Need help in Calculating Wavefunction Variance

[Jay R. Yablon]

Dear Friends:

I am attempting to calculate the variance of a non-Gaussian
wavefunction:

psi(x) = exp [-(1/2)Ax^2-Bx]

in the general situation where A and B are *interdependent*, i.e., dA/dB
<> 0. I can do this easily when dA/dB=0, but not for dA/dB <> 0
generally.

On one sheet, attached at:

http://jayryablon.files.wordpress.com/2008/04/non-gaussian-calculation.pdf ,

I have shown how far I am able to get. Does anyone have a clue how to
calculate through for the final terms in each of (2) and (3), presumably
by turning them into an ln expression but maybe via some other approach?

Please note, I have had trouble clicking the link above to open the
file, but have been able to right click and then download and view the
file. So, if you have trouble, that may be the solution.

Thanks.

Jay.
____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

[Roland Franzius]

Jay R. Yablon schrieb:
> Dear Friends:
>
> I am attempting to calculate the variance of a non-Gaussian
> wavefunction:
>
> psi(x) = exp [-(1/2)Ax^2-Bx]
>
> in the general situation where A and B are *interdependent*, i.e., dA/dB
> <> 0. I can do this easily when dA/dB=0, but not for dA/dB <> 0
> generally.


Gaussian integral are done by completion of squares and shifting the
domain of integration:

int_(-oo)^oo dx x^2 exp(-A x^2 - 2 B x) =
exp(B^2/A) int_(-oo)^oo dx x^2 exp(-A ( x^2 - 2 B/A x + (B/A)^2 )=

exp(B^2/A) int_(-oo)^oo dx x^2 exp(-A ( x^2 - B/A)^2 )=
exp(B^2/A) int_(-oo)^oo dy (y+B/A)^2 exp(-A y^2 )

So you get

var(x)=<x^2>-<x>^2
= int_(-oo)^oo dy (y+B/A)^2 exp(-A y^2 )/
int_(-oo)^oo dy exp(-A y^2 )
- { int_(-oo)^oo dy (y+B/A) exp(-A y^2 )/
int_(-oo)^oo dy exp(-A y^2 ) }^2

All you need is to calculate the integrals
<1>, <y> and <y^2> with distribution e^(-A y^2)

--

Roland Franzius