I have difficulty in understanding the path-integral formalism of SSB, and that of Effective Action.

Let's say a complex scalar field theory has the global $U(1)$ SSB, $$L(\phi)=(\partial^\mu \phi)^2-m^2\phi^\dagger\phi-\frac{1}{4}(\phi^\dagger\phi)^2$$

The proof of the masslessness of Goldstone bosons associated with this theory usually uses the $U(1)$ symmetry of the Effective action $\Gamma[\phi]$, which usually is proven by assuming that in the generating functional $$\tag{1} Z[J]=\int [D\phi D\phi^\dagger]e^{i\int d^4x[L(\phi)+J^\dagger\phi+\phi^\dagger J]}$$ the measure $[D\phi D\phi^\dagger]$ is also invariant under $\phi \rightarrow e^{i\theta}\phi$, and thereby proving the symmetry of $Z[J]$ under $J \rightarrow e^{i\theta}J$.

But the problem is, if we assume $[D\phi D\phi^\dagger]$ is invariant, we could easily get $$\langle\Omega|\phi|\Omega\rangle=e^{i\theta}\langle\Omega|\phi|\Omega\rangle$$, where $\Omega$ is the vacuum, and thus $$\tag{2} \langle\Omega|\phi|\Omega\rangle=0$$,which obviously contradicts SSB.

Where did I get wrong, or how to correctly prove that the effective action $\Gamma[\phi]$ indeed has the $U(1)$ symmetry?

let me make my question clearer.
The fact is that we don't really need the generating functional $Z[J]$ to calculate the VEV, we can simply go back to the functional-integral version of the VEV:
$$\tag{3} \text{VEV}=\int [D\phi D\phi^\dagger]\space \phi\space e^{iS[\phi]}$$ without doting about $J$.
Now the measure $[D\phi D\phi^\dagger]$ (and its region/asymptotic behavior) appearing in $(3)$ is the same as that in $(1)$, so if we assume its invariance to prove the symmetry of $Z[J]$, we can as well use it to prove $(2)$. The proof (if I'm right) goes like this:

$(3)$ just calculates VEV as a expect value of $\phi$, since everything is in the integral, the variables $\phi$ are dummy, we can just rename $\phi$ as $\phi^\prime$ without affecting anything: $$\tag{4} \text{VEV}=\int [D\phi^\prime D\phi^{\prime\dagger}]\space \phi^\prime\space e^{iS[\phi^\prime]}$$ next is the real stuff:
we make a variable transformation
$$\tag{5} \phi^\prime=e^{i\theta}\phi$$ and get $$\tag{6} \text{VEV}=\int [D(e^{i\theta}\phi) D(e^{-i\theta}\phi^\dagger)]\space e^{i\theta}\phi\space e^{iS[e^{i\theta}\phi]}$$ now according to our assumption $$\tag{7}\int[D(e^{i\theta}\phi) D(e^{-i\theta}\phi^\dagger)]=\int[D\phi D\phi^\dagger]$$ $$\text{and}\space\space\space\space\space\space S[e^{i\theta}\phi]=S[\phi]$$ we have, from $(6)$, $$\tag{8} \text{VEV}=\int [D\phi D\phi^\dagger]\space e^{i\theta}\phi\space e^{iS[\phi]}=e^{i\theta}\int [D\phi D\phi^\dagger]\space \phi\space e^{iS[\phi]}=e^{i\theta}\text{VEV}$$ thus we see that $$\text{VEV}=0$$ as in $(2)$.
Looking back, $(7)$ holds only if the functional integral is over an symmetrical region/class of field configurations $\phi(x)$, for example, if the integral is over all field configurations that vanish at infinity, $\phi(\infty)=0$.
So my conclusion is that, since $(8)$ contradicts SSB, the functional integration region shouldn't be invariant under $\phi \rightarrow e^{i\theta}\phi$, and thus picks up a particular/favored "orientation" in field's configurations, and thereby yields an nonzero VEV.
But if this is the case, I don't understand how the effective action $\Gamma[\phi]$ has the delightful property that it has the same symmetry as the classical action $S[\phi]$.

