# A question about the proof of Goldstone's theorem

I am studying the textbook An introduction to Quantum Field Theory by Peskin and Schroeder, and am not able to understand the discussions in the part Goldstone's Theorem Revisited in Ch. 11.6 on P.388. It shows why the Goldstone's theorem is also valid in the presence of quantum corrections.

There, the authors first state that the effective potential $V_{eff}(\phi_{cl})$ encapsulates the full solution to the theory by including all orders of quantum corrections in terms of 1PI propagators and vertices. At the same time, it is invariant to the symmetries of the theory. Therefore, the arguments presented before for classical theory on P.352 near Eq.(11.13) can be inherited to show that the matrix $$\frac{\partial^2 V_{eff}}{\partial \phi_{cl}\partial \phi_{cl}}$$ has a zero eigenvalue.

Now there is only one piece missing to complete the proof, namely, a zero eigenvalue of the above matrix implies the existence of a massless particle after the quantum correction is considered.

It is exactly where I don't really understand the discussions in the textbook. I can only "roughly" see how the arguments lead to the connection: the zero eigenvalue can be shown on the diagonal term by a proper transformation (by properly choosing the degrees of freedom of the theory), and the second order derivative of the effective potential is closely related the mass due to its physical interpretation. But I have not really made myself understood. To be specific, I am not sure/much confused about the following few points.

• I understand that $\phi_{cl}$ is a constant field independent of the coordinate $x$. This is implied in the textbook (while exception is briefly discussed in the textbook and related to soliton solution). Therefore, the difference between $\Gamma$ and $V_{eff}$ is a constant factor of the 4-volume as shown in Eq.(11.50), besides the fact that the former is a functional and the latter is a function.

• It states that for a theory involves several scalar fields $\phi^i$, Eq.(11.105) $$\int d^4x \, e ^{ip\cdot (x-y)}\frac{\delta^2 \Gamma}{\delta \phi^i\delta \phi^j}(x,y)=0 \ \ \ \ \ \ \ \ \ (1)$$ corresponds to the equation $$p^2-m_0^2-M(p^2)=0 ,$$ whose solution determines the particle mass (the pole in the propagator). This was discussed before near Eq(11.97) on P.383. A zero mass implies that $p^2-m^2=p^2=0$ is a solution of the above equation. In other words, to show the existence of a massless particle is to show $p^2=0$ is a solution of the above equation. $p=0$ is a even stronger condition, if $p=0$, then $p^2=0$ is implied.

• A zero eigenvalue implies something like $$\frac{\partial^2 V_{eff}}{\partial \phi^i_{cl}\partial \phi^j_{cl}} \Delta^j(\phi) =0$$ as in Eq.(11.13). After some proper linear combination of the field, one can diagonalize the matrix and transfer the relevant degree of freedom to the first line and first row of the matrix, namely, $$\frac{\partial^2 V_{eff}}{\partial \phi^1_{cl}\partial \phi^1_{cl}}=0.$$ But if this is true, the above Eq.(1) seems always satisfied for any $p$, which cannot be right...

What am I missing? Many thanks in advance for the explanation!

Edit

According to the additional comments in Adam's answer, now I understand that the functional derivative of effective action is not proportional to the derivative of effective potential: $$iD^{-1}(x,y)=\frac{\delta^2\Gamma}{\delta\phi^i(x)\delta\phi^j(y)}\bigg|_{\phi=\phi_{cl}} \neq \frac{\partial^2}{\partial \phi^i_{cl}\partial \phi^j_{cl}}\left(\Gamma[\phi]|_{\phi_{cl}}\right)=\Omega\frac{\partial^2}{\partial \phi^i_{cl}\partial \phi^j_{cl}}\left(V_{eff}(\phi_{cl})\right).$$ But, evaluating the inverse propagator in its momentum space at $p=0$ does. To be specific, $$i\tilde{D}^{-1}(p=0)=i\int d^4x e^{i\cdot 0\cdot (x-y)}D(x,y)=\int d^4x e^{i\cdot 0\cdot (x-y)}\frac{\delta^2\Gamma}{\delta\phi^i(x)\delta\phi^j(y)}\bigg|_{\phi=\phi_{cl}}=\int d^4x\frac{\delta^2\Gamma}{\delta\phi^i(x)\delta\phi^j(y)}\bigg|_{\phi=\phi_{cl}}=\Omega\frac{\partial^2}{\partial \phi^i_{cl}\partial \phi^j_{cl}}\left(V_{eff}(\phi_{cl})\right).$$ The functional Taylor expansion in Adam's proof does the trick!

