Can a self-adjoint operator have a continuous set of eigenvalues?

Question

This should be a trivial question for mathematicians but not for typical physicists.

I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" and "residual" parts [I gather that no boundedness assumption is needed but I could be wrong]. I further know that the point spectrum coincides with the set of eigenvalues of the operator. It seems from the terminology that the point spectrum is a discrete set of isolated point and that the eigenvalues cannot form a continuum. But I haven't been able to find a clear statement in a math reference about this.

Actually, I'm mostly interested in self-adjoint operators on a Hilbert space; so a simpler version of my question would be: Can a self-adjoint operator have a continuous set of eigenvalues? And if yes, under what conditions do the eigenvalues have to be discrete?

I appreciate any help.

Answer 1

Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal. In a separable Hilbert space, any orthogonal set is countable. So a self-adjoint operator on separable Hilbert space has only countably many eigenvalues. (As noted, this does not mean the spectrum is countable.)

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Proof for "Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal".
Let $\alpha \ne \beta$ be eigenvalues of the self-adjoint operator $A$. They are real. There exist nonzero eigenvectors $x, y$ so that $Ax=\alpha x, Ay=\beta y$. We must show $x \perp y$.

Compute $$ \alpha\langle x,y\rangle = \langle \alpha x, y\rangle = \langle Ax, y\rangle = \langle x, Ay\rangle = \langle x, \beta y\rangle = \beta\langle x,y\rangle $$ Then $$ (\alpha-\beta)\langle x,y\rangle = 0 $$ But $\alpha \ne \beta$, so $\alpha - \beta \ne 0$ and thus $$ \langle x,y\rangle = 0, $$ so $x \perp y$ as claimed.

Answer 2

You are confusing two notions. First of all, the point spectrum just means eigenvalues; there is no assumption that these form a discrete set. The shift operator is a simple example where the spectrum is "continuous".

The condition for the eigenvalues to be discrete is precicsely that the operator $A:H \to H$ is compact. It is however possible for non-compact self adjoint operators to have a discrete spectrum. The simplest example of this is the orthogonal projection operator $P:H \to Y$ where $Y$ is a closed subspace of the Hilbert space $H$. Here the spectrum is $0$ and $1$.

Bounded self adjoint operators have no residual spectrum but they do indeed have a continuous spectrum. Take any compact operator $A:H \to H$ where dim$H=+\infty$. Then $0$ belongs to the continuous spectrum because otherwise $A:H \to H$ would be invertible, implying that dim$H <\infty$. Continuous = "exists a set of approximate eigenvectors".

If you want a continuous range of spectrum take $Af = x f(x)$ on $L^2([0,1])$. Then the range of the spectrum is just $[0,1]$. There are no eigenvalues for this operator and moreoever since the residual spectrum is empty for self adjoint operator, $[0,1]$ is the spectrum and it is equivalent to the continuous spectrum.

So a final point, continuous just means that $R(A-\lambda I)$ is not dense but that $\lambda$ is not an eigenvalue. It has nothing to do with the actual spectrum being discrete or continuous.

So I think you were mixing up two notions but hopefully I've provided examples for both.

Answer 3

One version of the spectral theorem states that if $A$ is a selfadjoint operator on a separable Hilbert space $H$, then we can find a set $X$, a $\sigma$-finite measure on $X$, a unitary operator $U:H\to L^2(X)$, and a measurable function $a:X\to \mathbb{R}$, such that $A=U^*aU$. In other words, any selfadjoint operator, in essence, is nothing but 'multiplication by a real valued function', in suitable 'coordinates'. The spectrum of $A$ coincides with the essential range of the function $a$.

Now, an eigenvalue must be a real number $\lambda$ such that the set $a^{-1}(\lambda)$ has a positive measure. Since $X$ is $\sigma$-finite, the eigenvalues can be at most a countable set. But you can have a continuous spectrum of course, only the points will not be eigenvalues. You can play with real valued functions (and with measures on $X$; any $\sigma$-finite measure is ok) to train intuition.

Answer 4

I add this remark because it may be part of what the OP wants. Note that, as to the spectrum of a bounded linear self-adjoint operator on $\ell^2$, of course, it can be any compact set $K$ of $\mathbb{R}$. Just take a diagonal operator where the set of the diagonal elements (eigenvalues) is dense in $K$.

Answer 5

Recall that the spectrum of an operator $A$ on a Hilbert space is the set of vales $\lambda$ such that $A-\lambda I$ does not have a bounded inverse . So if $A$ is multiplication by a function in $L^2$ of a measure space, then any point of the essential range of the function is in the spectrum. So, for example (and this is the classic example) the spectrum of multiplication by $x$ on the real line is the whole line. However, a point of the spectrum is not necessarily an eigenvalue. In fact $\lambda$ is an eigenvalue (or in the point spectrum) iff $A- \lambda I$ has a non-trivial null space. (And as others have pointed out, on a separable Hilbert space, the point spectrum is countable.)

Answer 6

Just so we have the silly example here. Consider $\ell^2([0,1])$, meaning the set of SEQUENCES $u_{x}$ indexed by a number $x\in [0,1]$. So the scalar product is $$ \langle u, v \rangle = \sum_{x \in [0,1]} \overline{u_x} v_x. $$ In order for $u \in \ell^2([0,1])$, we must have that $u_x \neq 0$ for at most countably many $x$. So this is very different from $L^2([0,1])$.

Now consider the operator $$ (Au)_x = xu_x. $$ This operator is diagonal and it's eigenvalues are $[0,1]$. The eigenvector corresponding to $x \in [0,1]$ is $$ u_y = \begin{cases} 1, & x=y,\\\ 0, & \text{otherwise}.\end{cases} $$ Of course the key to this example and the question is that $\ell^2([0,1])$ is a NON-separable Hilbert space.

Answer 7

The resolvent set is the set of all $\zeta \in \mathbb{C}$ for which $T-\zeta$ is invertible (which means especially that the Range is all of $H$). The spectrum $\Sigma$ is the complement of the resolvent.

Normally, a $\lambda \in \Sigma$ is only called eigenvalue, if a $x \in H$ exists such that $Tx = \lambda x$. The set of those eigenvalues is countable as already argued but not necessarily discrete. For example: A compact operator on a Hilbert space has a countable spectrum, but not necessarily a discrete one; the spectrum has an accumulation point at zero, which may be an eigenvalue.

A noncompact bounded operator may have a countable spectrum consisting only of "true" eigenvalues, but then the eigenvalues -- counted with multiplicity -- have a nonzero accumulation point.

An unbounded operator whose resolvent is compact (for some element of the resolvent set) has a discrete spectrum; this criterion works on Laplacians on compact manifolds, for example. The Laplacian on $L^2(\mathbb{R}^n)$ with usual Lebesgue measure -- which is self-adjoint if $\mathrm{dom}(\nabla) = H^2$ -- however has as spectrum the whole positive real line, but no eigenvalue.

Answer 8

The hyperbolic Laplacian is self adjoint and its spectrum has a continuous component.