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In derivation of the BdG mean field Hamiltonian as follows, I have a confusion here in the second step:

$H_{MF-eff} = \int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})H_{E}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})+\int d^{3}r\psi_{\downarrow}^{\dagger}(\mathbf{r})H_{E}(\mathbf{r})\psi_{\downarrow}(\mathbf{r}) +\int d^{3}r\triangle^{\star}(\mathbf{r})\psi_{\downarrow}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})+\int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}^{\dagger}(\mathbf{r})\triangle(\mathbf{r})-\int d^{3}r\frac{|\triangle(\mathbf{r})|^{2}}{U}$ $ = \int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})H_{E}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})-\int d^{3}r\psi_{\downarrow}(\mathbf{r})H_{E}^{\star}(\mathbf{r})\psi_{\downarrow}^{\dagger}(\mathbf{r}) +\int d^{3}r\triangle^{\star}(\mathbf{r})\psi_{\downarrow}(\mathbf{r})\psi_{\uparrow}(\mathbf{r})+\int d^{3}r\psi_{\uparrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}^{\dagger}(\mathbf{r})\triangle(\mathbf{r})-\int d^{3}r\frac{|\triangle(\mathbf{r})|^{2}}{U}$ $= \int d^{3}r\left(\begin{array}{cc} \psi_{\uparrow}^{\dagger}(\mathbf{r}) & \psi_{\downarrow}(\mathbf{r})\end{array}\right)\left(\begin{array}{cc} H_{E}(\mathbf{r}) & \triangle(\mathbf{r})\\ \triangle^{\star}(\mathbf{r}) & -H_{E}^{\star}(\mathbf{r}) \end{array}\right)\left(\begin{array}{c} \psi_{\uparrow}(\mathbf{r})\\ \psi_{\downarrow}^{\dagger}(\mathbf{r}) \end{array}\right)+const. $

with

$H_{E}(\mathbf{r})=\frac{-\hbar^{2}}{2m}\nabla^{2}$

In the second step, we have taken

$\int d^{3}r\psi_{\downarrow}^{\dagger}(\mathbf{r})\nabla^{2}\psi_{\downarrow}(\mathbf{r}) = -\int d^{3}r\psi_{\downarrow}(\mathbf{r})\nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})$............(1).

I can prove (by integration by parts and putting the surface terms to 0) that $\int d^{3}r\psi_{\downarrow}^{\dagger}(\mathbf{r})\nabla^{2}\psi_{\downarrow}(\mathbf{r}) = \int d^{3}r \nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}(\mathbf{r})$

but how is it justified to now take

$\int d^{3}r \nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})\psi_{\downarrow}(\mathbf{r}) = - \int d^{3}r\psi_{\downarrow}(\mathbf{r})\nabla^{2}\psi_{\downarrow}^{\dagger}(\mathbf{r})$

in order to prove (1) ?

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    $\begingroup$ I think you have a sign error in your integration by parts formula; you have to integrate by parts twice since $\nabla^2$ is a second derivative, so you should get a net positive sign. $\endgroup$ – Andrew Mar 23 '14 at 6:03
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Write the differential operator $\nabla^2$ in terms of derivatives as $\nabla^2=\partial^2_x+\partial^2_y+\partial^2_z$. Write each derivative as a limit (i.e.: $\partial_x f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $). Rearrange the fields by putting $\psi_\downarrow (r')$ on the left, of course keep track of fermionic exchanges with a minus sign. Reinstate the limit as a derivative and then as a differential operator.

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Let's do a fourier transform of the field operator:

$$\Psi_\downarrow(r)=\frac{1}{\sqrt{N}}\sum_p\phi_p a_p$$

Where $\phi_p$ is the plane wave hence the eigenvector of $H_E$.

Now it is easy to show that:

$$\int d^3r\Psi_\downarrow^\dagger(r)H_E\Psi_\downarrow(r)+\int d^3r\Psi_\downarrow(r)H_E^*\Psi_\downarrow^\dagger(r)=\sum_p \frac{p^2}{2m}=const$$ I have used the anticommute relation $\{a_p,a_p^\dagger\}=1$.

Now you can absorb this const into the final equation

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The negative sign comes from the anti-commutation of the fermion operators $\psi$ and $\psi^\dagger$. That is $\psi^\dagger(r)\psi(r') = - \psi(r')\psi^\dagger(r) + \delta(r-r')$. The delta function just gives the some constant that will be absorbed into the $const$ term. Maybe it is bothering you that this term is formally divergent but this is removed by any regularization, such as putting the system on a lattice.

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