Distributional Derivatives

A distributional derivative extends differentiation to objects defined by their action on test functions. It makes precise statements such as “the derivative of a step function is a delta function” and explains why singular potentials create jump conditions.

The guiding rule is integration by parts with the derivative moved onto the test function.

Definition

Let $T$ be a distribution on a test-function space. Its distributional derivative $DT$ is defined by

$$\langle DT,\varphi\rangle = -\langle T,\varphi'\rangle$$

for every test function $\varphi$.

The second derivative is defined by applying the same rule twice:

$$\langle D^2T,\varphi\rangle = \langle T,\varphi''\rangle.$$

More generally,

$$\langle D^nT,\varphi\rangle = (-1)^n \langle T,\varphi^{(n)}\rangle.$$

This definition is chosen so that, when $T=T_f$ comes from a smooth ordinary function $f$, the distributional derivative agrees with the ordinary derivative.

Agreement with Ordinary Derivatives

Suppose $f$ is smooth and decays enough, or the test function has compact support. Then

$$\begin{aligned} \langle DT_f,\varphi\rangle &= -\int_{-\infty}^{\infty} f(x)\varphi'(x)\,dx \\ &= \int_{-\infty}^{\infty} f'(x)\varphi(x)\,dx \\ &= \langle T_{f'},\varphi\rangle. \end{aligned}$$

The boundary term vanishes because the test function is chosen to make integration by parts legitimate. Distributional differentiation is therefore not a different derivative for smooth functions; it is the extension of the same integration-by-parts identity to singular or nonsmooth objects.

Step Function

Let $H(x)$ be the Heaviside step function:

$$H(x) = \begin{cases} 0, & x<0,\\ 1, & x>0. \end{cases}$$

The value at $x=0$ does not affect the associated distribution. Its distributional derivative is

$$DH=\delta.$$

Indeed,

$$\begin{aligned} \langle DH,\varphi\rangle &= -\langle H,\varphi'\rangle \\ &= -\int_0^\infty \varphi'(x)\,dx \\ &= \varphi(0) \\ &= \langle \delta,\varphi\rangle. \end{aligned}$$

Thus the derivative is zero away from the jump, but the jump itself contributes a delta distribution.

Delta Derivatives

The derivative of a delta distribution is defined by

$$\langle D\delta_a,\varphi\rangle = -\varphi'(a).$$

This object is often written as $\delta'(x-a)$. It should be understood by its action on test functions:

$$\int_{-\infty}^{\infty} \delta'(x-a)\varphi(x)\,dx = -\varphi'(a).$$

The minus sign is not optional. It comes directly from the definition of distributional derivative.

Jump Formula

Let $f$ be smooth on the two sides of a point $a$, with one-sided limits $f(a^-)$ and $f(a^+)$. Define the jump

$$[f]_a = f(a^+)-f(a^-).$$

Then the distributional derivative is

$$Df = f'_{\mathrm{pw}} + [f]_a\,\delta_a,$$

where $f'_{\mathrm{pw}}$ is the ordinary derivative on each side of $a$.

If $f$ is continuous but $f'$ jumps, then $Df$ has no delta term, but the second derivative does:

$$D^2f = f''_{\mathrm{pw}} + [f']_a\,\delta_a.$$

If $f$ itself jumps, then

$$D^2f = f''_{\mathrm{pw}} + [f']_a\,\delta_a + [f]_a\,D\delta_a.$$

This hierarchy is the clean way to see which singular terms appear in a differential equation.

Example: Exponential Green Function

Let

$$g(x) = e^{-\kappa\lvert x\rvert}, \qquad \kappa>0.$$

The function is continuous at $0$, so $[g]_0=0$. Its derivative is

$$g'(x) = \begin{cases} \kappa e^{\kappa x}, & x<0,\\ -\kappa e^{-\kappa x}, & x>0, \end{cases}$$

so

$$[g']_0 = g'(0^+)-g'(0^-) = -2\kappa.$$

Away from $0$,

$$g''_{\mathrm{pw}}(x) = \kappa^2 e^{-\kappa\lvert x\rvert}.$$

Therefore

$$D^2g = \kappa^2 e^{-\kappa\lvert x\rvert} -2\kappa\delta(x).$$

Equivalently,

$$\left( -\frac{d^2}{dx^2} +\kappa^2 \right) \frac{e^{-\kappa\lvert x\rvert}}{2\kappa} = \delta(x),$$

which is the full-line Green function for $-d^2/dx^2+\kappa^2$. The same result appears in Green Functions.

