# Equivalence Theorem of the S-Matrix

as far as I know the equivalence theorem states, that the S-matrix is invariant under reparametrization of the field, so to say if I have an action $S(\phi)$ the canonical change of variable $\phi \to \phi+F(\phi)$ leaves the S-matrix invariant.

In Itzykson's book there is now an exercise in which you have to show, that the generating functional $$Z^\prime(j)=\int \mathcal{D}[\phi] \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j(x)(\phi +F(\phi)\}}$$ gives the same S-matrix as the ordinary generating functional with only $\phi$ coupled to the current due to the vanishing contact terms. He then writes, that this proves the equivalence theorem, which I do not fully understand.

Suppose I take this canonical change of variable, then I get a new action $S^\prime(\phi)=S(\phi+F(\phi))$ and a generating functional $$Z(j)=\int \mathcal{D}[\phi] \exp\{iS(\phi+F(\phi))+i\int d^4x\hspace{0.2cm} j(x)\phi\}$$ If I now "substitute" $\phi+F(\phi)=\chi$ I get $$Z(j)=\int \mathcal{D}[\chi] \det\left(\frac{\partial \phi}{\partial \chi}\right) \exp\{iS(\chi)+i\int d^4x\hspace{0.2cm} j(x)\phi(\chi)\}$$ with $\phi(\chi)=\chi + G(\chi)$ the inverse of $\chi(\phi)$. Therefore comparing $Z(j)$ and $Z^\prime(j)$ I get an extra jacobian determinant.

Where is my fallacy, or why should the determinant be 1?

• To clarify, the function $F(\phi)(x)$ is a function only of $\phi(x)$, or is it allowed to depend on local derivatives, or is it a general smooth functional? Jul 2, 2013 at 17:54
• Echoing @BebopButUnsteady's comment, what is your definition of a "canonical change of variables" in a Lagrangian theory? Jul 2, 2013 at 18:06
• I mean a change of variables, which is invertible and has therefore a nonvanishing jacobi determinant. It should further be of the kind $x \to x + F(x)$ so to say a point transformation. The function $F(\phi)$ should only depend on $\phi(x)$ here. Jul 2, 2013 at 18:18
• There is a discussion in Zee (page 68, Appendix 2 : Field Redefinition) Jul 3, 2013 at 8:49
• @gaugi: I agree with you that this at least somewhat more subtle than the texts imply. I will try to write an (incomplete) answer summarizing what I've figured out. Jul 3, 2013 at 14:55

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
$$Z[j]=\int \mathcal{D}[\phi] \exp{\{iS(\phi)+i\int d^4x\hspace{0.2cm} j(x)\phi(x) \}}$$

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)

• My second formula was from the viewpoint, that I have an action $S(\phi)$ for which I do not know, that it is a somehow transformed other action, e.g. for the free case. Then I naively write my generating functional as my second formula, as I only know the action $S(\phi)$. I have transformed nothing there. Why is it wrong then? If I now transform this action I wrote to the free case I get the third formula which coincides with the transformation law you gave, but not with the formula from Itzykson due to the missing determinant. Jul 4, 2013 at 12:49
• By the way, I do understand that terms like $j(t)G(x)$ do not contribute in the S-Matrix due to LSZ, as they are contact terms and that I must not compare Green´s functions. Jul 4, 2013 at 12:51
• @drake: I believe we are on the same page. I was essentially trying to explain the second paragraph of your answer, which claims that if we were handed the non linear $S$ we would know to right down the det in the measure. I understand this as being the fact that a Lagrangian does not unambiguously define correlators, because contact terms are singular. Only certain prescriptions for these terms will lead to a coherent theory. The det is a manifestation of the fact that our usual prescriptions are not coherent for this lagrangian. Jul 5, 2013 at 4:24
• @gaugi: I think the point is that writing simply $W[J] = \int\exp(i\int\mathcal{L} +J\chi)$ gives pathological results when there are derivatives in the interaction. You need to start from the Hamiltonian prescription. So if someone hands you a Lagrangian $\mathcal{L}$ you should get the Hamiltonian, write down the path integral and then integrate out the momenta, which will get you the determinant. Jul 5, 2013 at 13:54
• @BebopButUnsteady Thank you! If the Hamiltonian density is $T_{ij}(q)p_ip_j+W_i(q)p_i+V(q)$, then the integral over momenta gives $(\det (T(q)))^{-1/2}$. $T_{ij}$ is often (but not always) a constant and thus it does not have any implication. Jul 5, 2013 at 21:41

I) Ref. 1 never mentions explicitly by name the following two ingredients in its proof:

