# Proton decay lifetime prediction SU(5)?

1. Jul 25, 2015

### Anchovy

I'm trying to understand how the SU(5) prediction for the proton decay lifetime of $\tau_{p} \sim10^{31}$ yr has been arrived at. I keep seeing it stated that $\tau_{p} \sim \frac{4\pi}{g_{5}^{2}} \frac{M_{X}^{4}}{M_{p}^{5}}$ where $g_{5}$ is the SU(5) coupling and $M_{X,p}$ are the X boson and proton masses. However, I'd like to understand where this expression for comes from and see the full equation. Anyone know?

2. Jul 25, 2015

### fzero

Let's try to work out at least part of the computation. The presence of $M_X$ suggests that this lifetime corresponds to a process mediated by the $X$ gauge boson. In another post you have the parameterization for the gauge bosons, can you dig out a parameterization for one generation of quarks and leptons. Next, we'll want to work out the vertices for the interactions between $X$ and the quarks and leptons, so we'll want to look at the relevant parts of the GUT Lagrangian.

3. Jul 25, 2015

### Anchovy

Are you talking about the so-called $\overline{\textbf{5}}$ and $\textbf{10}$ representations here?

4. Jul 25, 2015

### fzero

Yes. The rules for how those representations decompose under the SM gauge group tell you how to fit the quarks and leptons into them.

5. Jul 25, 2015

### Anchovy

OK so we have
$$\overline{\textbf{5}} = \begin{pmatrix} d_{R} \\ d_{G} \\ d_{B} \\ e^{+}\\ \overline{\nu}_{e} \end{pmatrix}_{R}, \hspace{1 cm} \textbf{10} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & \overline{u}_{B} & -\overline{u}_{G} & u_{R} & d_{R} \\ -\overline{u}_{B} & 0 & \overline{u}_{R} & u_{G} & d_{G} \\ \overline{u}_{G} & -\overline{u}_{R} & 0 & u_{B} & d_{B} \\ -u_{R} & -u_{G} & -u_{B} & 0 & \overline{e} \\ -d_{R} & -d_{G} & -d_{B}& -\overline{e} & 0 \end{pmatrix}_{L}$$

6. Jul 25, 2015

### fzero

Right. The $\mathbf{10}$ is a bit complicated in color indices, but we can treat it in terms of the blocks. Can you work out how everything couples to $X$? It's probably best to organize in terms of expressions like $d^{\dagger a} X_{a\alpha} \ell^\alpha$, where $a,\ldots$ are color and $\alpha,\ldots$ are weak isospin.

Edit: Actually, the right-handed $d$ quarks should be in the $\bar{\mathbf{5}}$.

7. Jul 25, 2015

### Anchovy

At this point do I need
$$V_{\mu} = \begin{pmatrix} g_{11}-\frac{2B}{\sqrt{30}} & g_{12} & g_{13} & \overline{X}^{R} & \overline{Y}^{R} \\ g_{21} & g_{22}-\frac{2B}{\sqrt{30}} & g_{23}& \overline{X}^{G} & \overline{Y}^{G} \\ g_{31} & g_{32} & g_{33}-\frac{2B}{\sqrt{30}} & \overline{X}^{B} & \overline{Y}^{B} \\ X^{R} & X^{G} & X^{B} & \frac{W^{3}}{\sqrt{2}} + \frac{3B}{\sqrt{30}} & W^{+} \\ Y^{R} & Y^{G} & Y^{B}& W^{-} & -\frac{W^{3}}{\sqrt{2}} + \frac{3B}{\sqrt{30}} \end{pmatrix}$$ ?

8. Jul 25, 2015

### fzero

I realized after posting that the right-handed down quarks should be in the $\mathbf{5}$, so we have to be careful.

Yes, you need to know that the $X$ is on the off-diagonal there. I don't expect that you should have to explicitly compute the matrix products in components. You should be able to see where the blocks fit together and write expressions in terms of color and weak indices.

Also, I'm not going to try to get numerical factors right. So let's just make sure we keep track of the gauge coupling and other parameters, but not factors of 2 and $\pi$, at least on the first go.

