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After going through the questions and answers, I still have a question lingering in my mind.

So, an observable is defined as a Hermitian operator whose eigenvectors make up a basis for the state space.

But why is it so important that this basis covers the entire state space? Could it be because if it doesn't, we can't express $|\psi\rangle$ as a combination of all the basis elements? And then, we'd struggle to understand the quantum state of the system, right? I guess this also means we wouldn't be able to describe all the possible outcomes of observables and their probabilities accurately.

By the way, thanks and have a great day.

Vector spaces without a basis and observables in quantum theory

Not all self-adjoint operators are observables? Why do an observable's eigenstates always form a basis?

The state space is somehow defined by the observables?

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    $\begingroup$ What conditions should a well-defined probability distribution satisfy? Think about this and then compare to the axioms of QM. And BTW: an observable is represented by a hermitian operator; this automatically guarantees that the operator admits a complete orthonormal basis as eigenvectors. $\endgroup$
    – Tobias Fünke
    Commented Jan 27 at 12:27
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    $\begingroup$ It's an ensemble theory. For every system that goes in there has to be an outcome. That outcome has to be a real number because our measurements are all classical and therefor produce real numbers. These numbers are the eigenvalues of our operators. $\endgroup$
    – FlatterMann
    Commented Jan 27 at 12:31
  • $\begingroup$ @TobiasFünke Oh I think that I see what you mean. A distribution probability to be well defined must satisfy that the sum of the probabilities of all possible outcomes is equal to 1. So if you can not do this with the possible outcomes of your observable you can´t have a well-defined probability distribution, right? I was also thinking that the definition of a basis is that you need its elements to be orthonormal, that this enables you to write $|\psi\rangle$ in an unic form as a linear combination of all its elements and the clausure relation. $\endgroup$
    – user353399
    Commented Jan 27 at 13:06

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So, an observable is defined as a Hermitian operator whose eigenvectors make up a basis for the state space.

No, simply any Hermitian (or rather self-adjoint) operator defines, in principle an observable.

Usually, physical observables have an additional property, that of being local in space. However sometimes this requirement (arising from the laws of physics) doesn't apply. For example in quantum computing sometimes one can try to build observables with long range interactions.

So the most general definition of observable, without any restriction, is simply that of a self-adjoint operator.

Consider a single particle in 3D space (such as an electron-proton system in it's center of mass, i.e. an hydrogen atom). The momentum $P$ has purely continuous spectrum. Therefore no eigenvalue and no eigenvector.

Something similar as what you say anyway, still holds. Indeed one can decompose any wavefunction via it's Fourier decomposition:

$$ \psi(x) = \int_{\mathbb{R^3}} dp \, \frac{e^{ipx}}{(2\pi)^3}\hat{\psi}(p) $$

In the above equation $p$ is the wannabe eigenvalues of $P$ and $\frac{e^{ipx}}{(2\pi)^3}$ is the wannabe eigenvector. You will notice though, that this function is not normalizable. So it doesn't belong to the Hilbert space. But one can still build legitimate wavefunctions with them (via the equation above). In this specific case one speaks, for example, of building wave packets.

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  • $\begingroup$ Hmm, I think this is a bit tricky. I understand that the P representation has elements that are not square integrable; therefore, they are not elements of the Hilbert space. However, with plane waves, you can still construct a basis that enables you to represent $|\psi\rangle$ in a unique form. This form also satisfies closure relations and orthonormality. $\endgroup$
    – user353399
    Commented Jan 27 at 20:09
  • $\begingroup$ Simply stating that an operator is Hermitian isn't sufficient for me. I believe it's important to explicitly mention that in the continuous case, one needs to be able to construct a basis. Whether the elements are or are not part of the Hilbert space is a separate consideration. $\endgroup$
    – user353399
    Commented Jan 27 at 20:09
  • $\begingroup$ In the finite case as Cohen says When E is finite-dimensional, we have seen (§ D-1-b) that it is always possible to form a basis with the eigenvectors of a Hermitian operator. $\endgroup$
    – user353399
    Commented Jan 27 at 20:10
  • $\begingroup$ No. Your vague idea of constructing a basis with plane waves is actually the statement that any function in $L^2$ is the Fourier transform of some function (i.e. that the Fourier transform is unitary on $L^2$). The actual physical requirement is something else. Something that is mathematically implied by Hermitian operators. I will update my answer $\endgroup$
    – lcv
    Commented Jan 27 at 21:37
  • $\begingroup$ "Simply stating that an operator is Hermitian isn't sufficient for me." . Are you trying to say that momentum is not an observable? Or that position is not an observable? The mathematical theory of quantum mechanics is perfectly coherent in this respect. The only thing needed mathematically for an operator to be an observable is that of being self adjoint. Once again, I will update my answer. $\endgroup$
    – lcv
    Commented Jan 27 at 21:42

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