Functional Derivatives

A functional derivative measures how a functional changes when its input function is varied at each point. If $F[f]$ is a functional and $\eta(x)$ is a test variation, the derivative is defined by writing the first variation as

$$\delta F[f;\eta] = \left. \frac{d}{d\epsilon} F[f+\epsilon\eta] \right\rvert_{\epsilon=0} = \int \frac{\delta F}{\delta f(x)} \eta(x)\,dx.$$

For a real input function, $\delta F/\delta f(x)$ represents this linear response relative to the stated integration measure. It is an ordinary gradient when represented by a function in the chosen inner-product space; in other cases it is a distribution. Complex wavefunctions require the separate variations explained below.

Helpful background. Calculus of Variations supplies integration by parts and first variations; Delta Function explains the point-evaluation rule.

Functional Derivatives in Quantum Calculations

Functional derivatives are the local language behind several familiar calculations:

• deriving Euler–Lagrange equations from an action;
• extremizing energy functionals over wavefunctions;
• varying trial-state actions in time-dependent variational principles;
• generating insertions from source terms in path integrals;
• writing field equations compactly as $\delta S/\delta\phi(x)=0$.

The Calculus of Variations page explains stationary functionals and Euler–Lagrange equations. This page isolates the derivative notation and the delta-function identities that make functional calculations work.

Definition

Let $F[f]$ be a functional defined on a suitable space of functions over a region $\Omega$. The first variation in the direction $\eta$ is

$$\delta F[f;\eta] = \left. \frac{d}{d\epsilon} F[f+\epsilon\eta] \right\rvert_{\epsilon=0}.$$

If this first variation can be represented as

$$\delta F[f;\eta] = \int_\Omega g(x)\eta(x)\,d\mu(x)$$

for all allowed variations $\eta$, then

$$\frac{\delta F}{\delta f(x)} = g(x),$$

relative to the measure $d\mu(x)$.

This “relative to the measure” clause is not cosmetic. If the inner product is

$$\langle u,v\rangle = \int_\Omega u^*(x)v(x)\,w(x)\,dx,$$

then the gradient represented by a functional derivative depends on whether the variation is paired with $dx$ or with $w(x)\,dx$.

Delta-Function Rule

The basic identity is

$$\frac{\delta f(y)}{\delta f(x)} = \delta(y-x).$$

It is defined by its action under integration:

$$\delta f(y) = \int \frac{\delta f(y)}{\delta f(x)} \delta f(x)\,dx = \int \delta(y-x)\delta f(x)\,dx.$$

For a source term

$$F[J] = \int J(y)\phi(y)\,dy,$$

one has

$$\frac{\delta F}{\delta J(x)} = \phi(x),$$

because

$$\delta F = \int \delta J(y)\phi(y)\,dy.$$

This is the finite-dimensional rule $\partial(J_i\phi_i)/\partial J_j=\phi_j$ with sums replaced by integrals and Kronecker deltas replaced by Dirac deltas.

Local Functionals

For a local functional

$$F[f] = \int_\Omega V(f(x))\,dx,$$

the variation is

$$\delta F = \int_\Omega V'(f(x))\eta(x)\,dx.$$

Therefore

$$\frac{\delta F}{\delta f(x)} = V'(f(x)).$$

For example,

$$F[f] = \int_\Omega \frac{\lambda}{4}f(x)^4\,dx$$

gives

$$\frac{\delta F}{\delta f(x)} = \lambda f(x)^3.$$

This is the easiest case because no derivatives of $f$ appear.

Derivative Terms and Boundary Conditions

Consider

$$F[f] = \int_a^b \frac12 \left( \frac{df}{dx} \right)^2 dx.$$

Under $f\mapsto f+\epsilon\eta$,

$$\delta F = \int_a^b f'(x)\eta'(x)\,dx.$$

Integration by parts gives

$$\delta F = \left[f'(x)\eta(x)\right]_a^b - \int_a^b f''(x)\eta(x)\,dx.$$

For fixed-endpoint variations, $\eta(a)=\eta(b)=0$, so

$$\frac{\delta F}{\delta f(x)} = -f''(x).$$

The minus sign is important. It comes from moving the derivative off the variation. If the endpoints are not fixed, the boundary term is part of the variational problem and can impose natural boundary conditions.

Euler–Lagrange Form

For

$$S[\phi] = \int d^dx\, \mathcal L \left( \phi,\partial_\mu\phi,x \right),$$

the functional derivative is

$$\frac{\delta S}{\delta\phi(x)} = \frac{\partial\mathcal L}{\partial\phi} - \partial_\mu \left( \frac{\partial\mathcal L} {\partial(\partial_\mu\phi)} \right),$$

up to boundary terms determined by the allowed variations. The stationary-action equation is

$$\frac{\delta S}{\delta\phi(x)}=0.$$

For a single particle path $q(t)$, the same statement reads

$$\frac{\delta S}{\delta q(t)} = \frac{\partial L}{\partial q} - \frac{d}{dt} \left( \frac{\partial L}{\partial\dot q} \right).$$

Thus the Euler–Lagrange equation is a functional-derivative equation.

Complex Wavefunctions

Quantum variational calculations usually involve complex functions. One can vary the real and imaginary parts independently, or equivalently collect the coefficients of $\delta\psi$ and $\delta\psi^*$. The latter is called treating $\psi$ and $\psi^*$ as independent variables: it is a calculational shorthand, not a claim that a physical variation breaks complex conjugacy.

