Let $H$ be a Hilbert space and $T \in \mathcal{L}(H)$ be self-adjoint. I know that this is a basic question, but I do not understand the infinite dimensional case of linear algebra very well. My question is, when does there exist a Hilbert basis for $H$ consisting entirely of eigenvectors of $T$? In the finite dimensional case, we only had to show that the matrix representation of $T$ in any basis was similar to a diagonal matrix. But in the infinite dimensional case, it is not clear that matrix representations even make any sense.
Eigenbases in the infinite dimensional case
adjoint-operatorseigenvalues-eigenvectorshilbert-spaceslinear algebraoperator-theory
Related Solutions
Take any orthonormal basis $\{e_{n} \}_{n=1}^{\infty}$ of a Hilbert space $H$ and define $Lf = \sum_{n=1}^{\infty}n(f,e_{n})e_{n}$ on the domain $\mathcal{D}(L)$ consisting of all $f\in H$ for which $\sum_{n}n^{2}|(f,e_n)|^{2} < \infty$. The operator $L : \mathcal{D}(L)\subset H \rightarrow H$ is a densely-defined selfadjoint linear operator that has simple eigenvalues at $n=1,2,3,4,\cdots$.
There are lots of orthonormal bases for $L^{2}$. Start with any countable dense subset and perform Gram-Schmidt on the dense subset, discarding dependent elements along the way. Then choose any sequence $\{ a_{n} \}$ of distinct real numbers for which $|a_{n}|\rightarrow \infty$. Define $$ Lf = \sum_{n=1}^{\infty} a_n (f,e_n)e_n. $$ Then $L$ is selfadjoint with spectrum consisting only of simple eigenvalues $a_n$. This works whenever $L^{2}$ is separable. If it is not separable, then $L$ cannot exist because the eigenvectors of $L$ as you have described must be a complete orthogonal basis of $L^{2}$.
If you're working strictly within the context of an inner product space or Hilbert space, the statement you have learned is false. Even if you could identify certain "states" that you would call "eigenstates," those objects are not vectors in the space. They are some unknown, unspecified object in some unknown and extended space.
For example, start with the position operator $M$ on an interval $[0,1]$. This operator is multiplication by $x$, meaning $(Mf)(x)=xf(x)$. This is a perfectly legitimate selfadjoint linear operator on the Hilbert space $L^2[0,1]$, but it has no eigenvectors, meaning that there are no non-trivial functions $f \in L^2[0,1]$ for which $Mf = \lambda f$, regardless of the choice of $\lambda$. The operator $M$ has "continuous spectrum" but no eigenvectors or eigenvalues. There just are no such "eigenstate" objects.
Physicists typically introduce the $\delta$ function into such a space (which is logically inconsistent) and claim that $(M \delta_{y})(x) = x\delta(x-y)=y\delta_{y}$. For many reasons, on many levels, that is Mathematical non-sense, and cannot be salvaged within $L^2[0,1]$ in a logically consistent way. The axioms of Quantum use Hilbert spaces, but $\delta$ functions cannot live in spaces such as $L^2[0,1]$ because two functions that are equal a.e. in $L^2[0,1]$ are identical, which means that pointwise values have no meaning for elements of $L^2[0,1]$. Dirac knew this, but he liked the intuition.
John von Neumann, who was a contemporary of P.A.M. Dirac, created consistent ways of dealing with the selfadjoint operators of Physics through the Spectral Theorem, and this was available to Dirac at the time; but Dirac chose an intuitive presentation over logical consistency. And it wasn't because rigorous Mathematics was unavailable at the time to handle everything; the correct Math was there.
All of the Hilbert spaces of Quantum must be separable, meaning that the space contains a countable dense subset. The reason for this has to do with constructibility, and being able to bootstrap to general answers through finite approximations. $L^2[0,1]$ is a separable space. One of the consequences of having a separable space, is that an orthonormal basis of such a space is always finite or countably infinite. You cannot have mutually orthogonal objects that are indexed by an interval of the real line, for example. That's impossible in separable spaces. And, if you have a selfadjoint operator $M$ on a Hilbert space, and you have $Mf=\lambda f$ and $Mg = \mu g$ for $\lambda\ne \mu$, then $(f,g)=0$ must hold. So, a selfadjoint operator on a Hilbert space that is allowed in Quantum Mechanics cannot have more than a countable number of actual eigenvalues. That's a consequence of correct axiomatic systems for Quantum. You can have continuous spectrum, but that's different than having a continuum of eigenvalues. If you enlarge the space to allow for such things, then you lose the ability to approximate in a finite way, which disconnects the theory from a setting where finite approximation makes sense.
Sorry to disappoint you. The theorem you have--as stated--is not true. Correct spectral theory gives you essentially the same thing, but not exactly that.
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Best Answer
Example: let $H=L^2[0,1]$ and let $T:H \to H$ be defined by $(Tf)(t)=tf(t)$. Then $T$ is self-adjoint but has no eigenvalues.
If $A:H \to H$ is compact and self-adjoint, then there exists an orthonormal basis of $H$ consisting of eigenvectors of $A$. (See: https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space).