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?


1 Answer 1


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 $.)

  • $\begingroup$ 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. $\endgroup$
    – nuwe
    Commented Sep 27, 2020 at 18:37

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