# Why can a particle have both 0 and infinite momentum uncertainty?

I am a little confused by the meaning of this result. If I take a constant wave function $$\phi_0 = \frac{1}{\sqrt{2a}}$$ spaning over $$(-a,a)$$, I can convert this the momentum space:

$$\Phi(p) = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} \phi(x) e^{-ipx/\hbar}dx = \frac{1}{p\sqrt{\pi a \hbar}}\sin(\frac{pa}{\hbar})$$

If I calculate $$\Delta P = \sqrt{-

^2}$$

using integrals (example):

$$ = \int_{-\infty}^{\infty} (\frac{p}{\sqrt{\pi a \hbar}})^2\sin(\frac{pa}{\hbar})^2p dp \rightarrow\infty$$

Whereas when I do this with the coordinate space wavefunction, I will get zero since $$\partial_x \phi = 0$$ since it is a constant. Why is this happening? Shouldn't momentum uncertainty be the same regardless if you calculate it in momentum or coordinate space?

• I've not yet thought this through, but could it be you're missing issues of normalizability? the wave function you take is not square-integrable. Nov 27, 2022 at 15:12
• $\partial_x \phi(x) =(\delta(x+a)-\delta(x-a))/\sqrt{2a}$ solves the puzzle... Nov 27, 2022 at 15:18
• I think it's just a typo - your momentum space integral has $p^2$ in the numerator instead of the denominator. The integrand should go as $\sin^2 p / p$, which gives zero. Nov 27, 2022 at 15:24
• there’s a separate problem that $\hat p$ is not self-adjoint on the finite line. see the canonical paper by Bonneau G, Faraut J, Valent G. Self-adjoint extensions of operators and the teaching of quantum mechanics. American Journal of physics. 2001 Mar;69(3):322-31. Nov 27, 2022 at 15:33

1. Answer of the mathematician:

Consider the Hilbert space $$\mathcal{H}=L^2(R)$$ and the selfadjoint unbounded linear operator $$P$$ defined on the dense domain $$\mathcal{D}(P)=\{\psi \in \mathcal{H} | \psi \, \text{ is absolutely continuous and} \, \psi^\prime \in \mathcal{H} \}$$ with $$(P\psi)(x)=-i \hbar \psi^\prime(x)$$ (see e.g. W. Thirring, Quantum Mathematical Physics, Springer, ch. 2.5). The function $$\phi(x)=c_I(x)/\sqrt{2a}$$ (where $$c_I$$ is the characterisic function of the interval $$I=[-a,a]$$) is an element of $$\mathcal{H}$$, but NOT of the domain $$\mathcal{D}(P)$$, as it fails to be absolutely continuous (which is needed for partial integration). The fact that $$\phi^\prime(x)=0$$ almost everywhere (except for the set of measure zero consisting of the two points $$\pm a$$) is NOT sufficient for $$\phi$$ being an element of the domain of $$P$$.

For the mathematical purist this might already suffice as an answer. For those who whish to look behind the scenes, consider a sequence $$\{\phi_n\}_{n \in N}$$ of unit vectors in $$\mathcal{D}(P^2)$$ converging to $$\phi$$ in the $$L^2$$ norm (it is easy to construct such functions by just smearing out the edges of $$\phi$$ at $$\pm a$$). The expection values $$\langle \phi_n | P^2 \phi_n \rangle$$ are now finite but unbounded in the limit $$n \to \infty$$.

As the Fourier transform $$\mathcal{F}: \mathcal{H} \to \mathcal{H}$$ defined by $$\tilde{\psi}(p) = (\mathcal{F}\psi)(p)=\int_{-\infty}^{+\infty}dx\frac{e^{-ip x /\hbar}}{\sqrt{2 \pi \hbar}} \psi(x)$$ is a unitary transformation, all observations obtained in the "x-representation" must have a counterpart in the "p-representation": the "momentum operator" $$\tilde{P} = \mathcal{F} P \mathcal{F}^{-1}$$ is now simply a multiplication operator $$\tilde{P} \tilde{\psi}(p)=p \tilde{\psi}(p)$$ with domain $$\mathcal{D}(\tilde{P})=\{\tilde{\psi} \in \mathcal{H} | p \tilde{\psi}(p) \in \mathcal{H}\}$$.

The Fourier transform of $$\phi$$ is now given by $$\tilde{\phi}(p) =\sqrt{\frac{\hbar}{\pi a}} \frac{1}{p} \sin(p a/\hbar)$$, being square-integrable and thus an element of $$\mathcal{H}$$ (as to be expected as a result of a unitary transformation). On the other hand, $$\tilde{\phi}$$ is NOT in the domain of $$\tilde{P}$$ as $$p \tilde{\phi}(p)$$ is NOT square-integrable over $$R$$ and we recover all the observations of the previous discussion.

1. Answer of the theoretical physicist:

She/he is not frightened by leaving the cosy atmosphere of Hilbert space, remarking that $$\phi^\prime (x) = (\delta(x+a)-\delta(x-a))/ \sqrt{2 a}$$. A detailed discussion along these lines can be found in answers to related questions at Physics Stackexchange (see e.g. "Rectangular window...")

1. Answer of the mathematical physicist:

She/he secretly performs a calculation like 2 but publicly presents the result like 1. This impresses the uninitiated.