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

P. 2.41 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.

Griffiths' solution

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 $.)

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – holahola Sep 27 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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