# Spectral theorem asked by a physicist

In physics we use the fact that, given a self adjoint operator A, every element in Hilbert space can be written as "generalized linear combination" of the eigenstates of A:

$$\tag{1} \psi(x)=\sum a_k \psi_k(x) +\int c(\lambda)\psi_\lambda(x) d\lambda$$

Where the integral is over the continuous spectrum.

If the dimension of H is finite, the spectral theorem guarantees that this expression is valid for all self adjoint operators with no continuous part.

In the case of infinite dimension spectral theorem says that every A self adjoint is unitary equivalent to a multiplication operator, that is, via basis change, we have

$$UTU^{-1}(\psi)(x)= f(x)\psi(x)$$

In which way is this related to the possibility to write every state in the form $$(1)$$?

• I think you need not only "generalized linear combination" but "generalized eigenstates", because your $\psi_\lambda$ are unlikely to be actual elements of the Hilbert space. Jul 29, 2021 at 20:35
• (Specifically, the $\psi_k$'s in the discrete part belong to the Hilbert space, and those in the continuous part do not.) Jul 30, 2021 at 0:37

Okay. Every self-adjoint operator $$A$$ is unitarily equivalent to a multiplication operator on some measure space $$X$$. Let's assume $$\|A\| \leq 1$$. We can write $$X = X_0 \cup X_1$$ where $$X_0$$ is discrete and $$X_1$$ is atomless. The standard basis vectors in $$L^2(X_0)$$ are eigenvectors for any multiplication operator, so any element of $$L^2(X_0)$$ is an $$l^2$$ linear combination of these eigenvectors.

Now suppose $$X_1$$ is $$[-1,1]$$ equipped with some atomless measure $$\mu$$ and $$A$$ acts on $$L^2(X_1)$$ by multiplication by $$x$$. Then you may want to say that any $$f \in L^2(X_1)$$ has the form $$f = \int_{[-1,1]} f(x)\delta_x\, d\mu$$, where $$\delta_x$$ is the Dirac delta at $$x$$ is thought of as an eigenfunction of $$A$$. Whatever meaning you give to that expression is the answer to your question, because any self-adjoint multiplication operator on a separable Hilbert space is unitarily equivalent to the direct sum of a multiplication operator on a countable discrete space and multiplication by $$x$$ on countably many copies of $$[-1,1]$$, each equipped with an atomless measure. This is some version of the spectral theorem.

• Aside: for spectral theorem in nonspeparable Hilbert space, see the Halmos book Introduction to Hilbert Space and the Theory of Spectral Multiplicity Jul 30, 2021 at 0:43
• What is specifically "X"? Have I to imagine it as a set of infinitesimal (or finite) subsets of R^3 who i can associate a "volume"? What "discrete and continuous parts of X" are exactly? I'm having some troubles to understand your notations and context...where the integral appears in the relation $$A\psi(x) = f(x)\psi(x)$$, which is the definition of "multiplication operator"? Jul 30, 2021 at 12:52
• Oh, I'm afraid you'll need to learn some measure theory before you can really understand this stuff. Jul 30, 2021 at 13:13
• @GEdgar: for applications to physics, learning about nonseparable Hilbert spaces is not the best use of one's time. Jul 30, 2021 at 13:57
• Could you explain me where to start with? And, in a qualitative way, how the possibility to expand an operator as linear cmb. Of generalized eigenstates is related to the fact that every self adjoimt operator is a multiplication operator, apart from a unitary trasformation? Jul 30, 2021 at 20:15

Though this is an attractive formalism--and it is attractive--it can't really be made to work in any simple way. There are no generalized eigenstates for a general self-adjoint linear operator on a Hilbert space, because these generalized eigen-states would not lie in the Hilbert space; so it would make no sense to be able to form linear combinations through sums, let alone integral combinations with respect to a parameter. Almost all of the formalism arose out of Sturm-Liouville problems, and there you can do something reasonable. But not for the general self-adjoint operator on a Hilbert space. The ideas are so attractive that they caught on, and have been taught ever since Dirac because of the intuitive understanding that someone can gain from such idealizations.

• Ummh..so quantum mechanics fundamentals are based on mothing? After defining a physics operators, i can't know a priori that such an expansion is guaranteed? Aug 5, 2021 at 7:08
• @Rosario : The classical examples where separation of variables applies will allow Dirac type expansions. Dirac built it for that. Aug 6, 2021 at 1:59