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  Defining quantum effective action (Legendre transformation), existence of inverse (field - source)?

+ 3 like - 0 dislike

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?

This post imported from StackExchange Physics at 2014-04-13 14:08 (UCT), posted by SE-user Thomas
asked Apr 11, 2014 in Theoretical Physics by UnknownToSE (505 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-13 14:08 (UCT), posted by SE-user Adam
answered Apr 11, 2014 by Adam (115 points) [ no revision ]
+ 1 like - 0 dislike

The inverse (and the Legendre transform) does not exist in general, not even for QFTs in a space-time of finitely many points  (where everything is simple multivariater analysis; see https://en.wikipedia.org/wiki/Legendre_transform ).

One usually assumes silently that the second functional derivative is uniformly positive definite; this guarantees existence (and explains why some theories have broken symmetries).

answered Apr 13, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

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