A bit of everything

# Quantum chomodynamics and effective theories

Computing the properties of a single proton in quantum chromodynamics is quite an undertaking as it is, as we have to deal with the QCD Lagrangian (ignoring other interactions and assuming masses without the Higgs field)

$$\mathcal{L} = \sum_{f} \bar{\psi}_{fa} i \gamma^\mu [\delta_{ab} \partial_\mu + i g (G_{\mu}^\alpha t_{\alpha})_{ab}] \psi_{qa}$$

Due to being a non-Abelian eight-dimensional gauge group, this isn't the easiest theory to work out, especially when we're involving systems of many quarks. The idea of effective theories for QCD is to take the most general theory with the same symmetries as this Lagrangian, and work it out from there.

What are the symmetries here? Outside of the obvious $\mathrm{SU}(3)$ from the gauge group, we also have that each term is invariant under a Lorentz transform, which is simply going to be an $\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R$ transformation, for the left-handed and right-handed spinor. If we assume that $N$ quarks have an identical mass, there is also an $\mathrm{SU}(N)$ flavor symmetry, so that in total, our model will have the symmetry

$$G = \mathrm{SU}(3)_{QCD} \times \mathrm{SU}(2)_L \times \mathrm{SU}(2)_R \times \mathrm{SU}(N)$$

This is a group acting on the object $\psi_{af\alpha}$, a spinor with $N$ flavors and $3$ colors, as well as its conjugate, $\bar{\psi}_{af\alpha}$, for the antiquark. We're not quite interested in the behaviour of quarks, though, but more on systems of them. What we're looking for is then a product of them :

$$N \otimes \bar{M} = \psi_1 \otimes \psi_2 \otimes \ldots \otimes \psi_n \otimes \bar{\chi}_1 \otimes \bar{\chi}_2 \otimes \ldots \otimes \bar{\chi}_M$$

And more to the point, we'd like their irreducible representations. Our two important such products will be the mesons, inhabiting the product $1 \otimes \bar{1}$, and the baryons, in $1 \otimes 1 \otimes 1$.

## The pion model

For now, let's consider a simple case : we're only concerned about the symmetry group of two flavors, corresponding to the quark $u$ and $d$. With a respective mass of $2.01 \pm 0.14 \mathrm{MeV}$ and $4.79 \pm 0.16 \mathrm{MeV}$, those can be considered roughly similar (and in fact $\approx 0$) in the energy scales involved. We'll consider every other quantity fixed here : our two quarks are of the appropriate opposite colors, and are of a given spin at all times. Looking back at our Lagrangian, this would be

$$\mathcal{L} = (\bar{\psi}^u_{a}, \bar{\psi}^d_{a}) i \gamma^\mu [\delta_{ab} \partial_\mu + i g (G_{\mu}^\alpha t_{\alpha})_{ab}] \begin{pmatrix}\psi^u_{a}\\\psi^d_{a}\end{pmatrix}$$

Our system is therefore the associated vector space of $\mathrm{SU}(2)$, $\mathbb{C}^2$, and the antiparticle sector is the conjugate representation. In other words, given the standard matrix representation $\Pi$ of $\mathrm{SU}(2)$ over our vector space $V = \mathbb{C}^2$, the antiquark is in the complex conjugate vector space $\bar{V} = \mathbb{C}^2$, and the representation $\bar{Pi}$ is such that $\bar{\Pi}(g) = (\Pi(g))^*$. Their product $V \otimes \bar{V}$ is acted on by the representation

\begin{eqnarray} \Pi \otimes \bar{\Pi} : \mathrm{SU}(2) &\to& \mathrm{GL}(V \otimes \bar{V})\\ g &\mapsto& \Pi(g) \otimes (\Pi(g))^* \end{eqnarray}

So that our doublet $(\psi, \bar{\chi})$ is acted on by $(\Pi(g)\psi, (\Pi(g))^* \bar{\chi})$. In terms of components, this would be

$$\psi^{A} \bar{\chi}^{\dot{B}} \to M_A^{A'} \psi^{A} (M_{B}^{B'})^* \bar{\chi}^{\dot{B}}$$

Our symmetry group here is $\mathrm{SU}(2)$, so that we can write our transformation as

\begin{eqnarray} M &=& e^{i \alpha \theta_i \sigma^i}\\ &=& \sum_{n = 0}^\infty i^{n} \frac{(\alpha \theta_i \sigma^i)}{n!}\\ &=& \ldots\\ &=& I \cos(\alpha) + i \theta_i \sigma^i \sin(\alpha) \end{eqnarray}

We have the good luck here that the decomposition into irreducible representation is easy enough to do. Decompose the product as

$$\psi^{A} \bar{\chi}^{\dot{B}} = \psi^{(A} \bar{\chi}^{\dot{B})} + \psi^{[A} \bar{\chi}^{\dot{B}]}$$

Our basis of $\mathbb{C}^2 \times \mathbb{C}^2$ will be noted as ${e_u, e_d}$ for the first part, and ${e_\bar{u}, e_\bar{d}}$ for the second, with the product basis $e_{ij} = e_i \otimes e_j$, with basis $\{ e_{u \bar{u}}, e_{u \bar{d}}, e_{d \bar{u}}, e_{d \bar{d}} \}$. Our two vectors will be :

\begin{eqnarray} \psi &=& \psi_u e_u + \psi_d e_d\\ \bar{\chi} &=& \bar{\chi}_{\bar{u}} e_\bar{u} + \bar{\chi}_{\bar{d}} e_\bar{d} \end{eqnarray}

