How can I solve this quantum mechanical "paradox"?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.