I have the following definition of a general correlation function $$ \langle \Phi(x_1)\dots \Phi(x_n)\rangle = \frac{1}{Z} \int [d\Phi] \Phi(x_1)\dots\Phi(x_n)e^{-S[\Phi]} $$ I have only just started to learn about these functions, so could someone explain what this equation actually means? I see that parts are reminiscent of what you find in statistical mechanics, such as $Z$ denoting the partition function and $\exp(-S[\Phi])$ denoting the weight function or Boltzmann factor, and I think $[d\Phi]$ stands for the integration measure over a set or family of fields, so the bracketed notation emphasizes that, rather than an integration over points. But I can't make sense of all the pieces collectively.

My other question is to do with a derivation in the book by Di Francesco 'Conformal Field Theory' P.43. He defines the quantity $$ \langle X \rangle = \frac{1}{Z} \int [d\Phi'] (X + \delta X) e^{-S[\Phi] - \int d x \partial_{\mu}j^{\mu}_a \omega_a(x)}, $$ where $X$ is a collection of fields and $\delta X$ is its variation under the transformation. He then expands this result to first order in $\omega(x)$ to obtain $$\langle \delta X \rangle = \int d x \partial_{\mu}\langle j^{\mu}_a(x)X\rangle\omega_a(x)$$ and then identifies $$\delta X = -i\sum_{i=1}^{n} (\Phi(x_1)\dots G_a \Phi(x_i)\dots \Phi(x_n))\omega_a (x_i)$$ but I am not sure how he obtained these two equations.

Any help would be great, many thanks.


1 Answer 1


For any (local) observable $\mathcal{O}$, its expectation value is defined as

$$ \langle \mathcal{O} \rangle = \frac{1}{Z}\int \mathcal{D}\Phi\mathcal{O}\mathrm{e}^{-S_E[\Phi]}$$


$$ Z = \int \mathcal{D}\Phi\mathrm{e}^{-S_E[\Phi]}$$

is indeed called the partition function. My writing $S_E$ is intended to show that this is for a Euclidean theory (i.e. a theory on a Riemannian manifold), for a Minkowskian theory (i.e. a theory on a Lorentzian manifold), one would have to add some $\mathrm{i}$ by Wick rotation. I have used $\mathcal{D}\Phi$ instead of $[\mathrm{d}\Phi]$ to denote the measure on the space of field configurations, both notations are widely used. It is important to note that, in most cases, this measure cannot be rigorously constructed and is only defined in some regularization procedure, but naively you can think of it as any kind of integration over the space of all possible field configurations $\Phi : \Sigma \rightarrow \mathbb{R}$, if $\Sigma$ is your spacetime manifold.

Now, your intuition about this thing are basically correct - $\mathrm{e}^{-S_E[\Phi]}$ is indeed a Boltzmann-like weighting factor, that, in the QFT case, weighs different field configurations $\Phi$. It is obvious that this factor will be maximal at minimal $S_E[\Phi]$, so the classically realized field configuration $\Phi_0$ for which $S_E[\Phi_0]$ is a minimum of the action contributes the most to this integral. The expectation value of the observable is indeed defined as an exact analogon to classical statististical mechanics.

You write something about a transformation, and from the context of Ward identities, I am assuming that you are talking about a gauge transformation, which should be a symmetry of the theory. Explicitly, the transformation is infintesimally given by

$$ \Phi \mapsto (1 + \mathrm{i}\omega_a(x)T^a)\Phi \equiv \Phi' \tag{1}$$

where the $T^a$ are generators of the symmetry group. Being a symmetry of the theory means that no observable may change its value under the transformation. We denote quantities after transformation by primes. Looking at some observable $\mathcal{O}$ (your $X$), we must have that

$$ \langle \mathcal{O} \rangle = \frac{1}{Z}\int \mathcal{D}\Phi\mathcal{O}\mathrm{e}^{-S_E[\Phi]} \overset{!}{=} \frac{1}{Z'}\int \mathcal{D}\Phi' (\mathcal{O} + \delta\mathcal{O})\mathrm{e}^{-S_E[\Phi]-\int\mathrm{d}^d x \partial_\mu j^\mu_a \omega_a(x)}$$

