The Unfolding Universe of Quantum Field Theory: A Journey Through Exercise 1.3 Part 2

Quantum Field Theory (QFT) continues to be a fountain of bewitching complexity and awe-inspiring beauty. Our last foray into this realm dealt with functionals and their interaction with derivatives. Today, we venture further into the second part of Exercise 1.3, extending the discussion to functionals involving both \(f\) and \(f’\).

Table of Contents

The Exercise

For the functional

\[H[f] = \int g\left(y, f(y), f'(y)\right)\,\mathrm{d}y,\]

show that

\[\frac{\delta H[f]}{\delta f(x)} = \frac{\partial g}{\partial f}(x, f(x), f'(x)) – \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial g}{\partial f’}(x, f(x), f'(x)).\]

This is the familiar Euler-Lagrange expression for a functional whose integrand depends on a function and its first derivative.

Step 1: Setting the Stage

Start with

\[H[f] = \int g\left(y, f(y), f'(y)\right)\,\mathrm{d}y.\]

To compute the functional derivative, perturb \(f\) by a small test variation \(\epsilon\eta\), where \(\eta\) is smooth and vanishes at the boundary. Then

\[H[f+\epsilon\eta] = \int g\left(y, f(y)+\epsilon\eta(y), f'(y)+\epsilon\eta'(y)\right)\,\mathrm{d}y.\]

The first variation is

\[\delta H = \left.\frac{\mathrm{d}}{\mathrm{d}\epsilon}H[f+\epsilon\eta]\right|_{\epsilon=0}.\]

Step 2: Tracing the Derivatives

Differentiating under the integral gives

\[\delta H = \int \left( \frac{\partial g}{\partial f}\eta + \frac{\partial g}{\partial f’}\eta’ \right)\,\mathrm{d}y.\]

The second term contains \(\eta’\), so integrate it by parts:

\[\int \frac{\partial g}{\partial f’}\eta’\,\mathrm{d}y = \left[\frac{\partial g}{\partial f’}\eta\right]_{\text{boundary}} – \int \frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{\partial g}{\partial f’}\right)\eta\,\mathrm{d}y.\]

Because the variation is taken to vanish at the boundary, the boundary term is zero. Therefore

\[\delta H = \int \left( \frac{\partial g}{\partial f} – \frac{\mathrm{d}}{\mathrm{d}y}\frac{\partial g}{\partial f’} \right)\eta(y)\,\mathrm{d}y.\]

By the definition of the functional derivative,

\[\delta H = \int \frac{\delta H[f]}{\delta f(y)}\eta(y)\,\mathrm{d}y,\]

so we identify

\[\frac{\delta H[f]}{\delta f(x)} = \frac{\partial g}{\partial f}(x, f(x), f'(x)) – \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial g}{\partial f’}(x, f(x), f'(x)).\]

The Finale: The Euler-Lagrange Expression

The result is not only the answer to the exercise. It is the functional derivative form of the Euler-Lagrange equation, a central tool in classical mechanics, field theory, and quantum field theory. If \(H[f]\) is stationary under arbitrary variations \(\eta\), then \(\delta H = 0\), which implies

\[\frac{\partial g}{\partial f} – \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial g}{\partial f’} = 0.\]

Understanding the Delta-Function Derivation

The same result can also be derived by using a delta-function variation. Formally, one may perturb the function by

\[f(y) \mapsto f(y) + \epsilon\delta(y-x),\]

so that

\[f'(y) \mapsto f'(y) + \epsilon\frac{\partial}{\partial y}\delta(y-x).\]

Then the first-order change in \(H\) is

\[\frac{\delta H[f]}{\delta f(x)} = \int \left[ \frac{\partial g}{\partial f}\delta(y-x) + \frac{\partial g}{\partial f’}\frac{\partial}{\partial y}\delta(y-x) \right] \mathrm{d}y.\]

The first term is selected directly by the delta function:

\[\int \frac{\partial g}{\partial f}\delta(y-x)\,\mathrm{d}y = \frac{\partial g}{\partial f}(x, f(x), f'(x)).\]

For the second term, use the distribution identity

\[\int \phi(y)\frac{\partial}{\partial y}\delta(y-x)\,\mathrm{d}y = -\phi'(x),\]

assuming boundary terms vanish. With \(\phi(y)=\partial g/\partial f’\), this gives

\[\int \frac{\partial g}{\partial f’}\frac{\partial}{\partial y}\delta(y-x)\,\mathrm{d}y = -\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial g}{\partial f’}(x, f(x), f'(x)).\]

Putting both terms together recovers

\[\frac{\delta H[f]}{\delta f(x)} = \frac{\partial g}{\partial f} – \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial g}{\partial f’}.\]

A Note on \(\delta'(x-a)\)

The derivative of the Dirac delta function must be understood as a distribution, not as an ordinary function. Its defining action is

\[\int_{-\infty}^{\infty} f(x)\delta'(x-a)\,\mathrm{d}x = -f'(a),\]

provided the boundary terms vanish. This is why the sign in the Euler-Lagrange expression is negative. It is not correct to treat \(f(x)\delta'(x)\) as simply equal to \(f'(x)\) pointwise; the meaningful statement is the distributional identity under an integral.

Reflections

This exercise shows how functionals and derivatives fit together with remarkable precision. The formal delta-function method is compact, while the test-function variation makes the assumptions more visible: smooth variations, integration by parts, and vanishing boundary terms.

As we push deeper into Quantum Field Theory, these small technical details become essential. They are the grammar behind the larger physical statements, and getting them right gives the subject much of its power.