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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.

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How do you normalize, say $\sin(x)$ ? Are your coefficients functions too? –  user13838 Sep 12 '11 at 1:29
    
@percusse Normalizing such that $\int_{-\infty}^{\infty} |\psi(x)|^2 = 1$ (which you can't do with $sin$). The coefficients are constants. –  Nick T Sep 12 '11 at 1:32
    
@percusse: You don't normalize $\sin$ over $\mathbb{R}$, you do that over some finite interval with length a multiple of $2\pi$. –  anon Sep 12 '11 at 1:34
    
You've left out some hypotheses. The correct statement is that $\langle \psi_i, \psi_j \rangle = 0$ if $\psi_i$ and $\psi_j$ are eigenfunctions of a self-adjoint operator for different eigenvalues. –  Robert Israel Sep 12 '11 at 2:48

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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: $$\left\|\sum_{i=1}^n c_i\psi_i \right\|^2=\sum_{i=1}^n|c_i|^2 \text{ when }\langle\psi_i,\psi_j\rangle=\delta_{ij}.$$

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I was just starting to write it out, but with a heinous abuse of mathematical syntax (e.g. $|c_0\psi_0(x)+c_1\psi_1(x)|^2 = |c_0\psi_0(x)|^2 + 2|c_0\psi_0(x)||c_1\psi_1(x)| +|c_1\psi_1(x)|^2$) –  Nick T Sep 12 '11 at 1:41
    
@NickT: Ah, careful: that $|c_0\psi_0||c_1\psi_1|$ is actually incorrect. –  anon Sep 12 '11 at 1:42
    
Could you elaborate a touch; I'm out of my element :( Is it because they're not necessarily communicative? –  Nick T Sep 12 '11 at 1:47
    
@NickT: It's because $\langle a,b\rangle$ does not evaluate to $|a| |b|$. (Or were you asking about my answer in general?) –  anon Sep 12 '11 at 2:01

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