The eigenvectors associated to the continuous spectrum in Dirac formalism

Question

I am comfused about the definition of an observable, eigenvectors and the spectrum in the physics litterature. All what I did understand from Dirac's monograph is that the state space is a complex vector $\Psi$ space equipped with an inner product $\langle.,.\rangle$ and an observable is a hermitian operators $A: \Psi\rightarrow \Psi$ that have a complet set of eigenvectors. The set of all eigenvalues is called the spectrum and denoted by $\sigma(A)$. My question is the following : Let $A$ be an abservable such that $\sigma(A)=\mathbb{R}$ and let $\lambda\in \sigma(A)$. Do we have, according to Dirac, the existence of a vector $|\lambda\rangle\in \Psi$ such that $A|\lambda\rangle=\lambda |\lambda\rangle$ ?

Answer 1

Dirac is dead, we cannot ask him what he really meant by this. However, the pretension that these eigenstates exist in some sense is prevalent in many texts on quantum mechanics. There are several possibilities to deal with the claim of the existence of eigenstates for values in the spectrum outside the point spectrum.

1. Just ignore it. This is the approach taken by most introductory texts to quantum mechanics, and most non-mathematical physicists. The results come out right if you don't think too hard about the nature of these "states" and stay away from trying to produce paradoxes with them.

2. Interpret as shorthand. As in other cases (see this answer of mine about the $\delta$-"function"), the notation $A\lvert \lambda \rangle = \lambda \lvert \lambda\rangle$ may be interpreted as approximative shorthand - what is really meant is that there's a sequence of vectors $\lvert \lambda_i\rangle$ such that $\lim_{i\to\infty} \lvert\lvert A\lvert\lambda_i\rangle - \lambda\lvert \lambda_i\rangle\rvert\rvert = 0$, i.e. that we can get arbitrarily close to begin an eigenvector with that eigenvalue (but the limit $\lim_{i\to\infty} \lvert \lambda_i\rangle$ does not exist inside the Hilbert space in this case). If you replace every mention of the eigenvector with a statement about this limiting case, you get a statement that rigorous mathematics would prove.

3. Interpret via rigged Hilbert spaces. As mike stone's answer mentions, there's another formulation where we extend the Hilbert space to a Gel'fand triple with a larger space where the continuous "eigenstates" live. In essence, this is little more than adding the limits $\lim_{i\to\infty}\lvert \lambda_i\rangle$ to the original space: For instance, for the position operator in position representation, the $\lvert \psi_i\rangle$ are square-integrable nascent $\delta$-functions, and their limit in the rigged Hilbert space is the $\delta$ distribution itself. This is a convenient way to retroactively justify the physicists' claim about the existence of these states as actual vectors, but the rigged formalism is little more - in practice, most rigorous functional analysis texts use the limiting approach in my second point if they have to, and largely avoid talking about these approximate eigenstates as much as they can.

Answer 2

As long as you allow elements of the associated rigged Hilbert space the answer is "yes."