Momentum Kets

A momentum ket on the line is a plane-wave functional whose coefficient in a state is the state’s unitary Fourier transform. The individual ket is not square integrable, while integrating it against a square-integrable momentum amplitude reconstructs a normalizable state. The transform also makes momentum domains explicit: powers of momentum correspond to weighted square-integrability of that amplitude.

Required background. Position Kets explains weak resolutions of identity. Distributions as Generalized Vectors fixes the antidual momentum sign. The momentum label is $p$, its measure is $dp$, and $\hbar>0$.

The Fourier kernel defines a momentum eigenfunctional

Use the kernel

$$k_p(x)=\frac{e^{ipx/\hbar}}{\sqrt{2\pi\hbar}}.$$

For a Schwartz test vector, define

$$\begin{aligned} \langle\phi|p\rangle &=\int\overline{\phi(x)}k_p(x)\,dx,\\ \langle p|\phi\rangle &=\frac1{\sqrt{2\pi\hbar}} \int e^{-ipx/\hbar}\phi(x)\,dx. \end{aligned}$$

The first is antilinear in $\phi$, and the second is linear. The bound $|\langle\phi|p\rangle|\le(2\pi\hbar)^{-1/2}\|\phi\|_1$, together with the Schwartz decay seminorms, proves continuity. The kernel itself has constant nonzero magnitude and hence is not in $L^2(\mathbb R,dx)$.

Distributional differentiation gives $-i\hbar k_p'=pk_p$. By the antidual extension formula, this means $P^\times|p\rangle=p|p\rangle$. The calculation uses the domain-preserving action of $P$ on the test space; it does not place the plane wave in the Hilbert-space momentum domain.

A unitary transform turns generalized coefficients into states

Write $\widetilde\psi=\mathcal F_\hbar\psi$, where $\mathcal F_\hbar$ is the transform just defined. Fourier inversion and Plancherel give

$$\begin{gathered} \psi(x)=\frac1{\sqrt{2\pi\hbar}} \int e^{ipx/\hbar}\widetilde\psi(p)\,dp,\\ \|\psi\|_2=\|\widetilde\psi\|_2. \end{gathered}$$

For Schwartz states, the ordinary inverse integral is well defined pointwise. For a general $L^2$ state, the transform and inverse are defined by unitary completion; the formula need not be an absolutely convergent pointwise integral. A concrete reconstruction is

$$\begin{gathered} \psi_R=\mathcal F_\hbar^{-1} \bigl(\mathbf1_{[-R,R]}\widetilde\psi\bigr),\\ \psi_R\longrightarrow\psi\text{ in }L^2. \end{gathered}$$

This follows because the omitted momentum tail has vanishing square norm. It supplies an exact Hilbert-space meaning for the formal expansion $|\psi\rangle=\int|p\rangle\langle p|\psi\rangle dp$.

The unitary transform, rather than a hypothetical Hilbert norm of $|p\rangle$, establishes normalization of the reconstructed state. Delta Normalization treats the corresponding kernel identity.

Momentum moments encode operator domains

The transform intertwines the line momentum operator with multiplication by $p$:

$$\mathcal F_\hbar P\mathcal F_\hbar^{-1}=M_p.$$

Consequently, $\psi\in D(P)$ exactly when $p\widetilde\psi\in L^2(dp)$, and more generally $\psi\in D(P^n)$ exactly when $p^n\widetilde\psi\in L^2$. These conditions describe the self-adjoint operator domains, not merely formal derivatives.

A momentum-region probability instead needs no such moment:

$$\Pr_\psi(P\in\Delta) =\int_\Delta|\widetilde\psi(p)|^2\,dp.$$

Every normalized $L^2$ amplitude defines this law. The existence of a first absolute moment, a finite second moment, and membership in all power domains are progressively different regularity statements.

Finite momentum bandwidth does not control spatial tails

For $a>0$, choose the normalized amplitude

$$\widetilde\psi(p)=\frac{\mathbf1_{[-a,a]}(p)}{\sqrt{2a}}.$$

Its inverse is the square-integrable function

$$\psi(x)=\sqrt{\frac{\hbar}{\pi a}} \frac{\sin(ax/\hbar)}{x},$$

with the removable value at zero supplied by continuity. This is an actual state assembled from generalized momentum components. Its momentum support is compact, so all momentum moments exist. Nevertheless, its spatial tail is only of order $1/|x|$.

The example distinguishes momentum regularity from membership in a common position-and-momentum test space. A state can be in every $D(P^n)$ and still fail $D(X)$, hence cannot be Schwartz. Neither a convenient spectral expansion nor excellent control of one observable implies regularity for every other observable.

Exercise: the bandwidth state’s domains

For the state above, compute $\langle P\rangle$ and $\|P\psi\|^2$. Verify that $\psi\notin D(X)$ and explain why it is still a valid normalized state.

Solution

The symmetric uniform momentum density gives $\langle P\rangle=0$ and $\|P\psi\|^2=(2a)^{-1}\int_{-a}^ap^2dp=a^2/3$. But $|x\psi(x)|^2=(\hbar/\pi a)\sin^2(ax/\hbar)$ is not integrable on the line. Thus $X\psi$ is not an $L^2$ vector. Plancherel still gives $\|\psi\|=1$; the missing position second moment restricts that unbounded operator’s action, not the existence of the state or its position event probabilities.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — the Fourier realization of momentum and its domains.
• 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 transforms, spectral multipliers, and Sobolev spaces.