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I cannot find a definition for a symmetry of the generating functional in Quantum Field Theory: $$ Z[J] = \int \mathrm d \mu \, \exp\left\lbrace i S[J] \right\rbrace \, .$$ I know it's a simple question, but I cannot find a definition. What is the definition of a `symmetry' of this object? And what is its significance?

The generating functional $Z[J]$ is a function of $J$ that spits out a complex number. The only technically sound definition of a symmetry would be a map $f : J \mapsto J'$ which leaves $Z[J]$ invariant, i.e. $Z[J] = Z[f(J)]$. But that doesn't seem to have anything to do with the usual meaning of symmetry in this context.

Qmechanic
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Myridium
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  • Relevant: https://en.wikipedia.org/wiki/Ward%E2%80%93Takahashi_identity#Derivation_in_the_path_integral_formulation – Andrew Jun 14 '22 at 02:07
  • @Andrew - What relation does this bear to Ward-Takahashi identities? Those identities do not require any kind of symmetry at all in the generating functional. They are identities. You can write down an arbitrary field transformation and get the identity $1 = \left\langle \det ! \left( J \right) e^{iS[J]} \right\rangle$ where $J$ is the determinant of the Jacobian of the field transformation. It has no obvious relation to symmetry. No symmetry is required to write down these identities. – Myridium Jun 14 '22 at 06:53

3 Answers3

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Dual transformations : Let $\phi \to g\cdot\phi$ be a (non anomalous) symmetry of the theory in the usual way (with $g$ belonging to some group $G$): $$S[\phi] = S[g\cdot\phi] \quad \text{and}\quad \mathcal D\phi = \mathcal D(g\cdot \phi)$$

Then, we can find a transformation $J\to g\cdot J$ such that $\int J\cdot \phi = \int (g\cdot J) \cdot (g\cdot \phi)$. Performing the change of variable in the generating functional, we get : \begin{align} Z[J] &= \int\mathcal D\phi e^{iS[\phi] - i\int J\cdot \phi} \\ &= \int\mathcal D\phi e^{iS[g\cdot \phi] - i\int (g\cdot J) \cdot (g\cdot \phi)} \\ &= \int\mathcal D[g^{-1}\cdot \phi] e^{iS[\phi] - i\int(g\cdot J) \cdot\phi} \\ &= \int\mathcal D[\phi] e^{iS[\phi] - i\int(g\cdot J) \cdot\phi}\\ &=Z[g\cdot J] \end{align}

A few examples

  • for translation symmetry $\phi(x) \to \phi(x-a)$ the dual transformation is $J(x) \to J(x-a)$
  • for a $U(1)$ symmetry, $\phi(x) \to e^{i\alpha} \phi(x)$, the dual transformation is $J(x) \to e^{-i\alpha}J(x)$
  • for a $U(n)$ symmetry $\phi^a(x) \to {U^a}_b \phi^b(x)$, the dual transformation is $J_a(x) \to J_b(x) {(U^{-1})^b}_a$
SolubleFish
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  • What would be the point of transforming $J$ as well? – flippiefanus Jun 15 '22 at 07:51
  • When you compute $Z[J]$, $\phi$ is a dummy variable that is integrated over. If you perform the change of variable corresponding to the symmetry, you need to transform $J$ in order to express the result in terms of $Z$. – SolubleFish Jun 15 '22 at 10:03
  • But $J$ is a generating parameter, which is eventually set equal to zero. It has no physical representation. Therefore the transformation of $J$ has no physical significance. – flippiefanus Jun 15 '22 at 11:07
  • $J$ contains all the information about the fundamental fields (and maybe some composite operators). Although $J$ is set to $0$ at the end, derivative with respect to $J$ appear in correlation functions. The transformation of $J$ is the dual of that of the physical fields. – SolubleFish Jun 15 '22 at 11:09
  • No, it is not $J$ that contains the information. It is the generating function that contains the information. – flippiefanus Jun 15 '22 at 11:13
  • A specific $J$ is not physical. The space of all $J$s is part of the definition of the generating functional, together with is transformation laws. – SolubleFish Jun 15 '22 at 14:59
  • This is great, thank you. This may be a basic question, but how do you actually find the dual transformation? That was stopping me from actually being able to formulate the answer myself. I thought there might be a dual transformation as you call it, but didn't know where to begin proving that it exists or constructing it. – Myridium Jun 20 '22 at 22:18
  • @flippiefanus transforming the J is like changing coordinates in the expectation values. – Myridium Jun 20 '22 at 22:20
  • I suppose you could match the translation in the field with the translation in J, and make J transform in the opposite way within a gauge group. I can see that it's possible I suppose, but was having trouble phrasing it formally. – Myridium Jun 20 '22 at 22:24
  • I don't know if there is a more general way of phrasing this. It seems that what you are proposing (matching the space-time transformations and reversing the gauge/internal transformations) will work in most cases – SolubleFish Jun 21 '22 at 11:21
  • One other case would be conformal transformations : we need the integral $\int \text dx J\cdot \phi$ to be invariant. For conformal transformation, the jacobian is not $1$. If $\phi$ is a scaling operator with weight $\Delta$, then you would need to rescale $J$ with a weight $d-\Delta$. – SolubleFish Jun 21 '22 at 11:23
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  1. Yes, OP is right. Symmetries in terms of the underlying fields $\phi^k$ is in the generating functional/partition function/path integral $$Z[J]~=~\int \!{\cal D}\phi \exp\left(\frac{i}{\hbar}\left( S[\phi]+J_k\phi^k\right)\right) $$ transcripted into symmetries of the sources $J_k$.

