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

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?

• What do you mean? There is, in general, not only one orthonormal basis. And every orthonormal basis is an eigenbasis of some operator (the identity operator, for example). Commented Apr 17, 2023 at 19:47
• A Hilbert space is simply made up of functions that behave in many ways similar to vectors. What basis you chose is completely irrelevant for the Hilbert space. It may be relevant to whether you can actually build convergent series from those bases for all elements in the Hilbert space. In many cases we can't. But that is a problem for mathematicians to worry about. It has absolutely nothing to do with physics. Commented Apr 21, 2023 at 16:58

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.

• Actually, you didn't answer the question: is it an assumption or not? Also, H atom associated Laguerre polynomials are exact solutions to the Coulombic atom. They do form a basis, but it is weird to think of exact diagonalisations of the Hamiltonian operator as the starting point to solve for the wavefunctions... Commented Apr 17, 2023 at 17:32
• @naturallyInconsistent yeah I realized that Schroedinger's equation is an eigenvalue equation and therefore any solution is a eigenfunction as well as any linear combination of it. In that case the coefficients are not determined. They only take on a value upon a measurement. Certainly in most cases a ket-vector represents the eigenfunction of an operator. Commented Apr 18, 2023 at 8:55

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.

• Eignevectors of operators satisfying certain conditions span the whole Hilbert state. Commented Apr 17, 2023 at 12:45
• @RogerVadim and I was just saying that if they do not satisfy enough conditions, yet are meant to be observable, then we postulate that they also span the whole Hilbert space. Commented Apr 17, 2023 at 12:47
• eigenvectors of what? It is sloppy use of the term in the question, and I am merely suggesting to correct it in your answer. Commented Apr 17, 2023 at 12:51
• Oh, ok, that, sure! Commented Apr 17, 2023 at 12:52
• That the continuous case doesn't cause any problems is an enormous understatement and quantum field theory is asking you to hold its beer. :-) Commented Apr 17, 2023 at 16:25