  • 3
    $\begingroup$ Of course $\langle \phi \rangle\neq 0$ is not invariant, that's why it is called SSB... Not sure what's left to say... $\endgroup$
    – Adam
    Nov 5, 2014 at 13:12
  • $\begingroup$ Note that the measure is invariant, since the Jacobian is $1$. $\endgroup$
    – Adam
    Nov 5, 2014 at 13:21
  • $\begingroup$ @Adam, but there are constraints in the measure, for example, the field approaches a certain nonzero value when the spacetime goes to infinity. $\endgroup$
    – LYg
    Nov 5, 2014 at 13:56
  • $\begingroup$ That has nothing to do with the global transformation you are doing. Call $\phi'=e^{i\theta}\phi$, compute the Jacobian, and you'll see. $\endgroup$
    – Adam
    Nov 5, 2014 at 13:59
  • $\begingroup$ @Adam, sorry I didn't make myself clear. if you make transformation $\phi'=e^{i\theta}\phi$ , then the asymptotic behavior of these $\phi'$ is not the same as that of $\phi$. this amounts to integral over a different class of field configuration. though the Jacobian is trivial, the integral region has changed, and so the measure at the end is changed. I think the vacuum infomation is encoded in the measure in the path-integral formalism, the change of the measure integral region corresponds to selection of a different vaccum. $\endgroup$
    – LYg
    Nov 6, 2014 at 3:09

2 Answers 2


There are various general ways to see the point. Let me give you a simple one that is still general enough (see also last comment at the end). Imagine you have a lagrangian density $L$ that is invariant under the continuous group transformation that acts linearly on the fields as $$ \phi\rightarrow \phi^\prime=U\phi=e^{i\omega^a T^a}\phi=(1+i\omega^a T^a+\ldots)\phi\,. $$ If this symmetry is realized on the states of the system, that is the symmetry isn't broken spontaneously, it means that the correlation functions are invariant too, $$\delta^a\langle\phi(x_1)\phi(x_2)\ldots\ldots\rangle=\langle\delta^a\phi(x_1)\phi(x_2)\ldots\rangle+\langle\phi(x_1)\delta^a\phi(x_2)\ldots\rangle+\ldots=0$$ where $\delta^a\phi(x)=iT^a\phi(x)$ is the infinitesimal transformation. (In the case at hand you are interested to the variation of the one-point function $\delta^a\langle\phi\rangle$). This invariance is equivalence to the so called Ward identity for the current matrix elements $$ \partial_\mu \langle J^{a\,\mu}(x)\phi(x_1)\phi(x_2)\ldots\phi(x_n)\rangle=\sum_{m}\delta^4(x-x_m)\langle \phi(x_1)\ldots\delta^a\phi(x_m)\ldots\phi(x_n)\rangle $$ as it is clear by integrating with respect to $x$ both sides IF we can discard the surface term on the left-hand side. (Under the same IF, one can define the associated Noether charges by integrating the correlators above in a certain way (see e.g. the CFT book by Di Francesco et al, sec. 2.4 for more details).

But in fact, spontaneous symmetry breaking occurs when the currents (that is, their matrix elements) don't die at infinity fast enough. Think of the Goldstone bosons $\pi^a(x)$ that in this case are generated by the the spontaneously broken currents $$ \langle J^a_\mu(x)\pi^b(y)\rangle\sim \delta^{ab} \int \frac{d^4p}{(2\pi)^4}e^{-ip(x-y)}\frac{-ip_\mu}{p^2+i\epsilon} $$ where $\pi(x)$ is the field (generically composite, that is built out of products of $\phi$'s) that creates the Goldstone boson from the vacuum. The tilde means that I am neglecting an overal numerical factor and taking also the large $x-y$ limit (the latter means that only massless modes contributes, and since the GB's have interactions that vanish as $p\rightarrow0$ we can use in that limit the free particle correlator). Indeed, we know from Goldstone theorem that $$\langle 0|J^{a\,\mu}(0)|\pi^b(p)\rangle=\delta^{ab}\frac{f_\pi p^\mu}{(2\pi)^3 2|p|}$$ so that $$ J^{a\,\mu}=f_\pi\partial_\mu\pi^{a}+\ldots $$ where the $\ldots$ are terms that vanish on the vacuum. From the 2-point function above one can therefore see that $$\partial_\mu \langle J^a_\mu(x)\pi^b(y)\rangle\sim \delta(x-y)$$ that integrated over $d^4x$ is not vanishing. Hence, the integration of the left-hand side of the Ward identity can't be discarded. In other words, in the theory with Goldstone bosons, the correlation functions of the (spontaneously broken) currents $J^{a\, \mu}$ do not vanish fast enough and therefore the vacuum expectation values of the field do not enjoy the same symmetry than the dynamics.

