In the renormalization procedure of quantum field theories, say $\lambda \phi^4$ theory for simplicity, we use the physical mass $m$, the physical coupling constant $\lambda$ and the physical field $\phi$ to fix the bare quantities $m_0$, $\lambda_0$ and $\phi_0$.

It is clear that the physical mass $m$ and the physical coupling constant $\lambda$ can be found from experiments (and using the S-matrix). But how does one "measure" the physical field $\phi$?

Edit: I would like to (hopefully) clarify my question. My question is related to field renormalization $\phi_0(x)=\sqrt{Z_\phi}\phi(x)$, and how we decide what value $\phi$ should be. For the physical mass and coupling constant, we can perform experimental measurements to determine their fixed values but we don't seem to do this for the physical field (as far as I understand; I still need to properly interpret JeffDror's answer). Are we basically looking at the divergence of Feynman diagrams and choose $Z_\phi$ such that we can absorb this divergence? If this is true, then this would mean that the "physical" field is a badly chosen word and it would make more sense to call it the "renormalized" field?

  • $\begingroup$ Do you mean the expectation value of $\phi$ in vacuum, $\langle 0|\phi|0\rangle$? $\endgroup$ Feb 25, 2014 at 14:34
  • $\begingroup$ @IsidoreSeville I'm not sure to be honest... Is the VEV the way we renormalize the bare field $\phi_0$? $\endgroup$
    – Hunter
    Feb 25, 2014 at 14:37
  • 1
    $\begingroup$ I don't think @Hunter is referring to the VEV. The wavefunction renormalization arises from $\sqrt{ Z } = \left\langle k \right| \phi ( 0 ) \left| 0 \right\rangle $. This is equivalent to the residue of the propagator condition. $\endgroup$
    – JeffDror
    Feb 25, 2014 at 14:52
  • 1
    $\begingroup$ @Jeff can it be measured? only by turning off the vacuum? it can be calculated, if we know the cutoff, but i don't think that counts. if there is no cutoff it is infinite anyway so certainly not physical, i'd guess $\endgroup$
    – innisfree
    Feb 25, 2014 at 15:00
  • 1
    $\begingroup$ @Hunter Fair enough! $\endgroup$ Feb 25, 2014 at 18:16

1 Answer 1


To answer this we first need to be clear about why this wavefunction renormalization arises. For simplicity we focus on $\phi^4$ theory. For a free field we have, \begin{equation} \phi ( x ) \left| 0 \right\rangle = \int \frac{ d ^3 p }{ (2\pi)^3 } \frac{1}{ 2 E _{ {\mathbf{p}} } } e ^{ - i {\mathbf{p}} \cdot {\mathbf{x}} } \left| {\mathbf{p}} \right\rangle \end{equation} Apart from the factor of $ \frac{1}{ 2E _{ {\mathbf{p}} }} $ this is just equal to $ \left| {\mathbf{x}} \right\rangle $. Thus we can intereprate $ \phi ( {\mathbf{x}} ) $ acting on the vacuum as a field producing an $ {\mathbf{x}} $ eigenstate.

The equation above can be used to show that \begin{equation} \left\langle 0 \right| \phi ( {\mathbf{x}} ) \left| {\mathbf{p}} \right\rangle = e ^{ i {\mathbf{p}} \cdot {\mathbf{x}} } \end{equation} To continue to interpret $ \phi ( {\mathbf{x}} ) $ as the field that creates $ \left| {\mathbf{x}} \right\rangle $ eigenstates the renormalized field must obey this same relation. Thus we impose the renormalization condition: \begin{equation} \left\langle 0 \right| \phi _r ( 0 ) \left| {\mathbf{p}} \right\rangle = 1 \end{equation}
where $ \phi _r = \sqrt{ Z } \left( \phi - \left\langle \phi \right\rangle \right) $ is the renormalized field. If the wavefunction doesn't change under renormalization then $Z$ is just equal to $1$.

Now $ Z $ is not measurable for the same reason that the bare masses and couplings can't be measured. However, the renormalized field is measured in every process since if the condition above was not set to $1$ then we would be getting an extra factor for each external field line and another for each propagator in every diagram.

For example if we just forgot about the wavefunction renormalization we would get an extra $\sqrt{Z}$ for each external line and an extra $Z$ for each propagator.

  • $\begingroup$ +1 Thanks for your reply. I'm going to have to think about it a bit more and maybe ask you some questions, before I can accept your answer. $\endgroup$
    – Hunter
    Feb 25, 2014 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.