For a normalized energy functional, write schematically

$$\mathcal F[\psi,\psi^*] = \int dx\,\psi^*(x)(H\psi)(x) - E \left( \int dx\,\psi^*(x)\psi(x)-1 \right).$$

Varying with respect to $\psi^*(x)$ gives

$$\frac{\delta\mathcal F}{\delta\psi^*(x)} = (H\psi)(x)-E\psi(x).$$

Stationarity therefore gives

$$H\psi=E\psi.$$

This is the wavefunction version of the constrained variational argument for the eigenvalue equation. In an actual Hamiltonian problem, the domain of $H$ and the boundary terms still have to be checked.

Source Derivatives and Correlation Functions

Functional derivatives become especially useful when a source is coupled to a coordinate, operator, or field. In a schematic field-integral notation,

$$Z[J] = \int\mathcal D\phi\, \exp \left[ \frac{i}{\hbar} \left( S[\phi]+\int J(x)\phi(x)\,dx \right) \right].$$

Differentiating with respect to the source brings down a field insertion:

$$\frac{\delta Z[J]}{\delta J(x)} = \frac{i}{\hbar} \int\mathcal D\phi\, \phi(x) \exp \left[ \frac{i}{\hbar} \left( S[\phi]+\int J(y)\phi(y)\,dy \right) \right].$$

Higher source derivatives generate higher products of fields, with ordering and normalization depending on the precise real-time, Euclidean, operator, or time-ordered convention. This is one reason ordinary quantum-mechanical path integrals prepare readers for the field-theory language without replacing a QFT course.

Worked Example: Driven Quadratic Functional

Let

$$F[f] = \int_a^b \left[ \frac12(f')^2 + \frac12\omega^2f^2 - Jf \right]dx,$$

with fixed-endpoint variations. The variation is

$$\delta F = \int_a^b \left[ f'\eta' + \omega^2f\eta - J\eta \right]dx.$$

Integrating the derivative term by parts gives

$$\delta F = \int_a^b \left[ -f'' + \omega^2f - J \right]\eta\,dx.$$

Therefore

$$\frac{\delta F}{\delta f(x)} = -f''(x)+\omega^2f(x)-J(x).$$

The stationary equation is

$$\left( -\frac{d^2}{dx^2}+\omega^2 \right)f(x) = J(x).$$

This displays the bridge to Green Functions: the stationary function is the response of a linear operator to a source.

Common Mistakes

• Treating $\delta/\delta f(x)$ as an ordinary partial derivative without specifying the pairing integral.
• Dropping boundary terms before checking the allowed variations.
• Missing the minus sign produced by integration by parts.
• Forgetting that the functional derivative depends on the measure or inner product used to identify gradients.
• Varying $\psi$ but not $\psi^*$ in complex wavefunction problems, or varying them inconsistently.
• Confusing a derivative with respect to a variational parameter with a functional derivative with respect to a whole function.
• Assuming functional derivatives exist for nonsmooth, constrained, or distribution-valued functionals without checking the function space.
• Using source-derivative formulas in path integrals without stating the ordering and normalization convention.

Related Topics

• Why Path Integrals?
• Time-Dependent Variational Principle

Exercises

1. Compute the functional derivative of

$$F[f]=\int_a^b f(x)^3\,dx.$$

Solution

Vary $f\mapsto f+\epsilon\eta$. Then

$$\delta F = \int_a^b 3f(x)^2\eta(x)\,dx.$$

Therefore

$$\frac{\delta F}{\delta f(x)} = 3f(x)^2.$$

2. With fixed-endpoint variations, find the functional derivative of

$$F[f] = \int_a^b \left[ \frac12(f')^2+U(f) \right]dx.$$

Solution

The variation is

$$\delta F = \int_a^b \left[ f'\eta'+U'(f)\eta \right]dx.$$

After integration by parts and using $\eta(a)=\eta(b)=0$,

$$\delta F = \int_a^b \left[ -f''+U'(f) \right]\eta\,dx.$$

Hence

$$\frac{\delta F}{\delta f(x)} = -f''(x)+U'(f(x)).$$

3. Show that

$$\frac{\delta}{\delta J(x)} \int J(y)\phi(y)\,dy = \phi(x).$$

Solution

Vary $J$ by $\delta J$. Then

$$\delta F = \int \delta J(y)\phi(y)\,dy.$$

Comparing with

$$\delta F = \int \frac{\delta F}{\delta J(x)} \delta J(x)\,dx$$

gives

$$\frac{\delta F}{\delta J(x)} = \phi(x).$$

4. Why does the answer to a functional-derivative calculation change when the boundary variations change?

Solution

Derivative terms require integration by parts to express the variation as an integral against $\eta(x)$. That integration produces boundary terms. If the allowed variations make those terms vanish, the bulk functional derivative is enough. If not, stationarity also imposes boundary conditions or requires endpoint terms. Changing the allowed variations therefore changes the variational problem.

References

• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• I. M. Gelfand and S. V. Fomin, Calculus of Variations, Dover, 2000.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.
• J. Zinn-Justin, Path Integrals in Quantum Mechanics, Oxford University Press, 2005.