Justification for phi^4 potential

[joneall]

TL;DR Summary

Aside from the important fact that it provides insight, what is the justification for assuming a phi^4 potential?

My understanding is at the level of Griffiths's Introduction to quantum mechanics or Robinson's Symmetry and the standard model, i.e., using the phi^4 potential to explain the effects of global and local symmetry breaking, Goldstone and Higgs bosons. These books and others use a potential of type $$\frac{1}{4}\lambda (\phi^{*} \phi)^2$$. But nobody every explains why they take such a form. What is the justification for this, aside from the important fact that it works (as if that were not enough)?

 

[Gaussian97]

Well, first of all the Higgs potential is not a pure quadratic one, but rather
$$V(\phi)=\frac{\mu}{2}(\phi^\dagger\phi)^2+\frac{1}{4}(\phi^\dagger\phi)^4$$

In a lot of cases, the real answer on why this an not another option is: Many were tried, and that was the only one that fit the experiments.

Even though, we can try to give some reason. What if we want to construct a potential for a complex scalar doublet like $\phi$? What are the options if we also want to impose gauge invariance under $SU(2)\times U(1)$?
Well, as a complex doublet, this field has 4 degrees of freedom, we could try to write them as
$$\begin{pmatrix}\phi_1+i\phi_2 \\ \phi_3+i\phi_4\end{pmatrix}$$, but usually is better to use these other:
$$\phi, \qquad \phi^*, \qquad \phi_C=\frac{i}{2}\sigma^2\phi^*, \qquad \phi_C^*$$
The question now is to know how they transform under gauge symmetries, in fact under $U(1)$ is easy to see that $\phi$ and $\phi_C^*$ transform the same way, and the other two transforms with the inverse transformation, therefore only this combinations are $U(1)$ invariant
$$\phi\cdot \phi^*, \qquad \phi\cdot \phi_C$$
Now the question is about $SU(2)$, it is also easy to see that now $\phi$ and $\phi_C$ transform the same way, and $\phi^*$ and $\phi_C^*$ transform with the inverse matrix. So the only combination that is invariant under $U(1)$ and $SU(2)$ is
$$\phi\cdot \phi^* = \phi^\dagger \phi$$
And therefore any power of the form $(\phi^\dagger \phi)^n$ is also gauge invariant.

It is possible to have extra terms, like
$$|\phi^\dagger \phi_C|^2$$
but is also easy to see that this term is equal to zero.
Therefore the most general potential of a complex scalar doublet, up to 4-fields interaction is the SM Higgs potential.

 

[vanhees71]

Summary:: Aside from the important fact that it provides insight, what is the justification for assuming a phi^4 potential?

My understanding is at the level of Griffiths's Introduction to quantum mechanics or Robinson's Symmetry and the standard model, i.e., using the phi^4 potential to explain the effects of global and local symmetry breaking, Goldstone and Higgs bosons. These books and others use a potential of type $$\frac{1}{4}\lambda (\phi^{*} \phi)^2$$. But nobody every explains why they take such a form. What is the justification for this, aside from the important fact that it works (as if that were not enough)?

The various kinds of $\phi^4$ theories in textbooks are just pedagogical examples to introduce the calculational techniques you need to evaluate (perturbatively as well as beyond) relativistic QFTs.

On the fundamental level the fundamental interactions are all described by gauge theories (including gravity, but that's not yet successfully quantized). In the electroweak sector of the Standard Model you have some scalar fields (in the usual minimal version a flavor-doublet field) and then for Dyson-renormalizability you also necessarily need an interaction of the type $|\phi^{\dagger} \phi|^2$ as in any gauge theory with scalar bosons as "matter fields".

Then one should be aware that local gauge symmetries cannot be broken, but that unfortunately the slang of "broken local symmetries" is impossible to get rid of. It would be so easy to simply speak about the "Higgs mechanism".

 

[Demystifier]

Summary:: Aside from the important fact that it provides insight, what is the justification for assuming a phi^4 potential?

My understanding is at the level of Griffiths's Introduction to quantum mechanics or Robinson's Symmetry and the standard model, i.e., using the phi^4 potential to explain the effects of global and local symmetry breaking, Goldstone and Higgs bosons. These books and others use a potential of type $$\frac{1}{4}\lambda (\phi^{*} \phi)^2$$. But nobody every explains why they take such a form. What is the justification for this, aside from the important fact that it works (as if that were not enough)?

I think the best answer is that anything else would lead to problems. Powers of $\phi$ higher than 4 are non-renormalizable, 3rd power (and higher odd powers) lead to energy unbounded from below, 2nd power alone is too trivial (it produces no particle interactions and no symmetry breaking) and non-integer powers cannot be treated by the usual perturbative methods (Feynman diagrams).

Another argument could be based on Landau theory of phase transitions, where one just expands $V(\phi)$ compatible with a certain symmetry into a Taylor series and keeps only the lowest terms.

 

[vanhees71]

Of course you can have a $\phi^3$ term in addition to a $\phi^4$ term. The resulting theory is still (Dyson-) renormalizable and has an energy bounded from below. It's a theory without any intrinsic symmetries ($\phi^4$ theory has the "field reflection" $\mathbb{Z}_2$ symmetry, $\phi \rightarrow -\phi$ which makes $\phi^4$ theory renormalizable without a $\phi^3$ term).

 

[Demystifier]

Of course you can have a $\phi^3$ term in addition to a $\phi^4$ term.

Of course, I meant $\phi^3$ alone.

 

[dextercioby]

Isn't the simple textbook example of $\lambda \varphi^4$ theory in $1+3$ flat spacetime example of not fulfilling Wightman's axioms?

 

[Demystifier]

Isn't the simple textbook example of $\lambda \varphi^4$ theory in $1+3$ flat spacetime example of not fulfilling Wightman's axioms?

Yes, but it doesn't make it physically inadequate. By regularization and renormalization, field theories make physical sense even without fulfilling Wightman's axioms.

 

[vanhees71]

It's just a toy theory to develop perturbative calculations of the various types of Green's functions for a Dyson-renormalizable model. It's not much simpler than QED (if QED isn't simpler with its one tree-level three-point vertex from a diagrammatical point of view). The only thing you avoid is that you have to struggle with the problems from gauge invariance to derive manifestly covariant Feynman rules. I'm not sure, whether I'd put so much emphasis on it when I had to teach introductory QFT, because you use a lot of time with a model which has no real-world applications, and you have to explain the troubles of quantizing QED anyway.

 

[Demystifier]

a model which has no real-world applications

Higgs particle?

 

[vanhees71]

That's not simple $\phi^4$ theory. Of course you can argue with Ginzburg-Landau theory in statistical physics.

 

[joneall]

On the fundamental level the fundamental interactions are all described by gauge theories (including gravity, but that's not yet successfully quantized)

Interesting. Where is the gauge theory in GR? For my limited understanding, that would be thru the use of the covariant derivative with Christoffel symbols as connections. But that hardly looks like a new field to me.

 

[PeterDonis]

Where is the gauge theory in GR?

The gauge freedom in GR is the freedom to choose any coordinate chart you like on a given spacetime geometry. Or, to put it another way, the same spacetime geometry--the same physical configuration--can be described by multiple different coordinate charts.

 

[vanhees71]

Another point view is that you make global Poincare invariance local. This leads to a slightly generalized theory, Einstein-Cartan space-time manifolds, if spinor fields are involved. For details see

https://www.semanticscholar.org/pap...bble/58b4aeaa2de802b0d76130899d65dd2b21805651