Higgs Expectation Value with Classical vs Quantum Potential

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
jim burns
Messages
4
Reaction score
0
I'm having a hard time following the arguments of how the Higgs gives mass in the Standard Model. In particular, the textbook by Srednicki gives the Higgs potential as:

$$V(\phi)=\frac{\lambda}{4}(\phi^\dagger \phi-\frac{1}{2}\nu^2)^2 $$

and states that because of this, $$\langle 0 | \phi(x) |0 \rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}v \\ 0 \end{pmatrix}$$

However, I thought the proper procedure would be to write $$\phi=q+\tilde{\phi}$$, and substitute this

$$\tilde{V}(\tilde{\phi})=\frac{\lambda}{4}((q+\tilde{\phi})^\dagger (q+\tilde{\phi})-\frac{1}{2}\nu^2)^2 $$

Then you would calculate the minimum of the quantum effective potential $$\tilde{V}_{eff}(\tilde{\phi})$$, which will give you an equation that gives you 'q' in terms of 'ν' and your other couplings. You then plug this value into 'q' to get

$$\tilde{V}(\tilde{\phi})=\frac{\lambda}{4}((q(\nu,\lambda)+\tilde{\phi})^\dagger (q(\nu,\lambda)+\tilde{\phi})-\frac{1}{2}\nu^2)^2 $$

Moreover, any field M that multiplies the Higgs in an interaction now has a component

$$q(\nu,\lambda)*M$$

Did Srednicki skip all these steps? How can you just say that the vacuum expectation value of a field is the minimum of the classical potential? Is he renormalizing all his couplings so that the minimum of the classical potential is the minimum of the quantum effective potential?
 
jim burns said:
Did Srednicki skip all these steps? How can you just say that the vacuum expectation value of a field is the minimum of the classical potential? Is he renormalizing all his couplings so that the minimum of the classical potential is the minimum of the quantum effective potential?

I think this is an interesting question that may be a bit out of my scope, but I'll try to provide at least a qualitative answer. Basically, It is not exact, but an approximation that can be justified in a regime of really weak couplings. In reality the quantum-mechanical ground state would be some superposition of all the "classical" vacuum states that is invariant under the symmetries of the Lagrangian/Hamiltonian, but for really weak couplings, this is unstable and falls into one of the potential states the way a classical system works. Essentially, my reasoning is similar to that used to explain why macroscopic objects are not in rotationally invariant ground states despite being governed by a rotationally invariant Hamiltonian. The Higgs Field is then a first order perturbation about this semiclassical vacuum state to get a perturbative spectrum (spectrum refers to the masses of particles).

The validity of this assumption seems to be supported due to the couplings to particles in this model are proportional to M/v , where M is the particle's mass and v is the v.e.v, which are all fairly small using experimental values. The top quark being a potential exception as well as some of the Higgs self-couplings

I believe a more rigorous answer could be found in talking about Effective Actions and the tree approximation, which are explained in Weinberg's Theory of Quantum Fields, vol. 2,
 
  • Like
Likes   Reactions: vanhees71