A last few comments about the relation with the path integral derivation of the symmetry of the correlation functions. Let's promote for a second the parameters $\omega^a$ of the transformation to spacetime dependent functions, $\omega^a\rightarrow \omega^a(x)$. The lagrangian was invariant with $\omega$ constant, so that at for an infinitesimal transformation we have $$ \delta S[\phi]=\int d^4x \partial_\mu \omega^a(x) J^{a\,\mu}(x)\,. $$ Now, look at the partition function $Z[J]=\int d\phi e^{i S[\phi]+J\cdot\phi}$ and change variable $\phi=e^{-i\omega^a(x)T^a}\phi^\prime=U^{-1}\phi^\prime$ $$ Z[j]=\int d\phi^\prime e^{iS[\phi^\prime]+j\cdot\phi^\prime}\left[1-\delta S[\phi^\prime]-i\omega^a J\cdot T^{a}\phi^\prime\right] $$ (assuming that the integration measure $d\phi$ is invariant as usual for non-anomalous symmetries). Since $\phi^\prime$ is summed over, we rename it back $\phi$ and get that $$ \int d\phi e^{iS[\phi]+j\cdot\phi}\left[\int d^4x \partial_\mu \omega^a(x) J^{a\,\mu}(x)+i\omega^a(x) j\cdot T^{a}\phi\right]=0\,. $$ For $\omega$ constant this is just the statement that $Z$ is invariant under the inverse (global) transformations. But more information can be extracted by taking an arbitrary varying $\omega(x)$ (with $\omega$ vanishing fast enough at infinity) for which $Z$ isn't invariant: we integrate by parts the first term above $$ \int d\phi e^{iS[\phi]+j\cdot\phi}\left[\int d^4 x \omega^a(x) \left( \partial_\mu J^{a\,\mu}(x)-i j^T(x) T^{a}\phi\right)\right]=0\, $$ By taking the functional derivatives $\delta^{(n)}/\delta j(x_1)\ldots \delta j(x_n)$ wrt to sources $j(x)$ (and then setting the $j=0$), we recover the Ward identities that we used above (remember that $\omega(x)$ is basically arbitrary) $$ \partial_\mu \langle J^{a\,\mu}(x)\phi(x_1)\phi(x_2)\ldots\phi(x_n)\rangle=\sum_{m}\delta^4(x-x_m)\langle \phi(x_1)\ldots\delta^a\phi(x_m)\ldots\phi(x_n)\rangle $$ which we argued don't imply the invariance of the correlation functions unless the current matrix elements on the left-hand side vanish fast enough. Again, this is not the case for spontaneously broken symmetries giving rise to Goldstone bosons.

Note added after more informations have been added in the original question

One can wonder how the VEV can be non zero simply by looking at the path integral with one $\phi$ insertion and no sources, namely $$ \langle\phi\rangle= \int d\phi \phi e^{iS[\phi]}\,. $$ Well, this formula formally assumes that the vacuum is a collection of harmonic oscillators wrt $\phi$ variables. In particular, the wave function $\langle\phi(x)|0\rangle$ of such a vacuum gives rise to the unfamous $+i\epsilon$ prescription in the field propagator (needed to makes correlation functions convergent) if $\langle\phi(x)|0\rangle$ is Gaussian (see Weinberg volume I chapter 9.2). This is nothing but a way to set the boundary conditions for the fields on the vacuum state. But here is the catch: for a spontaneous symmetry the would be vacuum wave function associated to charged fields is not Gaussian, and doesn't die off at infinity (see e.g. the discussion on the Ward identity above). So, the actual vacuum and the proper correlation functions associated with it are defined by the path integral over the fields $\pi(x)$ with a better behavior at infinity, that for which the vacuum wave function is gaussian-like $$\langle\pi(x)|0\rangle=\int d^4x d^4y e^{-\frac{1}{2}\pi(x)K(x-y)\pi(y)}$$ (where the kernel function is a Hankel function, see again Weinberg volume I, chapter 9.2 for more details). These are basically the fields that you obtained by shifting the fields around that their VEV's. What is left anyway are the Ward identity that tells you when all of this happens: when the current at infinity doesn't go to zero fast enough over certain field configurations (the GB's) that dominates the path integral at large distances.

Let me add the last comment: I have been working with a continuous symmetry because it allows one to track down explicitly the breaking of the symmetry for the correlation functions from their Ward identities. This is nice and it also shows why in 1+1 the SSB can't happen for continuous symmetries: there is no GB in 1+1 (since the correlation function $\langle\pi\pi\rangle$ diverges logarithmically in the IR) and therefore the current will have matrix elements well behaved at infinity and the symmetry will be linearly realized. The drawback of this explanation for continuous symmetries is, however, that says very little about discrete symmetries (that don't deliver Ward identities). In such a case I think that the best general physical explanation is given by Weinberg in volume II chapter 19.1. He explains very well that the SSB mechanism require infinitely many degrees of freedom (along with causality and momentum conservation) that forbids the transitions among different degenerate vacua which instead would restore a symmetric vacuum state as it happens in QM.

