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Equivalence Theorem of the S-Matrix

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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?

Thank you in advance

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user gaugi
asked Jul 2, 2013 in Theoretical Physics by gaugi (70 points) [ no revision ]
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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?

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user BebopButUnsteady
Echoing @BebopButUnsteady's comment, what is your definition of a "canonical change of variables" in a Lagrangian theory?

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user Qmechanic
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.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user gaugi
There is a discussion in Zee (page 68, Appendix 2 : Field Redefinition)

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user Trimok
thank you, but this is exactly as in Itzykson´s book with a generating functional $Z^\prime$ and without this determinant. He then shows with LSZ that those contact terms I mentioned above vanish, what I understand. What I do not understand is why you can simply take this generating functional $Z^\prime$ instead of transforming the functional with the new action, as I did above.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user gaugi
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Weinberg talks about this rather unhelpfully in Vol. 1 Sec 9.3 and there references this PRD prd.aps.org/abstract/PRD/v3/i10/p2486_1 from 1971.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user BebopButUnsteady
@BebopButUnsteady I have tried to clarify your doubts in an edit of my answer.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user drake

2 Answers

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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
answered Jul 3, 2013 by drake (880 points) [ no revision ]
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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.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user gaugi
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.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user gaugi
@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.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user BebopButUnsteady
@BebopButUnsteady Are you talking about ordering prescriptions? If yes, these are always present and are related to the definition of the measure in the path integral. However, there is not always a determinant. I'm perhaps misunderstanding you (?).

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user drake
thank you again for your answer, and sorry if I should be slow on the uptake, but if I only had the action $S(\chi)$ and not the transformation $\chi+G(\chi)$ and wanted to get the generating functional $Z^\prime(\chi)$ I would not get the determinant you have in your $Z^\prime(\chi)$, as I really do not know about the transformation.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user gaugi
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@BebopButUnsteady: Thanks, that was the point I was missing. @ drake: Thank you very much for your long and enlightening answer.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user gaugi
@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.

This post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user drake
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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 post imported from StackExchange Physics at 2014-03-31 22:22 (UCT), posted by SE-user Qmechanic
answered Jul 5, 2013 by Qmechanic (2,580 points) [ no revision ]

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