Functional Derivative

We use the Hamiltonian Eq. (IV.2.1),

$$\hat{H} = \hat{\mathcal H}_0+\hat{U}\equiv \sum_{i=1}^N \hat{H}_0^{(i)}+ \frac{1}{2}\sum_{i\ne j}^N U_{ij}$$

$$\hat{H}_0^{(i)} = -\frac{\hbar^2}{2m}\Delta_i+ V({\bf r}_i),\quad U_{ij}= U(\xi_i,\xi_j).$$ (3.13)

The energy functional now depends on the $N$ wave functions $\psi_{\nu_i}({\bf r},\sigma)$, $i=1,...,N$,

$$F[\Psi]= F[\psi_{\nu_1},...,\psi_{\nu_N}]=F[\{\psi_{\nu_i}\}].$$ (3.14)

The definition of the functional derivative is not more complicated than in the one-component case,

$$\frac{\delta F[\Psi]}{\delta \Psi}\equiv \lim_{\varepsilon\to 0} ... ..._{\nu_i}+\varepsilon \cdot \delta \psi_i\}] -F[\{\psi_{\nu_i}\}]}{\varepsilon},$$ (3.15)

where we now have $i=1,...,N$ independent `deviations' $\delta \psi_i$ from the functions $\psi_i$. We furthermore want to ensure that all single particle states $\vert\nu_i\rangle$ are normalised. Therefore, we introduce our functional $F[\Psi]$ with $N$ Lagrange multipliers $\lambda_i$,

$$F[\Psi] \equiv \langle \nu_N,...,\nu_1 \vert\hat{H}\vert\nu_1,...,\nu_N\rangle_A + \sum_{i=1}^N \lambda_i[ \langle \nu_i\vert\nu_i\rangle -1 ].$$ (3.16)

We have calculated the energy expectation values already in Eq. (IV.2.5) and Eq. (IV.2.9),

$$F[\{\psi_{\nu_i}\}] = \sum_{i=1}^{N}\langle\nu_{i}\vert\hat{H}_0\vert\nu_{i}\rangle + \... ...j}\rangle -\langle\nu_{i}\nu_{j}\vert U_{ij} \vert\nu_{i}\nu_{j}\rangle \right]$$

$$+ \sum_{i=1}^N \lambda_i [ \langle \nu_i\vert\nu_i\rangle -1 ].$$ (3.17)

The individual terms are simply calculated:

$$\frac{\delta}{\delta \Psi} \sum_{i=1}^{N}\langle\nu_{i}\vert\hat{H}_0\vert\nu_{i}\rangle=$$

$$= \lim_{\varepsilon\to 0} \frac{1}{\varepsilon} \left[ \sum_{i=1}^{... ..._i\rangle- \sum_{i=1}^{N}\langle\nu_{i}\vert\hat{H}_0\vert\nu_{i}\rangle\right]$$

$$= \sum_{i=1}^{N} \left[\langle \delta\nu_i\vert\hat{H}_0\vert\nu_{i}\rangle + \langle \nu_i\vert\hat{H}_0\vert\delta\nu_{i}\rangle\right].$$ (3.18)

The term from the interaction $U$ yields

$$\frac{\delta}{\delta \Psi} \frac{1}{2}\sum_{ij}\left[\langle\nu_{... ...\nu_{j}\rangle -\langle\nu_{i}\nu_{j}\vert U \vert\nu_{i}\nu_{j}\rangle \right]$$ (3.19)

$$= \frac{1}{2} \sum_{ij}\Big[\langle \delta \nu_{j}\nu_{i}\vert U\ve... ...{j}\rangle + \langle\nu_{j}\nu_{i}\vert U\vert\nu_{i}\delta \nu_{j}\rangle\Big]$$

$$- \frac{1}{2} \sum_{ij}\Big[\langle \delta \nu_{i}\nu_{j}\vert U\ve... ...j}\rangle + \langle\nu_{i}\nu_{j}\vert U\vert\nu_{i}\delta \nu_{j}\rangle\Big].$$

We can use the symmetry property (Exercise: proof!)

$$\langle \nu_i \nu_k \vert U\vert\nu_l \nu_m \rangle = \langle \nu_k \nu_i \vert U\vert\nu_m \nu_l\rangle$$ (3.20)

to simplify things by using, e.g., $\langle\nu_{j} \delta \nu_{i}\vert U \vert\nu_{i}\nu_{j}\rangle = \langle \delta \nu_{i}\nu_{j}\vert U \vert\nu_{j}\nu_{i}\rangle$ and changing the summation indices $i$, $j$ such that

$$\frac{\delta}{\delta \Psi} \frac{1}{2}\sum_{ij}\left[\langle\nu_{... ...\nu_{j}\rangle -\langle\nu_{i}\nu_{j}\vert U \vert\nu_{i}\nu_{j}\rangle \right]$$ (3.21)

$$= \sum_{ij}\Big[\langle \delta \nu_{j}\nu_{i}\vert U\vert\nu_{i}\nu... ...j}\rangle - \langle\nu_{i}\nu_{j}\vert U\vert\nu_{i}\delta \nu_{j}\rangle \Big]$$

$$= \sum_{ij}\Big[\langle \delta \nu_{j}\nu_{i}\vert U\vert\nu_{i}\nu... ...- \langle \delta \nu_{j}\nu_{i}\vert U\vert\nu_{j}\nu_{i}\rangle + (H.c.)\Big],$$

where again in the ` $-$' term we have swapped indices, and $H.c$ means that there are two terms which are the hermitian conjugates of the two others.