2

I have some confusion about the vacuum expectation value of scalar field in QFT. I know that the one-point function $$ \langle \Omega | \phi(x) | \Omega \rangle = \langle \Omega | \phi(0) | \Omega \rangle =: v $$ is a constant, which follows immediately by translation properties of $\phi$, that is $$ \phi(x) = e^{i P x} \phi(0) e^{-iPx}. $$ Q1: If one would like to determine $v$, should one just calculate the one-point function (that means the tadpole diagrams) in perturbation theory? Or is there an easier (or exact) way to get it?

Q2: Also I have heard that one can set the one-point function to zero by making a field redefinition $\phi \rightarrow \phi - v$ (see e.g. Why can the $1$-point correlation function be made to vanish?) and therefore one can safely consider it zero while happily continuing with the same theory. I see that it sets $v \rightarrow 0$, but it also massively changes the theory, since is may generate additional interaction terms in the Lagrangian? I need some clarification.

jkb1603
  • 1,119

1 Answers1

4
  1. Yes, in order to determine $v$ in the original field's formulation you would need to calculate the $1$-point correlation function, and in the interacting theory this would mean summation over all diagrams.
  2. However, you are free to perform the field redefinition $\phi \rightarrow \phi - v$. Fields by themselves are not observable quantities, the choice of whether to use field $\phi$ or some other $\chi$ is purely a matter of convenience and convention. The physically observable is $S$-matrix, which encodes in itself amplitudes of all scattering processes $i \rightarrow f$, where $i$ is the initial state, and $f$ is the final.

Field redefinition can slightly change the action $S^{'}[\phi] = S[\phi + v]$, vertices with a certain power of the field may emerge of disappear, but this would be the only change since replacement $\phi \rightarrow \phi + v$ has a unit Jacobian and measure in the functional integration is preserved.

For some practical purposes, this redefinition can be inconvenient, however for theoretical purposes one can assume that $\langle \phi \rangle = 0$ without loss of generality.

spiridon_the_sun_rotator
  • 4,835
  • Thanks, I understand now! Is it then correct to say that the field variable $\phi$ where $\langle \phi \rangle = 0$ is in some sense special since it should be used as the variable in the quantum theory? I conclude this because the vanishing of the vaccum expectation value is required for the LSZ formula (see point 4. in the answer of https://physics.stackexchange.com/questions/311856/confusion-over-assumptions-made-in-the-lsz-reduction-formula). – jkb1603 Apr 26 '21 at 11:40
  • 1
    @jkb1603 I would that it is the most convenient choice, and required by the LSZ formula, but I would not say, that It is more physical. I suppose, if one takes $\phi$ with a non-vanishing expectation value, one can still get reasonable results. However, the LSZ formula would need to be modified in a certain way – spiridon_the_sun_rotator Apr 26 '21 at 14:19