# What is the equivalence theorem in quantum field theory?

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?

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.

• 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 $L\rightarrow L(\eta)$? Commented Aug 17, 2021 at 9:29
• 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. Commented Aug 18, 2021 at 10:31
• Ok Will edit my question 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.

• 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. Commented Aug 10, 2021 at 19:39
• @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. Commented Aug 10, 2021 at 20:37
• yeah, exactly. It is a nontrivial result about perturbative S-matrix. Commented Aug 10, 2021 at 20:39
• @Prof.Legolasov it can't be proven non-perturbatively? Commented Aug 10, 2021 at 20:40
• 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 Commented Aug 10, 2021 at 20:42