- #1
Karliski
- 9
- 0
Homework Statement
I have to calculate the expected value of the Hamilton operator (average energy) of a two, non interacting, identical particle system. Thus these particles can be bosons or fermions, but at the moment I just want to look at fermions.
Homework Equations
H=-\frac{h^{2}}{2m}(\frac{d^{2}}{dx_{1}^{2}}+\frac{d^{2}}{dx_{2}^{2}})
The localised wave functions (not eigenstates of the Hamilton operator) are given by:
<br /> \psi_{-}(x_{1}) & = & \exp[-\frac{\beta}{2}(x_{1}-a/2)^{2}]<br />
<br /> \psi_{+}(x_{2}) & = & \exp[-\frac{\beta}{2}(x_{2}+a/2)^{2}]
Thus for fermions we need to anti-symmetrise:
\psi(x_{1},x_{2})=\psi_{-}(x_{1})\psi_{+}(x_{2})-\psi_{-}(x_{2})\psi_{+}(x_{1})
Thus one particle is localised at x = a/2 and one at x = -a/2.
Thus to get the expected value of the Hamilton operator:
E(a)=\int_{\mathbb{R}}dx_{1}\int_{\mathbb{R}}dx_{2}\psi^{*}(x_{1},x_{2})H\psi(x_{1},x_{2})
One actually divides this expression by another function, but I solved that already and not relevant to my problem.
The Attempt at a Solution
So far I substituted the Hamilton operator into E(a) to get:
E(a)=\frac{h^{2}}{2m}\int_{\mathbb{R}}dx_{1}\int_{\mathbb{R}}dx_{2}\psi^{*}(x_{1},x_{2})(\frac{d^{2}}{dx_{1}^{2}}+\frac{d^{2}}{dx_{2}^{2}})\psi(x_{1},x_{2})
And then multiply it out to get two terms (both similar, only listing one):
\frac{h^{2}}{2m}\int_{\mathbb{R}}dx_{1}\int_{\mathbb{R}}dx_{2}\psi^{*}(x_{1},x_{2})\frac{d^{2}}{dx_{1}^{2}}\psi(x_{1},x_{2})
This is where the problem comes in, calculating that 2nd derivative creates a big mess, even when working with a computer algebra system. Is there a way to rewrite this in a better form?
[