Consider a real scalar field $\phi$ in a theory with a Lagrangian
$$
\mathcal{L}:=-\frac{1}{2}\partial _\mu \phi \partial ^\mu \phi -V(\phi ),
$$
where
$$
V(\phi ):= -\mu ^2\phi ^2+\frac{\lambda}{4!}\phi ^4,
$$
where both $\mu$ and $\lambda$ are positive real numbers.
We see that the potential has a couple of non-zero minima:
$$
V'(\phi )=-2\mu ^2\phi +\frac{\lambda}{3!}\phi ^3=0\Rightarrow \phi =0,\pm 2\mu \sqrt{\frac{3}{\lambda}}=:\pm V_0
$$
(It turns out that $\phi =0$ is a local max, and $\phi =\pm V_0$ are local mins; check the second derivative.)
As the usual story goes, we must define a new field $\psi :=\phi -V_0$ and re-write the theory in terms of this $\psi$ to get the appropriate Feynman rules of the quantum theory. If I did my algebra correctly (the details aren't exactly relevant here anyways), this substitution gives us
$$
\mathcal{L}=-\frac{1}{2}\partial _\mu \psi \partial ^\mu \psi -2\mu ^2\psi ^2+\mu \sqrt{\frac{\lambda}{3}}\psi ^3+\frac{\lambda}{4!}\psi ^4-6\frac{\mu ^4}{\lambda}.
$$
(Our Lagrangian no longer admits the symmetry $\psi \mapsto -\psi$, hence the term "symmetry breaking".)
Th question arises: Why is this substitution special? This form of the Lagrangian has some nice properties (namely that the potential has a local min at $0$), but surely there are some other substitutions that could given us some other nice properties as well. What about those?
My understanding of this was the following: The LSZ Reduction Formula, among other things, requires a priori that the fields one is working with have vanishing vacuum expectation value. Thus, when applying the LSZ formula, we must be working with $\psi$, not $\phi$, and so the appropriate Feynman rules can be read off only when the Lagrangian is written in terms of $\psi$. I have just recently discovered a problem with this explanation, however.
Before, I was under the impression that $\langle 0_\pm |\phi |0_\pm \rangle =\pm V_0$ (this theory evidently has two physical vacuums, whatever that precisely means), so that the definition of $\psi$ forces $\psi$ to have vanishing expectation value, so that the LSZ formula can be applied. However, I recently learned that $\pm V_0$ is only an approximation to $\langle 0_\pm |\phi |0_\pm \rangle$, which implies that $\psi$ only approximately has vanishing vacuum expectation value, which means that LSZ doesn't technically apply.
It seems that the proper substitution is in fact $\psi :=\phi -\langle 0_+|\phi|0_+\rangle$. There are several problems I see with this:
-
The Lagrangian re-written in terms of $\psi$ should have a small, but non-zero, linear term in $\psi$.
-
The Feynman rules I've been using all along that arise from the substitution $\psi :=\phi -V_0$ are only approximation.
-
The coefficients that arise from the 'proper' substitution $\psi :=\phi -\langle 0_+|\phi |0_+\rangle$ are going to be written in terms or something that can (to the best of my knowledge) only be calculated perturbatively (namely $\langle 0_+|\phi |0_+\rangle$), but we need to know these coefficients to obtain the Feynman rules to begin with (resulting in a 'circularity' problem).
How does one go about resolving all these issues?
(Disclaimer: I asked a very similar question here not quite a year ago, but my understanding of the situation has improved since then, and as is usual, my improvement of understanding has only brought forth many more questions regarding this, so I felt it was appropriate to address the issue once again.)
Best Answer
First of all, classically (neglecting loop corrections), we obviously want to expand around the true minima that are found in the vacuum, which means around one of the states $|0_\pm\rangle$ – these two are really equivalent to one another due to the gauge symmetry. The state around $\phi=0$ is a maximum of energy, not a minimum, so Nature doesn't spend much time over there before it rolls down to the true minimum.
If we expanded around a shifted scalar field, there would be, assuming that the Lagrangian is Taylor-expanded, also first-order terms in the scalar field. They would produce Feynman "vertices" with one external line connected to a "cross" (in which a Feynman diagram ends like an external line but isn't associated with an actual external particle). These simple decorations of Feynman diagrams could be resummed and their effect would be simple – effective shift the scalar field back to the minimum.
When loop corrections are taken into account, the true minimum isn't exactly given by the classical approximation and the vev isn't exactly as the location of the minimum, too. All these things get quantum corrections – suppressed formally by positive powers of $\hbar$ and more quantitatively by the small dimensionless values of the coupling constants.
The calculations in a quantum field theory shift the scalar field to the new, true minimum (incorporating quantum corrections) automatically. Do you remember I took about one-external-leg vertices of Feynman diagrams that we could eliminate by a clever shift? Well, when you consider loop diagrams, there is a similar subtlety – tadpole diagrams such as this one:
It's a one-loop diagram and the loop at the end plays the same role as the small "cross" indicating the direct Feynman diagram vertex from the beginning of the discussion. But even if we eliminate the linear terms from the action we start with, quantum loops effectively generate their own tadpoles that may be attached through other vertices to any Feynman diagram and that effectively shift the scalar field to the right minimum corrected by the corrections suppressed by powers of $\hbar$.
The LSZ formula doesn't break at all. If you correctly add all the Feynman diagrams, there will also be the Feynman diagrams with the tadpoles attached. They fixed the "internal" part of the Feynman diagram so that it knows about the true minimum. The external part of the LSZ formula is also fixed automatically because the particles in LSZ are connected with the creation action by a near-mass-shall mode of the scalar fields. And it doesn't matter whether you pick a Fourier component of $\phi$ or $\phi-c$ for a constant $c$ – only the actual field $c$ will get magnified near $k^2=m^2$ while the constant term is annihilated by $(k^2-m^2)$, anyway.
So:
So you shouldn't worry about any of these things. The calculation has intermediate steps that are not unique but if you carefully follow your conventions, you don't have to do anything special and the summation over all the Feynman diagrams produces the right physical answers without any extra "cures".