Let $L(\phi)$ be a Lagrangian and $\phi$ a quantum field. The equivalence theorem says that the $S$ matrix remains invariant under field redefinition.

Let us take for example the Lagrangian $$L=\partial_\mu\phi\partial^\mu\phi-m^2\phi^2\tag 1$$ The canonically conjugate field is $$ \pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi} $$ which leads to the Hamilton density $$ \mathcal{H}(\phi,\pi)=\pi \dot{\phi}-\mathcal{L}=\frac{1}{2}\left(\pi^{2}+\left(\nabla \phi\right)^{2}+m^{2} \phi^{2}\right) . $$ Suppose we make the field redefinition $\phi=F(\eta)$, then since $\phi$ and $\pi$ are independent we should have $$ \mathcal{H}(\phi,\pi)=\mathcal{H}(F(\eta),\pi)=\mathcal{H'}(\eta,\pi)=\frac{1}{2}\left(\pi^{2}+\left(\nabla F(\eta)\right)^{2}+m^{2} F(\eta)^{2}\right) . $$ Now since the Hamiltonian does not change the $S$ matrix should not change also. But this result is so trivial that I am not understanding why bother to make it a theorem.

On the other hand if we make the transformation $\phi \rightarrow \eta=F(\phi)$ then we would have $$\mathcal{H}(\phi,\pi) \rightarrow\mathcal{H'}(\phi,\pi)=\frac{1}{2}\left(\pi^{2}+(\nabla F(\phi))^{2}+m^{2} F(\phi)^{2}\right) . $$

Here the notation $\phi \rightarrow \eta=F(\phi)$ means replace $\phi $ by $\eta$ as is showed above

In order for the scattering amplitude to be invariant we should have $\mathcal{H}=U^\dagger\mathcal{H'}U$ where $U$ is unitary operator.

So is this redefinition $\phi=F(\eta)$ or the redefinition $\phi \rightarrow \eta=F(\phi)$ that is stated in this theorem?


2 Answers 2


The theorem only holds for invertible redefinitions, so both options $\phi=F(\eta)$ and $\eta=F'(\phi)$ are equivalent. The two options are related via $F'=F^{-1}$.

The theorem is obviously false for non-invertible definitions, e.g. you can choose the constant functional $F(\eta)=1$. The "change of variables" $\phi=1$ clearly does not leave the theory invariant.

  • $\begingroup$ What I am asking is how the substitution is made. For example is the Lagrangian transformed Like this $L\rightarrow L(F(\eta))$ or like this $L\rightarrow L(\eta)$? $\endgroup$ Commented Aug 17, 2021 at 9:29
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    $\begingroup$ I don't know what the symbol $\to$ means. The arrow notation is very imprecise and only leads to confusion. Try to formulate the question without using $\to$'s. $\endgroup$ Commented Aug 18, 2021 at 10:31
  • $\begingroup$ Ok Will edit my question $\endgroup$ Commented Aug 18, 2021 at 16:44

Both are equivalent.

  • In the first, we change field variables from $\phi$ to $\eta$ where $\phi = F(\eta)$. Simply plug in $F(\eta )$ wherever you see $\phi$.
  • In the second, we simply replace the symbol $\phi$ with $\eta$ (does nothing). Then we state that $\eta= F(\phi )$. This is the exact same as before, but with the symbols reversed.

Edit: I removed a bit about the equivalence theorem following trivially from the path integral. The equivalence theorem is not so trivial.

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    $\begingroup$ The equivalence theorem is not a simple corollary of the path integral. Changes of variables also change the path integral measure nontrivially. In fact, variable changes like this do change the QFT! It is only the S-matrix that remains unchanged. So these can not be corollaries of the path integral. $\endgroup$ Commented Aug 10, 2021 at 19:39
  • $\begingroup$ @Prof.Legolasov I deleted my previous comment. I see what you're saying now. I understand that the path integral measure changes under field redefinitions, however I misunderstood what the equivalence theorem was. I thought it was that, in the path integral, if you change variables the overall result doesn't change. What it's actually saying is that, starting with the $\phi$ theory, if you calculate S-matrix elements but using the action $S[F[\phi]]$ rather than $S[\phi]$, you'll get the same results. $\endgroup$ Commented Aug 10, 2021 at 20:37
  • $\begingroup$ yeah, exactly. It is a nontrivial result about perturbative S-matrix. $\endgroup$ Commented Aug 10, 2021 at 20:39
  • $\begingroup$ @Prof.Legolasov it can't be proven non-perturbatively? $\endgroup$ Commented Aug 10, 2021 at 20:40
  • $\begingroup$ I actually am not sure, the definition of S-matrix is a bit different non-perturbatively. You’ll need Haag-Ruelle scattering theory. My guess is that it is possible to prove for individual models $\endgroup$ Commented Aug 10, 2021 at 20:42

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