Distributions as Generalized Vectors

A distribution becomes a generalized quantum vector only after its test space and pairing convention have been specified. The usual distribution dual is linear in a test function, while a ket paired with a testing vector is antilinear in that vector. An explicit conjugation connects the two descriptions and prevents sign errors when differential operators act.

Required background. Distribution Theory Minimum defines distributional derivatives. Gelfand Triples defines the antidual and its operator extension. Here the test space is $\mathcal S(\mathbb R)$, with its standard conjugation $\phi\mapsto\overline\phi$.

One function defines two useful functionals

For a suitable function $f$, the ordinary linear distribution and its generalized ket are

$$\begin{aligned} T_f(\phi)&=\int f(x)\phi(x)\,dx,\\ F_f(\phi)&=\int\overline{\phi(x)}f(x)\,dx. \end{aligned}$$

Thus $F_f(\phi)=T_f(\overline\phi)$. More generally, every tempered distribution $T$ defines an antilinear continuous functional $F_T$ by this same formula. Conjugation preserves the Schwartz topology, so continuity is preserved.

When $f\in L^2$, $F_f$ agrees with the Hilbert embedding $Jf$. When $f$ is a polynomial or an oscillatory plane wave, $T_f$ can still be tempered even though $f\notin L^2$. The functional remains meaningful; the Hilbert norm does not become available by extending the notation.

For example, $T_1(\phi)=\int\phi(x)dx$ is continuous on Schwartz space because sufficiently many powers of spatial decay make the integral absolutely bounded. The corresponding constant-amplitude ket is a functional, not a normalized state spread uniformly over the whole line.

Momentum tests the conjugation convention

On Schwartz functions, $P=-i\hbar\,d/dx$ is symmetric and preserves the test space. Its antidual extension satisfies $P^\times F(\phi)=F(P\phi)$. For smooth $f$ such that both $f$ and $f'$ have at most polynomial growth, integration by parts gives

$$\begin{aligned} (P^\times F_f)(\phi) &=i\hbar\int\overline{\phi'(x)}f(x)\,dx\\ &=-i\hbar\int\overline{\phi(x)}f'(x)\,dx\\ &=F_{-i\hbar f'}(\phi). \end{aligned}$$

The plus sign in the first line comes from conjugating $-i$. The minus sign in the second comes from the derivative. Together they recover the physical momentum action. The identity extends distributionally when $f'$ is interpreted as a weak derivative.

By contrast, the plain transpose on the linear dual, $P'T(\phi)=T(P\phi)$, gives $P'T_f=T_{+i\hbar f'}$. This is a transpose formula, not the same Hilbert-to-ket extension. A transpose, a Hilbert adjoint, and an antidual extension encode different conventions. The dictionary must be carried along when moving between them.

Testing defines convergence, not a norm of generalized states

Weak convergence in the antidual means

$$F_n(\phi)\longrightarrow F(\phi) \quad\text{for every }\phi\in\mathcal S.$$

It does not require Hilbert-norm convergence and does not supply a norm on all of $\mathcal S^\times$. Distributional approximations to a delta can have diverging $L^2$ norms. Conversely, a sequence of normalized oscillating vectors can have a zero weak antidual limit. The preliminary distribution example and the Gelfand-triple exercise show these two mechanisms.

The physical probability law is attached to a normalized Hilbert vector or an appropriate state functional on an observable algebra. A generic distribution does not become such a state merely because it is called a generalized vector.

Not every Hilbert-space operation extends

Multiplication of a distribution by a smooth function is defined by moving that function onto the test function, provided the multiplier preserves the chosen test space. Multiplication by an arbitrary bounded measurable function need not be defined. Differentiation is always defined on distributions, but the derivative need not correspond to an $L^2$ vector.

There is no canonical inner product of two arbitrary elements of $\mathcal S^\times$. Pairings involve a test vector and a functional. Expressions involving two generalized kets can sometimes denote a kernel identity after integration, but that meaning must be established separately.

Likewise, the formal outer product $|x\rangle\langle x|$ is not a bounded rank-one projection on $L^2(dx)$. The bounded position projector belongs to a measurable region and is supplied by its PVM. Squaring a delta distribution does not construct the missing singleton projector.

Exercise: an extended relation outside Hilbert space

Define $F_0(\phi)=\overline{\phi(0)}$ and $G(\phi)=-\overline{\phi'(0)}$. Show that $X^\times F_0=0$ and $X^\times G=-F_0$. Would a self-adjoint Hilbert-space operator admit nonzero vectors obeying the corresponding relations?

Solution

Testing gives $F_0(x\phi)=0$ and $G(x\phi)=-\overline{\phi(0)}$. These continuous functionals therefore satisfy the displayed relations. If Hilbert vectors $f,g$ instead obeyed $Xf=0$ and $Xg=-f$ on the appropriate domains, symmetry would give $-\|f\|^2=\langle f,Xg\rangle=\langle Xf,g\rangle=0$. Hence $f$ would vanish. There is no contradiction: the antidual functionals lack the Hilbert inner product used in that argument. Algebraic relations in a larger distribution space cannot automatically be interpreted as Hilbert-space spectral statements.

References

• de la Madrid, Rafael, “The role of the rigged Hilbert space in quantum mechanics,” European Journal of Physics 26, 287–312 (2005), doi:10.1088/0143-0807/26/2/008, arXiv:quant-ph/0502053 — the distinction between linear bras and antilinear ket functionals.
• Gel’fand, I. M., and N. Ya. Vilenkin, Generalized Functions, Vol. 4: Applications of Harmonic Analysis, translated by Amiel Feinstein, Academic Press, 1964, publisher edition — generalized function spaces and operator transposes.
• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — differential operator domains and Hilbert-space spectral statements.