Basis in quantum mechanics

Question

My quantum mechanics textbook (Primer of Quantum Mechanics, by Marvin Chester) says that both the momentum space and the position space are basis spaces. It also says that the momentum space is quantized while the position space is a continuum. I have two questions:

1. Is the fact that the momentum space and the position space are basis spaces an experimental result or something that can be derived from more basic laws?

2. I learn from my linear algebra course that any basis of a vector space must have the same number of elements. But the momentum space is countable and the position space is uncountable (for a particle confined in a ring shaped track). How can they both be bases of the same vector space?

Answer 1

First, the term "basis spaces" isn't standard in quantum physics but let us assume that we understand what the sentences approximately mean.

Second, the momentum is continuous (not quantized) if the position space is noncompact (infinite). The momentum only becomes quantized if the position space is compact (or periodic), and indeed, it's been experimentally verified that the momentum is quantized e.g. in the potential well.

Third, the same Hilbert space (more precisely, rigged Hilbert space etc.) may have both countable and uncountable (labeled by continuous numbers) bases. There is no contradiction because at this level of accuracy, the cardinality of the basis doesn't have an impact on the "size" of the Hilbert space as long as it is infinite-dimensional. See e.g.

http://motls.blogspot.com/2014/02/cardinality-of-bases-doesnt-matter-for.html?m=1

So for example, a particle on the line is described by the Hilbert space of complex-valued functions $\psi(x)$ which may be constructed out of the continuous "bases" of position eigenstates; or as superpositions of the harmonic oscillator energy eigenstates (this basis is countable). From the viewpoint of a mathematician who likes to think about cardinality of sets, the bases may be "differently large". But from the viewpoint of physics, they are equally large.

It is completely common and normal in quantum mechanics that some operators have continuous spectra, other operators have discrete spectra, and other operators have mixed (discrete plus continuous) spectra. They may still have bases of eigenstates – which are exactly sufficient to write every vector as a linear superposition. For the discrete bases, the linear superposition is written as a sum; for the continuous bases, the linear combination is written as an integral (and has to be supported by some extended axiomatic system of "rigged Hilbert spaces" etc. to remain rigorous); for mixed bases, the superposition is the sum of the sum and an integral.

Operators with continuous, discrete, and mixed spectra must be considered "equally good operators" from the physics viewpoint. Indeed, the Hamiltonian – the operator of energy that also governs the time evolution – may have discrete, continuous, or mixed spectra and the answer often requires complicated dynamical calculations: it is in no way determined "a priori" whether the Hamiltonian should have a discrete part of the spectrum. Its spectrum has a discrete part (the spectrum is discrete or mixed) if the Hamiltonian admits "bound states" and it usually can't be guessed "immediately" whether such bound states exist.

Answer 2

Answer to question one : The Principle of Quantum Mechanics by R. Shankar page 149 reads "Barring a few exceptions, the schrodinger equation is always solved in a particular basis. Although all basis are equal mathematically, some are more equal that others. First of all, since H = H(X,P) the X and P basis recommend themselves....The choice between the two depends on the Hamiltonian." Basically choice of the basis depends on form on the Hamiltonian. In the basis in which the Hamiltonian has simpler form is chosen for ease of calculations. Some problem are equally easy to solve in multiple basis e.g. the Harmonic Oscillator problem is equally easy to solve in the X and P basis.