Show that a wave function fits the Schrödinger's equation. (Harmonic oscillator)

[Asphyxion]

Homework Statement

The wave function $$\psi_0 (x) = A e^{- \dfrac{x^2}{2L^2}}$$
represents the ground-state of a harmonic oscillator. (a) Show that $$\psi_1 (x) = L \dfrac{d}{dx} \psi_0 (x)$$ is also a solution of Schrödinger's equation. (b) What is the energy of this new state? (c) From a look at the nodes of this wave function, how would you classify this excited state?

Homework Equations

$$E_n = (n + 1/2) \hbar \omega$$

SE:
$$-\frac{\hbar ^2}{2m} \dfrac{\partial}{\partial x} \psi + \frac{1}{2} m \omega ^2 \psi = E \psi$$

The Attempt at a Solution

$$\dfrac{\partial}{\partial x} \psi_0 = -\dfrac{xA}{L^2} e^{- \dfrac{x^2}{2L^2}}$$

$$\dfrac{\partial}{\partial x} \psi_1 = - \dfrac{A}{L} e^{- \dfrac{x^2}{2L^2}} + \dfrac{x^2A}{L^3} e^{- \dfrac{x^2}{2L^2}}$$

$$\dfrac{\partial ^2}{\partial x^2} \psi_1 = (\dfrac{xA}{L^3} \dfrac{x^2}{2L^2}} + \dfrac{2xA}{L^3} \dfrac{x^2}{2L^2}} - \dfrac{x^3A}{L^5} \dfrac{x^2}{2L^2}})e^{- \dfrac{x^2}{2L^2}}$$

$$\dfrac{\partial ^2}{\partial x^2} \psi_1 = (\dfrac{3xA}{L^3} \dfrac{x^2}{2L^2}} - \dfrac{x^3A}{L^5} \dfrac{x^2}{2L^2}})e^{- \dfrac{x^2}{2L^2}}$$
Put into SE this gives me:
$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 = E$$
This is where I'm not getting any further. I just can't see how this is supposed to match the energy state of a harmonic oscillator. Am I thinking completely wrong, have I done the maths wrong. Or both? Any pointers would be greatly appreciated! (First time trying to post my LaTeX work on a forum, so be forgiving!)

 

[Asphyxion]

[STRIKE]Wait with reading this, this is total gibberish :)[/STRIKE]
Okay, I think my chain of thoughts should be understandable now.

 

[Mindscrape]

Okay, two hints:

Hint #1:
$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 = E$$
should actually be
$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 x^2 = E$$

Hint #2
Should energy have an x dependence? In other words, you know this has to satisfy the SE, so what can you do that will make it do that?

 

[Asphyxion]

Yeah, I've tried to set x=0 which gives me $$E_1= 3/2 \dfrac{\hbar ^2}{2mL^2}$$ which i don't find satisfy the $$E_1 = 3/2 \hbar \omega$$.
And on my paper i ofcourse had the x^2 part of the portential.

 

[Mindscrape]

You can't just set x=0, you have to get rid of it in another way. Look at the text below if you get super stuck.

You have to collect the x^2 terms and use the constraint of the constants to take away the x^2 terms. You will be able to find out what L^2 is, and it will give you the E1 energy level.

 

[Asphyxion]

I'm afraid I'm completely stuck even with the hint! I cannot find any example of anything similar to this in my book. And since you were hiding that hint, I can only assume it's supposed to be a rather short and easy step in the proof.

 

[Mindscrape]

Okay here's how it goes:

$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 x^2 = E$$

collect the x^2 terms

$$x^2(-\dfrac{\hbar ^2}{2mL^4}+\frac{1}{2} m \omega ^2) + \dfrac{3}{L^2}\dfrac{\hbar ^2}{2m} = E$$

set the constants multiplying the x^2 to 0 so that the x^2 term goes away
$$-\dfrac{\hbar ^2}{2mL^4}+\frac{1}{2} m \omega ^2 = 0$$

Solve for L^2 and plug it back into the new SE you will get the $3/2 \hbar \omega$ you are looking for.

 

[Asphyxion]

Thanks a bunch! I see in retrospect that I would've never figured this out. I hope that's not a bad sign for me as a student of physics :)