How can I solve this quantum mechanical "paradox"?

Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?

• why must sigma_x be finite for a plane wave? electron6.phys.utk.edu/qm1/modules/m1/free_particle.htm Commented Feb 2, 2016 at 15:49
• @annav This problem is about a particle in a ring of circumference $a$. Commented Feb 2, 2016 at 22:01

This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space.

Let's set $$a=1$$ for convenience. The operator $$p =-\mathrm{i}\hbar\partial_x$$ acting on wavefunctions with periodic boundary conditions defined on $$D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$$ is self-adjoint, that is, on the domain of definition of $$p$$, we have $$p=p^\dagger$$, and $$p^\dagger$$ admits the same domain of definition. The self-adjointness of $$p$$ follows from the periodic boundary conditions killing the surface terms that appear in the $$L^2$$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $$\psi\in D(p)$$ and every $$\phi\in D(p^\dagger) = D(p)$$, but not for $$\phi$$ with $$\phi(0)\neq\phi(1)$$.

Now, for the question of the commutator: the multiplication operator $$x$$ is defined on the entire Hilbert space, since for $$\psi\in L^2([0,1])$$ $$x\psi$$ is also square-integrable. For the product of two operators $$A,B$$, we have the rule $$D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$ and $$D(A+B) = D(A)\cap D(B)$$ so we obtain \begin{align} D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\ D(xp) & = D(p) \end{align} and $$x\psi\in D(p)$$ means $$0\cdot \psi(0) = 1\cdot\psi(1)$$, that is, $$\psi(1) = 0$$. Hence we have $$D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$ and finally $$D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$ meaning the plane waves $$\psi_{p_0}$$ do not belong to the domain of definition of the commutator $$[x,p]$$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $$A,B$$, you may rewrite the uncertainty principle as $$\sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$ where the r.h.s. and l.h.s. are now both defined on $$D(A)\cap D(B)$$. Applying this version to the plane waves yields no contradiction.

• But how do you recover Heisenberg's uncertainty principle then? Commented Feb 3, 2016 at 8:53
• @AndreaDiBiagio: It is the last equation in my post. For plane waves, the r.h.s. reads zero, so $\sigma_p = 0$ is not a contradiction. For functions in $D([x,p])$, it is the usual $\sigma_p\sigma_x\geq\hbar/2$. I'm not sure what you think there needs to be recovered. Commented Feb 3, 2016 at 14:18
• @ACuriousMind You answer is detailed and very informative. However, it seems to me to miss OP's question's spirit, which is about the HUP understood colloquially as the product of the uncertainty about the position and momentum of a particle is always larger than some non-zero bound. To better you answer you could perhaps consider adding a less mathematical punch-line. Commented Feb 3, 2016 at 15:40

Notice that $\psi(x)$ is defined on a circle of circumference $a$. Multiplying $x$ on this circle is really multiplying a periodic extension of $x$, i.e., the sawtooth function $x - a\lfloor x/a\rfloor$, where $\lfloor y\rfloor$ means the largest integer not greater than $y$. So, the commutator of the position and momentum operators involves the derivative of not only $x$ but also the discontinuous part $-a\lfloor x/a\rfloor$. Therefore, $$\sigma_{x} \sigma_p \geq \frac{1}{2}\Big|\langle \psi|\,[\hat{x},\hat{p}]\,|\psi\rangle\Big| = \frac{\hbar}{2}\Bigg|\Big\langle\psi\,\Big|\frac{d}{dx}\big(x - a\lfloor x/a\rfloor\big)\Big|\,\psi\Big\rangle\Bigg| = \frac{\hbar}{2}\Big|1-a|\psi(0)|^{2}\Big|.$$ For a plane wave $\psi(x) = e^{ikx}/\sqrt{a}$, the above reduces to $\sigma_{x} \sigma_p\ge0$, as desired.

• This is a nonrigorous version of more mathematical answers other users might provide. Commented Feb 2, 2016 at 9:38
• You don't need to modify the position operator, it is perfectly self-adjoint without the sawtooth function added to it. What you need to do is think about the domain of validity of the uncertainty principle, in particular the commutator. Commented Feb 2, 2016 at 15:01
• @ACuriousMind You are right, but my purpose is to keep the discussion at the level of undergraduate QM. Considering domains of operators and distinguishing between Hermiticity and self-adjointness are somewhat too mathematically involved for many physicists including myself. Commented Feb 2, 2016 at 15:32
• @higgsss from what you wrote, doesn't the expression in the absolute value read $1-a/|\sqrt{a}|^2 = 0$ for a plane wave? Commented Feb 2, 2016 at 20:21
• @AndreaDiBiagio Yes it does. So we have $\sigma_x \sigma_p\ge 0$ for a plane wave. Commented Feb 2, 2016 at 21:04

There are two ways to interpret the boundary conditions you are imposing.

The first case is that of a system which is infinite in extent, but has a periodic regularity. This is like an electron in an idealised 1D crystal, where the periodic boundary condition is imposed by the presence of nuclei regularly spaced. In this case, the plane wave solution has $\sigma_p$ = 0 but $\sigma_x$ is infinite.

The second case, is that of a particle in a ring. In this case, you can imagine the particle as being constrained within the ring by a infinitely deep potential well. The system is not actually 1D, it is 2D. Now you have to consider both $\sigma_x \sigma_{p_x}$ and $\sigma_y \sigma_{p_y}$, and even though $\sigma_x = \sigma_y \sim a$, the uncertainty in momentum will be imposed by the thickness of the ring. The plane wave solution will in fact represent angular momentum eigenstates.

• If one embeds the ring in 2D (realized by a ring-shaped potential well), the "plane wave along the ring" will become an angular momentum eigenstate. The "paradox" will persist because in an angular momentum eigenstate, the uncertainty of the angle is $2\pi$, while that of the angular momentum is 0. Commented Feb 3, 2016 at 19:22
• But $2\pi$ corresponds to total uncertainty, so the "colloquial" HUP holds. @higgsss Commented Feb 3, 2016 at 22:48
• "Colloquially", we have $\Delta L_z \Delta\varphi \ge \hbar/2$, and this doesn't hold for angular momentum eigenstates. Commented Feb 4, 2016 at 2:52