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.

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    $\begingroup$ Is your action linear? $\endgroup$ Commented Apr 25, 2021 at 18:34
  • $\begingroup$ To add to @Richard Myers comment: $S[\phi+\phi_c]\ne S[\phi]+S[\phi_c]$ even for non interacting theories. $\endgroup$
    – mike stone
    Commented Apr 25, 2021 at 19:52
  • $\begingroup$ Right, I agree. Somehow I thougth that this was true. I will fix it in the post. $\endgroup$
    – jkb1603
    Commented Apr 25, 2021 at 19:56
  • $\begingroup$ Just a caution that the transformation property you quoted is not sufficient to conclude that the vev is constant. That is an operator relation (true for any state), and there are certainly states where the one-point function of the field is not a constant. $\endgroup$
    – Andrew
    Commented Apr 25, 2021 at 20:17
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    $\begingroup$ related: physics.stackexchange.com/q/311856/84967 $\endgroup$ Commented Apr 25, 2021 at 20:21

1 Answer 1

  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.

  • $\begingroup$ 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 physics.stackexchange.com/questions/311856/…). $\endgroup$
    – jkb1603
    Commented Apr 26, 2021 at 11:40
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    $\begingroup$ @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 $\endgroup$ Commented Apr 26, 2021 at 14:19

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