9. Jul 25, 2015

### Anchovy

OK so am I now supposed to be finding the relevant bits of $\overline{\textbf{5}} V_{\mu} \textbf{10}$ ?

10. Jul 25, 2015

### fzero

Not quite. Let's use $\psi^i$ to stand for the $\bar{\mathbb{5}}$ and $\psi_{ij}$ for the $\mathbb{10}$. The kinetic terms are something like
$$i \bar{\psi}_i ({\not \! D})^i_j \psi^i + i \bar{\psi}^{ij} ({\not \! D})_j^k \psi_{ki},$$
so we don't get a direct vertex between the $\bar{\mathbb{5}}$ and $\mathbb{10}$ from these. The GUT gauge boson couples the representation to its conjugate.

11. Jul 25, 2015

### Anchovy

Ah ok, I'm having trouble finding out what ${\not \! D}$ should look like, am I right in thinking capital D is referring to covariant derivative $D_{\mu} = \partial_{\mu} - ig_{5}T^{\alpha}V_{\mu}^{\alpha}$ ? So... ${\not \! D} = {\not \! \partial_{\mu}} - ig_{5}T^{\alpha}V_{\mu}^{\alpha}$ ?

12. Jul 25, 2015

### fzero

Yes, but I'm using lowercase for $\mathbf{5}$ indices and uppercase for $\bar{\mathbf{5}}$ ones. The adjoint representation appears in $\mathbf{5} \otimes \bar{\mathbf{5}}$, so we can write an adjoint matrix with one index up and one down. So I would write
$$(D_{\mu})^i_j = \delta^i_j \partial_{\mu} - ig_{5} (V_{\mu})^i_j.$$

13. Jul 25, 2015

### Anchovy

OK, so,
$$-ig_{5} (V_{\mu})^i_j \psi^{i} = -ig_{5} \begin{pmatrix} g_{11}-\frac{2B}{\sqrt{30}} & g_{12} & g_{13} & \overline{X}^{R} & \overline{Y}^{R} \\ g_{21} & g_{22}-\frac{2B}{\sqrt{30}} & g_{23}& \overline{X}^{G} & \overline{Y}^{G} \\ g_{31} & g_{32} & g_{33}-\frac{2B}{\sqrt{30}} & \overline{X}^{B} & \overline{Y}^{B} \\ X^{R} & X^{G} & X^{B} & \frac{W^{3}}{\sqrt{2}} + \frac{3B}{\sqrt{30}} & W^{+} \\ Y^{R} & Y^{G} & Y^{B}& W^{-} & -\frac{W^{3}}{\sqrt{2}} + \frac{3B}{\sqrt{30}} \end{pmatrix} \begin{pmatrix} d_{R} \\ d_{G} \\ d_{B} \\ e^{+}\\ \overline{\nu}_{e} \end{pmatrix}_{R}$$

$$= -ig_{5} \begin{pmatrix} ... + \overline{X}^{R}e^{+} + \overline{Y}^{R}\nu_{e}\\ ... + \overline{X}^{G}e^{+} + \overline{Y}^{G}\nu_{e} \\ ... + \overline{X}^{B}e^{+} + \overline{Y}^{B}\nu_{e}\\ X^{R}d_{R} + X^{G}d_{G} + X^{B}d_{B} + ...\\ Y^{R}d_{R} + Y^{G}d_{G} + Y^{B}d_{B} + ...\end{pmatrix}_{R}$$

and

$$\delta^i_j \partial_{\mu} \psi^{i} = \partial_{\mu} \psi^{j}$$
(not sure what happens next with that bit...)

