3.3 The Pati-Salam Model

Next we turn to a unified theory that is not so `grand': its gauge group is not a simple Lie group, as it was for the ${\rm SU}(5)$ and ${\rm Spin}(10)$ theories. This theory is called the Pati-Salam model, after its inventors [28]; it has gauge group ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$, which is merely semisimple.

We might imagine the ${\rm SU}(5)$ theory as an answer to this question:

Why are the hypercharges in the Standard Model what they are?

The answer it provides is something like this:

Because ${\rm SU}(5)$ is the actual gauge group of the world, acting on the representation $\Lambda {\mathbb{C}}^5$.

But there are other intriguing patterns in the Standard Model that ${\rm SU}(5)$ does not explain--and these lead us in different directions.

First, there is a strange similarity between quarks and leptons. Each generation of fermions in the Standard Model has two quarks and two leptons. For example, in the first generation we have the quarks $u$ and $d$, and the leptons $\nu$ and $e^-$. The quarks come in three `colors': this is a picturesque way of saying that they transform in the fundamental representation of ${\rm SU}(3)$ on ${\mathbb{C}}^3$. The leptons, on the other hand, are `white': they transform in the trivial representation of ${\rm SU}(3)$ on ${\mathbb{C}}$.

Representations of ${\rm SU}(3)$
Particle	Representation
Quark	${\mathbb{C}}^3$
Lepton	${\mathbb{C}}$

Could the lepton secretly be a fourth color of quark? Maybe it could in a theory where the ${\rm SU}(3)$ color symmetry of the Standard Model is extended to ${\rm SU}(4)$. Of course this larger symmetry would need to be broken to explain the very real difference between leptons and quarks.

Second, there is a strange difference between left- and right-handed fermions. The left-handed ones participate in the weak force governed by ${\rm SU}(2)$, while the right-handed ones do not. Mathematically speaking, the left-handed ones live in a nontrivial representation of ${\rm SU}(2)$, while the right-handed ones live in a trivial one. The nontrivial one is ${\mathbb{C}}^2$, while the trivial one is ${\mathbb{C}}\oplus {\mathbb{C}}$:

Representations of ${\rm SU}(2)$
Particle	Representation
Left-handed fermion	${\mathbb{C}}^2$
Right-handed fermion	${\mathbb{C}}\oplus {\mathbb{C}}$

But there is a suspicious similarity between ${\mathbb{C}}^2$ and ${\mathbb{C}}\oplus {\mathbb{C}}$. Could there be another copy of ${\rm SU}(2)$ that acts on the right-handed particles? Again, this `right-handed' ${\rm SU}(2)$ would need to be broken, to explain why we do not see a `right-handed' version of the weak force that acts on right-handed particles.

Following Pati and Salam, let us try to sculpt a theory that makes these ideas precise. In the last two sections, we saw some of the ingredients we need to make a grand unified theory: we need to extend the symmetry group ${G_{\mbox{\rm SM}}}$ to a larger group $G$ using an inclusion

$${G_{\mbox{\rm SM}}}\hookrightarrow G$$

(up to some discrete kernel), and we need a representation $V$ of $G$ which reduces to the Standard Model representation when restricted to ${G_{\mbox{\rm SM}}}$:

$$F \oplus F^* \cong V.$$

We can put all these ingredients together into a diagram

$$\xymatrix{ {G_{\mbox{\rm SM}}}\ar[r] \ar[d] & G \ar[d] \\ {\rm U}(F \oplus F^*) \ar[r]^-\sim & {\rm U}(V) \\ }$$

which commutes only when our $G$ theory works out.

We now use the same methods to chip away at our current challenge. We asked if leptons correspond to a fourth color. We already know that every quark comes in three colors, $r$, $g$, and $b$, which form a basis for the vector space ${\mathbb{C}}^3$. This is the fundamental representation of ${\rm SU}(3)$, the color symmetry group of the Standard Model. If leptons correspond to a fourth color, say `white', then we should use the colors $r$, $g$, $b$ and $w$, as a basis for the vector space ${\mathbb{C}}^4$. This is the fundamental representation of ${\rm SU}(4)$, so let us take that group to describe color symmetries in our new GUT.

