quantum mechanics - Does the Hilbert space include states that are not solutions of the Hamiltonian?

I've studied Quantum Mechanics and I know the usual answer "The dimension of the Hilbert Space is the maximum number of linear independent states the system can be found in". There is something about this statement that bothers me, let me try to explain it.

Imagine a particle whose dynamics satisfy Schrodinger equation. Before we give it a hamiltonian, in principle the particle can have any Square-integrable continuous function as a state. When we write a particular Hamiltonian we find the actual eigenstates of the particle and then every possible state is a linear combination of that eigenstates. Now, acording to the first definition of the hilbert space, it has all the eigenstates of the hamiltonian. Now, several questions arise:

1) Does the Hamiltonian determine the Hilbert Space?

2) What if I make two particles with diferent Hamiltonians interact? Do they live in diferent Hilbert spaces?

3) What about perturbation theory? Do I change the Hilbert space each time I add a new term in the Hamiltonian?

Now I tend to think that the Hilbert Space contains every possible state of the particle whether it is a solution of the Schrodinger equation or not. Please help me sort this problem out.

Answer

There is a certain subtlety to your question.

For quantum systems with a finite number of degrees of freedom, as commonly dealt with in intro QM, things are relatively simple:

1. 
   

   Yes and no: the Hamiltonian certainly determines a basis for the Hilbert space of states, but the working Hilbert space depends on the domain of definition of the problem and on associated boundary conditions. See particle in a 3D-box vs. free particle on the entire 3D-space, as well as particle in a box with Dirichlet b.c.-s vs. particle in a box with periodic b.c.-s, etc. Alternatively, the Hilbert space is determined by the algebra of system observables, as pointed out in user1620696's answer, but the two descriptions are eventually equivalent. Moreover, there exists an even deeper equivalence of Hilbert spaces, see point(3) below.

   
   


2. 
   

   Each particle lives in its own Hilbert space, but the combined interacting system lives in the direct product of individual Hilbert spaces. Again, see relation to algebra of observables as in user1620696's answer.

   
   


3. 
   

   Leaving aside spin and the spin interactions mentioned by Hosein, generally no, for a finite number of degrees of freedom the Hilbert space does not change under perturbations. According to the Stone-von Neumann theorem, in this case all possible Hilbert spaces are isomorphic to one another (or equivalently, there is a unique irreducible representation of the canonical commutation relations), hence distinguishing one over the other makes no formal difference. At most, the total Hilbert space decomposes into a direct sum of multiple isomorphic copies.

   
   
   

For systems with an infinite number of degrees of freedom, which are the domain of Quantum Field Theory, points (1) and (2) above remain largely valid, but the situation does change drastically in regards to point (3).

The Stone-von Neumann theorem does not hold for quantum fields, and one may find that certain unitary transformations defined on one Hilbert space, constructed around a given Hamiltonian, produce states that are orthogonal to that whole Hilbert space and living in an entirely new, inequivalent state space. This is the case of inequivalent vacua of many QFT Hamiltonians, from condensed matter ones (see boson condensation, superconductivity, etc.) to QCD.

Further, the nature of such inequivalent vacua (or better say, unitarily inequivalent representations of the dynamics) is determined by the nature of interactions between free-fields described by some free-particle Hamiltonian and corresponding state space.

For an idea of what is going on, see for instance Sec. 1.2 of this review on Canonical Transformations in Quantum Field Theory.