Is it necessary that the Hilbert space basis is made up of the eigenvectors/eigenfunctions of the operator under consideration?

Question

When using Dirac bra-ket notation to make some statement regarding an operator acting on the vectors of a Hilbert space, is it necessary that the basis of the Hilbert space is made up of the eigenvectors/eigenfunctions of the operator? Or doesn't it matter?

Answer 1

It is usually assumed that the solution of the Schroedinger equation which governs large part of Quantum mechanics is $\in L_2(\Omega)$ (typically 1-particle QM, in the many particle case is the space is more complicated) where $\Omega$ is the domain where the solution $u$ is sought.

$L_2(\Omega)$ has together with a scalar product

$$ \langle u | v \rangle = \int_\Omega dx u(x)\bar{v(x)}$$

Hilbert space structure, in particular as $L_2(\Omega)$ is complete using the $L_2$-norm:

$$ || u||_{L_2} = \int_\Omega dx u(x)\bar{u(x)} = \int_\Omega |u|^2 dx$$

A Hilbert space has many properties which have also finite-dimensional vector spaces. The completeness of Hilbert space in $L_2(\Omega)$-norm guarantees that every element of the Hilbert space can be expanded in (typically an infinite number of) base vectors $e_i(x)$ (But finite-dim Hilbert spaces do exist too).

$$u(x) = \sum_{i=1}^\infty a_i e_i(x)$$

So if $u(x)$ is known we can know the coefficients $a_i$:

$$ a_i = \langle u | e_i\rangle = \int_\Omega dx u(x) \bar{e_i(x)}$$

A good example to demonstrate this are Fourier-series. Any periodic function defined on $\Omega =[0,2\pi]$ can be developed in series of trigonometric functions which serve in that case as base vectors of this particular Hilbert space. Actually, the base vectors can be chosen arbitrarily as long as they form a complete basis, i.e. $$||u(x)-\lim_{n\rightarrow\infty}\sum_{i=0}^n\langle u| e_i\rangle e_i(x)||_{L_2(\Omega)} \rightarrow 0$$

Fourier series are very good example since apart from basis formed by trigonometric functions we can easily change the basis and use exponential function as base vectors instead.

In application of physics, for instance Quantum mechanics, can be very useful to choose indeed a basis that is eigenvector of a (series of) operator(s) which has to be complete in order to form a complete basis as it is required above. In many cases this is the appropiate way to work in a Hilbert space. But as in a vector space of 3D the basis of a space can be chosen freely.

Bonus: Usually we do not know $u(x)$ as it is the solution of a complicated partial differential equation -- Schroedinger's equation. One technique to find the solution is to choose a basis: for instance Laguerre-polynomials in the H-atom case and then get an equation for the coefficients $a_i$ (a priori unknown as we don't know $u$ yet) which can then be solved and thereby the solution of $u$ can be found.

Answer 2

It can be proved for any finite Hilbert space that eigenvectors of observable's operators span the Hilbert space. That it is true for infinitely many, especially the continuous case (the eigenvectors of which live in Rigged Hilbert space, not normalisable, not suitable for quantum states since they must be normalisable and live in Hilbert space, but still can be used to expand any quantum state).

Because we cannot prove it in the bigger case, we postulate it. It has never been the source of any problems, though.