Now ${\rm SU}(3)$ has an obvious inclusion into ${\rm SU}(4)$, using block diagonal matrices:

$$g \mapsto \left( \begin{array}{cc} g & 0 \\ 0 & 1 \end{array}\right)$$

When restricted to this subgroup, the fundamental representation ${\mathbb{C}}^4$ breaks into a direct sum of irreps:

$${\mathbb{C}}^4 \cong {\mathbb{C}}^3 \oplus {\mathbb{C}}.$$

These are precisely the irreps of ${\rm SU}(3)$ that describe quarks and leptons. For antiquarks and antileptons we can use

$${\mathbb{C}}^{4*} \cong {\mathbb{C}}^{3*} \oplus {\mathbb{C}}.$$

It looks like we are on the right track.

We can do even better if we start with the splitting

$${\mathbb{C}}^4 \cong {\mathbb{C}}^3 \oplus {\mathbb{C}}.$$

Remember that when we studied ${\rm SU}(5)$, the splitting

$${\mathbb{C}}^5 \cong {\mathbb{C}}^2 \oplus {\mathbb{C}}^3$$

had the remarkable effect of introducing ${\rm U}(1)$, and thus hypercharge, into ${\rm SU}(5)$ theory. This was because the subgroup of ${\rm SU}(5)$ that preserves this splitting is larger than ${\rm SU}(2) \times {\rm SU}(3)$, roughly by a factor of ${\rm U}(1)$:

$$({\rm U}(1) \times {\rm SU}(2) \times {\rm SU}(3)) / {\mathbb{Z}}_6 \cong \S ({\rm U}(2) \times {\rm U}(3))$$

It was this factor of ${\rm U}(1)$ that made ${\rm SU}(5)$ theory so fruitful.

So, if we choose a splitting ${\mathbb{C}}^4 \cong {\mathbb{C}}^3 \oplus {\mathbb{C}},$ we should again look at the subgroup that preserves this splitting. Namely:

$$\S ({\rm U}(3) \times {\rm U}(1)) \subseteq {\rm SU}(4) .$$

Just as in the ${\rm SU}(5)$ case, this group is bigger than ${\rm SU}(3) \times {\rm SU}(1)$, roughly by a factor of ${\rm U}(1)$. And again, this factor of ${\rm U}(1)$ is related to hypercharge!

This works very much as it did for ${\rm SU}(5)$. We want a map

$${\rm U}(1) \times {\rm SU}(3) \to {\rm SU}(4)$$

and we already have one that works for the ${\rm SU}(3)$ part:

$$\begin{array}{ccc} {\rm SU}(3) &\to & {\rm SU}(4) \\ h &\... ...gin{array}{cc} h & 0 \\ 0 & 1 \end{array}\right) \end{array}$$

So, we just need to include a factor of ${\rm U}(1)$ that commutes with everything in the ${\rm SU}(3)$ subgroup. Elements of ${\rm SU}(4)$ that do this are of the form

$$\left( \begin{array}{c c} \alpha & 0 \\ 0 & \beta \end{array}\right)$$

where $\alpha$ stands for the $3 \times 3$ identity matrix times the complex number $\alpha \in {\rm U}(1)$, and similarly for $\beta$ in the $1 \times 1$ block. For the above matrix to lie in ${\rm SU}(4)$, it must have determinant 1, so $\alpha^3 \beta = 1$. Thus we must include ${\rm U}(1)$ using matrices of this form:

$$\left( \begin{array}{c c} \alpha & 0 \\ 0 & \alpha^{-3} \end{array}\right).$$

This gives our map:

$$\begin{array}{ccc} {\rm U}(1) \times {\rm SU}(3) &\to& {\rm... ...lpha h & 0 \\ 0 & \alpha^{-3} \end{array}\right) . \end{array}$$