We now assume that our symmetry is non-anomalous, i.e. $\mathcal{D}\Phi' = \mathcal{D}\Phi$. Then, for the equality to hold, we have:

$$\int \mathcal{D}\Phi\mathcal{O}\mathrm{e}^{-S_E[\Phi]} \overset{!}{=} \int \mathcal{D}\Phi (\mathcal{O} + \delta\mathcal{O})\mathrm{e}^{-S_E[\Phi]-\int\mathrm{d}^d x \partial_\mu j^\mu_a \omega_a(x)}$$

If you now expand $\mathrm{e}^{-\int\mathrm{d}^d x \partial_\mu j^\mu_a \omega_a(x)}$ to first order in $\omega_a$, you will get $\langle \mathcal{O} \rangle$ on both sides of the equation, which cancels, and the remaining terms are precisely the Ward identities, i.e. the first equation you asked about.

The second equation follows from doing the explicit transformation on $\mathcal{O}$: (Assuming, as you said, that the $G_a$ are the generators $T^a$ of the transformation): Look again at $(1)$. Since you said that $\mathcal{O}$ is a collection of fields, it is

$$ \mathcal{O} = \prod_{i=1}^n \Phi(x_i)$$

for some $n$. Now, carry out $(1)$ on every field $\Phi$ and keep only the terms at most first order in $\omega_a$. This is exactly your $\delta \mathcal{O}$.

(This is "exact" since we consider an infinitesimal gauge transformation, anyways. There are rigorous foundations in Lie theory for this "throwing away" of higher order terms. Nevertheless, it is quite important to carry these tricks the physicist likes so much oneself, since the answer coming out is correct.)

  • $\begingroup$ Hi ACuriousMind, upon expanding, I am left with $$\langle \delta X \rangle = \int d^d x (\partial_{\mu}j^{\mu}_a) \omega_a(x) \cdot \frac{1}{Z}\int [d\Phi](X+\delta X)e^{-S[\Phi]}$$ but I don't see how the result follows from this. Also, in my notation, it looks like my $G_a \rightarrow T_a$ in yours. Thanks. $\endgroup$
    – CAF
    Commented Jul 15, 2014 at 13:58
  • $\begingroup$ @CAF Move the $\partial_\mu j_a^\mu$ into the $\Phi$ integral, this will directly yield $\partial_\mu \langle j_a^\mu (X + \delta X) \rangle$. And the $\delta X$ you can throw away, since $\omega_a \delta X$ is second order in $\omega_a$ (the change $\delta X$ under transformation must be at least of order 1 in the transformation parameter). $\endgroup$
    – ACuriousMind
    Commented Jul 15, 2014 at 14:07
  • $\begingroup$ Thanks, but wouldn't writing something like $\partial_{\mu} \langle j^{\mu}_a X \rangle$ suggest that the partial acts on everything in the brackets $\langle \dots \rangle$? $\endgroup$
    – CAF
    Commented Jul 15, 2014 at 14:16
  • $\begingroup$ @CAF: Ah, the eternal struggle with notation. Yes and no. $\partial_\mu$ acts upon precisely one $x_i$ coordinate - and here, it's the one $j_a^\mu$ depends on, so the $X$ is left untouched by it. I agree it's a weird way to write it, but that's how it's customary. (by the way, let me also know if the way to get your second equation I indicated in my updated answer satisfies you) $\endgroup$
    – ACuriousMind
    Commented Jul 15, 2014 at 14:20
  • $\begingroup$ I see, so writing that out more explicitly $\partial_{\mu}\langle j^{\mu}_a X\rangle = \partial_{\mu}\langle j^{\mu}_a \Phi(x_1)\dots \Phi(x_n)\rangle$ and all the $\Phi(x_i)$ are untouched by the $\partial_{\mu}$? Before I understand your updated answer, what does the quantity $\langle \Phi(x_1)\dots \Phi(x_n) \rangle$ represent? As I read it, we are multiplying together the value of the field at different positions $x_i$ defined on some base manifold and then taking the average of this quantity over some domain containing all points $x_i$? $\endgroup$
    – CAF
    Commented Jul 15, 2014 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.