Consider a quantum system defined on the interval $[0,L]$. Suppose we take the periodic boundary condition,

$$ \psi(0) = \psi(L ) . $$

It is know that for such a system, the position operator is not well defined. It is often said that even if $\psi (x)\in \mathcal{H }$, generally $x \psi(x) \notin \mathcal{H}$.

But are the states in the Hilbert space $\mathcal{H }$ dense in the space $L^2 (0,L)$?

If so, $x \psi(x)$ can be approximated by some state $\phi(x)$ in $\mathcal{H }$. Then possibly we can define the position operator as

$$ x \psi (x) = \phi(x) .$$

Can this idea work?

  • 1
    $\begingroup$ How is this different from setting up an angle operator on a circle? Similar to the azimuthal angle in polar coordinates. $\endgroup$ – Sean E. Lake Nov 6 '16 at 10:57

The issue is much easier than you suppose. If you define $X$ as multiplicative operator in the Hilbert space $L^2([0,L], dx)$, differently to what happens in $L^2(\mathbb R, dx)$, its domain is the whole space. If $\psi \in L^2([0,L], dx)$ then $$||X\psi||^2 = \int_0^L |x \psi(x)|^2 dx \leq L^2 \int_0^L |\psi(x)|^2 dx \leq L^2 ||\psi||^2 < +\infty\:.$$ This operator is self-adjoint, bounded and defined on the whole Hilbert space. It is possible to prove that $||X||=L$ and $\sigma(X)= \sigma_c(X)= [0,L]$.

Boundary conditions are completely irrelevant here, since the Hilbert space $L^2(\mathbb R, dx)$ is defined independently from them. However, if you try to define $$(X_0\psi)(x) := x \psi(x)\quad \psi \in L^2([0,L], dx) \quad \psi(0) = \psi(L)$$ this definition does not make sense, as $0$ and $L$ have zero Lebesgue measure and $\psi$ is defined up to zero-measure sets.

You may artificially add some condition to make sensible the last requirement like this $$(X_0\psi)(x) := x \psi(x)\quad \psi \in L^2([0,L], dx)\:, \mbox{$\psi$ continuous,} \quad \psi(0) = \psi(L)\:.$$ With this definition the self-adjointness condition $X_0^* =X_0$ fails, just because it turns out that $$X_0^* = X$$ where $X$ is the self-adjoint operator I initially introduced which is defined in the whole Hilbert space. In other words, the adjoint of $X_0$ has a larger domain than $X_0$ which, in fact, coincides with the whole Hilbert space. To obtain a self-adjoint operator from $X_0$ one may maximally extend it from its initial domain, made of continuous functions, to the closure of this domain which is the whole $L^2([0,L], dx)$ since the initial domain is dense therein. This extension is possible because $X_0$ is continuous. This way, as a matter of fact, we come back to my initial definition.

It is however possible to prove that, since $X_0$ is symmetric and its adjoint is self-adjoint, $X$ is the only possible self-adjoint linear extension of $X_0$, obtained by continuity or not.

  • $\begingroup$ And what does it say about quantum mechanics of particle confined to a circle? $\endgroup$ – Blazej Nov 6 '16 at 14:25
  • $\begingroup$ The fact that we are dealing with a circle rather than a segment is decided by the the group of symmetries we fix. The circle admits $\mathbb S^1$ as isometry group and thus the theory on the circle supports an unitary representaiton of this group physically meaningful: the generator is the quasi-momentum. If the theory is in $[0,L]$ this group representation has no physical meaning. $\endgroup$ – Valter Moretti Nov 6 '16 at 14:47
  • $\begingroup$ After all $L^2(\mathbb R^3, dx)$ is the Hilbert space of either a classic and relativistic (massive) particle. The distinction is done when looking at the fundamental group of symmetries we adopt in that Hilbert space: a unitary representation of Galilei's group or a unitary representation of Poincaré group. $\endgroup$ – Valter Moretti Nov 6 '16 at 14:51

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