Simple Harmonic Oscillator (time independant Schrodingers)

[indie452]

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-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

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 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)

 

[lanedance]

so shrodingers:

\frac{-hbar^2}{2m}*[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?

 

[indie452]

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

 

[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...

 

[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.

 

[indie452]

thats what I am 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 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?

 

[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

 

[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}

 

[lanedance]

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

 

[lanedance]

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

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

 

[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}