Quantum Field Theory/QFT Schwinger-Dyson

In quantum field theory, action is given by the functional S of field configurations (which only depends locally on the fields), then the time ordered vacuum expectation value of polynomially bounded functional F, , is given by


 * $$=\frac{\int \mathcal{D}\phi F[\phi]e^{iS[\phi]}}{\int\mathcal{D}\phi e^{iS[\phi]}}$$

In fact, on shell equations for the classical case usually have their quantum analog because, in a hand wavy way, when integrating over regions of the configuration space which are significantly off shell, the rapidly oscillating phases would tend to produce "destructive interference" wheareas for regions close on shell, we tend to have "constructive interference".

For example, what is the analog of the on shell Euler-Lagrange equations, $$\frac{\delta}{\delta \phi}S[\phi]=0$$?

If the functional measure $$\mathcal{D}\phi$$ turns out to be translationally invariant (we'll assume this for the rest of this article, although this does not hold for, let's say nonlinear sigma models) and if we assume that after a Wick rotation


 * $$e^{iS[\phi]}$$,

which now becomes


 * $$e^{-H[\phi]}$$

for some H, goes to zero faster than any reciprocal of any polynomial for large values of &phi;, integrate by parts (after a Wick rotation, followed by a Wick rotation back) to get the following Schwinger-Dyson equations:


 * $$<\frac{\delta}{\delta \phi}F[\phi]>=-i$$

for any polynomially bounded functional F.

These equations are the analog of the on shell EL equations.

If J (called the source field) is an element of the dual space of the field configurations (which has at least an affine structure because of the assumption of the translational invariance for the functional measure then, the generating functional Z of the source fields is defined to be:

$$Z[J]=\int \mathcal{D}\phi e^{i(S[\phi]+)}$$

Note that

$$\frac{\delta^n Z}{\delta J(x_1) \cdots \delta J(x_n)}[J]=i^n Z[J] <\phi(x_1)\cdots \phi(x_n)>_J$$

where

$$_J=\frac{\int \mathcal{D}\phi F[\phi]e^{i(S[\phi]+)}}{\int\mathcal{D}\phi e^{i(S[\phi]+)}}$$

Basically, if $$\mathcal{D}\phi e^{iS[\phi]}$$ is viewed as a functional distribution (this shouldn't be taken too literally as an interpretation of QFT, unlike it's Wick rotated statistical mechanics analogue, because we have time ordering complications here!), then $$<\phi(x_1)\cdots \phi(x_n)>$$ are its moments and Z is its Fourier transform.

If F is a functional of &phi;, then for an operator K, F[K] is defined to be the operator which substitutes K for &phi;. For example, if $$F[\phi]=\frac{\partial^{k_1}}{\partial x_1^{k_1}}\phi(x_1)\cdots \frac{\partial^{k_n}}{\partial x_n^{k_n}}\phi(x_n)$$ and G is a functional of J, then $$F[-i\frac{\delta}{\delta J}]G[J]=(-i)^n \frac{\partial^{k_1}}{\partial x_1^{k_1}}\frac{\delta}{\delta J(x_1)} \cdots \frac{\partial^{k_n}}{\partial x_n^{k_n}}\frac{\delta}{\delta J(x_n)} G[J]$$.

Then, from the properties of the functional integrals, we get the "master" Schwinger-Dyson equation:

$$\frac{\delta S}{\delta \phi(x)}[-i \frac{\delta}{\delta J}]Z[J]+J(x)Z[J]=0$$

If the functional measure is not translationally invariant, it might be possible to express it as the product $$M[\phi]\mathcal{D}\phi$$ where M is a functional and $$\mathcal{D}\phi$$ is a translationally invariant measure. This is true, for example, for nonlinear sigma models where the target space is diffeomorphic to Rn. However, if the target manifold is some topologically nontrivial space, the concept of a translation does not even make any sense.

In that case, we would have to replace the S in this equation by another functional $$\hat{S}=S-i\ln(M)$$

If we expand this equation as a Taylor series about J=0, we get the entire set of Schwinger-Dyson equations.

