Simple Harmonic Oscillator (time independant Schrodingers)

1. Nov 10, 2009

indie452

1. The problem statement, all variables and given/known data

Particle mass m is confined by a one dimensional simple harmonic oscillator potential V(x)=Cx2, where x is the displaecment from equilibrium and C is a constant

By substitution into time-independant schrodingers with the potential show that
$$\psi$$(x)=Axe-ax2
is a possible spatial wavefunction for this particle provided constant a has a certain value.
Find a in term os C, m, h bar (\h)
What the corresponding energy eigen value

3. The attempt at a solution

$$\psi$$(x)= Axe-ax2
d/dx $$\psi$$(x)= [A - 2Aax2]*e-ax2
d2/d2x $$\psi$$(x)= [4a2Ax3 - 6aAx]*e-ax2

so shrodingers:

$$\frac{-hbar^2}{2m}$$*[4a2x2 - 6a] + Cx2 = E

Ive gotten to this bit but i dont understand what to do, i have read the 4 page solution in Eisburg&Resnick and spoke to my lecturer, but my lecturer has said that i dont need to do the long winded complete solution and that i is also not necessary to evaluate the normalisation constant A.
I just dont know how to show that the psi above is a possible wavefunction

help or advice on how to do this problem would be much appreciated

ps I have tried looking at and doing some other example but they all had the potential as V(x)= 1/2 Cx2, this seemed to be the standard example. I also tried using the substitution w2= (c/m) and this would give a potential V(x)= mw2x2, but i didnt know if this was right as the question ask for a in terms of C (and m and hbar)

Last edited: Nov 10, 2009
2. Nov 10, 2009

lanedance

so i think you're pretty much there, so assuming you have subbed into the TISE correctly (which looks ok to me...), then how can you "choose" a, so that the above equation i satisfied?

3. Nov 10, 2009

indie452

well the thing is i dont know when it would be satisfied other than the LHS=RHS but i dont know how i would get E on the LHS, unless i made the sub that E=hbar*w

4. Nov 10, 2009

lanedance

I don't see a w anywhere... but there should be one value of a to satisfy TISE for all x, and what if the TISE is telling you the energy of the given state....

5. Nov 10, 2009

jimmy neutron

As lanedance said, you're nearly there. You do know that E is a constant (but don't know it yet), i.e. it is independent of x.

6. Nov 10, 2009

indie452

thats what im having trouble with - knowing when it is satisfied

is the LHS supposed to equal the RHS?
i tried a=$$\sqrt{\frac{mc}{hbar}}$$

and i got

Cx2(1-2hbar) + $$\frac{3x(hbar)\sqrt{hbar*mC}}{m}$$ = E

but i dont know what to do with it...is my choice for a even right?
How do you choose the value, is it a guess? or is there some method?

7. Nov 10, 2009

lanedance

no, i don't think so, the Energy of a eigenstate in the TISE, should be a constant scalar value (independent of x), this should help you choose a to remove any x dependence

Last edited: Nov 10, 2009
8. Nov 10, 2009

indie452

ok i tried this

$$\frac{(hbar)^2}{2m}$$*[6a - 4ax2 + $$\frac{2m}{(hbar)^2}$$Cx2] = E

6a + x2*[$$\frac{2m}{(hbar)^2}$$Cx2 - 4a] = $$\frac{2m}{(hbar^2}$$Cx^2]E

i tried a= mC/2(hbar)2

this gave me

$$\frac{6mC}{2(hbar)^2}$$ + x2[$$\frac{2mC}{(hbar)^2}$$ - $$\frac{4mC}{2(hbar)^2}$$] = $$\frac{2m}{(hbar)^2}$$Cx^2]E

=$$\frac{6mC}{2(hbar)^2}$$ = $$\frac{2m}{(hbar)^2}$$Cx^2]E

E=$$\frac{3C}{2}$$

9. Nov 10, 2009

lanedance

i can't follow your equations i the last post... but i think you've got the idea

10. Nov 10, 2009

lanedance

also do you know you can write longer equations in tex? eg.

$$\frac{\hbar^2}{2m}(4a^2x^2 - 6a) + Cx^2 = E$$

11. Nov 10, 2009

indie452

sorry made a mistake there, this is what i meant

$$\frac{(hbar)^2}{2m}$$*[6a - 4ax2 + $$\frac{2m}{(hbar)^2}$$Cx2] = E

6a + x2*[$$\frac{2m}{(hbar)^2}$$Cx2 - 4a] = $$\frac{2m}{(hbar)^2}$$E

i tried a= mC/2(hbar)2

this gave me

$$\frac{6mC}{2(hbar)^2}$$ + x2[$$\frac{2mC}{(hbar)^2}$$ - $$\frac{4mC}{2(hbar)^2}$$] = $$\frac{2m}{(hbar)^2}$$E

=$$\frac{6mC}{2(hbar)^2}$$ = $$\frac{2m}{(hbar)^2}$$E

E=$$\frac{3C}{2}$$