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I'm reading some notes about the symmetry breaking and the standard model. From what I understand, typically one has a symmetry group of transformations (the gauge group) acting on vectors of fields, then writes the most general invariant potential and tries to find it's minimum configurations (vevs).

Leaving the theory of symmetry breaking aside, my doubt is specifically about the computation of finding such minima. In my notes I have an example that I cannot understand:

the potential

$$\quad V(\Phi_\,\Phi_2)=-\mu_1^2\Phi_1^\dagger\Phi_1-\mu_2^2\Phi_2^\dagger\Phi_2+\lambda_1(\Phi_1^\dagger\Phi_1)^2+\lambda_2(\Phi_2^\dagger\Phi_2)^2+\lambda_3(\Phi_1^\dagger\Phi_1)(\Phi_2^\dagger\Phi_2)+\lambda_4(\Phi_1^\dagger\Phi_2)(\Phi_2^\dagger\Phi_1)+\lambda_5\left[(\Phi_1^\dagger\Phi_2)^2+(\Phi_2^\dagger\Phi_1)^2\right] \tag{1}$$

is given as the most general potential preserving some symmetries, where $\Phi_{1,2}$ are two doublets containing complex fields. Now, in my notes I have written that such potential attains minimum at

$$\left<\phi_0^1\right> = v_1e^{i\pi/2}\quad\left<\phi_0^2\right> = v_2.$$

however I can't understand how to reproduce this. Should I write out all the fields explicitly, take the derivative w.r.t to all of them ( I believe there are 8 fields) and then set the derivatives equal to zero? I'm sure it's something easier that I'm not seeing right now.

I've see other, easier examples, where $V$ only depends on a single field, like $$V(\Phi^\dagger\Phi)=a\Phi^\dagger\Phi+b(\Phi^\dagger\Phi)^2,$$ but these are easier because since $\Phi^\dagger\Phi$ is an invariant number, I can just call it $\phi_0$ and then take the derivative ${{\partial V}/{\partial\phi_0}}=0.$ This is easier, but I don't see how to extend this technique to the case of the potential (1) because I don't know how to treat the mixed terms $$\Phi_i^\dagger\Phi_j\quad(i \neq j)$$ that appear in $\lambda_4$ and $\lambda_5$ .

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    $\begingroup$ I think you are right. There is also a term V0 that "does not depend on the relative phases of the $\Phi_1$ with $\Phi_2$" which I thought that was simply a constant. $\endgroup$
    – geodesic
    Jan 21, 2023 at 20:21
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    $\begingroup$ I believe this is called the "two Higgs doublet model" (?). I also found this reference link where they find the same minima in equation (302). Also Peskin writes the same potential at the end of chapter 20. (Exercise 20.5 in my copy) $\endgroup$
    – geodesic
    Jan 21, 2023 at 20:32
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    $\begingroup$ but I still don't see how one minimizes V $\endgroup$
    – geodesic
    Jan 21, 2023 at 23:15
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    $\begingroup$ You should edit your question giving the full potential you wish to minimize. The quartic piece of the potential you wrote down is not bounded from below and a minimization of this incomplete part alone does not make sense. $\endgroup$
    – Hyperon
    Jan 21, 2023 at 23:27
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    $\begingroup$ Thank you @Hyperon to point that out. I have now corrected V $\endgroup$
    – geodesic
    Jan 22, 2023 at 9:04

1 Answer 1

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Assuming the usual charge assignment $\Phi_i^\dagger = \left( \phi_i^- , (\phi_i^0)^\ast\right)$ (with $i=1,2$) in an $\rm SU(2)$ doublet and the requirement that $\rm U(1)_{em}$ should remain unbroken, we have $\langle \Phi_i^\dagger \rangle = \left(0, \langle \phi_i^0 \rangle^\ast \right)$. A convenient parametrization for the determination of the (classical) minimum of the potential is thus given by $\Phi_1^\dagger \Phi_1=x \ge 0$, $\Phi_2^\dagger \Phi_2= y \ge 0$, $\Phi_1^\dagger \Phi_2 = \sqrt{xy} \, e^{i\alpha/2}$ with a real phase angle $\alpha$. Expressed in terms of these variables, the potential (for the uncharged fields) takes the simple form

$$V(x,y,\alpha) =-\mu_1^2 \, x -\mu_2^2 \, y+\lambda_1 x^2 +\lambda_2 \, y^2+(\lambda_3+\lambda_4)\, x y +2 \lambda_5 \, x y \cos \alpha. $$

It is now a relatively easy task (a) to find conditions on the parameters $\mu_i^2, \lambda_i$ such that the potential admits local minima and (b) to justify the value of the phase angle $\alpha$ (at the minimum of the potential) given in the question. As presenting full solutions of homework-type problems is off-topic in PSE, these final steps are left as an exercise.

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