# 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?

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 multplication 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? – Andrea Feb 3 '16 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. – ACuriousMind Feb 3 '16 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. – Andrea Feb 3 '16 at 15:40
• @ACuriousMind When you have defined the domain for $p$, you have included the condition $\Psi(0)=\Psi(1)$, which is the specified boundary condition. But you do not seem to have included that condition in the domain of $x$. Is that right? If so, why? – Curiosa Aug 5 '16 at 18:52
• @Curiosa I took $L^2([0,1])$ as the Hilbert space we're dealing with, i.e. a particle confined to the interval $[0,1]$. If I imposed the condition from the start, i.e. on all wavefunctions, I would be dealing with $L^2(S^1)$, i.e. a particle on a ring, whose quantum theory has different subtleties (a basic variant of the Aharonov-Bohm effect, for instance). It's not actually meaningful to prescribe "boundary conditions" for the wavefunction - those tend to arise naturally as in the case of the momentum operator - you can only get it self-adjoint if you restrict its domain to that conditions. – ACuriousMind Aug 8 '16 at 20:00

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, \begin{equation} \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|. \end{equation} 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. – higgsss Feb 2 '16 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. – ACuriousMind Feb 2 '16 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. – higgsss Feb 2 '16 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? – Andrea Feb 2 '16 at 20:21
• @AndreaDiBiagio Yes it does. So we have $\sigma_x \sigma_p\ge 0$ for a plane wave. – higgsss Feb 2 '16 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. – higgsss Feb 3 '16 at 19:22
• But $2\pi$ corresponds to total uncertainty, so the "colloquial" HUP holds. @higgsss – Andrea Feb 3 '16 at 22:48
• "Colloquially", we have $\Delta L_z \Delta\varphi \ge \hbar/2$, and this doesn't hold for angular momentum eigenstates. – higgsss Feb 4 '16 at 2:52