If we let ${\rm U}(1) \times {\rm SU}(3)$ act on ${\mathbb{C}}^4 \cong {\mathbb{C}}^3 \oplus {\mathbb{C}}$ via this map, the `quark part' ${\mathbb{C}}^3$ transforms as though it has hypercharge $\frac{1}{3}$: that is, it gets multiplied by a factor of $\alpha$. Meanwhile, the `lepton part' ${\mathbb{C}}$ transforms as though it has hypercharge $-1$, getting multiplied by a factor of $\alpha^{-3}$. So, as a representation of ${\rm U}(1) \times {\rm SU}(3)$, we have

$${\mathbb{C}}^4 \quad \cong \quad {\mathbb{C}}_{\frac{1}{3}} \... ...}}^3 \quad \oplus \quad {\mathbb{C}}_{-1} \otimes {\mathbb{C}}.$$

A peek at Table 1 reveals something nice. This exactly how the left-handed quarks and leptons in the Standard Model transform under ${\rm U}(1) \times {\rm SU}(3)$!

The right-handed leptons do not work this way. That is a problem we need to address. But this brings us to our second question, which was about the strange difference between left- and right-handed particles.

Remember that in the Standard Model, the left-handed particles live in the fundamental rep of ${\rm SU}(2)$ on ${\mathbb{C}}^2$, while the right-handed ones live in the trivial rep on ${\mathbb{C}}\oplus {\mathbb{C}}$. Physicists write this by grouping left-handed particles into `doublets', while leaving the right-handed particles as `singlets':

$$\left( \! \begin{array}{c} \nu_L \\ e^-_L \end{array} \! \right)\quad \quad \begin{array}{c} \nu_R \\ e^-_R \end{array}.$$

But there is a suspicious similarity between ${\mathbb{C}}^2$ and ${\mathbb{C}}\oplus {\mathbb{C}}$. Could there be another copy of ${\rm SU}(2)$ that acts on the right-handed particles? Physically speaking, this would mean that the left- and right-handed particles both form doublets:

$$\left( \! \begin{array}{c} \nu_L \\ e^-_L \end{array} \! \rig... ...t( \! \begin{array}{c} \nu_R \\ e^-_R \end{array}\! \right)$$

but under the actions of different ${\rm SU}(2)$'s. Mathematically, this would amount to extending the representations of the `left-handed' ${\rm SU}(2)$:

$${\mathbb{C}}^2 \quad \quad {\mathbb{C}}\oplus {\mathbb{C}}$$

to representations of ${\rm SU}(2) \times {\rm SU}(2)$:

$${\mathbb{C}}^2 \otimes {\mathbb{C}}\quad \quad {\mathbb{C}}\otimes {\mathbb{C}}^2$$

where the first copy of ${\rm SU}(2)$ acts on the first factor in these tensor products, while the second copy acts on the second factor. The first copy of ${\rm SU}(2)$ is the `left-handed' one familiar from the Standard Model. The second copy is a new `right-handed' one.

If we restrict these representations to the `left-handed' ${\rm SU}(2)$ subgroup, we obtain:

$$\begin{eqnarray*} {\mathbb{C}}^2 \otimes {\mathbb{C}}& \cong & {\mathbb{C}}^2 \... ...otimes {\mathbb{C}}^2 & \cong & {\mathbb{C}}\oplus {\mathbb{C}}. \end{eqnarray*}$$

These are exactly the representations of ${\rm SU}(2)$ that appear in the Standard Model. It looks like we are on the right track!

Now let us try to combine these ideas into a theory with symmetry group ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$. We have seen that letting ${\rm SU}(4)$ act on ${\mathbb{C}}^4$ is a good way to unify our treatment of color for all the left-handed fermions. Similarly, the dual representation on ${\mathbb{C}}^{4*}$ is good for their antiparticles. So, we will tackle color by letting ${\rm SU}(4)$ act on the direct sum ${\mathbb{C}}^4 \oplus {\mathbb{C}}^{4*}$. This space is 8-dimensional. We have also seen that letting ${\rm SU}(2) \times {\rm SU}(2)$ act on ${\mathbb{C}}^2 \otimes {\mathbb{C}}\oplus {\mathbb{C}}\otimes {\mathbb{C}}^2$ is a good way to unify our treatment of isospin for left- and right-handed fermions. This space is 4-dimensional.

