# Expected Value of the Hamilton operator

1. Apr 16, 2008

### Karliski

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

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.

2. Relevant 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:

$$\psi_{-}(x_{1}) & = & \exp[-\frac{\beta}{2}(x_{1}-a/2)^{2}]$$

$$\psi_{+}(x_{2}) & = & \exp[-\frac{\beta}{2}(x_{2}+a/2)^{2}]$$

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.

3. 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?
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2. Apr 17, 2008

### genneth

Try working in the momentum basis.