Simple Harmonic Oscillator (time independant Schrodingers)

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indie452
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Homework Statement



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-independent schrodingers with the potential show that
[tex]\psi[/tex](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

The Attempt at a Solution



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

so shrodingers:


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

Ive gotten to this bit but i don't 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 don't need to do the long winded complete solution and that i is also not necessary to evaluate the normalisation constant A.
I just don't 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)
 
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indie452 said:
so shrodingers:

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

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?
 
well the thing is i don't know when it would be satisfied other than the LHS=RHS but i don't know how i would get E on the LHS, unless i made the sub that E=hbar*w
 
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...
 
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.
 
thats what I am having trouble with - knowing when it is satisfied

is the LHS supposed to equal the RHS?
i tried a=[tex]\sqrt{\frac{mc}{hbar}}[/tex]

and i got

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

but i don't 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?
 
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
 
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ok i tried this

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

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

i tried a= mC/2(hbar)2

this gave me

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

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

E=[tex]\frac{3C}{2}[/tex]
 
sorry made a mistake there, this is what i meant

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

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

i tried a= mC/2(hbar)2

this gave me

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

=[tex]\frac{6mC}{2(hbar)^2}[/tex] = [tex]\frac{2m}{(hbar)^2}[/tex]E

E=[tex]\frac{3C}{2}[/tex]