1
$\begingroup$

I am a physicist learning functional analysis because of its fundamental role in quantum mechanics.

There are so many bizarre facts.

One is, there are densely defined operators which seem cannot be defined everywhere.

Any simple example?

$\endgroup$

3 Answers 3

5
$\begingroup$

Not true. Any linear operator defined on a linear subspace can be extended to the whole space, but

  1. You need some form of the Axiom of Choice to do this.
  2. You lose any nice properties of the operator (e.g. if the operator is symmetric, the extension won't be).
$\endgroup$
1
  • $\begingroup$ To use a quantum mechanical language, I would say that such extensions are unphysical. $\endgroup$ Dec 8, 2014 at 22:13
4
$\begingroup$

Let's take the simplest possible system: a one-dimensional spinless particle. We can describe it in terms of wave functions $\psi\in L^2(\mathbb{R})$. The fundamental observable is "position" $x$, which must be represented as self-adjoint operator acting on wave functions. Quantum mechanics tells us that the right way of doing so is multiplying by $x$. Problem is, if we take an arbitrary wave function $\psi$, it is NOT granted that $x\psi(x)$ is again a wave function, because it needs not belong to $L^2(\mathbb{R})$ space.

Hence the necessity of considering operators that are not defined on the whole Hilbert space.

Moreover, the mapping $$ \psi \mapsto x\psi(x) $$ is NOT continuous with respect to the $L^2$ norm. To wit, take a peak shaped wave function, concentrated around the origin. Have it travel to infinity: $$ \psi_n(x)=\psi(x- n). $$ You are not altering the $L^2$ norm, of course: $\lVert \psi_n\rVert_2 = \lVert \psi\rVert_2$. However, after applying the position operator, you have $\lVert x\psi_n\rVert_2 \to \infty$, because multiplying by $x$ you amplify the travelling peak as it goes away. No continuous linear operator can act this way.

Hence the necessity to introduce the machinery of unbounded (=discontinuous) linear operators.

$\endgroup$
0
$\begingroup$

Let $X=C[0,1]$ with the uniform metric, $Y=C^1[0,1]$ and $D\colon Y\to X$ the derivative operator: $Df=f'$ for all $f\in Y$. $Y$ is dense in $X$ and $D$ cannot be extended to $X$ as a continuous operator.

$\endgroup$
4
  • 1
    $\begingroup$ It's already not continuous, so this doesn't really say anything. $\endgroup$ Dec 8, 2014 at 21:56
  • $\begingroup$ Continuous operators densely defined can always be extended. $\endgroup$ Dec 8, 2014 at 22:00
  • $\begingroup$ But the OP is not talking about continuous operators. $\endgroup$ Dec 8, 2014 at 23:12
  • $\begingroup$ I interpret that the OP is talking about unbounded operators defined on a dense subset, like the derivative or the example in Giuseppe Nero's answer. $\endgroup$ Dec 9, 2014 at 10:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .