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 ${\mathbb{C}}$.

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 ${\mathbb{C}}\oplus {\mathbb{C}}$:

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

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

We can put all these ingredients together into a diagram

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:

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

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

It looks like we are on the right track.

We can do even better if we start with the splitting

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

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 ${\rm U}(1)$:

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:

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

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

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

where $\alpha$ stands for the 3ドル \times 3$ identity matrix times the complex number $\alpha \in {\rm U}(1)$, and similarly for $\beta$ in the 1ドル \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:

This gives our map:

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

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':

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:

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

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

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:

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

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:

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:

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



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

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,

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

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

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 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:

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:

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

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

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 $c= r, g, b$:

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

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

This table defines an isomorphism of Hilbert spaces

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:

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.

2010年01月11日