Since $8 \times 4 = 32$, and the Standard Model representation is 32-dimensional, let us take the tensor product

$$V = \big(({\mathbb{C}}^2 \otimes {\mathbb{C}}) \; \oplus \; (... ...es \; \big({\mathbb{C}}^4 \; \oplus \; {\mathbb{C}}^{4*}\big).$$

This becomes a representation of ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$, which we call the Pati-Salam representation. To obtain a theory that extends the Standard Model, we also need a way to map ${G_{\mbox{\rm SM}}}$ to ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$, such that pulling back $V$ to a representation of ${G_{\mbox{\rm SM}}}$ gives the Standard model representation.

How can we map ${G_{\mbox{\rm SM}}}$ to ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$? There are several possibilities. Our work so far suggests this option:

$$\begin{array}{ccl} {\rm U}(1) \times {\rm SU}(2) \times {\rm ... ... 0 \\ 0 & \alpha^{-3} \end{array}\right) \right) \end{array}$$

Let us see what this gives. The Pati-Salam representation of ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$ is a direct sum of four irreducibles:

$$V \quad \cong \quad {\mathbb{C}}^2 \otimes {\mathbb{C}}\otim... ... {\mathbb{C}}\otimes {\mathbb{C}}^2 \otimes {\mathbb{C}}^{4*}.$$

We hope the first two will describe left- and right-handed fermions, so let us give them names that suggest this:

$$F_L = {\mathbb{C}}^2 \otimes {\mathbb{C}}\otimes {\mathbb{C}}^4,$$

$$F_R = {\mathbb{C}}\otimes {\mathbb{C}}^2 \otimes {\mathbb{C}}^4 .$$

The other two are the duals of the first two, since the 2-dimensional irrep of ${\rm SU}(2)$ is its own dual:

$$F_L^* = {\mathbb{C}}^2 \otimes {\mathbb{C}}\otimes {\mathbb{C}}^{4*},$$

$$F_R^* = {\mathbb{C}}\otimes {\mathbb{C}}^2 \otimes {\mathbb{C}}^{4*} .$$

Given our chosen map from ${G_{\mbox{\rm SM}}}$ to ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$, we can work out which representations of the ${G_{\mbox{\rm SM}}}$ these four spaces give. For example, consider $F_L$. We have already seen that under our chosen map,

$${\mathbb{C}}^4 \quad \cong \quad {\mathbb{C}}_{\frac{1}{3}} \... ...C}}^3 \quad \oplus \quad {\mathbb{C}}_{-1} \otimes {\mathbb{C}}$$

as representations of ${\rm U}(1) \times {\rm SU}(3)$, while

$${\mathbb{C}}^2 \otimes {\mathbb{C}}\cong {\mathbb{C}}^2$$

as representations of the left-handed ${\rm SU}(2)$. So, as representations of ${G_{\mbox{\rm SM}}}$ we have

$$F_L \quad \cong \quad {\mathbb{C}}_{\frac{1}{3}} \otimes {\m... ... {\mathbb{C}}_{-1} \otimes {\mathbb{C}}^2 \otimes {\mathbb{C}}.$$

Table 1 shows that these indeed match the left-handed fermions.

If we go ahead and do the other four cases, we see that everything works except for the hypercharges of the right-handed particles--and their antiparticles. Here we just show results for the particles:

The Pati-Salam Model -- First Try
Particle	Hypercharge: predicted	Hypercharge: actual
$\left( \! \begin{array}{c} \nu_L \\ e^-_L \end{array} \! \right)$	$-1$	$-1$
$\left( \! \begin{array}{c} u_L \\ d_L \end{array} \! \right)$	$\frac{1}{3}$	$\frac{1}{3}$
$\nu_R$	$-1$	$0$
$e^-_R$	$-1$	$-2$
$u_R$	$\frac{1}{3}$	$\frac{4}{3}$
$d_R$	$\frac{1}{3}$	$-\frac{2}{3}$