Delta Potentials

Consider the stationary Schrödinger equation with a delta potential:

$$-\frac{\hbar^2}{2m}D^2\psi(x) +\lambda\delta(x)\psi(x) = E\psi(x).$$

For an ordinary delta potential, the wavefunction is taken continuous at the origin:

$$[\psi]_0=0.$$

Then $D^2\psi$ contains a delta term if the derivative jumps:

$$D^2\psi = \psi''_{\mathrm{pw}} + [\psi']_0\delta(x).$$

The product $\delta(x)\psi(x)$ is interpreted as $\psi(0)\delta(x)$ when $\psi$ is continuous. Comparing the coefficient of $\delta(x)$ in the Schrödinger equation gives

$$-\frac{\hbar^2}{2m}[\psi']_0 +\lambda\psi(0) = 0.$$

Thus

$$\psi'(0^+)-\psi'(0^-) = \frac{2m\lambda}{\hbar^2}\psi(0).$$

This is the derivative jump condition. If $\psi$ itself had a jump, $D^2\psi$ would contain a $D\delta$ term, which is not present in the ordinary delta-potential equation. That is why continuity is part of the matching rule for this model.

For the wave-mechanics boundary-condition statement, see Boundary Conditions.

Fourier-Space Rule

Distributional derivatives also preserve the familiar Fourier derivative rule. With the ordinary $k$ convention,

$$\mathcal F[Df](k) = ik\,\mathcal F[f](k),$$

interpreted distributionally. This remains meaningful even when $f$ has jumps or singular parts.

With the wavefunction $p$ convention,

$$\mathcal F_p[D\psi](p) = \frac{ip}{\hbar}\phi(p),$$

whenever the pairing is interpreted in the distributional sense. To use this as a Hilbert-space momentum operator, one needs more: the derivative must be represented by an $L^2$ function and the state must lie in the operator domain.

This distinction is explained in Domains of Operators.

Common Mistakes

• Thinking a distributional derivative must be an ordinary function.
• Forgetting the minus sign in $\langle D\delta_a,\varphi\rangle=-\varphi'(a)$.
• Treating a step function as having derivative zero everywhere and missing the jump delta.
• Applying the momentum operator to every $L^2$ state just because every distribution has a derivative.
• Multiplying a delta distribution by a discontinuous function without specifying a convention or model.
• Requiring derivative continuity across a delta potential.
• Ignoring the $D\delta$ term created by a discontinuity in the wavefunction.

Cross-Links

• Distributions
• Delta Function
• Principal Value Distributions
• Real Analysis Essentials
• Boundary Conditions
• Green Functions
• Domains of Operators
• Unbounded Operators
• Wave-Mechanics Boundary Conditions

References

• I. M. Gel’fand and G. E. Shilov, Generalized Functions, Volume 1, Academic Press, 1964.
• L. Schwartz, Théorie des distributions, Hermann, 1966.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.

Exercises

1. Derive $DH=\delta$ for the Heaviside function.

Solution

For every test function $\varphi$,

$$\begin{aligned} \langle DH,\varphi\rangle &= -\langle H,\varphi'\rangle \\ &= -\int_0^\infty \varphi'(x)\,dx \\ &= -\left[ 0-\varphi(0) \right] \\ &= \varphi(0) \\ &= \langle\delta,\varphi\rangle. \end{aligned}$$

Therefore $DH=\delta$ distributionally.

2. Let $\operatorname{sgn}(x)=-1$ for $x\lt0$ and $\operatorname{sgn}(x)=1$ for $x\gt0$. Find its distributional derivative.

Solution

The sign function can be written as

$$\operatorname{sgn}(x) = 2H(x)-1.$$

Since $D1=0$ and $DH=\delta$,

$$D\operatorname{sgn} = 2\delta.$$

Equivalently, the jump is $1-(-1)=2$, so the jump formula gives $2\delta$.

3. Compute $D^2e^{-\kappa\lvert x\rvert}$ distributionally.

Solution

The function is continuous, so there is no $D\delta$ term. Away from zero,

$$g''_{\mathrm{pw}}(x) = \kappa^2e^{-\kappa\lvert x\rvert}.$$

The derivative jumps from $g'(0^-)=\kappa$ to $g'(0^+)=-\kappa$, so

$$[g']_0=-2\kappa.$$

Therefore

$$D^2e^{-\kappa\lvert x\rvert} = \kappa^2e^{-\kappa\lvert x\rvert} -2\kappa\delta(x).$$

4. Use the distributional second derivative to derive the delta-potential jump condition.

Solution

For a continuous wavefunction with a possible derivative jump at $0$,

$$D^2\psi = \psi''_{\mathrm{pw}} + [\psi']_0\delta(x).$$

Substitute this into

$$-\frac{\hbar^2}{2m}D^2\psi +\lambda\delta(x)\psi(x) = E\psi(x).$$

The regular terms hold away from the origin. The coefficient of $\delta(x)$ must vanish:

$$-\frac{\hbar^2}{2m}[\psi']_0 +\lambda\psi(0) = 0.$$

Hence

$$\psi'(0^+)-\psi'(0^-) = \frac{2m\lambda}{\hbar^2}\psi(0).$$