# Normalising a wavefunction where $\psi$ is equal to a sum of functions [closed]

The wavefunction $\psi(x)$ = $\phi_1(x)$ + $2\phi_2(x)$ + $3\phi_3(x)$ is to be normalised. The functions $\phi_1(x)$, $\phi_2(x)$, $\phi_3(x)$ are normalised eigenfunctions of a Hermitian operator $\hat{O}$ with eigenvalues $\lambda_1=1$, $\lambda_2=5$, $\lambda_3=9$.

I know that to normalise a wavefunction you do:

$\int$ $|\psi(x)|^2 dx = 1$

But substituting for $\psi(x)$ gives a long sum of a combination of the $\phi(x)$s and I don't see how you can integrate that.

## closed as off-topic by Emilio Pisanty, tpg2114♦, Abhimanyu Pallavi Sudhir, Waffle's Crazy Peanut, Qmechanic♦Nov 6 '13 at 9:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, tpg2114, Waffle's Crazy Peanut, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

## 2 Answers

First, if you want to normalize it the wave function needs to have some free constant, so $\psi(x)=A\,(\phi_1(x)+2\phi_2(x)+3\phi_3(x))$. Then normalize as you said: $$\int\left|\psi(x)\right|^2dx = A^2\;(\int\left|\phi_1(x)\right|^2dx + 4\int\left|\phi_2(x)\right|^2dx + 9\int\left|\phi_3(x)\right|^2dx + \text{cross terms})=1$$ The eigenfunctions of a Hermitian operator are orthogonal, so all the cross terms are zero. $$A^2 = \frac{1}{14}$$

• So you are saying that the integrals of $|\phi|^2$ = 1 and that's why you get A^2 = 1/14 ? – turnip Nov 6 '13 at 14:18
• It makes sense, I just didn't think it's this simple. I thought you'd have to integrate $\phi$ somehow... – turnip Nov 6 '13 at 14:20

The integral is a linear operation, it gets "distributed" over sums and multiplications. And given that the $\phi$'s are the eigenfunctions, they're orthogonal to each other (they're non degenerate).