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

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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. $$A^2 = \frac{1}{14}$$

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


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