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Exchange potential operator?

by ismaili
Tags: exchange, operator, potential
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ismaili
#1
Jul1-08, 03:38 AM
P: 161
Dear all,

I read from a book discussing about the scattering that if we consider the scattering of spin-1/2 particles (two particle scattering), we can describe the effect of spin in terms of the "exchange potential operator", which satisfies
[tex]V_e^{\text{op}}\psi(\mathbf{r}) = V_e(r)\psi(-\mathbf{r})[/tex]
where [tex]V_e(r)[/tex] is a c-number and we assumed the potential to be central. In this way, the radial Schrodinger equation in the center of mass frame would be
[tex] \big[\nabla^2 + k^2 - U_d(r)\big]\psi(\mathbf{r}) - U_e(r)\psi(-\mathbf{r}) = 0[/tex] where [tex]U_d(r) = 2m V_d/\hbar^2[/tex], [tex] U_e(r) = 2m V_e/\hbar^2 [/tex] and [tex]V_d(r)[/tex] is the "direct potential" that unrelated to spin.
My question is, how can we find the "exchange potential"? and why we have to introduce an "exchange potential operator" first? Thanks for any instructions!
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