Quantum Harmonic Oscillator Problem

[Crush1986]

Homework Statement

Substitute $$\psi = Ne^{-ax^2}$$ into the position-space energy eigenvalue equation and determine the value of the constant a that makes this function an eigenfunction. What is the corresponding energy eigenvalue?

Homework Equations

$$\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \langle x | E \rangle + \frac{1}{2} m \omega^2 x^2 \langle x| E \rangle = E \langle x | E \rangle$$

The Attempt at a Solution

So, initially I tried to solve for a by plugging in $$\psi$$ but I got a nasty quadratic $$\frac{2 \hbar^2 x^2}{m} a^2 - \frac{\hbar^2}{m} a + \left( E - \frac{1}{2} m \omega^2 x^2 \right) = 0$$that didn't really seem right. I then did some research and found a similar problem where the book stopped at a similar quadratic equation (the problem was for the first excited state) and said that "the x^2 terms must cancel.

Why is that? I guess I haven't seen the energy eigenstates depend on x before... So I suppose that gives reason to believe that the terms with x will negate each other?

Following that recipe I arrived at the same value of a as they do $$a = \frac{m \omega}{2 \hbar}$$ and I also arrive at the expected energy (Since this given psi is of the form of the ground state harmonic oscillator) $$E = \frac{\hbar \omega}{2}$$.

Thanks for any help with understanding this deeper.

 

[blue_leaf77]

Why is that?

Because that quadratic equation must hold for any value of $x$.

 

[Crush1986]

Ok! I think I see. the rest of the terms are constant, so if x varies the zero on the right must still obviously be a 0. Only way this happens is if the two x^2 terms are equal, right?

 

[blue_leaf77]

Yes.

 

[Crush1986]

Thank you so much! That is kind of a subtle thing to notice! Makes total sense now though, thanks again!