# All Eigenfunctions can be Represented by a Linear Superposition

1. Jun 22, 2014

### kq6up

1. The problem statement, all variables and given/known data

The time-independent wave function $\psi (x)$ can always be taken to be real (unlike $\Psi (x,t)$, which is necessarily complex). This doesn't mean that every solution to the time-independent Schödinger equation is real; what it says is that if you've got one that is not, it can always be expressed as a linear combination of solutions (with the same energy) that are. So you might as well stick to $\psi$'s that are real. Hint: If $\psi (x)$ satisfies Equation 2.5, for a given E, So too does its complex conjugate. and hence also the real linear combinations $(\psi +\psi^*)$ and $i(\psi - \psi^*)$.

2. Relevant equations

N/A

3. The attempt at a solution

Does not any two $\psi$'s of the same energy level form a subspace of the total general function $\Psi$ that covers all of that subspace? That is if you put a real coefficient in front of one, and an imaginary coefficient in front of the other, you have the ability to make a vector that touches *every* point of that subspace. This seems self evident to me, and I don't understand why there needs to any kind of formal proof that involves plugging this in to Schrödingers equation.

Agreed, or am I missing something?

Chris

2. Jun 22, 2014

### Simon Bridge

... well isn;t that most of what you have to prove?

Note: the wavefunctions cannot be just any old functions, they have to be solutons to the schrodinger equation - so what is true for any old functions may not be true for psi's.

You have been thinking about this problem backwards - of course if you start by defining the eigenfunctons as a subspace of the overall function then the answer is self evident - to do a proof you cannot assume that.
You have to start from the definition of the eigenfunctions and of the general function, and demonstrate that they have this relationship.