Let's go through step by step. The rough picture is correct, but there are important subtleties missing.
fog37 said:
Thanks dextercioby.
Let me summarize what I think I know so you don't have to repeat yourself:
- Every observable (position, momentum, angular momentum, energy, spin, etc. did I forget any?) is represented by a Hermitian linear operator in quantum mechanics. The operator's spectrum can be continuous, discrete or hybrid depending on the type of spatial domain over which the system exists and the forces involved.
Hermitecity is not sufficient, it must be an essentially self-adjoint operator, i.e., the operator must be defined on a dense subspace of Hilbert space. In general this subspace is a proper subspace of the Hilbert space, and it's co-domain must be also in this subspace, i.e., if ##|\psi \rangle## is in the domain of the operator ##\hat{A}##, representing an observable, also ##\hat{A} |\psi \rangle## should be in this subspace.
A generic abstract state of a system is given by ##| \Psi>## is a state vector that can be expressed as a weighted superposition of eigenstates of a certain operator. That means that we can represent ##| \Psi>## in many different ways according to the operator and its associated basis. But the basis of eigenvectors of each different operator is associated to different linear vector spaces... So in which vector Hilbert space does ##| \Psi>## live in? Is there a different ##| \Psi>## for each different vector space associated to each different operator? For instance, the spin operator has a basis of only two eigenvectors (2-dimensional) while the energy basis generally contains an infinity (discrete or continuous) of eigenvectors. It does not seem possible to represent the same state ##| \Psi>## that is superposition of many energy eigenstates using only two spin eigenstates.
A pure state is not represented by ##|\Psi \rangle## itself but the entire ray defined by this Hilbert-space vector. That's important since it enables us to describe, e.g., particles with half-integer spin, and the surrounding matter around us is made up of such particles (quarks and leptons).
The most general description, valid for both pure and mixed states, is the statistical operator. A state is pure if and only if the representing statistical operator is a projection operator, ##\hat{\rho}=|\Psi \rangle \langle \Psi |## with a normalized Hilbert-space vector ##|\Psi \rangle##.
b) What about a "total" vector state ##| \Psi>## that contains information about all the observables?
Is don't know what you mean by that. The state vector contains of course probabilistic information about all possible observables through Born's rule, i.e., if ##\hat{A}## represents the observable ##A## and if ##|a,\beta \rangle## denotes all (generalized) eigenvectors of ##\hat{A}## with eigenvalue ##a## (which is a possible value when this observable is measured), then the probability (distribution) to measure this value ##a## if the system is prepared in a state represented by ##|\Psi \rangle##
$$P(a)=\sum_{\beta} |\langle a,\beta|\Psi \rangle|^2.$$
Of course ##\beta## can also be in a continuous set, and then the sum over ##\beta## becomes and integral (or you can have the case that ##\beta## takes both discrete and continuous values, in which case you have both a sum and an integral).
c) In the particle in the box example, spin is not considered. Only the energy and position observables are discussed. Why? The particle trapped in the box could be an electron which has a nonzero spin.
There's nothing that hinders you to consider also spin. In the non-relativistic case, it's just another label on the eigenstates, i.e., for an electron ##\sigma_z=\pm 1/2## in addition to energy (there's no momentum observable in this case, BTW!).
d) Based on what said above, how does the tensor product between two (or more?) linear vector spaces fit it? Is it used when the total wavefunction is expressed in terms of eigenvectors that include multiple observables? In this thread, it was said that spin is an additional degree of freedom. Aren't position, momentum, energy, etc. also degrees of freedom?
Let me stop here. Thank you for any help.
Yes, if you have a composite system the Hilbert space of this system is the tensor product of the Hilbert spaces of the composites. Formally, if you have a single particle with spin ##s##, a generalized basis is the momentum-spin basis, ##|\vec{p},\sigma_z \rangle##, where ##\vec{p} \in \mathbb{R}^3## and ##\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}##. Formally you can build up this basis as a tensor product
$$|\vec{p},\sigma_z \rangle=|p_x \rangle \otimes |p_y \rangle \otimes |p_x \rangle \otimes \sigma_z \rangle.$$