# Spectrum of the Laplacian for free space versus spectrum of Laplacian with boundary conditions?

In this question the spectrum of the Laplacian in free space is defined as $]-\infty, 0]$. So it seems the spectrum can be determined without regard to boundary conditions? If instead we consider the Laplacian with Dirichlet (or Neumann) boundary conditions, would you expect this to effect the spectrum? If so, why would boundary conditions affect the spectrum?

• this is the basis of quantum physic : a differential operator in a bounded space will have a discrete spectrum, in opposition to a free space, at first you just have to try it and see it happens. then you can see the bounded space as a separable Hilbert space ($H^1_0(\Omega)$ for the Dirichlet conditions) – reuns Jul 19 '16 at 10:39
• The Laplacian on free space is essentially selfadjoint when considered on the domain $C_c^{\infty}(\mathbb{R}^n)$ consisting of compactly-supported infinitely-differentiable functions. That is, the restriction of $\Delta$ to such a domain has a selfadjoint closure. So there are no boundary conditions required (or possible) at $\infty$. Can you explain how you are imposing conditions? – DisintegratingByParts Jul 19 '16 at 12:47

The spectrum of an operator, in this case an unbounded operator, greatly depends on the domain of definition and the surrounding Banach or Hilbert space. For example on the space $C^2[0,1]$ of twice differentiable functions with the usual Banach norm, the Laplacian is a bounded operator, and hence its spectrum is bounded.
On the other hand, if we take the $X=\{ f \in C^2[0,1]\bigm| f(0) = 0\}$, then the spectrum is just $\{0\}$.
• So the Laplacian isn't invertible if we set one of the boundary conditions to $0$? Is it the case that setting that boundary condition means that the Laplacian will map every $u \in C^2([0, 1])$ to $0$? – csss Jul 21 '16 at 8:17