- #1
kq6up
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Homework Statement
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^*)##.
Homework Equations
N/A
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 into Schrödingers equation.
Agreed, or am I missing something?
Chris