Gauge boson layout in $\rm SU(5)$ GUT unification

Question

It is known that the theory of the Great Unification $\rm SU(5)$ GUT encapsulates the gauge groups of the standard model to one large group $\rm SU(5)$. But is it possible to show schematically how to extract a massless photon and Z boson from it?

The problem i see is that in standard (SM) notation there are a few operators that create them. $A_μ = W_{11}sinθ_{w} + B_μcosθ_{w}$ and $Z_μ = W_{11}cosθ_{w} - B_μsinθ_{w}$ corresponding to 2x2 $\rm SU(2)$ block in SM. But it seems that $\rm SU(5)$ has a different structure, in the 2x2 $\rm SU(2)$ block the signs are incorrect and the constants near the fields $W$ and $B$ do not correspond to weak mixing angle.

So what is the correct way to look at all this?

$V_μ = \begin{pmatrix} G_1^1-\frac{2}{\sqrt{30}}B & G_2^1 & G_3^1 & \bar{X^1} & \bar{Y^1}\\ G_1^2 & G_2^2-\frac{2}{\sqrt{30}}B & G_3^2 & \bar{X^2} & \bar{Y^2}\\ G_1^3 & G_2^3 & G_3^3-\frac{2}{\sqrt{30}}B & \bar{X^3} & \bar{Y^3}\\ X^1 & X^2 & X^3 & \frac{1}{\sqrt{2}}W^3+\frac{3}{\sqrt{30}}B & W^{+}\\ Y^1 & Y^2 & Y^3 & W^{-} & -\frac{1}{\sqrt{2}}W^3+\frac{3}{\sqrt{30}}B\\ \end{pmatrix}$

Answer 1

I am rewriting this to avoid sign confusions, which I originally tried to coddle you with by switching the location of the v.e.v.. The mass matrix which you are trying to diagonalize arises out of the 55 entry of the square of your vector field matrix element surviving action on the 5 Higgs field v.e.v., $$ \left (\sqrt{\frac{3}{5}}B-W^3\right )^2 , $$ where I have taken out a common normalization of to be incorporated into your embedding normalizations and does not affect the mixing. Now, define $\tan^2\theta = 3/5$.

The above mass term then, up to normalization, may be re-expressed as $$ (B,W^3) \begin{pmatrix} \sin^2 \theta & -\sin \theta\cos \theta \\ -\sin \theta\cos \theta & \cos^2 \theta \end{pmatrix} \begin{pmatrix} B \\ W^3 \end{pmatrix}, $$ whose evident symmetric (zero-determinant) mass-matrix eigenvectors are $(\cos\theta,\sin\theta)^T$ and $(-\sin\theta,\cos\theta)^T$, respectively, orthogonal to each other, of course. The first one has vanishing eigenvalue, so zero mass.

The matrix then collapses to $$ \begin{pmatrix} \cos\theta& -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} \begin{pmatrix} \cos\theta& \sin\theta \\ - \sin\theta & \cos\theta \end{pmatrix}. $$

Mindful of irrelevant common factors to be incorporated in the couplings, the massless eigenvector combination amounts to $$ \cos\theta ~B +\sin\theta ~W^3 \mapsto \sqrt{5/8} ~B + \sqrt{3/8} ~W^3 , $$ and the massive one to $$ \mapsto - \sqrt{3/8} B + \sqrt{5/8} W^3, $$ identifiable with the γ and the Z, respectively, the routine expressions you wrote. I don't see any sign discrepancies!

The square of the sine of the mixing angle at the GUT scale is then $$ \sin^2\theta = {\tan^2\theta\over 1+\tan^2\theta}= 3/8 \approx 0.38, $$ to be compared to the physical 0.23.

In fact, this is sufficiently close (and dealt with renormalizing down from the GUT scale to the SM SSB scale) that it counted as an early encouraging success of the SU(5) model.