If wavefunction is $\phi$, we know that $|\phi (x)|^2$ represents probability of finding a particle at $x$.

Now let us talk about some pedagogical example in spontaneous symmetry breaking in QFT, Mexican hat potential.

$V(\phi,\phi^\dagger) = -\frac{1}{2}\mu^2|\phi|^2 + \frac{\lambda^2}{4}|\phi|^4$

But I don't really know what would be physical meaning behind $|\phi|^2$. It is definitely different from $|\phi (x)|^2$ as that would refer to particular space $x$. What exactly is it?


You must first realize that a priori there doesn't need to be a relationship between the 2 things you're comparing here:

A. The former, $|\phi(x)|^2= |\langle x|\phi \rangle|^2$ is the square of the position space amplitude of a state $|\phi\rangle$ living in Hilbert space

B. The latter has two possible meanings depending on context:

  1. If we view $V(\phi)$ as the potential of a full, UV-complete theory, then $\phi$ is an operator in Hilbert space, and $|\phi(x)|^2 \equiv\phi(x)^\dagger \phi(x)$.
  2. If we view $V(\phi)$ as an effective potential and $\phi$ as a background scalar field, as often is the case when studying spontaneous symmetry breaking, then we can view $\phi$ as a complex classical scalar field, and $|\phi(x)|^2 = \phi(x)^*\phi(x)$ where $*$ denotes complex conjugation.

More physically, $\phi$ in situation A describes the physical state that a system is in, while in B $\phi$ is the dynamical variable in some field theory: you can obtain physical states in this theory by, for example, quantizing this theory, deriving creation/annnihilation operators and acting on a vaccuum state, but the field $\phi$ never has the interpretation of a physical state. Therefore as stated, there is no reason a priori to expect a connection between the physical interpretation of the two objects - they're just different objects to which you have given the same name.

We can now try to ask what is the physical interpretation of $|\phi|^2$. The quick answer is that you know $\phi$ has the interpretation of a quantum field, the dynamical variable in a field theory. $|\phi|^2$ is just the way that it happens to appear in the Lagrangians of this field theory. In the specific context of symmetry breaking, you can see that in order for the theory to have the desired $U(1)$ symmetry the action can only be a function of $|\phi|^2$ and not of $\phi$ and $\phi^\dagger$ individually.

Additionally if we go back to viewing $V$ as an effective potential, $\phi$ can be viewed as a classical background field, you can take the parameters $\mu$ and $\lambda$ to be functions of some renormalization scale $M$, it is often convenient to choose $M=\phi$, and when this choice is made $\phi$ has the additional interpretation of a renormalization scale. Most often in literature when you see the standard model potential graphed to a good degree of accuracy, they make this choice of renormalization scale described here. But once again, $|\phi|^2$ does not have any additional physical meaning other than those that $\phi$ already has.


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