• What is $\Delta^{j}$ in Your notation? – Name YYY Jan 19 '17 at 16:06
• @Name YYY Thx for asking, $\Delta^j$ is a nonzero vector which is the eigenvector corresponds to the zero eigenvalue. It is non-zero due to the fact that the vacuum state is denerate and changes continusously with respect to the symmetry, see the discussions near Eq.(11.13). – gamebm Jan 20 '17 at 3:40
• @Cosmas Zachos, thx for the comments. I think my confusion comes mostly from the part related to the quantum corrections. The classical picture of spontaneous symmetric breaking is fully explained in the textbook near Eq.(11.13) on P.352. If any, I am sorry to bring up a quite specific question concerning a specific textbook. – gamebm Jan 20 '17 at 3:47
• I've posted a question concerning (11.105). – GotchaP Aug 17 '17 at 2:20
• I think that's because of the uncertainty principle. When $\tilde{\phi}(p) \sim \delta(p)$, one gets $\phi(x) \sim \int d^4p \, e^{-ip \dot x} \delta(p)=1$ – GotchaP Aug 17 '17 at 2:51

I will assume that the physical value of the classical field (the value that minimizes the effective action) is homogeneous, as it is the case in P&S. I'll note this value $\bar \phi_i$. Let's expand the effective action around $\bar \phi_i$ : $$\Gamma[\phi]=\Gamma[\bar \phi]+\frac12\int_{x,y} \Gamma^{(2)}_{ij}(x,y)(\phi_i(x)-\bar\phi_i)(\phi_j(y)-\bar\phi_j)+\cdots,$$ where $$\Gamma^{(2)}_{ij}(x,y)=\frac{\delta^2\Gamma}{\delta\phi_i(x)\delta\phi_j(y)}\bigg|_{\phi=\bar\phi}$$ is the physical value of the inverse propagator (i.e. the 2-point vertex function evaluated at the minimum of the potential).
If we evaluate $\Gamma[\phi]$ in constant field, we obtain by definition the effective potential $\Gamma[\phi]\big|_{\phi\, {\rm cst.}}=\Omega V(\phi)$, ($\Omega$ is the (space-time) volume) and thus $$V(\phi)=V(\bar \phi)+\frac12 \left(\frac1\Omega\int_{x,y}\Gamma^{(2)}_{ij}(x,y)\right)(\phi_i-\bar\phi_i)(\phi_j-\bar\phi_j)+\cdots$$ which gives us $$\frac{\partial^2 V}{\partial \phi_i\partial\phi_j}\bigg|_{\bar\phi}=\frac1\Omega\int_{x,y}\Gamma^{(2)}_{ij}(x,y)=\Gamma^{(2)}_{ij}(p=0),$$ which is the inverse propagator at zero momentum.
Edit : Note that functional derivation and evaluation at constant field do not commute, that is $$\frac{\delta^2\Gamma}{\delta\phi_i(x)\delta\phi_j(y)}\bigg|_{\phi=\varphi} \neq \frac{\partial^2}{\partial \varphi_i\partial \varphi_j}\left(\Gamma[\phi]|_{\varphi}\right),$$ where $\varphi$ is a constant field. This is most clearly seen in the following example of a one component field with $$\Gamma[\phi]=\int_x \left(\frac{(\nabla \phi(x))^2}{2}+U(\phi(x))\right).$$ We have $$\frac{\delta^2\Gamma}{\delta\phi(x)\delta\phi(y)}\bigg|_{\phi=\varphi}= \delta(x-y)(-\triangle+U''(\varphi)),$$ whereas $\Gamma[\phi]|_{\varphi}=\Omega U(\varphi)$ and thus $$\frac{\partial^2}{\partial \varphi\partial \varphi}\left(\Gamma[\phi]|_{\varphi}\right)=\Omega U''(\varphi)$$
• Thanks for the answer, I misunderstood that p=0 is related to that $\phi_{cl}$ is a constant field in space. In fact, after reading the textbook again, and also your answer, the latter is a simplified assumption. The former $p=0$ should satisfy the equality, Eq.(11.105), if $m=0$. – gamebm Sep 13 '17 at 3:01
• No, $p=0$ is not necessarily related to a constant field. It is an external parameter coming from the fourier transform. It is just that the effective potential is by definition evaluated at constant field, and that usually the physical value of the classical field is constant. However, I don't understand your last sentence, so maybe we agree anyway. – Adam Sep 13 '17 at 5:06