# On the 1D Quantum Mechanics Harmonic Oscillator

I was solving the P. 2.41 of Griffiths' Introduction to Quantum Mechanics.

Nothing really new until I read a proposed solution (from Griffiths' himself) for the problem in which it states that I can write a function $$\Psi$$ which is a linear combination of the first three states $$\Psi_0$$, $$\Psi_1$$ and $$\Psi_2$$ as

$$\Psi(x,t) = \sum\limits_{n=0}^2 c_n \Psi_n e^{-iwt(n+1/2)}$$

where the $$c_n$$'s and $$\Psi_n$$'s are known.

Question: Why can I represent the $$\Psi$$ wavefunction in the harmonic oscillator like this?

The eigenstates $$\Psi_n$$ of the Harmonic Oscillator are the Hermite polynomials (with a Gaussian weight) which form a complete set, and so you can write any initial state $$\Psi(x,0) = \sum_{n=0}^\infty c_n \Psi_n(x),$$ and solve for the $$c_n$$s. Once you've done this, you just need to tack on the time dependence for each eigenstate (of the form $$e^{-iE_n/\hbar t }$$) which gives you the complete solution.
The interesting question is why we don't need more than three $$\Psi_n$$s to actually write $$\Psi(x,0)$$. The reason for that is because $$\Psi(x,0)$$ contains a polynomial of degree 2. As you can see, no power of $$x$$ greater than $$x^2$$ exists in it. As a result, it must be expressable as a linear combination of all the Hermite polynomials with degree up to 2, i.e. $$\Psi_0(x), \Psi_1(x)$$, and $$\Psi_2(x)$$!
(This is very similar to saying that any polynomial of degree $$m$$ can be expressed as a linear sum of all $$x^n$$, where $$n=0,\dots m$$.)
• I appreciate a lot your answer! so it was the form of the time dependent part of the equation what I was missing. As $E_n$ takes the form of $\hbar \omega (n+1/2)$ the issue is solved. – holahola Sep 27 at 18:37