Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.