proving a step in this field-theoretic derivation of the BdG equations

Question

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) ?

This post imported from StackExchange Physics at 2014-03-23 12:23 (UCT), posted by SE-user user38579

Comments

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.

This post imported from StackExchange Physics at 2014-03-23 12:23 (UCT), posted by SE-user Andrew

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@Andrew : hey, thanks. I made the correction.

This post imported from StackExchange Physics at 2014-03-23 12:23 (UCT), posted by SE-user user38579

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If both integrals are true, then one would expect \(\int\mathrm{d}^3r\left(\psi^\dagger_\downarrow(\mathbf{r})\nabla^2\psi_\downarrow(\mathbf{r})+\psi_\downarrow(\mathbf{r})\nabla^2\psi^\dagger_\downarrow(\mathbf{r})\right)=0\\ \implies \int\mathrm{d}^3r\nabla^2\left(\psi^\dagger_\downarrow(\mathbf{r})\psi_\downarrow(\mathbf{r})\right)=0\\ \implies \nabla^2\left(\psi^\dagger_\downarrow(\mathbf{r})\psi_\downarrow(\mathbf{r})\right)=0\\ \implies \psi^\dagger_\downarrow(\mathbf{r})\psi_\downarrow(\mathbf{r})\neq \psi^\dagger_\downarrow(x,y,z)\psi_\downarrow(x,y,z)\)

However, \(\mathbf{r}=(x,y,z)\), and hence I am led to believe that the substitution (1) is unjustified.