# [Physics] How to solve this quantum mechanical “paradox”

heisenberg-uncertainty-principlemomentumquantum mechanicswavefunction

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?

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.