# Spectral theorem and diagonal expansion for self adjoint operators

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 $$A$$:

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

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 A self adjoint 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, ossia, 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?

• what does "ossia" mean?
– YCor
Jul 29 at 21:44
• Cross post over at math.SE. Jul 29 at 21:56
• @YCor in music ossia means an alternative, usually simpler way to play a passage. I think the poster means something similar here, an alternative and simpler expression. It comes from the Italian o sia meaning "or be it." Jul 29 at 22:21

This is easier to explain on the finite dimensional example (though the idea is the same). The spectral theorem says that for every self-adjoint (same as Hermitian, in finite dimensional case) operator there is an orthogonal basis of the whole space consisting of eigenvectors. In terms of matrices, this means that every Hermitian matrix $$A$$ can be diagonalized by a unitary matrix $$U$$: $$U^{-1}AU=D,$$ where $$D$$ is a diagonal matrix (consisting of eigenvalues of $$A$$ standing on the main diagonal of $$D$$). The columns of unitary matrix $$U$$ are eigenvectors of $$A$$. Since eigenvectors make a basis, every vector can be expanded in terms of them, as your first formula says.
This formula can be thought as a change of coordinates: if $$x=Uv,$$ then the coordinates $$v_k$$ are coordinates of $$x$$ in the new basis consisting of the columns of $$U$$.
On the other hand, $$D$$ is really a multiplication operator: it acts on a vector by the rule $$Dv=(\lambda_1v_1,\ldots,\lambda_nv_n)^T$$. To see that this is the same as multiplication on an independent variable, arrange $$\lambda_k$$ in the increasing order, and think of vectors $$v$$ as functions of $$\lambda$$ which take the value $$v_k$$ at the point $$\lambda_k$$ (zero at all other points). Then $$D$$ is the "multiplication on the independent variable", namely it multiplies such functions on $$\lambda$$.