Anyway then we have
$$i\overline{\psi}_{i}(-ig_{5})(V_{\mu})^i_j \psi^{i} = g_{5} \overline{\psi}_{i} (V_{\mu})^i_j \psi^{i} = g_{5} (\overline{d}_{R}, \overline{d}_{G}, \overline{d}_{B}, e^{-}, \nu_{e})_{L} \begin{pmatrix} ... + \overline{X}^{R}e^{+} + \overline{Y}^{R}\nu_{e}\\ ... + \overline{X}^{G}e^{+} + \overline{Y}^{G}\nu_{e} \\ ... + \overline{X}^{B}e^{+} + \overline{Y}^{B}\nu_{e}\\ X^{R}d_{R} + X^{G}d_{G} + X^{B}d_{B} + ...\\ Y^{R}d_{R} + Y^{G}d_{G} + Y^{B}d_{B} + ...\end{pmatrix}_{R} \\ = g_{5} (... + \overline{d}_{R_{L}} \overline{X}^{R}e^{+}_{R} + \overline{d}_{R_{L}}\overline{Y}^{R}\nu_{e_{R}} \\ + ... + \overline{d}_{G_{L}} \overline{X}^{G}e^{+}_{R} + \overline{d}_{G_{L}}\overline{Y}^{G}\nu_{e_{R}} \\ + ... + \overline{d}_{B_{L}} \overline{X}^{B}e^{+}_{R} + \overline{d}_{B_{L}}\overline{Y}^{B}\nu_{e_{R}} \\ + e^{-}_{L}X^{R}d_{R_{R}} + e^{-}_{L}X^{G}d_{G_{R}} + e^{-}_{L}X^{B}d_{B_{R}} + ... \\ + \nu_{e_{L}}Y^{R}d_{R_{R}} + \nu_{e_{L}}Y^{G}d_{G_{R}} + \nu_{e_{L}}Y^{B}d_{B_{R}} + ...)$$

This is taking me ages, before I go any further am I OK up to here? Do I need color indices seeing as the 'right handed' is clashing with the 'red'? And have I correctly changed $\psi \rightarrow \overline{\psi}$ by switching particles for antiparticles and right handed for left?

14. Jul 25, 2015

### fzero

Yes, I didn't want to to get hung up writing all of the components out explicitly. But you are seeing that the $X$ pairs the upper block of the matter fields with the lower block. So maybe we can use a more typical notation here where we use $q_{a\alpha}$ to stand for the left-handed quark doublets, $u_R^a,d_R^a$ for the RH quark singlets, $\ell_\alpha$ for the lepton doublet and $e_R$ for the RH singlet. Then we have something like
$$\begin{split} & \psi^i = \begin{pmatrix} d_R^a \\ \ell_\alpha\end{pmatrix}, \\ & \psi_{ij} = \begin{pmatrix} f_{abc} u_R^c & q_{a\alpha} \\ -q_{a\alpha} & \epsilon_{\alpha\beta} e_R \end{pmatrix}. \end{split}$$
From the parameterization of the gauge boson, it looks like $X$ is defined with the color index down $X_{a\alpha}$, but the complex conjugate would be $\bar{X}^a_\alpha$.

Edit: Actually $X = V_{a1}$, while $Y=V_{a2}$.

Last edited: Jul 25, 2015
15. Jul 29, 2015

### Anchovy

16. Jul 30, 2015

### Anchovy

OK I have attached what I've got for $$i \bar{\psi}_i ({\not \! D})^i_j \psi^i + i \bar{\psi}^{ij} ({\not \! D})_j^k \psi_{ki}$$ but I'm not well practiced with this stuff and the last bit is a step or two away from being complete - I got confused on the last line I've shown here. Also I wasn't sure if the way I've correctly written out $\overline{\psi}_{i}, \overline{\psi}_{ij}$ given $\psi_{i}, \psi_{ij}$ (I've shown what I did for each in the top right corner)?

