# Eigenfunctions in a harmonic oscillator

This assignment is about the one dimensional harmonic oscillator (HO). The hamiltonian is just as you know from the HO, same goes for the energies, but I get that the wavefunction of the particle, at $t = 0$, is given by:

$\psi \left( x,\,\,t=0 \right)=A\left( {{\phi }_{0}}\left( x \right)+2i{{\phi }_{2}}\left( x \right) \right)$

Now I have to calculate A, so that $\psi$ is normalized.

My question is, what is ${{\phi }_{0}}\left( x \right)$ and ${{\phi }_{2}}\left( x \right)$ given by ?

Otherwise I guess is just doing the usual:

$\int_{-\infty }^{\infty }{{{\left| \Psi \left( x,\,\,t \right) \right|}^{2}}}dx=1$

To figure out A.

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The explicit eigenfunctions of the Harmonic oscillator hamiltonian are given here, but I would highly discourage you from explicitly doing an integral using these expressions to determine $A$. It is significantly easier to use the fact that the eigenfunctions are orthogonal; $$\int_{-\infty}^\infty dx\,\phi_m^*(x)\phi_n(x) = \delta_{mn}$$ If you use this fact, then the integral on the left hand side of the $t=0$ normalization condition you wrote down will be very easy. Try this out, and if you still have trouble we can give you more guidance since this is a homework question.
Ahhh, so simple. So basically you do the $\int_{-\infty }^{\infty }{{{\left| \Psi \left( x,\,\,t \right) \right|}^{2}}}dx$ and see that it will give: ${{A}^{2}}{{\phi }_{0}}\left( x \right)+4{{A}^{2}}{{\phi }_{2}}\left( x \right)$ And then just use your statement, and you get that: $A=\sqrt{\frac{1}{5}}$ Thank you very much :D –  Denver Dang May 5 '13 at 20:18
@DenverDang ; it will actually give $\int (A^2 |\phi_0|^2 + 4A^2|\phi_2|^2)$ but yea; you get the picture. –  joshphysics May 5 '13 at 20:33
@DenverDang No prob. Btw there's one other interesting subtlety here; $A$ can actually be complex in which case the normalization condition gives $|A|^2 = 1$. This shows that in your case you can just as well set $A = e^{i\theta}\sqrt{1/5}$ for any real number $\theta$. This just reflects the fact that states which defer by multiplication by a phase are considered physically equivalent in quantum, so there is a freedom to multiply any normalization you choose by a phase. In particular, you have just chosen a phase to make $A$ real, which is often the simplest choice. –  joshphysics May 5 '13 at 20:42