# Why should Ward identities only be used with the effective action (as opposed to the generating functional for connected diagrams)?

My question is about the derivation of Ward identities. I will sketch it here in the case of an O(N) symmetric model and point out what it bothering me when I am done. I am being very sloppy with the notation. Please ask in the comments if you don't understand it. I assume that you know about the generating functional of connected correlation functions $W[j] = \ln[Z[j]]$ and its Legendre transform $\Gamma[\phi]$.

Consider an action that depends on an $N$ component field, $\phi_a$, and is invariant under rotations in this space, $$\phi_a \rightarrow \phi_a = U(\theta_\alpha)_{ab} \phi_b' = \left[\text{e}^{i \theta_\alpha J^\alpha}\right]_{ab} \phi_b' \, .$$ $J^\alpha$ are the generators of the rotations and $\theta_\alpha$ the corresponding "angles".

The ward identities are derived by making a change of variables in the generating functional $$Z[j^a] = \int D\phi \, \text{e}^{S\left[ \phi_a \right]+\int j^a \phi_a} = \int D \phi' \, \text{e}^{i S[U(\theta)\phi']+\int j^a U(\theta)^{ab} \phi_b'} \, .$$ After renaming $\phi'\rightarrow \phi$ and exploiting the symmetry of the action $S[U(\theta)\phi']=S[\phi']$ we write $$Z[j^a] = \int D \phi \, \text{e}^{i S[\phi]+\int j^a U(\theta)^{ab} \phi_b} \, .$$ Finally we expand to linear order in $\theta$, $$Z[j^a] = \int D \phi \, \text{e}^{i S[\phi]+\int j^a \phi_a} \left(1 + i \theta _\alpha \int j^a \left[J^{\alpha}\right]_{ab}\phi^b\right) \, .$$ We conclude that $$\int j^a \left[J^{\alpha}\right]_{ab} \langle \phi^b \rangle = 0 \, . \tag{1}$$ $\langle \phi^b \rangle$ is the expectation value of $\phi^b$ with the source $j$, $$\langle \phi^b \rangle = \frac{\delta Z}{\delta j^b} \, .$$

Next this is expressed in terms of the effective action, $\Gamma[\phi] = -\log[Z[j]] + \int j^a \phi_a \, ,$

$$\int \frac{\delta \Gamma}{\delta \phi_a} \left[J^{\alpha}\right]_{ab} \phi^b = 0 \, . \tag{2}$$

Taking one field derivative of the last equation and evaluation at physical solutions, $j=\delta\Gamma/\delta \phi = 0$ leads to $$\int \frac{\delta^2 \Gamma}{\delta \phi_c \delta \phi_a} \left[J^{\alpha}\right]_{ab} \phi^b = 0 \, .$$ When the fourier transform of this equation is taken we find Goldstones' theorem $$\frac{\delta^2 \Gamma}{\delta \phi_c \delta \phi_a}(p=0) \left[J^{\alpha}\right]_{ab} \phi^b = 0\, .$$ I.e. for each generator of the symmetry there is one mode with zero mass.

My question is the following: As I understand it, Eq. (1) is valid for any choice of the source $j$. However if I use it to constrain correlation functions, I get (after one derivative with respect to $j^c$ and evaluating at $j=0$, $$\int \left[J^{\alpha}\right]_{cb} \langle \phi^b \rangle = 0 \, .\tag{3}$$ I find that the correlation functions are symmetric, $\langle \phi^b \rangle = 0$. This simply tells me that there is no symmetry breaking. What went wrong here? Why can I use the ward identity in terms of $\Gamma$, Eq. (2) and not the one which is written in terms of $Z$, Eq. (1)?

