Continuum States in QM

Question

In the Hilbert space of QM, in the finite dimensional case, for a complete orthonormal set of basis vectors, one writes the generic state vector as: $\psi=\sum_j(\phi_j,\psi)\phi_j$. When the complete orthogonal states form a continuum, most of the authors of books that I am reading (including Dirac, Weinberg and many others) replace the previous expression with an integral of the sort: $\psi=\int(\phi_\xi,\psi)\phi_\xi d\xi$ , which is justified with some kind of generic argument about taking a large number of discrete state and then pushing the number to infinity. Despite having some basic knowledge of the Riemann and Lebesgue integration and even of integration with respect to a projection-valued measures, I still cannot make sense of that integral expression. To me it looks merely formal. The same as saying that: $f(\xi)=\int \delta(\xi - \xi')f(\xi') d\xi'$ , which is just kind of hand waving, knowing what $\delta$ actually is, i.e. a distribution. Is there any way to assign such an integral expression a meaningful interpretation?

Answer 1

In infinite-dimensional Hilbert spaces, one may not be lucky enough to have valid eigenvectors of operators. As an example, an eigenvector of the position operator on $L^{2}([0,1])$ is $|{x}\rangle$ which is not normalizable, and so cannot belong to the Hilbert space.

However, we can look for the next best thing, which is an approximate eigenvector. In the context of our example, we may look for a subspace of vectors supported on a tiny interval $[x,x+ \epsilon] \subset [0,1]$ such that $X|{x}\rangle \approx x |{x}\rangle$. For bounded observables, we may collect such subspaces into a projection-valued measure $\{E_{x}\}$.

By the spectral theorem for bounded self-adjoint operators, we may write the observable in terms of an integral over this projection-valued measure $X= \int_{\sigma(X)} x dE_{x}$, where $\sigma(X)$ is the spectrum of $X$. Comparison between this formula and standard quantum mechanics textbooks identifies $dE_{x}$ as $|x\rangle \langle x|dx$.

Thus, the "formal" looking integrations you quote result from inserting a resolution of the identity $I = \int_{\sigma(X)} dE_{x} = \int_{\sigma(X)} |x\rangle \langle x|dx$ so that $|\psi\rangle = \int dE_{x} |\psi\rangle = \int \langle x | \psi \rangle |x\rangle dx$. In terms of interpretation, each $dE_{x}$ projects the state $|\psi\rangle$ onto a subspace of approximate eigenvectors for $X$, and so integrating over all such subspaces recovers the total state.