The problem is that the right-handed particles are getting the same hypercharges as their left-handed brethren. To fix this problem, we need a more clever map from ${G_{\mbox{\rm SM}}}$ to ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$. This map must behave differently on the ${\rm U}(1)$ factor of ${G_{\mbox{\rm SM}}}$, so the hypercharges come out differently. And it must take advantage of the right-handed copy of ${\rm SU}(2)$, which acts nontrivially only on the right-handed particles. For example, we can try this map:

$$\begin{array}{ccc} {\rm U}(1) \times {\rm SU}(2) \times {\rm ... ... 0 \\ 0 & \alpha^{-3} \end{array}\right) \right) \end{array}$$

for any integer $k$. This will not affect the above table except for the hypercharges of right-handed particles. It will add $k/3$ to the hypercharges of the `up' particles in right-handed doublets ($\nu_R$ and $u_R$), and subtract $k/3$ from the `down' ones ($e^-_R$ and $d_R$). So, we obtain these results:

The Pati-Salam Model -- Second Try
Particle	Hypercharge: predicted	Hypercharge: actual
$\left( \! \begin{array}{c} \nu_L \\ e^-_L \end{array} \! \right)$	$-1$	$-1$
$\left( \! \begin{array}{c} u_L \\ d_L \end{array} \! \right)$	$\frac{1}{3}$	$\frac{1}{3}$
$\nu_R$	$-1 + \frac{k}{3}$	$0$
$e^-_R$	$-1 - \frac{k}{3}$	$-2$
$u_R$	$\frac{1}{3}+ \frac{k}{3}$	$\frac{4}{3}$
$d_R$	$\frac{1}{3}- \frac{k}{3}$	$-\frac{2}{3}$

Miraculously, all the hypercharges match if we choose $k = 3$. So, let us use this map:

$$\begin{array}{ccc} \beta \colon {\rm U}(1) \times {\rm SU}(2)... ...0 \\ 0 & \alpha^{-3} \end{array}\right) \right). \end{array}$$

When we take the Pati-Salam representation of ${\rm SU}(2) \times {\rm SU}(2) \times {\rm SU}(4)$ and pull it back along this map $\beta$, we obtain the Standard Model representation. As in Section 3.1, we use complete reducibility to see this, but we can be more concrete. We saw in Table 4 how we can specify the intertwining map precisely by using a specific basis, which for $\Lambda {\mathbb{C}}^5$ results in the binary code.

Similarly, we can create a kind of `Pati-Salam code' to specify an isomorphism of Hilbert spaces

$$\ell \colon F \oplus F^* \to \big(({\mathbb{C}}^2 \otimes {\m... ...es \; \big({\mathbb{C}}^4 \; \oplus \; {\mathbb{C}}^{4*}\big),$$

and doing this provides a nice summary of the ideas behind in the Pati-Salam model. We take the space ${\mathbb{C}}^2 \otimes {\mathbb{C}}$ to be spanned by $u_L$ and $d_L$, the left-isospin up and left-isospin down states. Similarly, the space ${\mathbb{C}}\otimes {\mathbb{C}}^2$ has basis $u_R$ and $d_R$, called right-isospin up and right-isospin down. Take care not to confuse these with the similarly named quarks. These have no color, and only correspond to isospin.

The color comes from ${\mathbb{C}}^4$ of course, which we already decreed to be spanned by $r$, $g$, $b$ and $w$. For antiparticles, we also require anticolors, which we take to do be the dual basis ${\overline{r}}$, ${\overline{g}}$, ${\overline{b}}$ and ${\overline{w}}$, spanning ${\mathbb{C}}^{4*}$.

It is now easy, with our knowledge of how the Pati-Salam model is to work, to construct this code. Naturally, the left-handed quark doublet corresponds to the left-isospin up and down states, which come in all three colors $c= r, g, b$:

$$u^c_L = u_L \otimes c \quad \quad d^c_L = d_L \otimes c.$$

The corresponding doublet of left-handed leptons is just the `white' version of this:

$$\nu_L = u_L \otimes w \quad \quad e^-_L = d_L \otimes w .$$

The right-handed fermions are the same, but with $R$'s instead of $L$'s. Thus we get the Pati-Salam code for the fermions:

$F_L$	$F_R$
$\nu_L = u_L \otimes w$	$\nu_R = u_R \otimes w$
$e^-_L = d_L \otimes w$	$e^-_R = d_R \otimes w$
$u^c_L = u_L \otimes c$	$u^c_R = u_R \otimes c$
$d^c_L = d_L \otimes c$	$d^c_R = d_R \otimes c$

