# Dimensionality of the quantum space of states

I don't understand what is the dimension of the space of the states because it looks different dependently on the base that I choose, for example:

If I use the position representation (the base are the states with a determinate position) the generic state is represented by its wave function which value can be seen as the components of the vector:

$$|s⟩=\int\psi(x)|x⟩dx$$

$$|s⟩$$ is identified by $$\psi(x)$$

If I use as base the eigenstates of the hamiltonian of spherically symmetric potential the generic states is represented by a discrete set of numbers.

$$\psi=\sum_{nml} a_{nml} Y^{nml}$$

$$\psi$$ is identified by the numbers $$a_{nml}$$

In the first case there is a non discrete set of elements to constitute the generic state in the second it's a discrete set of elements. Am i misunderstanding something?

• The position eigenstates are not normalizable and therefore not actually part of the Hilbert space, so they don't actually form a basis. – Javier Jun 3 '19 at 16:00
• – SRS Jun 3 '19 at 16:59

To answer your question, you might like to translate it first to a similar question in a more familiar area, namely the classical physics of waves and the wave equation. Consider the wave equation on an infinite line. You can use Fourier analysis, and you will need a continuous set of states (Fourier components) to cover every possibility. However, as soon as the problem involves only a finite length of the line, then you can use a discrete set of states (a discrete Fourier series). Similar statements apply in 3 dimensions, and similar statements apply in quantum mechanics.

The move between discrete and continuous can require some thought, and I understand that Dirac chose to develop both possibilities alongside one another in his textbook. Other authors make other choices in the order of ideas while developing the subject. I would say you would be well advised to stick to discrete sets of wavefunctions in the first instance, and then go to the continuous limit when you need to. This is how more complex calculations are often set up in practice, for example by the use of periodic boundary conditions and things like that.

In short, the Hilbert space of infinite dimensions can be treated as a limiting case of a Hilbert space of a finite number of dimensions as far as I am aware, though I admit there may be technical mathematical reasons why this statement does not capture all possibilities.

The Hilbert space associated to a quantum theory (on which the algebra of observables is (irreducibly) represented) is, in almost all concrete cases, separable. In other words, it has a countable dense subset and a basis (complete set of orthonormal vectors) of cardinality $$\aleph_0$$ (the cardinality of natural numbers).

The space of normal quantum states with respect to a given representation (i.e. the set of density matrices on the aforementioned Hilbert space) is also separable, provided the Hilbert space is such. This is a cone in a Banach space, and therefore the definition using a basis is not really meaningful, still there is a countable dense subset.

The algebra of quantum observables is, however, always non-separable, unless it is a trivial algebra. As a consequence, there are examples in which the space of all quantum states is non-separable (especially in quantum field theory).

• This answer uses a lot of fancy words and is technically correct, but it doesn't address OP's confusion. – Javier Jun 3 '19 at 15:59
• I must agree with @Javier – gented Jun 3 '19 at 16:48