Separable differential equations?
Separable differential equations?
A lot of times, for example when solving the 2D or 3D schroedinger equation, we assume that our solutions will be separable. My question is, when, and why can we assume that a solution is separable? is it just a guessing thing? and then we accept it since it satisfies our original DE when we plug our solution in? Or is there some other reason for the assumption?

 Posts: 42
 Joined: Sat Nov 12, 2011 5:01 pm
Re: Separable differential equations?
In principle, separating the PDE is an act of total desperation. We're only going to get solutions whose functions are simple products of singlevalue functions, so we're throwing away an enormous number of possible solutions.
However, when we do this, we usually get functions that look like sin/cos of every possible frequency. Then something amazing happens  it turns out that there is a theorem in Fourier analysis claiming that EVERY function on a bounded interval that isn't too pathological (a fourier series converges in a certain sense) can be represented by these solutions. So we can use our limited infinity of functions to model any arbitrary function, so although we're limited to separated solutions, they have a form so that we can fit them to any function (sin/cos form a basis for functons). Some PDE's have some other set of solutions, e.g. Hermite/Laguerre/Legendre polynomials and the such  these often also span the space of all functions. If the PDE + BC's have a uniqueness theorem, we even know that our form of the solution is equal to any other form you can find.
I believe that in quantum mechanics, you need to take it as an axiom of the theory that the eigenfunctions of the Hamiltonian always form a basis for all possible states. I know that this is true for a finite dimensional Hermitian matrix, but I think there is no proof for an arbitrary operator on a Hilbert space, so we need to assume completedness (every Hamiltonian I've seen does satisfy this property).
However, when we do this, we usually get functions that look like sin/cos of every possible frequency. Then something amazing happens  it turns out that there is a theorem in Fourier analysis claiming that EVERY function on a bounded interval that isn't too pathological (a fourier series converges in a certain sense) can be represented by these solutions. So we can use our limited infinity of functions to model any arbitrary function, so although we're limited to separated solutions, they have a form so that we can fit them to any function (sin/cos form a basis for functons). Some PDE's have some other set of solutions, e.g. Hermite/Laguerre/Legendre polynomials and the such  these often also span the space of all functions. If the PDE + BC's have a uniqueness theorem, we even know that our form of the solution is equal to any other form you can find.
I believe that in quantum mechanics, you need to take it as an axiom of the theory that the eigenfunctions of the Hamiltonian always form a basis for all possible states. I know that this is true for a finite dimensional Hermitian matrix, but I think there is no proof for an arbitrary operator on a Hilbert space, so we need to assume completedness (every Hamiltonian I've seen does satisfy this property).

 Posts: 381
 Joined: Mon May 24, 2010 11:34 pm
Re: Separable differential equations?
Given the gradient operator (momentum in the Schroedinger equation), there's about six coordinate systems where it separates. In all others you're out of luck. So it really amounts to our hoping that the problem's boundary conditions are compatible with one of those coordinate systems. (Fortunately, the people who create homework problems are very lucky.)rhenley wrote:My question is, when, and why can we assume that a solution is separable?