Momentum on the Real Line

Whole-line momentum is the self-adjoint weak derivative $P=-i\hbar\,d/dx$ on $H^1(\mathbb R)$. Its resolvent is an exponentially decaying integral operator, and its spectrum is the entire real line with absolutely continuous multiplicity one. The Schrödinger representation establishes the construction and common-core theorem; here the resolvent gives a concrete test of that realization.

Required background. Resolvent Operators defines $(P-z)^{-1}$ and its spectral role. Momentum Kets fixes the unitary Fourier normalization.

The maximal weak-derivative domain fixes the operator

On $L^2(\mathbb R,dx)$,

$$D(P)=H^1(\mathbb R),\qquad P\psi=-i\hbar\psi'.$$

The Fourier transform identifies this operator with multiplication by the real momentum variable:

$$\mathcal F_\hbar P\mathcal F_\hbar^{-1}=M_p,\qquad D(M_p)=\{f:pf\in L^2(dp)\}.$$

This establishes self-adjointness and the maximal domain. Smooth compactly supported functions are an operator core, so starting with the same derivative on $C_c^\infty(\mathbb R)$ has this unique self-adjoint closure.

The derivative is weak rather than pointwise. A state may have a kink and still belong to $H^1$; a jump generally introduces a delta in its distributional derivative and prevents membership. The pointwise appearance of a graph is less informative than the Sobolev condition.

There is no endpoint phase on the whole line. Introducing an interval, half-line, or excluded point changes the seed operator and requires a new extension analysis.

The upper-half-plane resolvent has a one-sided kernel

Let $\operatorname{Im}z>0$. Define

$$G_z(s)=\frac{i}{\hbar}\,\mathbf1_{s>0}e^{izs/\hbar}.$$

Distributional differentiation gives $(P-z)G_z=\delta_0$: the smooth part cancels, while differentiating the step contributes $-i\hbar(i/\hbar)\delta_0=\delta_0$.

Since the exponential decays for positive $s$,

$$\|G_z\|_{L^1}=\frac1{\operatorname{Im}z}.$$

Therefore convolution gives a bounded map on $L^2$,

$$((P-z)^{-1}f)(x) =\frac{i}{\hbar}\int_{-\infty}^{x} e^{iz(x-y)/\hbar}f(y)\,dy.$$

The equation implies $\psi'=(i/\hbar)(f+z\psi)\in L^2$, so the output lies in the correct domain. A homogeneous solution $Ce^{izx/\hbar}$ is not square integrable at both ends unless $C=0$, proving uniqueness.

For $\operatorname{Im}z<0$, the decaying direction reverses:

$$((P-z)^{-1}f)(x) =-\frac{i}{\hbar}\int_x^\infty e^{iz(x-y)/\hbar}f(y)\,dy.$$

These opposite spatial directions express square-integrability of the resolvent solution. They are not a choice of physical forward or backward time evolution.

The resolvent norm detects distance to the real spectrum

In Fourier space the resolvent multiplies by $(p-z)^{-1}$. Hence

$$\|(P-z)^{-1}\| =\operatorname*{ess\,sup}_{p\in\mathbb R}\frac1{|p-z|} =\frac1{|\operatorname{Im}z|}.$$

The convolution estimate is therefore sharp. It also explains why the resolvent does not remain bounded as $z$ approaches a real momentum.

Every real number lies in the spectrum because the multiplier $p-z$ can be arbitrarily small on sets of positive measure. None is an eigenvalue: a square-integrable function supported only at a single momentum is zero. For every state the scalar spectral measure is

$$\mu_\psi(\Delta) =\int_\Delta|\widetilde\psi(p)|^2\,dp,$$

so the spectrum is purely absolutely continuous. The multiplication variable has one scalar component, giving spectral multiplicity one. This differs from the free energy operator, which identifies the two momentum values $\pm p$.

Resolvents regularize once, while translations preserve regularity

The resolvent sends $L^2$ boundedly into $H^1$ with its graph norm because $P(P-z)^{-1}=I+z(P-z)^{-1}$. It need not map every input into $H^2$.

Momentum evolution instead translates a state:

$$e^{-iaP/\hbar}\psi(x)=\psi(x-a).$$

It preserves each Sobolev regularity class and cannot repair a missing weak derivative. Resolvent regularization and unitary propagation are different operations, despite both being defined by the same spectral calculus.

Exercise: test the adjoint relation of the kernel

Verify directly that the integral kernels above satisfy $R_z^*=R_{\overline z}$.

Solution

For $\operatorname{Im}z>0$, the kernel is $R_z(x,y)=(i/\hbar)\mathbf1_{x>y}e^{iz(x-y)/\hbar}$. Its adjoint kernel is $\overline{R_z(y,x)}=(-i/\hbar)\mathbf1_{y>x}e^{i\overline z(x-y)/\hbar}$, which is exactly the lower-half-plane formula. Both the sign and the direction of the integration interval are necessary for the self-adjoint resolvent identity.

References

• Bonneau, Guy, Jacques Faraut, and Galliano Valent, “Self-adjoint extensions of operators and the teaching of quantum mechanics,” American Journal of Physics 69, 322–331 (2001), doi:10.1119/1.1328351, arXiv:quant-ph/0103153 — whole-line momentum and its contrast with boundary problems.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapters 7–8 — Fourier domains and momentum spectral calculus.