Quantum Field Theory for the Gifted Amateur Exercise 1.6 Hunt the Papertiger

Craving a deep dive into the intricate world of Quantum Field Theory (QFT)? Follow me as we dissect a seemingly complex functional derivative and arrive at a compact result.

Table of Contents

The Genesis: Our Original Equation

In the realm of QFT, Exercise 1.6 presents us with a useful challenge. It begins with a functional, $Z_0[J]$, described by the equation:

\[Z_0[J]=\exp \left(-\frac{1}{2} \int \mathrm{d}^4 x\, \mathrm{d}^4 y\, J(x) \Delta(x-y) J(y)\right).\]

It adds an important symmetry condition:

\[\Delta(x)=\Delta(-x).\]

Our goal is to prove that

\[\frac{\delta Z_0[J]}{\delta J(z_1)}=-\left[\int \mathrm{d}^4 y\, \Delta(z_1-y) J(y)\right] Z_0[J].\]

The Journey Begins: Unwrapping the Functional

The Derivative: Starting with First Principles

The first step is to use the definition of the functional derivative. Let the source be shifted by a delta-function probe at $z_1$:

\[\frac{\delta Z_0[J]}{\delta J(z_1)} = \lim_{\epsilon \rightarrow 0} \frac{Z_0[J + \epsilon \delta(z_1,x)] – Z_0[J]}{\epsilon}.\]

Equivalently, the variation is $J(x) \mapsto J(x)+\epsilon \delta^{(4)}(x-z_1)$. The notation in the exercise uses $\delta(z_1,x)$ for the same delta function.

Digging into the Expansion

Now consider what $Z_0[J + \epsilon \delta(z_1,x)]$ means. Starting from

\[Z_0[J] = \exp \left( -\frac{1}{2} \int \mathrm{d}^4 x\, \mathrm{d}^4 y\, J(x) \Delta(x-y) J(y) \right),\]

we substitute the shifted source:

\[Z_0[J + \epsilon \delta] = \exp \left( -\frac{1}{2} \int \mathrm{d}^4 x\, \mathrm{d}^4 y\, [J(x)+ \epsilon \delta(z_1,x)] \Delta(x-y) [J(y)+ \epsilon \delta(z_1,y)] \right).\]

Expanding the exponent gives the original quadratic term, two terms linear in $\epsilon$, and one term of order $\epsilon^2$:

\[\begin{aligned} &-\frac{1}{2} \int \mathrm{d}^4 x\, \mathrm{d}^4 y\, [J(x)+ \epsilon \delta(z_1,x)] \Delta(x-y) [J(y)+ \epsilon \delta(z_1,y)] \\ &= -\frac{1}{2} \int \mathrm{d}^4 x\, \mathrm{d}^4 y\, J(x)\Delta(x-y)J(y) \\ &\quad -\frac{\epsilon}{2}\int \mathrm{d}^4 x\, \mathrm{d}^4 y\, \delta(z_1,x)\Delta(x-y)J(y) \\ &\quad -\frac{\epsilon}{2}\int \mathrm{d}^4 x\, \mathrm{d}^4 y\, J(x)\Delta(x-y)\delta(z_1,y) +O(\epsilon^2). \end{aligned}\]

The first linear term is

\[\int \mathrm{d}^4 x\, \mathrm{d}^4 y\, \delta(z_1,x)\Delta(x-y)J(y) = \int \mathrm{d}^4 y\, \Delta(z_1-y)J(y).\]

For the second linear term, integrate over $y$ first:

\[\int \mathrm{d}^4 x\, \mathrm{d}^4 y\, J(x)\Delta(x-y)\delta(z_1,y) = \int \mathrm{d}^4 x\, J(x)\Delta(x-z_1).\]

Using $\Delta(x)=\Delta(-x)$, we have $\Delta(x-z_1)=\Delta(z_1-x)$, so after renaming the dummy variable $x$ to $y$, this becomes

\[\int \mathrm{d}^4 y\, \Delta(z_1-y)J(y).\]

Thus the two equal linear terms combine with the factor $-1/2$ to give

\[Z_0[J + \epsilon \delta] = Z_0[J]\exp\left(-\epsilon \int \mathrm{d}^4 y\, \Delta(z_1-y)J(y)+O(\epsilon^2)\right).\]

To first order in $\epsilon$,

\[Z_0[J + \epsilon \delta] = Z_0[J]\left(1-\epsilon \int \mathrm{d}^4 y\, \Delta(z_1-y)J(y)+O(\epsilon^2)\right).\]

Zeroing In on the Solution

Substitute this expansion into the definition of the functional derivative:

\[\begin{aligned} \frac{\delta Z_0[J]}{\delta J(z_1)} &= \lim_{\epsilon \rightarrow 0} \frac{Z_0[J+\epsilon \delta(z_1,x)] – Z_0[J]}{\epsilon} \\ &= \lim_{\epsilon \rightarrow 0} \frac{Z_0[J]\left(1-\epsilon \int \mathrm{d}^4 y\, \Delta(z_1-y)J(y)+O(\epsilon^2)\right)-Z_0[J]}{\epsilon} \\ &= -\left[\int \mathrm{d}^4 y\, \Delta(z_1-y)J(y)\right] Z_0[J]. \end{aligned}\]

That proves the required identity.

Why This Matters

This exercise is a small but important example of how Gaussian functionals behave under functional differentiation. In free-field generating functionals, differentiating with respect to the source $J$ pulls down factors involving the propagator $\Delta$. This is the mechanism behind extracting correlation functions from a generating functional, which is one of the standard computational tools in quantum field theory.