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On the 1D Quantum Mechanics Harmonic Oscillator

Physics Asked on May 20, 2021

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

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

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?

One Answer

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 $.)

Correct answer by Philip on May 20, 2021

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