First, the equivalence theorem refers to S-matrix elements rather than off-shell n-point functions, or their generator $Z[j]$, which are generally different. What you have to study is the LSZ formula that gives the relation between S-matrix elements and expectation values of time-ordered product of fields (off-shell n-point functions, what one gets after taking derivatives of $Z[j]$ and setting $j=0$). You will see that even thought these time-ordered products are different, the S-matrix elements are equal just because the residues of these products in the relevant poles are "equal" (they are strictly equal if the matrix elements of the fields between vacuum and one-particle states ( $\langle p|\phi|0\rangle$) are equal, if they are not equal, but both of them are different from zero, one can trivially adapt the LSZ formula to give the same results).
Second, the generating functional
\begin{equation}
Z[j]=\int \mathcal{D}[\phi] \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j(x)\phi(x) \}}
\end{equation}
is not valid for all actions functionals $S$. I will illustrate this with a quantum-mechanical example—the generalization to quantum field theory is trivial. The key point is to notice that the "fundamental" path integral is the phase-space or Hamiltonian path integral, that is, the path integral before integrating out momenta.
Suppose an action $S[q]=\int L (q, \dot q) \, dt=\int {\dot q^2\over 2}-V(q)\, dt$, then the generating of n-point functions is:
$$Z[j]\sim\int \mathcal{D}[q] \exp{\{iS(q)+i\int dt\hspace{0.2cm} j(t)q(t) \}}$$
The Hamiltonian that is connected with the action above is $H(p,q)={p^2\over 2}+V(q)$ and the phase-space path integral is:
$$Z[j]\sim \int \mathcal{D}[q]\mathcal{D}[ p] \exp{\{i\int p\dot q - H(p,q)\;dt+i\int dt\hspace{0.2cm} j(t)q(t) \}}$$
Now, if one performs a change of coordinates $q=x+G(x)$ in the Lagrangian:
$$\tilde L(x,\dot x)=L(x+G(x), \dot x(1+G(x)))={1\over 2}\dot x^2 (1+G'(x))^2-V(x+G(x))$$
the Hamiltonian is: $$\tilde H={\tilde p^2\over 2(1+G'(x))}+V\left( x+G(x)\right)$$
where the momentum is $\tilde p={d\tilde L\over d\dot x }=\dot x \; (1+G'(x))^2$. A change of coordinates implies a change in the canonical momentum and the Hamiltonian. And now the phase-space path integral is:
$$W[j]\sim \int \mathcal{D}[x]\mathcal{D}[\tilde p] \exp{\{i\int \tilde p\dot x - \tilde H(\tilde p,x)\;dt+i\int dt\hspace{0.2cm} j(t)x(t) \}}\,,$$
as you were probably expecting. However, when one integrates the momentum, one obtains the Langrangian version of the path integral:
$$W[j]\sim\int \mathcal{D}[x]\;(1+G'(x)) \exp{\{iS[x+G(x)]+i\int dt\hspace{0.2cm} j(t)x(t) \}}$$
where $(1+G'(x))$ is just $\det {dq\over dx}$. Thus, your second equation is wrong (if one assumes that the starting kinetic term is the standard one) since the previous determinant is missing. This determinant cancels the determinant in your last equation. Nonetheless, $Z[j]\neq W[j]$, since changing the integration variable in the first equation of this answer
$$Z[j]\sim\int \mathcal{D}[x]\;(1+G'(x)) \exp{\{iS[x+G(x)]+i\int dt\hspace{0.2cm} j(t)(x(t)+G(x)) \}}$$
which does not agree with $W[j]$ due to the term $j(t)(x(t)+G(x))$. So that, both generating functional of n-point functions are different (but the difference is not the Jacobian), although they give the same S-matrix elements as I wrote in the first paragraph.
Edit: I will clarify the questions in the comments
Let $I=S(\phi)$ be the action functional in Lagrangian form and let's assume that the Lagrangian generating functional is given by
$$Z[j]=\int \mathcal{D}[\phi] \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j\phi \}}$$
Obviously, we may change the integration variable $\phi$ without changing the integral. So that, if $\phi\equiv \chi + G(\chi)$, one obtains:
$$Z[j]=\int \mathcal{D}[\chi]\,\det(1+G'(\chi)) \exp{\{iS(\chi +G(\chi))+i\int d^4x\hspace{0.2cm} j(\chi + G(\chi))\}}$$
If we want to use this generating functional in terms of the field variable $\chi$, the determinant is crucial. If we had started with the action $S'(\chi)=S(\chi +G(\chi))=I$ — without knowing the existence of the field variable $\phi$ —, we would had derived the following Lagrangian version of the generating functional:
$$Z'[j]=\int \mathcal{D}[\chi]\,\det(1+G'(\chi)) \exp{\{iS'(\chi )+i\int d^4x\hspace{0.2cm}j \chi\}}$$
Note that $Z'[j]\neq Z[j]$ (but $Z[j=0]=Z'[j=0]$) and therefore the off-shell n-point functions are different. If we want to see if these generating functional give rise the same S-matrix elements, we can, as always, perform a change of integration variable without changing the functional integral. Let's make the inverse change, that is, $\chi\equiv\phi+F(\phi)$:
$$Z'[j]=\int \mathcal{D}[\phi]\, \det(1+F'(\phi)) \det(1+G'(\chi)) \exp{\{iS'(\phi+F(\phi) )+i\int d^4x\hspace{0.2cm} j(\phi + F(\phi))\}}=\int \mathcal{D}[\phi]\, \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j(\phi + F(\phi))\}}$$
So that, one has to introduce the n-point functions connected with $Z[j]$ and $Z'[j]$ in the LSZ formula and analyze if they give rise to same S-matrix elements, even though they are different n-point functions.
(Related question: Scalar Field Redefinition and Scattering Amplitude)
This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user drake
Q&A (4871)
