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We have the Hartree wave function of N particles: $ \psi_{H} ( 1, ...., N) := \phi_{1}(1) \cdot \phi_{2}(2) ... \phi_{N}(N) $ where $\phi_{j}(I)$ is the one particle wave function of the I-th particle in state j. In order to find the differential equation to solve, one needs to minimize the average of the Hamiltonian under the condition that the one particle wave functions are normalized, more precisely we have to minimize the quantity: $X := \left\langle \psi_{H} | \mathcal{H} |\psi_{H}\right\rangle - \sum_{j=0}^{N} \epsilon_j \cdot \big[\int |\phi_j(\vec{r})|^2 d^3r - 1\big]$, where $|\phi_j(\vec{r})|^2 = \phi_j(\vec{r}) \cdot \phi_j(\vec{r})^*$ and $\epsilon$ the Lagrange muliplier. In oder to minimize X, so is told in literature, we have to calculate $ \frac{\delta X}{\delta \phi_{j}^*} = 0$, where $\frac{\delta\phi_{i}(x^´)}{\delta\phi_{j}(x)} = \delta_{ij} \delta(x - x')$. This is the step I did not understand properly. Why is it that the deinition of such a functional derivative is like the one above? And more important than that, why should we only consider the complex conjugate of the wave function in the process of minimizing? Why do we consider the wave function and its complex conjugate to be two fully independent entities. By that I mean for example the fact that: $\frac{\delta\phi_{i}(x^´)}{\delta\phi_{j}(x)^*} = 0$

Let me elaborate the derivation further: Considering a Hamiltonian of N interacting electrons: $\mathcal{H} = \sum_{j=0}^{N} \frac{p_{i}^2}{2m} - \frac{Ze^2}{4\pi\epsilon_{0}r_{i}} + \frac{1}{2} \sum_{i,j , i\neq j } \frac{1}{4\pi\epsilon_{0}}\frac{1}{|\vec{r}_{i} - \vec{r}_{j}|}$

That means for X = $ \int dr^3_{1}...dr^3_{N} \bigg \{\phi_{1}^*(\vec{r}_{1}) ...\phi_{N}^*(\vec{r}_{N})\bigg[ \sum_{i}\frac{-h^2}{2m}\Delta_{i} - \frac{Ze^2}{4\pi\epsilon_{0}r_{i}} + \frac{1}{2} \sum_{i,j , i\neq j } \frac{1}{4\pi\epsilon_{0}}\frac{1}{|\vec{r}_{i} - \vec{r}_{j}|} \bigg] \phi_{1}(\vec{r}_{1}) ...\phi_{N}(\vec{r}_{N}) \bigg\} - \sum_{j=0}^{N} \epsilon_j \cdot \big[\int |\phi_j(\vec{r})|^2 d^3r - 1\big]$.

If we perform the derivative as above we get the Hartree equations:

$\bigg[ \frac{-h^2}{2m}\Delta_{i} - \frac{Ze^2}{4\pi\epsilon_{0}r_{i}} \bigg] \phi_{i}(\vec{r}_{i}) + \sum_{i\neq j} \int d^3r_{j} \phi_{j}^*(\vec{r}_{j})\frac{e^2}{4\pi\epsilon_{0}|\vec{r}_{i} - \vec{r}_{j}|}\phi_{j}(\vec{r}_{j})\phi_{i}(\vec{r}_{i}) = \epsilon_{i}\phi_{i}(\vec{r}_{i}) $. The integral describes the average potential.

  • The Imaginary part of a function is not the same as the complex conjugate of a function. The complex conjugate still contains the real part of the function. The functional derivative with respect to the complex conjugate function holds the same information as the derivative with respect to the "normal" function in this case while taking the derivative with respect to only the imaginary part would be a wholly different thing. – Hans Wurst May 17 '21 at 20:07
  • Sorry, you are right, I mixed them up. What i ment was of course the complex conjugate. I agree that the functional derivative with respect to the complex conjugate function holds the same information as the derivative with respect to the "normal" function, but in this case we consider the wave function and its complex conjugate to be two independent entities. – ibroketheinternet May 17 '21 at 20:15
  • I'll leave this link here as this question is related and the answers given might help, https://physics.stackexchange.com/questions/89002/why-treat-complex-scalar-field-and-its-complex-conjugate-as-two-different-fields – Hans Wurst May 17 '21 at 20:49

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