Prof. Dr. Stefan Floerchinger

Researcher and professor for quantum field theory.

Grand unification

SU\((5)\) unification

We now discuss a proposed extension of the Standard Model which leads to a unification of the gauge groups into SU\((5)\). This has been proposed by Howard Georgi and Sheldon Glashow in 1974.

Note that the SU\((3)\) and SU\((2)\) generators naturally fit into SU\((5)\) generators and similar for the spinors \[\begin{pmatrix} \Bigg ( 3 \times 3 \Bigg )_{\text{SU}(3)} & \\ & \bigg ( 2 \times 2 \bigg )_{\text{SU}(2)} \end{pmatrix} \begin{pmatrix} \psi^1 \\ \psi^2 \\ \psi^3 \\ \psi^4 \\ \psi^5 \end{pmatrix} .\] There are \(5^2 - 1 = 24\) generators of SU\((5)\) corresponding to the hermitian traceless \(5 \times 5\) matrices. Out of them, eight generate SU\((3)\), while three generate SU\((2)\).

Moreover, within SU\((5)\) there is one hermitian traceless matrix \[\frac{1}{2} Y = \begin{pmatrix} - \frac{1}{3} & & & & \\ & - \frac{1}{3} & & & \\ & & - \frac{1}{3} & & \\ & & & \frac{1}{2} & \\ & & & & \frac{1}{2} \end{pmatrix}.\] That generates a U\((1)\) subgroup which actually gives U\((1)_Y\). The remaining generators correspond to additional gauge bosons not present in the Standard Model so they are supposedly very heavy or confined. We find the embedding \[\text{SU}(5) \to \text{SU}(3) \otimes \text{SU}(2) \otimes \text{U}(1).\]

Fundamental representation \(\mathbf{5}\)

Now let us consider representations. Take the fundamental representation of SU\((5)\) the spinor \(\psi^m\). From the above illustration one sees that it decomposes like \[\mathbf{5} = \bigg ( \mathbf{3} , \ \mathbf{1} , \ - \frac{1}{3} \bigg ) \oplus \bigg ( \mathbf{1}, \ \mathbf{2}, \ \frac{1}{2} \bigg ),\] in a natural way. The conjugate decomposes \[\mathbf{5^*} = \bigg ( \mathbf{3^*} , \ \mathbf{1} , \ \frac{1}{3} \bigg ) \oplus \bigg ( \mathbf{1} , \ \mathbf{2} , \ - \frac{1}{2} \bigg ).\] Indeed these could be the representations for the right-handed down quark and the anti-lepton doublet, \[d_R, \quad\quad\quad \begin{pmatrix} \bar{\nu}_L & \bar{e}_L \end{pmatrix},\] and their anti-particles \[\bar{d}_R, \quad\quad\quad \begin{pmatrix} \nu_L \\ e_L \end{pmatrix},\] respectively.

Note that the hypercharges indeed conspire to be such that the generator \(\frac{1}{2}Y\) is indeed traceless. It can therefore be one of the generators of SU\((5)\).

Antisymmetric tensor representation \(\mathbf{10}\)

So what about the other representations? The next smallest representation is the anti-symmetric tensor \(\psi^{mn}\) with dimension ten. We still need \[\bigg ( \mathbf{3} , \ \mathbf{2} , \ \frac{1}{6} \bigg ), \quad\quad\quad \bigg ( \mathbf{3^*} , \ \mathbf{1} , \ - \frac{2}{3} \bigg ) , \quad\quad\quad \bigg ( \mathbf{1} , \ \mathbf{1} , \ 1 \bigg ),\] and the corresponding anti-fields. These are ten fields indeed. Now \(\psi^{mn}\) decomposes into irreducible representations according to \[\begin{split} \bigg ( \mathbf{3} , \ \mathbf{1} , \ - \frac{1}{3} \bigg ) \otimes_A \bigg ( \mathbf{3} , \ \mathbf{1} , \ - \frac{1}{3} \bigg ) &= \bigg ( \mathbf{3^*} , \ \mathbf{1} , \ - \frac{2}{3} \bigg ), \\ \bigg ( \mathbf{3} , \ \mathbf{1} , \ - \frac{1}{3} \bigg ) \otimes_A \bigg ( \mathbf{1} , \ \mathbf{2} , \ \frac{1}{2} \bigg ) &= \bigg ( \mathbf{3} , \ \mathbf{2} , \ \frac{1}{6} \bigg ), \\ \bigg ( \mathbf{1} , \ \mathbf{2} , \ \frac{1}{2} \bigg ) \otimes_A \bigg ( \mathbf{1} , \ \mathbf{2} , \ \frac{1}{2} \bigg ) &= \bigg ( \mathbf{1} , \ \mathbf{1} , \ 1 \bigg ) . \end{split}\] This matches indeed to \(\bar{u}_R\), the left-handed quark doublet \((u_L, d_L)\) and \(\bar e_R\), respectively.

Note that we have used here tensor product decomposition relations discussed before such as for SU\((3)\) \[\mathbf{3} \otimes \mathbf{3} = \mathbf{3}^*_A \oplus \mathbf{6}_S,\] or for SU\((2)\) \[\mathbf{2} \otimes \mathbf{2} = \mathbf{1}_A \oplus \mathbf{3}_S.\] The U\((1)\) charges are simply added. Indeed things work out! Also in this sector one finds that the hypercharges add up to zero, \(3\times 1\times (- \frac{2}{3})+3\times 2 \times \frac{1}{6} +1\times 1 \times1 =0\).

