Distribution Theory Minimum

Distribution theory interprets expressions such as a position ket or the delta kernel by their action on test functions. It does not turn every formal wavefunction into a normalizable quantum state. The practical minimum is to identify the test space, say what convergence means after testing, and keep distributional differentiation separate from membership in a quantum operator domain.

Helpful background. Distributions and Distributional Derivatives introduce the general definitions. Fourier Transform supplies the transform conventions used to change representation.

Test functions determine what a generalized object means

On $\mathbb R$, $\mathcal D=C_c^\infty(\mathbb R)$ consists of smooth compactly supported functions. A distribution is a continuous linear functional on this space with its test-function topology. The Schwartz space $\mathcal S(\mathbb R)$ consists of smooth functions for which every seminorm

$$\begin{gathered} p_{a,b}(\phi)=\sup_x|x^a\phi^{(b)}(x)|,\\ a,b=0,1,2,\ldots, \end{gathered}$$

is finite. Its continuous linear dual $\mathcal S'$ is the space of tempered distributions. The topology involves all these seminorms, not merely an $L^2$ norm. Fourier transformation preserves $\mathcal S$ and extends by duality to $\mathcal S'$.

Here distributional pairings are linear in the test function: $T(\phi)$. A locally integrable function gives $T_f(\phi)=\int f(x)\phi(x)dx$. This pairing has no complex conjugation. A Hilbert inner product has one. When translating to Dirac bras and kets, introduce conjugation explicitly rather than identifying the linear dual and the antidual silently.

The delta distribution is $\delta_{x_0}(\phi)=\phi(x_0)$. Evaluation is continuous in the test-function topology. It is neither well defined on $L^2$ equivalence classes nor bounded in their norm. Thus the notation $|x_0\rangle$ belongs to a generalized-vector construction; it is not a unit vector in the state Hilbert space.

Approximate delta functions illustrate two different limits

Let $g\in\mathcal S$ be nonnegative with $\int g=1$, and define $g_\epsilon(x)=\epsilon^{-1}g(x/\epsilon)$. For every test function,

$$\begin{aligned} \int g_\epsilon(x)\phi(x)dx &=\int g(u)\phi(\epsilon u)du\\ &\longrightarrow\phi(0). \end{aligned}$$

Dominated convergence justifies the limit, so $g_\epsilon\to\delta_0$ distributionally. But

$$\|g_\epsilon\|_2=\epsilon^{-1/2}\|g\|_2,$$

which diverges. There is no $L^2$ limit.

A normalized narrow wavepacket instead scales as $\psi_\epsilon(x)=\epsilon^{-1/2}h(x/\epsilon)$ with $\|h\|_2=1$. Its probability density can converge to a delta measure while its vector has no normalized Hilbert-space limit. The amplitude and the probability density are different objects and require different normalization powers. The operational meaning of localization belongs to probabilities for finite detector regions.

Weak derivatives do not supply missing operator domains

The derivative of a distribution is defined by

$$T'(\phi)=-T(\phi').$$

Every distribution has such a derivative. However, the momentum operator $P=-i\hbar\,d/dx$ on the line requires the weak derivative of its input to lie in $L^2$. Its self-adjoint domain is $H^1(\mathbb R)$, a strict subset of $L^2(\mathbb R)$.

For example, the indicator $u=\mathbf1_{(0,1)}$ is square integrable, but

$$u'=\delta_0-\delta_1$$

as a distribution. This derivative is not an $L^2$ function, so $u\notin D(P)$. Formally applying a derivative and obtaining an answer in $\mathcal D'$ does not show that the Hilbert-space operator can act on the state. Boundary conditions introduce further restrictions when the configuration space is an interval.

Kernel identities must be read under testing

The expression $I(x,y)=\delta(x-y)$ means that integrating the kernel against suitable test functions reproduces their pairing. It does not assign a finite value to $\delta(0)$. Likewise,

$$\frac{1}{2\pi\hbar}\int_{\mathbb R} e^{ip(x-y)/\hbar}\,dp=\delta(x-y)$$

is a distributional Fourier identity. One may insert a Gaussian regulator, evaluate the ordinary integral, test against a Schwartz function, and then remove the regulator. Interchanging these operations without a limiting statement loses the meaning of the equation.

Products of distributions are not generally defined. In particular, neither $\delta^2$ nor a norm computed by multiplying two delta distributions becomes legitimate by calling the calculation formal. Extra structure or a specified regularization and convergence theorem is necessary. The Rigged Hilbert Spaces chapter develops generalized eigenvectors beyond these preliminary tests.

Exercise: a distributional eigen-equation

Show that $x\delta_{x_0}=x_0\delta_{x_0}$. Explain why this does not give an eigenvector of multiplication by $x$ in $L^2(\mathbb R,dx)$.

Solution

Testing gives $(x\delta_{x_0})(\phi)=\delta_{x_0}(x\phi)=x_0\phi(x_0)$. An $L^2$ eigenvector would satisfy $(x-x_0)\psi(x)=0$ almost everywhere, hence would be supported on a Lebesgue-null singleton and equal zero as an $L^2$ class. The nonzero distributional solution exists in a larger space with a different topology; it does not contradict the absence of point spectrum.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — continuous spectral representations and momentum.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s edition — Fourier transformation, weak derivatives, and Sobolev operator domains.