Now how about the on shell Noether's theorem for the classical case? Does it have a quantum analog as well? Yes, but with a caveat. The functional measure would have to be invariant under the one parameter group of symmetry transformation as well.

Let's see how it goes. Let's just assume for simplicity here that the symmetry in question is local (I don't mean local in the gauge sense. I mean local in the sense that the transformed value of the field at any given point under an infinitesimal transformation would only depend on the field configuration over an arbitrarily small neighborhood of the point in question.). Let's also assume that the action is local in the sense that it is the integral over spacetime of a Lagrangian and that $$Q[\mathcal{L}(x)]=\partial_\mu f^\mu (x)$$ for some function f where f only depends locally on &phi; (and possibly the spacetime position). If we don't assume any special boundary conditions, this would not be a "true" symmetry in the true sense of the term in general unless f=0 or something. Here, Q is a derivation which generates the one parameter group in question. We could have antiderivations as well, like for example BRST and supersymmetry. Let's also assume

$$\int \mathcal{D}\phi Q[F][\phi]=0$$ for any polynomially bounded functional F. This property is called the invariance of the measure. And this does not hold in general. See anomaly (physics) for more details.

Then,

$$\int \mathcal{D}\phi Q[F e^{iS}][\phi]=0$$, which implies

$$+i=0$$ where the integral is over the boundary. This is the quantum analog.

Now, let's assume even further that Q is a local integral $$Q=\int d^dx q(x)$$ where q(x)[&phi;(y)]=&delta;(d)(x-y)Q[&phi;(y)] so that $$q(x)[S]=\partial_\mu j^\mu (x)$$ where $$j^{\mu}(x)=f^\mu(x)-\frac{\partial}{\partial (\partial_\mu \phi)}\mathcal{L}(x) Q[\phi]$$ (this is assuming the Lagrangian only depends on &phi; and its first partial derivatives! More general Lagrangians would require a modification to this definition!). Note that we're NOT insisting that q(x) is the generator of a symmetry (i.e. we're NOT insisting upon the gauge principle), but just that Q is. And let's also assume the even stronger assumption that the functional measure is locally invariant:

$$\int \mathcal{D}\phi q(x)[F][\phi]=0$$.

Then, we'd have

$$+i=+i=0$$

Alternatively,

$$q(x)[S][-i \frac{\delta}{\delta J}]Z[J]+J(x)Q[\phi(x)][-i \frac{\delta}{\delta J}]Z[J]=\partial_\mu j^\mu(x)[-i \frac{\delta}{\delta J}]Z[J]+J(x)Q[\phi(x)][-i \frac{\delta}{\delta J}]Z[J]=0$$

The above two equations are the Ward-Takahashi identities.

Now for the case where f=0, we can forget about all the boundary conditions and locality assumptions. We'd simply have

=0.

Alternatively,

$$\int d^dx J(x)Q[\phi(x)][-i \frac{\delta}{\delta J}]Z[J]=0$$

An example: &phi;4
To give an example, suppose

$$S[\phi]=\int d^dx \left (\frac{1}{2} \partial^\mu \phi(x) \partial_\mu \phi(x) -\frac{1}{2}m^2\phi(x)^2 -\frac{\lambda}{4!}\phi(x)^4\right )$$

for a real field &phi;.

Then,

$$\frac{\delta S}{\delta \phi(x)}=-\partial_\mu \partial^\mu \phi(x) -m^2 \phi(x) - \frac{\lambda}{3!}\phi(x)^3$$.

The Schwinger-Dyson equation for this particular example is:

$$i\partial_\mu \partial^\mu \frac{\delta}{\delta J(x)}Z[J]+im^2\frac{\delta}{\delta J(x)}Z[J]-\frac{i\lambda}{3!}\frac{\delta^3}{\delta J(x)^3}Z[J]+J(x)Z[J]=0$$

Note that since $$\frac{\delta^3}{\delta J(x)^3}$$ is not well-defined ($$\frac{\delta^3}{\delta J(x_1)\delta J(x_2) \delta J(x_3)}Z[J]$$ is a distribution in x1, x2 and x3), this equation needs to be regularized!