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In the book "Piers Coleman - Introduction to Many-Body Physics (2016, Cambridge University Press)" in page 491

When an electron pair is created, electrons can only be added above the Fermi surface, so that $$ |\Psi\rangle = \Lambda^\dagger|FS\rangle = \sum_{|\mathbf{k}|>k_F} \phi_\mathbf{k}|\mathbf{k}_P\rangle, \tag{14.15}$$ where $|\mathbf{k}_P\rangle \equiv |\mathbf{k}\uparrow,−\mathbf{k}\downarrow\rangle = c^\dagger_{\mathbf{k}\uparrow}c^\dagger_{−\mathbf{k}\downarrow}|FS\rangle$. Now suppose that the Hamiltonian has the form $$ H = \sum_\mathbf{k} \epsilon_\mathbf{k} c^\dagger_{\mathbf{k}\sigma}c_{\mathbf{k}\sigma} + \hat{V}, \tag{14.16}$$ where $V$ contains the details of the electron–electron interaction; if $|\Psi\rangle$ is an eigenstate with energy $E$, then $$ H|\Psi\rangle = \sum_{|\mathbf{k}|>k_F} 2\epsilon_\mathbf{k}\phi_\mathbf{k}|\mathbf{k}_P\rangle + \sum_{|\mathbf{k}|,|\mathbf{k}'|>k_F} |\mathbf{k}_P\rangle\langle \mathbf{k}_P|\hat{V}|\mathbf{k}'_P\rangle\phi_{\mathbf{k}'}. \tag{14.17}$$ Identifying this with $E|\Psi\rangle = E \sum_\mathbf{k} \phi_\mathbf{k}|\mathbf{k}_P\rangle$, so comparing the amplitudes to be in the state $|\mathbf{k}_P\rangle$, $$ E\phi_\mathbf{k} = 2 \epsilon_\mathbf{k}\phi_\mathbf{k} + \sum_{|\mathbf{k}'|>k_F} \langle \mathbf{k}_P|\hat{V}|\mathbf{k}'_{P}\rangle\phi_{\mathbf{k}'}. \tag{14.18}$$

can anyone tell me how the equation 14.17 is coming? I am specifically confused how does $2\epsilon$ is coming in the second quantized kinetic energy part?

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    $\begingroup$ It's currently unclear what exactly this question is asking without clicking on the link you provided. To make questions more accessible and guard against link rot, please include all relevant information, such as the explanation of notation or specific terminology used, in your question. $\endgroup$ – ACuriousMind Jun 17 at 17:36
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We are talking about creating pairs of electrons, so as to not disturb the zero momentum of the state.

The $2\epsilon_\mathbf{k}$ term is actually $\epsilon_{\mathbf k} +\epsilon_{-\mathbf k}$: you get the single-particle energy of both electrons summed together independently. With an assumption of isotropy these should be the same energy.

The term on the right is somewhat harder to interpret but it presumably comes from some sort of completeness relation of the form, $$\sum_{\mathbf k} |\mathbf k_P\rangle \langle \mathbf k_P| = 1.$$

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I think you are confused about 2$\epsilon_\textbf{k}$ because equation (14.16) is wrong. To correct it, replace the first sum $\sum_{\textbf{k}}$ for $\sum_{\textbf{k}\sigma}$ (some books assume this is implicit). The electrons from the two spin-states must be taken into account to the total kinect energy operator. Note that $\epsilon_\textbf{k}$ are kinect energies only for a "free electron gas", but the answer remains in any case satisfying $\epsilon_\textbf{k}=\epsilon_{-\textbf{k}}$.

Nevertheless, if you still dont get (14.17), I can show some clarifying calculations.

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  • $\begingroup$ Can you show me some clear calculations? $\endgroup$ – Saptarshi Biswas Jun 18 at 7:23

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