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Consider the potential scattering problem described by the Hamiltonian

$$H=K+V$$

where $K$ is the kinetic energy operator (e.g. $K=-\frac{1}{2} \nabla^2$) and $V$ is a position dependent potential.

Let $G_0^\pm$ be the "free" Green function

$$G_0^\pm = (\frac{q^2}{2}-K \pm i\eta)^{-1}$$

Then the operator identity

$$KG_0^\pm = G_0^\pm K = \frac{q^2}{2}G_0^\pm-1$$

holds.

Question

I don't really get this identity. Appart from that I can't show it, it also confuses me because if I take an inner product with some free eigenstates $|k\rangle$ defined by $K|k\rangle=\frac{k^2}{2}|k\rangle$ the above identity gives

$$\langle k'|G_0^\pm K|k\rangle = \frac{q^2}{2}\langle k'|G_0^\pm|k\rangle - \langle k'|k\rangle.$$

while I should be able to just apply $K$ to the ket and that would give me

$$\langle k'|G_0^\pm K|k\rangle = \frac{k^2}{2}\langle k'|G_0^\pm|k\rangle.$$

What happened to the last term and the $q$? My suspicion is that there is some problem connected to the states $|k\rangle$ being non-normalizable.

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1 Answer 1

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Using the fact that the operator $G_0^\pm$ is diagonal in the $|k'\rangle$ basis, one immediately gets that $$G_0^\pm|k'\rangle=\frac{1}{q^2/2-k^{'2}/2\pm i\eta}|k'\rangle.$$

With this identity, one shows easily that $$KG_0^\pm = G_0^\pm K = \frac{q^2}{2}G_0^\pm-1,$$ for all states $|k'\rangle$ (up to a vanishingly small contribution proportional to $\eta$).

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