  • $\begingroup$ Thank you! Your derivation of the Ward identity indeed does't depend on the boundary behavior of $\phi$ in the integration measure, since you have assumed that $\omega$ vanishes fast enough at infinity. Can you please demonstrate the asymptotic behavior of $\langle J^{a\,\mu}(x)\phi(x_1)\rangle$ which is relevant to my question? $\endgroup$
    – LYg
    Nov 12, 2014 at 3:07
  • $\begingroup$ I added PS to my original question, please review it. $\endgroup$
    – LYg
    Nov 12, 2014 at 3:10
  • $\begingroup$ @LYg I have added extra lines to address your new part in the question. As for the the 2-poin function $\langle J^{a\,\mu}\pi^b\rangle$ I have revised the text to make the derivation more clear and consistent. $\endgroup$
    – TwoBs
    Nov 12, 2014 at 8:38

Since the conceptual problem is not really coming from field theory itself, let's look at the 0D case. This case is too naive (no SSB here), but the discussion is still correct qualitatively. The partition function is $$Z(j,j^*)=\int dz dz^* e^{-V(|z|^2)+ z^*j + j^* z}, $$ where $dzdz^*$ really means $d\Re z\,d\Im z$ and $V(x)=r\, x+x^2$ where $r$ can be positive or negative.

If we change the phase of $j$ and $j^*$ such that $j^{(*)}{}'=e^{(-)i\theta}j$, we find $$Z(j',j^{*}{}')=\int dz dz^* e^{-V(|z|^2)+ e^{i\theta}z^*j + e^{-i\theta}j^* z}, $$ and using $z'=e^{-i\theta}z$, and the fact that the jacobian and $V$ are invariant, we obtain that $$Z(j',j^{*}{}')=Z(j,j^*)=e^{W(j,j^*)}$$ is a function of $|j| {}^2$ only.

One can wonder how the SSB phase ($r<0$) could be obtained since $\bar z =\langle z\rangle=\frac{\partial W}{\partial j}(0,0)$ is expected to be zero by the symmetry discussed above. The reason is that in the SSB phase, $W$ is not analytic at small $j$. Typically, we have $$ W(j,j^*)=W(0,0)+2a\sqrt{j\, j^*}+\cdots,$$ and we obtain $\bar z = a\, e^{-i\arg j}$ ($a$ is a constant that depends on the system) which depends explicitly on the phase of $j$ and thus on the way we send $j$ to zero as expected. One can easily verify that by computing $W$ numerically, for example with mathematica.

EDIT: more details, concerning the OP's post scriptum.

Of course, if one compute $\bar z$ naively one find $$ \bar z = \int dz dz^* \, z\, e^{-V(|z|^2)}/Z(0,0)=0,$$ for all $r$, even in the SSB phase, by symmetry. But the solution of this problem is well known, one should compute $\bar z$ at finite sources $j=Je^{-i\Theta}$ and then send the sources to zero, i.e. $J\to 0$.

In the symmetric phase $\bar z (j,j^*)\propto j j^*=J^2 $ and one finds $\bar z(0^+,0^+)=0$, whereas in the SSB phase, one finds (see above) $\bar z (j,j^*)= a e^{i\Theta}$ as $J\to 0$, see above. Of course, since we are in the SSB phase, $\bar z$ depends on the phase of the source, and changing the phase of $j$ will change that of $\bar z$.

The same thing happens in the paragmagnetic-ferromagnetic transition in the Ising model where the sign of the magnetisation depends on the sign of the magnetic field $h$. I invite the OP to start with that case.

  • $\begingroup$ Thanks, could you please explain what $a$ is, and how you get that typical small $j$ dependence of $W(j,j^*)$? BTW, please see my PS, since I have further questions even if yours is the right answer to one aspect. $\endgroup$
    – LYg
    Nov 12, 2014 at 2:35
  • $\begingroup$ @LYg: see my edit. You should start with the easy case of Ising.You can also have a look at the appendix A of arXiv:1106.5585, where the same analysis is done for a simple quantum system that exhibits SSB. $\endgroup$
    – Adam
    Nov 12, 2014 at 13:15

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