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Expected Value of the Hamilton operator

  • Thread starter Karliski
  • Start date
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1. 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.

2. Homework Equations

[tex]H=-\frac{h^{2}}{2m}(\frac{d^{2}}{dx_{1}^{2}}+\frac{d^{2}}{dx_{2}^{2}})[/tex]

The localised wave functions (not eigenstates of the Hamilton operator) are given by:

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

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

Thus for fermions we need to anti-symmetrise:

[tex]\psi(x_{1},x_{2})=\psi_{-}(x_{1})\psi_{+}(x_{2})-\psi_{-}(x_{2})\psi_{+}(x_{1})[/tex]

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:

[tex]E(a)=\int_{\mathbb{R}}dx_{1}\int_{\mathbb{R}}dx_{2}\psi^{*}(x_{1},x_{2})H\psi(x_{1},x_{2})[/tex]

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:

[tex]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})[/tex]

And then multiply it out to get two terms (both similar, only listing one):

[tex]\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})[/tex]

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?
[
 

Answers and Replies

979
1
Try working in the momentum basis.
 

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