Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a Quantum field theory, for a scalar field $\phi$ with generic Action $S[\phi]$, we have the generating functional $$Z[J] = e^{iW[J]} = \frac{\int \mathcal{D}\phi e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi e^{iS[\phi]}}.$$

The one-point function in the presence of a source $J$ is.

$$\phi_{cl}(x) = \langle \Omega | \phi(x) | \Omega \rangle_J = {\delta\over\delta J}W[J] = \frac{\int \mathcal{D}\phi \ \phi(x)e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi \ e^{i(S[\phi]+\int d^4x J(x)\phi(x))}}.$$

The effective Action is defined as the Legendre transform of $W$

$$\Gamma[\phi_{cl}]= W[J] -\int d^4y J(y)\phi_{cl}(y),$$ where $J$ is understood as a function of $\phi_{cl}$.

That means we have to invert the relation $$\phi_{cl}(x) = {\delta\over\delta J}W[J]$$ to $J = J(\phi_{cl})$.

How do we know that the inverse $J = J(\phi_{cl})$ exists? And does the inverse exist for every $\phi_{cl}$? Why?

share|cite|improve this question

This is an interesting question, and although I don't know a rigorous answer, we can discuss some typical cases.

Usually, the inverse exists, but the cases where this inverse does not exist are not necessarily pathological (sound models can have the problem that the inverse does not exist).

For standard field theories (say, $\phi^4$, O(N) models, classical spins models, ...), generically the inverse exists, and this can be shown order by order in a loop expansion (I don't know if this has been proven at all order, but in standard textbooks, this is shown to order 1 or 2). However, the inverse will not exist necessarily for all $\phi_{cl}$, especially in broken symmetry phases. Indeed, an ordered phase is characterized by $$\bar\phi_{cl}=\lim_{J\to 0 } \phi_{cl}[J]=\lim_{J\to 0 }W'[J]\neq 0 ,$$ where $\bar \phi_{cl}$ is the equilibrium value of the order parameter. Therefore, you cannot inverse the relationship $\phi_{cl}[J]$ for $\phi_{cl}\in [0,\bar \phi_{cl}]$ ($\phi_{cl}[J]$ generically increases when $J$ increases).

Furthermore, there are cases where the inverse is simply not defined, because $\phi_{cl}[J]={\rm const}$ for all $J$. This is usually the case when the field has no independent dynamics without a source. For instance, if you take a single quantum spin at zero temperature, the only dynamics is given by external magnetic field (here in the $z$ direction) $$\hat H= -h.\sigma_z.$$ With $h>0$, the ground state is always $|+\rangle$, and the "classical field" $\phi_{cl}(h)=\langle \sigma_z\rangle=1/2$ for all $h$, and the Gibbs free energy (the Legendre transform of the free energy with respect to $h$, which is essentially the effective action) does not exist.

share|cite|improve this answer

If we treat the generating functional $W_c[J]$ for connected diagrams as a formal power series in the sources $J_i$, and if the connected propagator$^1$

$$ W_c^{ij} ~=~\langle \phi^i \phi^j \rangle~=~\left. \frac{\delta^2 W_c[J]}{\delta J_i\delta J_j }\right|_{J=0} $$

is invertible, then the effective action

$$ \Gamma[\phi_{\rm cl}]~=~W_c[J]-J_i \phi^i_{\rm cl} $$

exists as a formal power series in the Legendre transformed variable $\phi_{\rm cl}$. In particular, the inversion of the formal power series

$$\phi^i_{\rm cl}~=~\frac{\delta W_c[J]}{\delta J_i} $$

then follows from a multi-variable generalization of Lagrange inversion theorem.


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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