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17. Jul 30, 2015

### fzero

OK, I realize I had a mistake where $f_{abc}$ should be $\epsilon_{abc}$ in $\psi_{ij}$, so
$$\begin{split} & \psi^i = \begin{pmatrix} d_R^a \\ \ell_\alpha\end{pmatrix}, \\ & \psi_{ij} = \begin{pmatrix} \epsilon_{abc} u_R^c & q_{a\alpha} \\ -q_{a\alpha} & \epsilon_{\alpha\beta} e_R \end{pmatrix}. \end{split}$$
You are on the right track and will get there eventually, but let me try to quickly get the terms that give the $X$ and $Y$ couplings. These are
$$i \bar{\psi}_i ({\not \! D})^i_j \psi^i + i \bar{\psi}^{ij} ({\not \! D})_j^k \psi_{ki}\rightarrow i \bar{\psi}_\alpha ({\not \! V})^\alpha_a \psi^a + i \bar{\psi}^{ia} ({\not \! V})_a^\alpha \psi_{\alpha i} + \text{h.c.}$$
The first term gives
$$i \bar{\psi}_\alpha ({\not \! V})^\alpha_a \psi^a = i \bar{\ell}_\alpha ({\not \! V})^\alpha_a d^a_R,~~~(1)$$
while the 2nd gives
$$\begin{split} i \bar{\psi}^{ia} ({\not \! V})_a^\alpha \psi_{\alpha i} &= i \bar{\psi}^{ba} ({\not \! V})_a^\alpha \psi_{\alpha b} + i \bar{\psi}^{\beta a} ({\not \! V})_a^\alpha \psi_{\alpha \beta} \\ &= i \epsilon^{bac} \bar{u}_{Rc} ({\not \! V})_a^\alpha (-q_{\alpha b}) + i \bar{q}^{\beta a} ({\not \! V})_a^\alpha \epsilon_{\alpha \beta} e_R.~~~(2)\end{split}$$

Since $X_a = V_a^1$, the terms with $X$ are
$$i \bar{e}_L ({\not \! X})_a d^a_R + i \epsilon^{bac} \bar{u}_{Rc} ({\not \! X})_a(-u_{L b}) + i \bar{d}_L^a ({\not \! X})_a e_R +\text{h.c.}.$$

See if this makes sense to you. I want to check a couple of things to make sure the 2nd term makes physical sense.

Last edited: Jul 30, 2015
18. Jul 30, 2015

### Anchovy

First of all for (1) I'm not sure why you got $i \bar{\ell}_\alpha ({\not \! V})^\alpha_a d^a_R$ (ie. quark and lepton) when I had terms that only had quark or only had lepton?

Also I keep seeing fields with that superscript c all over the place ($u^{c}_{R}, e^{c}_{R}$, the attached diagram etc.), what does that indicate? Charge conjugation?

And for (2), on the first line I don't understand what happened there - the first term on the right hand side is the same as what appears on the left hand side, just with $b$ replacing $i$, what's the difference caused by simply changing the label? Clearly there is one because a similar-looking term appears on the right hand side too, this time with $\beta$ replacing $i$. I can see that $b, \beta$ are both indices on the $\epsilon$'s but I don't understand what happened there.

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19. Jul 30, 2015

### fzero

I'm sorry, I didn't notice that. Looking back at your computation at the last line of (i), You have $V \psi$ and expand to have $X$ in the upper block. But $X$ is an off-diagonal block of $V$, so it should appear in the lower SU(2) block in that part of the computation

If you look at certain references, a $c$ index is used to indicate the right-handed fields. I am using $a,b,c$ for SU(3) color indices, so I used $R$ to label the RH fields.

Let me review my index conventions. $i,j,\ldots = 1,2,3,4,5$ are SU(5) indices. $a,b,\ldots=1,2,3$ are SU(3) color and $\alpha,\beta=1,2$ are SU(2) weak isospin labels. The way we're parameterizing the upper blocks as color, $i=1,2,3$ are the same as $a=1,2,3$, while $i=4,5$ are identified with $\alpha=1,2$. I think this convention is less confusing than making $\alpha,\beta =4,5$, but it doesn't take some time to get comfortable with.

When I wrote (1), I used the fact that $X,Y$ are the off-diagonal blocks of the SU(5) gauge field so they have one $a$ color index and one $\alpha$ weak index. Then we see that the upper block on $\psi_i$ gets paired by $X$ with the lower block on the conjugate.

I am replacing a sum on $i$ with a pair of sums: one over the color part $a$ and a second over the weak part $\alpha$. So $A^i B_i = A^a B_a + A^\alpha B_\alpha$ with my conventions.

20. Jul 31, 2015

### Anchovy

OK I understand your (1) now and also the first line of (2) but I can't quite work out how to get the second line of (2) ?

Looking at the first term of the second line of (2) I'm not sure why $\overline{\psi}^{ba}$ has been replaced with $\epsilon_{bac}\overline{u}_{Rc}$ and $\overline{\psi}_{\alpha b}$ has been replaced with $-q_{a\alpha}$

Last edited: Jul 31, 2015