Proving solution for Schrodinger's Simple Harmonic Oscillator

[PFCJeff]

1. The problem statement, all variables and given/known data
Hi guys. I've been working on this problem for a while, it's starting to frustrate me.

"Show that the function of Ѱ=e^(-bx^2) with b=mw/2ħ is a solution and that the corresponding energy is ħw/2."


2. Relevant equations
Schrodinger Eqn: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/imgqua/hosc2.gif
Ѱ=e^(-bx^2)
b=mw/2ħ

3. The attempt at a solution
What I first did is rewrote the entire exponential function to include b, then tried to differentiate it (problems with that). I'm thinking you have to differentiate that exponential function and insert it into schrodingers and solve for E (having to be ħw/2) ...I hope I'm not missing something or am completely off here, but I'd appreciate any help! :)

[ideasrule]

You just need to show it satisfies Schrodinger's equation, so plug in Ѱ=e^(-bx^2) on the left side, E=ħw/2 on the right side, and show that the two sides are equal.

[PFCJeff]

right, ok. thats what i thought...
but im having trouble on the d^2/dx^2 part once i plug in Ѱ=e^(-bx^2)

[ideasrule]

Why are you having trouble deriving it? Do you know the chain rule?

[PFCJeff]

i know how to do the chain rule, but i'm having issues with deriving the exponential function (e), PLUS the division in the exponent....it's throwing me off completely, i don't even think I can formulate a proper derivation.

[ideasrule]

What division in the exponent? To derive e^(-bx^2), you first derive with respect to (-bx^2) by treating the whole thing as one variable. You get e^(-bx^2). Then you derive (-bx^2) with respect to x and multiply the two together. The final answer should be e^(-bx^2)*-2bx. Now you just need to derive it again to get the second derivative.

[PFCJeff]

but b=mw/2ħ...
or am i allowed to just "plug" that in after? that was the division i was talking about.

[PFCJeff]

after my second derivation, i get 2be^(-bx^2) * (2bx^2 - 1)

then i can sub in that equation for b? to me, it doesn't seem right, but i could be wrong...

[vela]

Remember b is just a constant. It doesn't complicate the differentiation at all.

[ideasrule]

after my second derivation, i get 2be^(-bx^2) * (2bx^2 - 1)

then i can sub in that equation for b? to me, it doesn't seem right, but i could be wrong...

Yes, that's right. As vela said, b is just a constant. The rules of calculus couldn't care less whether you choose to call the constant "b", "mw/2h", or "abcdefg/hijklkm"; no matter what name you give it, it's still just a constant.