With these components, our $2 \times 2$ matrix corresponds to

\begin{eqnarray} \psi^{(A} \bar{\chi}^{\dot{B})} &=& \begin{pmatrix} \psi_u \bar{\chi}_{\bar{u}} & \psi_u \bar{\chi}_{\bar{d}} + \psi_d \bar{\chi}_{\bar{u}} \\ \psi_u \bar{\chi}_{\bar{d}} + \psi_d \bar{\chi}_{\bar{u}} & \psi_d \bar{\chi}_{\bar{d}} \end{pmatrix} \\ \psi^{[A} \bar{\chi}^{\dot{B}]} &=& (\psi_u \bar{\chi}_{\bar{d}} - \psi_d \bar{\chi}_{\bar{u}}) \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \end{eqnarray}

From our original $4$-dimensional $\mathbb{C}^4$, we can see that the first term is three dimensional, made from the span of $u \bar{u}$, $d \bar{d}$, and a new vector $u \bar{d} + d \bar{u}$, while our second term is one-dimensional, spanned by $u \bar{d} - d \bar{u}$. Let's see how $\mathrm{SU}(2)$ act on them :

$$M^{A'}_{A}\psi^{A} (M^{B'}_{B})^* \bar{\chi}^{\dot{B}} = M^{A'}_{A} (M^{B'}_{B})^* \psi^{(A} \bar{\chi}^{\dot{B})} + M^{A'}_{A} (M^{B'}_{B})^* \psi^{[A} \bar{\chi}^{\dot{B}]}$$

Working out for each component, let's consider first the action on the vectors themselves :

\begin{eqnarray} M^{A'}_{A} \psi^{A} &=& \begin{pmatrix} \cos(\alpha) + i \theta_3 \sin(\alpha) & i \theta_1 \sin(\alpha) + \theta_2 \sin(\alpha) \\ i \theta_1 \sin(\alpha) - \theta_2 \sin(\alpha) & \cos(\alpha) - i \theta_3 \sin(\alpha) \end{pmatrix} \begin{pmatrix} u \\ d \end{pmatrix} \\ &=& \begin{pmatrix} (\cos(\alpha) + i \theta_3 \sin(\alpha)) u + (i \theta_1 \sin(\alpha) + \theta_2 \sin(\alpha)) d\\ ( i \theta_1 \sin(\alpha) - \theta_2 \sin(\alpha)) u + (\cos(\alpha) - i \theta_3 \sin(\alpha)) d \end{pmatrix}\\ M^{B'}_{B} \bar{\chi}^{A} &=& \begin{pmatrix} \cos(\alpha) + i \theta_3 \sin(\alpha) & i \theta_1 \sin(\alpha) + \theta_2 \sin(\alpha) \\ i \theta_1 \sin(\alpha) - \theta_2 \sin(\alpha) & \cos(\alpha) - i \theta_3 \sin(\alpha) \end{pmatrix} \begin{pmatrix} u \\ d \end{pmatrix} \\ &=& \begin{pmatrix} (\cos(\alpha) + i \theta_3 \sin(\alpha)) u + (i \theta_1 \sin(\alpha) + \theta_2 \sin(\alpha)) d\\ ( i \theta_1 \sin(\alpha) - \theta_2 \sin(\alpha)) u + (\cos(\alpha) - i \theta_3 \sin(\alpha)) d \end{pmatrix} \end{eqnarray} \begin{eqnarray} \begin{pmatrix} \cos(\alpha) + i \theta_3 \sin(\alpha) & i \theta_1 \sin(\alpha) + \theta_2 \sin(\alpha) \\ i \theta_1 \sin(\alpha) - \theta_2 \sin(\alpha) & \cos(\alpha) - i \theta_3 \sin(\alpha) \end{pmatrix} M^{A'}_{A} (M^{B'}_{B})^* \psi^{[A} \bar{\chi}^{\dot{B}]} \end{eqnarray} \begin{eqnarray} M^{A'}_{A} (M^{B'}_{B})^* \psi^{(A} \bar{\chi}^{\dot{B})} &=& (I^{A'}_A \cos(\alpha) + i \theta_i (\sigma^i)^{A'}_A \sin(\alpha)) (I^{A'}_A \cos(\alpha) + i \theta_i (\sigma^i)^{A'*}_A \sin(\alpha)) \psi^{(A} \bar{\chi}^{\dot{B})} \\ &=& (I^{A'}_A \cos(\alpha) + i \theta_i (\sigma^i)^{A'}_A \sin(\alpha)) (I^{A'}_A \cos(\alpha) + i \theta_i (\sigma^i)^{A'*}_A \sin(\alpha)) \psi^{(A} \bar{\chi}^{\dot{B})} \end{eqnarray}

If we wish for the irreducible representations, we can consider the Young tableau of our vector space.

### Young tableaux

Last updated : 2020-02-19 14:33:59