The result is very similar for the antifermions in $F^*_L$ and $F^*_R$, but watch out: taking antiparticles swaps up and down, and also swaps left and right, so the particles in $F^*_L$ are right-handed, despite the subscript $L$, while those in $F^*_R$ are left-handed. This is because it is the right-handed antiparticles that feel the weak force, which in terms of representation theory means they are nontrivial under the left ${\rm SU}(2)$. So, the Pati-Salam code for the antifermions is this:

$F^*_L$	$F^*_R$
$e^+_R = u_L \otimes {\overline{w}}$	$e^+_L = u_R \otimes {\overline{w}}$
$\overline{\nu}_R = d_L \otimes {\overline{w}}$	$\overline{\nu}_L = d_R \otimes {\overline{w}}$
$\overline{d}^{\overline{c}}_R = u_L \otimes {\overline{c}}$	$\overline{d}^{\overline{c}}_L = u_R \otimes {\overline{c}}$
$\overline{u}^{\overline{c}}_R = d_L \otimes {\overline{c}}$	$\overline{u}^{\overline{c}}_L = d_R \otimes {\overline{c}}$

Putting these together we get the full Pati-Salam code:

Table 5: Pati-Salam code for first-generation fermions, where $c= r, g, b$ and ${\overline {c}}= {\overline {r}}, {\overline {b}}, {\overline {g}}$.

The Pati-Salam Code
$F_L$	$F_R$	$F^*_L$	$F^*_R$
$\nu_L = u_L \otimes w$	$\nu_R = u_R \otimes w$	$e^+_R = u_L \otimes {\overline{w}}$	$e^+_L = u_R \otimes {\overline{w}}$
$e^-_L = d_L \otimes w$	$e^-_R = d_R \otimes w$	$\overline{\nu}_R = d_L \otimes {\overline{w}}$	$\overline{\nu}_L = d_R \otimes {\overline{w}}$
$u^c_L = u_L \otimes c$	$u^c_R = u_R \otimes c$	$\overline{d}^{\overline{c}}_R = u_L \otimes {\overline{c}}$	$\overline{d}^{\overline{c}}_L = u_R \otimes {\overline{c}}$
$d^c_L = d_L \otimes c$	$d^c_R = d_R \otimes c$	$\overline{u}^{\overline{c}}_R = d_L \otimes {\overline{c}}$	$\overline{u}^{\overline{c}}_L = d_R \otimes {\overline{c}}$

This table defines an isomorphism of Hilbert spaces

$$\ell \colon F \oplus F^* \to \big(({\mathbb{C}}^2 \otimes {\m... ...mes \; \big({\mathbb{C}}^4 \; \oplus \; {\mathbb{C}}^{4*}\big)$$

so where it says, for example, $\nu_L = u_L \otimes w$, that is just short for $\ell(\nu_L) = u_L \otimes w$. This map $\ell$ is also an isomorphism between representations of ${G_{\mbox{\rm SM}}}$. It tells us how these representations are the `same', just as the map $h$ did for $F \oplus F^*$ and $\Lambda {\mathbb{C}}^5$ at the end of Section 3.1.

As with ${\rm SU}(5)$ and ${\rm Spin}(10)$, we can summarize all the results of this section in a commutative square:

Theorem 3   . The following square commutes:

$$\xymatrix{ {G_{\mbox{\rm SM}}}\ar[r]^-{\beta} \ar[d] & {\rm ... ...s \big({\mathbb C}^4 \oplus {\mathbb C}^{4*}\big)\right) \\ }$$

where the left vertical arrow is the Standard Model representation and the right one is the Pati-Salam representation.

The Pati-Salam representation and especially the homomorphism $\beta$ look less natural than the representation of ${\rm SU}(5)$ on $\Lambda {\mathbb{C}}^5$ and the homomorphism $\phi \colon {G_{\mbox{\rm SM}}}\to {\rm SU}(5)$. But appearances can be deceiving: in the next section we shall see a more elegant way to describe them.