1. The pivotal role of the Lehmann-Symanzik-Zimmermann (LSZ) reduction formula $$\left[ \prod_{i=1}^n \int \! d^4 x_i e^{ip_i\cdot x_i} \right] \left[ \prod_{j=1}^m \int \! d^4 y_j e^{-ik_j\cdot y_i} \right] \langle \Omega | T\left\{ \phi(x_1)\ldots \phi(x_n)\phi(y_1)\ldots \phi(y_m )\right\}|\Omega \rangle$$ $$~\sim~\left[ \prod_{i=1}^n \frac{i\langle \Omega |\phi(0)|\vec{\bf p}_i\rangle }{p_i^2-m^2+i\epsilon}\right] \left[ \prod_{j=1}^m \frac{i\langle \vec{\bf k}_j |\phi(0)|\Omega\rangle }{k_j^2-m^2+i\epsilon}\right] \langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$$ $$\tag{A} +\text{non-singular terms}$$ $$\text{for each} \quad p_i^0~\to~ E_{\vec{\bf p}_i} , \quad k_j^0~\to~ E_{\vec{\bf k}_j}, \quad i~\in~\{1, \ldots,n\}, \quad j~\in~\{1, \ldots,m\}.$$ [Here we have for simplicity assumed that spacetime is $\mathbb{R}^4$; that interactions take place in a compact spacetime region; that asymptotic states are well-defined; that there is just a single type of scalar bosonic field $\phi$ with physical mass $m$.]

2. That eqs. (9-102), (9-103), (9-104a) and (9-104b) on p.447 are just various versions of the Schwinger-Dyson (SD) equations. [The SD equations can be proved either via integration by part, or equivalently, via an infinitesimal changes in integration variables, in the path integral. The latter method is used in Ref. 1.]

On the middle of p. 447, Ref. 1 refers to a field redefinition $\varphi\to \chi$ as canonical if

[...] the relation $\varphi\to \chi$ may be inverted (as a formal power series).

This is certainly not standard terminology. Also it is a somewhat pointless definition, since any reader would have implicitly assumed without being told that field redefinitions are invertible. Note in particular, that Ref. 1 does not imply a Hamiltonian formulation with the word canonical.

II) The Equivalence Theorem states that the $S$-matrix $\langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$ [calculated via the LSZ reduction formula (A)] is invariant under local field redefinitions/reparametrizations.

In this answer, we will mainly be interested in displaying the main mechanism behind the Equivalence Theorem at the level of correlation functions (as opposed to carefully tracing the steps of Ref. 1 at the level of the partition function).

In the LSZ formula (A), let us consider an infinitesimal, local field redefinition

$$\tag{B} \phi~ \longrightarrow ~\phi^{\prime}~=~ \phi +\delta \phi$$

without explicit space-time dependences; i.e., the transformation

$$\tag{C} \delta \phi(x)~=~ f\left(\phi(x), \partial\phi(x), \ldots, \partial^N\phi(x)\right)$$

at the spacetime point $x$ depends on the fields (and their spacetime derivatives to a finite order $N$), all evaluated at the same spacetime point $x$. [If $N=0$, the transformation (C) is called ultra-local.]

One may now argue that near the single particle poles, this will only lead to a multiplicative rescaling on both sides of the LSZ formula (A) with the same multiplicative constant, i.e., the $S$-matrix is invariant. This multiplicative rescaling is known as wave function renormalization or as field-strength renormalization in Ref. 3.

III) Finally, let us mention that Vilkovisky devised an approach, where $1$-particle-irreducible (1PI) correlation functions are invariant off-shell under field reparametrizations, cf. Ref. 4.

References:

1. C. Itzykson and J-B. Zuber, QFT, (1985) Section 9.2, p. 447-448.

2. A. Zee, QFT in a Nutshell, 2nd ed. (2010), Chapter 1, Appendix B, p. 68-69. (Hat tip: Trimok.)

3. M.E. Peskin and D.V Schroeder, (1995) An Introduction to QFT, Section 7.2.

4. G.A. Vilkovisky, The Unique Effective Action in QFT, Nucl. Phys. B234 (1984) 125.

• This is an old answer but I had a couple of questions: 1. Aren't field redefinitions involving derivatives non-invertible? 2. Is the assumption that field redefinition be local necessary? Jul 19, 2017 at 2:25
• @Qmechanic How to proof your statement "One may now argue that near the single particle poles, this will only lead to a multiplicative rescaling on both sides of the LSZ formula (A) with the same multiplicative constant" ? Oct 10, 2021 at 9:10