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What's wrong with this general solution of the Lippmann-Schwinger equation: $$ |\psi_k \rangle=|\phi_k \rangle+G_k V|\psi_k \rangle $$ Taking the inner product with $\langle\phi_{k'}|$ \begin{align} \langle \phi_{k'}|\psi_k \rangle&=\langle \phi_{k'} |\phi_k \rangle+\langle \phi_{k'} |G_k V|\psi_k \rangle \\ &=\delta(k'-k)+g(E_k-E_{k'})\langle \phi_{k'} |V|\psi_k \rangle\\ &=\delta(k'-k)+g(E_k-E_{k'})\langle \phi_{k'} |H-H_0|\psi_k \rangle\\ &=\delta(k'-k)+g(E_k-E_{k'})\langle \phi_{k'} |E_k-E_{k'}|\psi_k \rangle \end{align} Basic algebra then gives us $$\langle \phi_{k'}|\psi_k \rangle[1-g(E_k-E_{k'})(E_k-E_{k'})]=\delta(k'-k)$$ and isolating the inner product on the left hand side gives us $$\langle \phi_{k'}|\psi_k \rangle=\delta(k'-k) [1-g(E_k-E_{k'})(E_k-E_{k'})]^{-1}$$ where $g(E_k-E_{k'})$ is the eigenvalue of $G_k$ resulting from hitting $\langle \phi_{k'} |$ on the left?

This seems to suggest that the eigenstate $|\psi_k \rangle $ of the full Hamiltonian $H$ is the same as the eigenstate $|\phi_k \rangle$ of $H_0$, up to a proportionality factor, since $|\psi_k \rangle $ has no amplitude to be in $|\phi_{k'\neq k} \rangle$.

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It seems to me that the passage from the 4th line to the 5th cannot be right because $[E_k]_k$ stands for the spectrum of the free hamiltonian $H_0$. And what you've done here is : $$\langle \phi_k|H=E_k \langle \phi_k| $$

which is wrong. $[E_k-E_{k'}]_{(k,k')}$ cannot be the spectrum of the potential $V$.

If you want the solution of this equation, you must compute an expansion on the Green operator $\hat{G}$ following the Dyson equation : $$\hat{G}=\hat{G_0}+\hat{G_0}\hat{V}\hat{G}$$ with $\hat{G_0}$ the free Green operator associated to $H_0$ ; and stop it at the order in $V$ you feel your calcultaions will be consistent.

For instance, at first order :

$$\hat{G}=\hat{G_0}+\hat{G_0}\hat{V}\hat{G_0}$$ i.e. with your notations : $|\psi_k\rangle=|\phi_k\rangle+\hat{G}_k\hat{V}|\phi_k\rangle$.

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  • $\begingroup$ In the equation: $\langle \phi_{k'} |H-H_0|\psi_k \rangle$, the $E_{k'}$ comes from acting on the ket with $H$, and the $E_k$ comes from acting on the bra with $H_0$. In scattering I think it's assumed that $H$ and $H_0$ have the same spectrum, and there is a 1:1 map between the eigenstates $\phi_k \rangle$ of the unperturbed Hamiltonian $H_0$ and the eigenstates (in-states) $|\psi_k \rangle$ of the full Hamiltonian $H$. $\endgroup$ – Chris L. Apr 29 '14 at 13:31
  • $\begingroup$ Yes, but still that you cannot know how the ket $|\psi_k\rangle$ is actually acting on $H-H_0$ since you dont know what is $|\psi_k\rangle$ either the spectrum of $H$ (which cannot be $[E_k]_k$). $\endgroup$ – dolun Apr 29 '14 at 13:40
  • $\begingroup$ I don't know what is $|\psi_k\rangle$ (trying to solve for it), but I do know that it's an eigenvector of $H$, so regardless of its actual form, $H$ operating on it should give its eigenvalue times itself. As for the spectrum of $H$, shouldn't it be the same as $H_0$ in the scattering sector? Nevertheless, even if we don't know the spectrum of $H$, we still know that $|\psi_k\rangle$ is an eigenvector of $H$, so we know the eigenvalue $E_k$ comes out even if we don't know its value. Hence the above solution to the Lipp-Schwing equation has an unknown eigenvalue in it, but is still a solution? $\endgroup$ – Chris L. Apr 29 '14 at 21:02
  • $\begingroup$ Ok so first there is a notation problem : to avoid any confusion you should note $[\varepsilon_k]_k$ the eigenvalues of $H$ rather than $[E_k]_k$. $H$ and $H_0$ can't have the same eigenvalues since scattering events introduce shifts in energy (depending on the potential amplitude). Actually, your solution is not a real solution since you don't know the $[\varepsilon_k]_k$. $\endgroup$ – dolun Apr 30 '14 at 9:25
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$\langle \phi_{k'} \rvert$ is not an eigenfunction of $G_k$, hence the second line of your derivation is not correct. The eigenfunctions are $\langle \psi_{k} \rvert$, and that is exactly what we usually solve Lippmann-Schwinger equation for.

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