter and antimatter in both spin up and down configurations: $A,B= \{1,2,3,4\} \equiv \{\uparrow_{\mathrm{matter}}, \downarrow_{\mathrm{matter}},\uparrow_{\mathrm{antimatter}}, \downarrow_{\mathrm{antimatter}} \} $.

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$[1]$ https://www2.physics.ox.ac.uk/sites/default/files/2014-03-31/qcdgrad_rojo_oxford_tt14_2_basics_pdf_40958.pdf - Page 7

$[2]$ https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group (The unitarian trick section)

$[3]$ https://www.math.uchicago.edu/~may/VIGRE/VIGRE2007/REUPapers/FINALFULL/Thvedt.pdf Page 5, above Theorem 4.4

Answer 1

Yes, the quark fields carry both spinor and color indices, both of which are often suppressed when people write down the QCD lagrangian. To be precise, the quark field lives in a representation of the 'spin-color' group $SU(2)_L \times SU(2)_R \times SU(3)$ (remember that the Lorentz algebra can be decomposed into left and right subgroups), where the Lorentz Group acts on the dirac index, and the colour group $SU(3)$ acts on the colour index.

In fact, in QCD often people consider even more kinds of symmetries. For example, in the chiral (massless) limit, you have a $SU(N_f)_L \times SU(N_f)_R$ symmetry group that rotates the left-handed massless quarks into each other, and the right-handed massless quarks into each other. This symmetry group acts on the flavor indices you wrote (usually restricted to the first three quark flavors, since they are the lightest). Usually this lie algebra is decomposed as $\mathfrak{su}(N_F)_V \oplus \mathfrak{su}(N_F)_A$, because the axial component undergoes spontaneous symmetry breaking, and this explains a lot about the light spectrum of QCD.

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These topics are covered for example in Schwarz in the following places:

• Chapter 10 discusses spinor indices, and how to build up representations of the Lorentz group.
• Section 25.1.1 and onwards discusses how quarks live in the fundamental of the $SU(N_c)$ gauge group.
• Section 30.5.2 discusses the anomalous breaking of the flavor symmetry of QCD quark fields

Answer 2

Right, the gamma matrices are a collection of matrices and therefore have two spinorial indices for matrices. For all representations of "Dirac matrices", A,B runs from 1 to 4. This is true in the Weyl basis, Majorana basis, and Dirac basis. However, in the Weyl basis, all of the gamma matrices can be represented in terms of 2x2 blocks of 2x2 matrices, which include the identity matrix and Pauli matrices. It was not precisely clear to me what was meant by "Pauli-Dirac" representation, so I'd clarify that there are 2x2 Pauli matrices that relate to Cl(3) and 4x4 Dirac matrices that relate to Cl(3,1). A Dirac spinor can be comprised of two Weyl spinors.

Eq. (5) as written is correct for A, B = 1, ... 4 as spinor indices. Therefore, quarks can also be written with a Dirac spinor index as well. Although, it is common to include 2-component Weyl spinors in the standard model. Nevertheless, projection operators can be added to project on chiral fermions, which can be hidden inside the covariant derivative to work with Dirac spinors in the standard model with chiral interactions.

Note that at times, researchers may be sloppy about this and write things like a 2-component spinor equaling a projection operator multiplied by a 4-component spinor. Technically, that would be a 4-component (bi)spinor with one of the 2-component spinors being zero. There are 2-component spinor indices for complex (chiral) Weyl spinors in D=3+1, while Majorana spinors in D=3+1 are 4-component real spinors, and Dirac bispinors are 4-component complex spinors. Dirac matrices should have 4-component spinor indices in D=3+1 because Dirac matrices are for Dirac spinors. Otherwise, reference to Pauli matrices or matrix representations of Clifford algebras might be more appropriate for spinors with other number of components.