Coefficients of a linear combination of eigenfunctions I'm trying to show that given $\psi(x)=c_0\psi_0(x)+c_1\psi_1(x)$, where all functions are normalized and additionally that $\psi_0$ and $\psi_1$ are eigenfunctions of an arbitrary operator, that $|c_0|^2+|c_1|^2 = 1$.
From what I understand, the eigenfunctions of a given operator should all be orthogonal to one  another (that is $\int_{-\infty}^{\infty}\psi_i(x)^\star\psi_j(x) = 0$, if $i \neq j$), so it makes sense on some level that a linear combination of them that is still normalized in and of itself (specified in the problem) would have a 'magnitude' of 1, so the length of the vector determined by the coefficients $c_0$ and $c_1$ would be 1 (or is my logic totally faulty?).
I'm not sure how to actually demonstrate this, however.
 A: You demonstrate it via $$1=\langle\psi,\psi\rangle=\langle c_0\psi_0+c_1\psi_1,c_0\psi_0+c_1\psi_1\rangle$$ $$=\langle c_0\psi_0,c_0\psi_0\rangle+\langle c_1\psi_1,c_0\psi_0\rangle+\langle c_0\psi_0,c_1\psi_1\rangle+\langle c_1\psi_1,c_1\psi_1\rangle$$
$$=c_0c_0^*\langle\psi_0,\psi_0\rangle+c_1c_0^*\langle\psi_1,\psi_0\rangle+c_0c_1^*\langle\psi_0,\psi_1\rangle+c_1c_1^*\langle\psi_1,\psi_1\rangle$$
$$=|c_0|^2+|c_1|^2.$$
Above we used the facts $\langle\psi_0,\psi_0\rangle=\langle\psi_1,\psi_1\rangle=1$ and $\langle\psi_1,\psi_0\rangle=\langle\psi_0,\psi_1\rangle=0$, as well as basic features of the inner product (linearity and conjugate symmetry):
$$\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle$$
$$\langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle$$
$$\langle au,bv\rangle=ab^*\langle u,v\rangle$$
with $a$ and $b$ scalars and $z^*$ denoting complex conjugation.
This generalizes to arbitrary linear combinations of normalized orthogonal vectors:
$$\left\|\sum_{i=1}^n c_i\psi_i \right\|^2=\sum_{i=1}^n|c_i|^2 \text{  when }\langle\psi_i,\psi_j\rangle=\delta_{ij}.$$