  2. More generally, the same thing happens for correlation functions $$\langle F \rangle_J ~:=~ \frac{1}{Z[J]}\int \! {\cal D}\phi~\exp\left(\frac{i}{\hbar}\left(S[\phi] +J_k\phi^k \right)\right)F[\phi] .$$

  3. Note that since the fields $\phi^k$ might not transform as tensorial objects, the $\phi^k$-symmetries may get encoded in non-trivial manners in the $J$-picture. For this reason, one seldomly tries to classify symmetries of a theory starting from the $J$-picture alone, which seems to be at the core of OP's question. Symmetries are typically identified in the $\phi$-picture, and then transcribed to the $J$-picture. Also keep in mind that sources $J_k$ are usually put to zero at the end of the calculations.

  4. Similarly, the $J_k$-symmetries get in turn transcripted into symmetries of the classical fields $\phi^k_{\rm cl}$ in the 1-PI effective action $\Gamma[\phi_{\rm cl}]$.

  5. For an example see e.g. my Phys.SE answer here.

  6. There might also be symmetries involving external parameters.

Qmechanic
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  • Thanks for your response but I can't understand this as an answer to my question. I basically have no idea what you're saying. – Myridium Jun 14 '22 at 06:50
  • I would find it really helpful if you could use the gauge and/or chiral transformation of the QED Lagrangian as an example. The transformation of fields has an associated Ward-Takahashi identity, fine. But every transformation has a Ward-Takahashi identity. It's effectively a change of coordinates. I don't know what is meant by a quantum symmetry. Perhaps you could show how there is a derived relationship on the expectations of source terms $J$? And what kind of relation constitutes a symmetry and the motivation for calling it that? – Myridium Jun 14 '22 at 06:58
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The generating function $Z[J]$ produces all the correlation functions of theory. The correlation functions contain all the symmetries of theory. However, these symmetries may be hidden in the generating function.

Since the correlation functions are produced by the number of functional derivatives, we can represent the generating function function crudely as $$ Z[J] = Z[0]+J\langle \phi \rangle +\tfrac{1}{2}J^2\langle \phi^2 \rangle +\tfrac{1}{3!}J^3\langle \phi^3 \rangle + ... , $$ with some integral contractions between the $J$'s and correlation functions that are not shown and also normalizations not shown. The symmetries are inside the correlation functions $\langle \phi^n \rangle$. The $J$'s are just tags with which we can extract the appropriate correlation function. Their transformation would therefore not have any role to play. Apart from the formal role of $J$'s to serve as a device to extract the appropriate correlation function, they have no physical significance and therefore do not take part in any symmetry transformations.

flippiefanus
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