When undergraduate physics students are trying to learn quantum mechanics, one of the most difficult obstacles is that abstract infinite dimensional vector spaces are needed, which makes it impossible for a person to mentally “see” the geometry in their minds in the way a three-dimensional space can be imagined. The fact that the coordinates of the points in that vector space are complex numbers certainly doesn’t make it any easier. When I was studying that myself in the early undergraduate studies, it took me over a year to get a hold of the concept that any self-adjoint operator acting on vectors in the quantum state space has a set of eigenvectors that form a basis of that space.
Sometimes, however, a quantum mechanics problem can be handled by approximating the state space with a finite-dimensional vector space. In fact, several physics Nobel prizes have been awarded for the study of two-state quantum systems. One example of a situation, where a quantum system is effectively a two-state system, is something where the ground state (n=0) and the first excited state (n=1) have energies that are close to each other, and the energy gap between the first and second (n = 2) excited states is unusually wide. Then, if the system is at low temperature, the Boltzmann factors of the lowest-lying two states,
are much larger numbers that the Boltzmann factors of any other states, and it is unlikely that a measurement of total energy will give a result other than one of the two lowest energy eigenvalues.
How, then, does one know that a quantum state space actually has to be infinite-dimensional? I once saw, in a mathematical physics textbook, a practice problem where this had to be proved, and I was proud for almost immediately finding the correct answer.
A cornerstone of nonrelativistic quantum mechanics is the position-momentum commutation relation
Now, if the quantum state space were finite-dimensional with dimension , then these operators and would be square matrices, and the constant on the RHS would be a diagonal matrix where all the diagonal elements have value
Actually, this equation can’t hold in the finite-dimensional case. To see this, consider the calculation rules for taking the trace of a matrix sum or a matrix product:
Now, if you take the trace of both sides of the commutation relation equation, and use these rules, you will get the result
Which clearly isn’t possible. So, we have done a false assumption somewhere in the derivation of the contradictory equation above, and the false assumption must be the finite-dimensionality of the space where and are defined. An infinite dimensional operator doesn’t always have a trace at all, and this is the case with the operator .
An interested reader may try to form finite-dimensional matrix representations for the raising and lowering operators of a quantum harmonic oscillator, use those those to form matrices for operators and , and see what happens when you calculate their commutator.