• I'm confused: 1. Where is there a Ward identity here, which would be $\partial_\mu \langle j^\mu \prod_i\phi_i \rangle = - \mathrm{i}\sum_j \langle \prod_{i\ne q j} \phi_i \delta \phi_j \delta(x-x_j)\rangle$ for a Noether current $j^\mu$? 2. "Goldstone's theorem" seems not to be a statement. 3. Where did you break the symmetry? – ACuriousMind Apr 18 '15 at 22:27
• Thanks for the "Goldstone" comment. I corrected it. By Ward identity, I mean the equations that follow from eq (2) and when evaluated at zero source. Sorry, if I use the wrong terminology. The symmetry is broken if $\langle \phi^b \rangle \neq 0$. This implies $\delta^2\Gamma/\delta^2 \phi (p=0) = 0$. – Steven Mathey Apr 18 '15 at 23:15

I) OP is asking about the case where the infinitesimal symmetry transformations

$$\delta\phi^a~=~t^a{}_b \phi^a \tag{A}$$

are linear in the fields in the path integral$^1$

$$Z[J] ~=~\exp\left[\frac{i}{\hbar}W_c[J]\right]~=~\int \! {\cal D}\phi~\exp\left[\frac{i}{\hbar}\left( S[\phi]+J_a\phi^a \right)\right]. \tag{B}$$

Recall the Legendre transformation between the generating functional $W_c[J]$ for connected diagrams and effective action

$$\Gamma[\phi_{\rm cl}]~=~W_c[J]-J_a \phi^a_{\rm cl}.\tag{C}$$

The classical fields becomes equal to the quantum average

$$\phi^a_{\rm cl}~=~\langle \phi^a \rangle_J. \tag{D}$$

$$\forall J:~~ J_a \langle \delta \phi^a \rangle_J ~=~ 0, \tag{E}$$

cf. Ref. 1. Combining eqs. (A), (C), (D) and (E), we get

$$\forall \phi_{\rm cl}:~~ \frac{\delta\Gamma[\phi_{\rm cl}]}{\delta\phi^a_{\rm cl}} t^a{}_b ~\phi^b_{\rm cl}~=~ 0. \tag{F}$$

Differentiation of eq. (F) wrt. $\phi_{\rm cl}$ yields

$$\forall \phi_{\rm cl}:~~\Delta^{-1}_{ba}~t^a{}_c ~\phi^c_{\rm cl}~=~ \frac{\delta\Gamma[\phi_{\rm cl}]}{\delta\phi^a_{\rm cl}} t^a{}_b ,\qquad \Delta^{-1}_{ab}~=~-\frac{\delta\Gamma[\phi_{\rm cl}]}{\delta\phi^a_{\rm cl}\delta\phi^b_{\rm cl}},\tag{G}$$

from which we (via standard arguments given in Ref. 1) can deduce for $J=0$ that non-zero VEV leads to zeromodes for the matrix $\Delta^{-1}$, so that the inverse matrix (traditionally denoted $\Delta$) does not exist.

III) On the other hand, combining eqs. (A) and (E), we get

$$\forall J:~~ J_a ~t^a{}_b \langle \phi^b \rangle_J ~=~ 0, \tag{H}$$

which is OP's eq. (1). Differentiation of eq. (H) wrt. $J$ yields

$$\forall J:~~ t^a{}_b ~\langle \phi^b \rangle_J +J_c ~t^c{}_b ~\Delta^{ba}~=~ 0,\qquad \Delta^{ab}~=~\frac{\delta^2 W_c[J]}{\delta J_a \delta J_b}\tag{I}.$$

The problem with eq. (I) is that $\Delta$ might become singular for $J=0$, so that one can not drop the second term in eq. (I) to derive OP's eq. (3).

References:

1. S. Weinberg, Quantum Theory of Fields, Vol. 2; Sections 16.1, 16.4, and 19.2.

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$^1$ We use DeWitt condensed notation to not clutter the notation.

• Thanks, that's very helpful. I have a question though: Is it possible to recover Eq (G) from Eq (I) in the limit $J \rightarrow 0$ ? I see what I am doing wrong when I use Eq (I), but I have no idea how to fix it. – Steven Mathey Apr 20 '15 at 8:12
• Yes, multiply with $\Delta^{-1}$ on both sides of eq. (I) to get eq. (G). – Qmechanic Apr 20 '15 at 8:20