Position Kets

The position ket $|x_0\rangle$ on the line is a continuous antilinear evaluation functional on Schwartz test vectors. It is a generalized eigenvector of position in the antidual, and it reconstructs ordinary region probabilities through weak integrals. It is not an $L^2$ vector or a projection onto a physical state at one point.

Required background. Gelfand Triples fixes the linear and antilinear pairing conventions. Projection-Valued Measures defines the bounded event operators reconstructed below. The configuration measure is Lebesgue measure $dx$.

Evaluation is continuous on the test space

For $\phi\in\mathcal S(\mathbb R)$, define the ket functional and its associated bra by

$$\langle\phi|x_0\rangle=\overline{\phi(x_0)}, \qquad \langle x_0|\phi\rangle=\phi(x_0).$$

Their magnitudes are bounded by the Schwartz seminorm $p_{0,0}(\phi)=\sup_x|\phi(x)|$. Thus both are continuous in the test topology, with the respective antilinearity and linearity.

Evaluation is not bounded in the Hilbert norm. To see this, choose $g\in\mathcal S$ with $g(0)=1$ and put $g_\epsilon(x)=g((x-x_0)/\epsilon)$. Then

$$g_\epsilon(x_0)=1, \qquad \|g_\epsilon\|_2=\sqrt\epsilon\,\|g\|_2\longrightarrow0.$$

No estimate $|\phi(x_0)|\le C\|\phi\|_2$ can hold. Consequently, the Riesz representation theorem cannot represent this evaluation by a Hilbert vector. Moreover, an arbitrary $L^2$ class has no preferred point value at all.

The notation is therefore more than an informal infinite norm. It denotes a functional continuous in one topology and discontinuous in the weaker topology used to complete the state space.

The position eigen-equation is an antidual identity

The self-adjoint position operator is multiplication by $x$, with domain $D(X)=\{\psi\in L^2:x\psi\in L^2\}$. It preserves Schwartz space. Writing $F_{x_0}(\phi)=\overline{\phi(x_0)}$, its antidual extension gives

$$\begin{aligned} (X^\times F_{x_0})(\phi) &=F_{x_0}(X\phi)\\ &=x_0\overline{\phi(x_0)} =x_0F_{x_0}(\phi). \end{aligned}$$

This is the precise sense of $X|x_0\rangle=x_0|x_0\rangle$. It does not assert membership in $D(X)$, because the functional does not belong to the underlying Hilbert space.

The obstruction is not simply that $X$ is unbounded. On $L^2(0,1)$, multiplication by $x$ is bounded and still has no nonzero point eigenvectors: a vector supported on a singleton vanishes as a Lebesgue class. Continuous-spectrum eigenkets require generalized interpretation even for this bounded observable.

A weak integral reconstructs the identity

For test vectors $\phi,\psi$, ordinary integration gives

$$\int_{\mathbb R} \langle\phi|x\rangle\langle x|\psi\rangle\,dx =\langle\phi,\psi\rangle.$$

Thus the notation $I=\int|x\rangle\langle x|dx$ means equality after testing with these vectors. For fixed $\psi\in\mathcal S$, the integral defines the antidual functional $J\psi$. It is not a norm-convergent integral of bounded rank-one projections: its individual ket factors are not Hilbert vectors.

Restricting the integral to a Borel set $\Delta$ instead gives

$$\int_\Delta\overline{\phi(x)}\psi(x)\,dx =\langle\phi,E_X(\Delta)\psi\rangle.$$

The right-hand side is bounded by $\|\phi\|\,\|\psi\|$, so it extends uniquely to all Hilbert vectors. It identifies the familiar multiplication projector $E_X(\Delta)$. This bounded extension is the rigorous event operator. For a singleton, the integral and projector are zero; there is no nonzero singleton rank-one projection hidden in the ket notation.

A wavefunction value and an event probability have different status

For a test state, $\langle x|\psi\rangle=\psi(x)$ is an actual evaluation. For a general $L^2$ state, the same notation denotes an almost-everywhere representative of its position spectral transform. The probability of a region is independent of that representative, while the value at an individually chosen point need not be.

Similarly, $|\psi(x)|^2$ is a probability density relative to $dx$, not the probability of obtaining precisely $x$. Changing the coordinate measure changes the density and ket normalization. Delta Normalization treats that measure dependence.

Exercise: a sharp cutoff can destroy momentum regularity

Let $\psi\in\mathcal S$ and let $\Delta=(a,b)$, with at least one of $\psi(a),\psi(b)$ nonzero. Show that $E_X(\Delta)\psi$ is a valid Hilbert vector but is not in the line momentum domain $H^1(\mathbb R)$.

Solution

Multiplication by the indicator is bounded on $L^2$. Its distributional derivative is

$$\mathbf1_{(a,b)}\psi' +\psi(a)\delta_a-\psi(b)\delta_b.$$

A nonzero endpoint delta is not an $L^2$ function, so the result has no square-integrable weak derivative. The position event remains well defined, but its ideal sharp cutoff can leave the momentum operator domain. A bounded measurement effect need not preserve every unbounded observable’s domain.

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 — generalized position functionals and weak expansions.
• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — multiplication operators, spectral measures, and continuous representations.
• 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 — weak derivatives and Sobolev domains.