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28
votes
3answers
860 views

Not all self-adjoint operators are observables?

The WP article on the density matrix has this remark: It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable.[...
18
votes
2answers
1k views

What's the deal with momentum in the infinite square well?

Every now and then a question comes up about the status of the momentum operator in the infinite square well, and while we have two good answers on the topic here and here, I'm generally not satisfied ...
6
votes
2answers
308 views

Symmetric, (essentially) self-adjoint operators and the spectral theorem

At the moment I got a bit confused about the notation in some QM textbooks. Some say the operators should be symmetric, some say they should be essentially self-adjoint, others that they has to be ...
5
votes
2answers
363 views

Unbounded operators defined only on dense subdomain of Hilbert space in QM?

I am relatively new to quantum mechanics. In a set of notes I am using, the following is a description of an aspect of some operators corresponding to observables. The notes state the following: "...
19
votes
7answers
2k views

Why is a Hermitian operator a “quantum random variable”?

To me, as a stupid mathematician, a random variable is a measurable function from some probability space $(\Omega, \sigma, \mu)$ to $(\Bbb{R}, B(\Bbb{R}))$. This makes sense. You have outcomes, events,...
4
votes
1answer
886 views

Conservation of momentum in infinite square well

This is inspired by Griffiths QM section 2.2, on the infinite square well, which is about how far I've gotten (so, sorry if this is addressed later in the book). For any given starting wavefunction, ...
3
votes
2answers
570 views

What exactly implies the need of quantum mechanics for self-adjoint and not only symmetric operators? [duplicate]

We know that quantum mechanics requires self-adjoint operators, not only symmetric. Can we say that this follows ONLY from the two following axioms of quantum mechanics, namely that each observable $...
2
votes
2answers
493 views

Why is quantum mechancis is not content with symmetric operators, but wants self-adjoint operators?

A symmetric operator has only real eigenvalues and different eigenvectors corresponding to different eigenvalues are orthogonal. These are exactly what we want for a physical observable. I think ...
2
votes
1answer
178 views

Making an Incomplete Set of Observables Complete

In quantum mechanics, it seems a standard procedure that if you have an incomplete set of observables, then one can make this set complete by adding more commuting observables until the set becomes ...
0
votes
2answers
228 views

Eigenstates of an observable

Can we use eigenstates of ANY observable as base of the Hilbert space? If we can, is this equal to the statement that those eigenstates are orthogonal to each other and normalizable?
10
votes
1answer
2k views

“An operator is hermitian”. Implications?

Alastair Rae states that there are 4 postulates of Quantum Mechanics in his text on the subject matter. The first part of his second postulate can be stated as: Every dynamical variable may be ...
22
votes
2answers
5k views

Intuitive meaning of Hilbert Space formalism

I am totally confused about the Hilbert Space formalism of Quantum Mechanics. Can somebody please elaborate on the following points: The observables are given by self-adjoint operators on the ...