All fermions

The fermion fields of a single generation in the Standard Model can be organised into the \(SU(5)\) representations \[\mathbf{5}^*: \quad \bar{d}_R , \ \begin{pmatrix} \nu_L \\ e_L \end{pmatrix},\] and \[\mathbf{10}: \quad \bar{u}_R , \ \bar{e}_R , \ \begin{pmatrix} u_L \\ d_L \end{pmatrix},\] as well as the corresponding anti-fields. There is no space here for a right handed neutrino, it would have to be a singlet \(\mathbf{1}\) under SU\((5)\).

The scalar Higgs field could be part of a \(\mathbf{5}\) scalar representation but the corresponding field with quantum numbers \[\bigg ( \mathbf{3} , \ \mathbf{1} , \ - \frac{2}{3} \bigg )\] is not present in the Standard Model and must be very heavy or otherwise suppressed.

Gauge bosons

The gauge bosons of SU\((5)\) can be found from decomposing \(\mathbf{5} \otimes \mathbf{5}^* = \mathbf{24} + \mathbf{1}\). In terms of SU\((3)\otimes \text{SU}(2)\otimes \text{U}(1)\) the \(\mathbf{24}\) decomposes into \[\mathbf{24} = \bigg ( \mathbf{1}, \ \mathbf{3} , \ 0 \bigg ) \oplus \bigg ( \mathbf{8}, \ \mathbf{1} , \ 0 \bigg ) \oplus \bigg ( \mathbf{1}, \ \mathbf{1} , \ 0 \bigg ) \oplus \bigg ( \mathbf{3}, \ \mathbf{2} , \ \frac{2}{3} \bigg ) \oplus \bigg ( \mathbf{3}^*, \ \mathbf{2} , \ -\frac{2}{3} \bigg ).\] We recognize the \(W\) boson triplet, the gluons, the hypercharge photon and two more gauge bosons that transform under both SU\((3)_\text{color}\) and the electroweak group SU\((2)\times \text{U}(1)\).

Proton decay?

The latter type of gauge bosons could in principle induce transitions of the type \[\begin{aligned} d &\to e^+, \\ u &\to \bar{u}, \end{aligned}\] and thus \(u + d \to \bar{u} + e^+\) causing \[\begin{aligned} uud &\to u \bar{u} + e^+, \\ p &\to \pi^0 + e^+ . \end{aligned}\] The proton could therefore decay! This is actually one of the main experimental signatures for such grand unified theories.

Proton decay has not been observed so the transition rate must be very small. This also implies that the unification scale where the three forces SU\((3)_\text{color}\), SU\((2)\) and U\((1)_Y\) unite, must be very high. The latest experimental constraint is that the proton half-life time must be at least \(1.6 \times 10^{34}\) years [Super-Kamiokande, PRD 95, 012004 (2017)]. If the decay rate goes like \[\Gamma \approx \frac{m_p^5}{M_\text{GUT}^4},\] one can estimate for the unification scale \(M_\text{GUT} > 10^{16}\) GeV.

The Georgi-Glashow model we discussed so far is not very realistic, in some sense it is already ruled out. For example it predicts massless neutrinos, which is in conflict with the observation of neutrino oscillations. Also the unification of renormalization group trajectories to a single SU\((5)\) coupling constant at the scale \(M_\text{GUT}\) does not seem to work as it should.

Charge quantization

Besides the nice matching of the representations, there is another theoretical reason that speaks for a unified gauge theory. In the standard model it is not explained by electric charge to be a multiple of the electron charge (with fractional charges 1/3 for the quarks). In the SU\((5)\) model the U\((1)_Y\) generator is part of SU\((5)\) and it is naturally explained why the charges have the values they have.

SO\((10)\) unification

There are further possibilities to construct unified theories. The Pati-Salam model for example has the gauge group \(\text{SU}(4)\times \text{SU}(2)_L \times \text{SU}(2)_R\).

Both the Georgi-Glashow and the Pati-Salam model can be further unified and embedded into the group SO\((10)\). The unified gauge theory based on SO\((10)\) is particularly elegant, but we will not discuss it in detail here. Let us just mention that there is a spinorial representation (similar to the left-handed or right-handed spinor representations of SO\((1,3)\) for chiral fermions) \(\mathbf{16}\), that decomposes in terms of SU\((5)\) representations as \[\mathbf{16} = \mathbf{10} \oplus \mathbf{5}^* \oplus \mathbf{1}.\] This contains all the representations we need for the Georgi-Glashow model and therefore the standard model fermions of one generation as well as one additional fermion that has the quantum numbers of the right-handed neutrino! The latter is anyway needed for the seesaw mechanism to give mass to the observable neutrinos. For this mechanism to work, the right-handed neutrino is supposed to be very heavy. On the other side, in the SO\((10)\) model, it is part of the \(\mathbf{16}\) representation together with all the other fermions, so it is supposed to be massless. There must be some mechanism that breaks SO\((10)\) at some high energy or mass scale and this mechanism needs to give the right-handed neutrino its mass. One can infer that the scale of SO\((10)\) breaking shows up (albeit somewhat indirectly) through the seesaw mechanism in the observable neutrino masses and more directly in the right-handed neutrino mass, which is also a candidate for dark matter.

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