0

I'm reading Quantum Mechanics by Cohen-tannoudji and I find some parts not clear.

  1. a hermitian operator is observable if there exists an orthonormal complete basis or if every orthonormal system of the vector is a complete basis.
  2. it's linked to point 1. The book tries to demonstrate that when two observable A and B commute there exists a complete basis formed by a common eigenvector. Thus, it takes a basis {$u_i$} for A, there are eigenspaces $U_n$, and it's simple to demonstrate that $U_n$ are subspaces invariant for B. But then it concludes the proof saying that B restricted to $U_n$ is still hermitian okay and then there exists a basis of $U_n$ in which B is diagonalizable. Why? I think there are some hypotheses omitted: if every $U_n$ is finite dimensional okay for spectral theorem, but for infinite dimensional eigenspaces? it seems like B observable is never used or it has omitted the proof that if exists a basis of eigenvector then it exists also for restrictions
Kasiéobì ùdumágà
  • 177
  • 5
Alessandro Vagni
  • 43

1 Answers1

0

The 1. does not make any sense.

The second part goes like you say, when the whole Hilbert space is finite-dimensional. For infinite-dimensional Hilbert spaces, the problem is much more difficult (even to state: if the operators are unbounded, then the usual commutation requirement can sometimes not be meaningful, and when it is, might not be enough; moreover, the spectrum can be quite complicated and the notion of eigenbasis might ot be general enough) and functional analysis is required. You will find some (if not all) answers in the book of Valter Moretti, Spectral Theory and Quantum Mechanics.

Plop
  • 2,647