Say we have a GUT e.g. $SU(5)$, if the Higgs field breaks this it can break it down to $SU(3)\times SU(2)\times U(1)$ if the Higgs field acquires values proportional to the weak hypercharge. But can the same theory break down in a different way such as to $SU(4)\times U(1)^2$ if the Higgs acquires different values?

i.e. are there certain values a Higgs field can take (such as certain local minima in it's potential) that break down the gauge group into different ways? Or is there always a lowest minima that it always takes on. Do people design the shape of the Higgs potential so that it always breaks down the gauge group in a certain way?

If the Higgs field could take on different values, what happens if the Universe finds itself in a local minima which isn't the lowest minima, could the Universe suddenly jump and the gauge group of the Universe change?

  • $\begingroup$ I'm not sure why you are taking on an extra U(1) in the latter case. I'd expect just SU(4)xU(1), instead, if the coefficient of the "true" quartic in the potential is negative. Do you want the technical details? Yes, people "design" the potential with that coefficient positive (and other conditions). If nature upended those, there'd be hell to pay. $\endgroup$ – Cosmas Zachos Apr 8 '20 at 14:13

Do people design the shape of the Higgs potential so that it always breaks down the gauge group in a certain way?

Yes they do. I'll review the Georgi-Glashow breaking of 12 generators of SU(5) to SU(3)×SU(2)×U(1); point out the discriminant parameter in the Higgs potential that dictates breaking only 8 generators to SU(4)×U(1) instead; and justify the "how/why" by a summary "seat of the pants" argument (the Fermi/Feynman style), leaving you to replicate the epic Ling-Fong Li analysis for this particular group.

But I 'll desist from bogus cosmological scenario speculation, and how the flip of that parameter of the potential I'll discuss upends the hypothetical SU(5) breaking into a freak universe.

The classic renormalizable and field-parity-symmetric quartic potential for the adjoint (24) Higgs $\Phi=\phi^a \lambda^a/\sqrt{2}$ is $$ V(\Phi)= -\frac{\mu^2}{2}\operatorname{Tr} \Phi^2 + \frac{a}{4}(\operatorname{Tr} \Phi^2)^2 + \frac{b}{2}\operatorname{Tr} \Phi^4 . $$ This looks just like the Higgs potential of the SM, except, crucially, for the extra quartic term with coefficient b. (For SU(2), such a term would be proportional to the a term, since the symmetric d coefficient of SU(2) vanishes, so the trace collapses to a square of traces: can check this with Pauli matrices. Not so for any other SU(N).)

It is a fact, then, justified below, that, for b>0 and 15 a +7 b>0 the adjoint Higgs vev at the minimum is the one proportional to the SM hypercharge (think action on the 5), and traceless (identifiable with a generator of SU(5)), $$ \langle \Phi\rangle= v \begin{pmatrix} 1&0 &0 &0 &0\\ 0&1 &0 &0 &0 \\0& 0&1 &0 &0 \\0 &0 &0 &-3/2 &0 \\ 0&0 &0 &0 &-3/2 \end{pmatrix}, $$ and thus symmetric under SU(3)×SU(2)×U(1).

Plugging it into the potential yields $$ V(\langle\Phi\rangle)= -\frac{\mu^2}{2} v^2 \frac{15}{2} + \frac{ a}{ 4} v^4 \left (\frac{15}{2}\right )^2 + \frac{ b}{2} v^4 \frac{105}{8} . $$ Solution of the full problem specifies that this is minimized for the $v^2$ whose variation yields zero, $$ {\partial V \over \partial v^2 } =0 = -\frac{\mu^2}{2} \frac{15}{2} + \frac{ a}{ 2} v^2 \left (\frac{15}{2}\right )^2 + \frac{ b}{2} v^2 \frac{105}{4} \leadsto \\ \mu^2 = v^2 (15a+7b)/2 ~ > 0. $$ The positivity of the parenthesis ensures the quartic term is increasing without a bound for large v.

Now you are effectively asking: "What conditions lead you, instead, to a traceless vev, $$ \langle \Phi\rangle= v \begin{pmatrix} 1&0 &0 &0 &0\\ 0&1 &0 &0 &0 \\0& 0&1 &0 &0 \\0 &0 &0 &1 &0 \\ 0&0 &0 &0 &-4 \end{pmatrix}, $$ symmetric under SU(4)×U(1) ?"

The answer is b<0 , for positive a, see below. Now $$ V(\langle\Phi\rangle)/10= -\mu^2 v^2 + 10 a v^4 + 13 b v^4 , $$ and variation yields the minimum value $$ \mu^2 = v^2 (20a+26b), $$ which yields the new condition 10a+13b>0. So a> -1.2b> 0.

Is it possible to appreciate this bifurcation summarily? Consider this interpolating traceless vev, $$ \langle \Phi\rangle= v \begin{pmatrix} 1&0 &0 &0 &0\\ 0&1 &0 &0 &0 \\0& 0&1 &0 &0 \\0 &0 &0 &1-x &0 \\ 0&0 &0 &0 &-4+x \end{pmatrix}, $$ that is, we've reinserted a new d.o.f. parameter x, which yields the first v.e.v. for x=5/2 and the second one for x=0, respectively.

Can we see how the sign of b prejudices the respective values? $$ V(\langle\Phi\rangle)= \\ = -\frac{\mu^2}{2} v^2 ( 3+ (x-1)^2 + (x-4)^2 ) + \frac{ a}{ 4} v^4 ( 3+ (x-1)^2 + (x-4)^2)^2 + \frac{ b}{2} v^4 (3+ (x-1)^4+(x-4)^4 ) \\ = -\frac{\mu^4}{4a} + a v^4 \left ( x(x-5/2)+10 -\frac{\mu^2}{2av^2}\right )^2 + b v^4 130 (1-x+O(x^2)) \\ =-\frac{\mu^4}{4a}+av^4 \left ( x(x-5/2)+10 -\frac{\mu^2}{2av^2}\right )^2+b v^4 ( 105/16+ 27(x-5/2)^2/2+(x-5/2)^4 ). $$ You then see that for b=0, the "conventional" potential terms cannot distinguish between the two values of x, and has a minimum for either value. Moreover, a negative b favors the x=0 value, since increasing x from it increases V (penultimate line). Likewise, for positive b, any excursion from 5/2 increases V, whose minimum is under investigation. You can be guided by this principle as you do the full minimization for the full 24-tuplet, as in Ling-Fong Li's 1974 paper.

  • $\begingroup$ Thanks, you certainly know your stuff. Seems like we wouldn't like to live in a Universe where b=0. $\endgroup$ – zooby Apr 8 '20 at 22:03
  • $\begingroup$ Also, it seems like we would need additional terms in the Higgs potential for higher $SU(N)$ otherwise there would only be a maximum of 2 ways to break the symmetry. $b<0$ and $b>0$ unless it operates in ranges. $\endgroup$ – zooby Apr 8 '20 at 22:12
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    $\begingroup$ Hm... yeah, for the adjoint, "these" are the only options. But with more/different reps, the sky is the limit... $\endgroup$ – Cosmas Zachos Apr 8 '20 at 22:15

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