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I’ve been studying the KKR method from the original Kohn and Kostoker’s paper (https://journals.aps.org/pr/abstract/10.1103/PhysRev.94.1111). On the text, they use variational calculus for dealing with an integral equation. Namely, the paper states that the integral equation (2.14):

$\psi(\vec{r})=\int G(\vec{r},\vec{r}_0)V(\vec{r}_0)\psi(\vec{r}_0)d^3r_0$

Is equivalent to:

$\delta\Lambda=0$

Where:

$\Lambda=\int \psi^*(\vec{r})V(\vec{r})\psi(\vec{r})d^3r-\int \psi^*(\vec{r})V(\vec{r})G(\vec{r},\vec{r}_0)V(\vec{r}_0)\psi(\vec{r}_0)d^3rd^3r_0$

However, my understanding of variational techniques is that you are supposed to minimize (or extremize) an action integral of a lagrangean. And that the result turns out to be that the integral path for which the action is an extreme is a solution to an Euler-Lagrange equation.

But I’m not able to see how this applies to an integral equation like this, or how the expression for $\Lambda$ appears. The fact that the authors proceed to extremize $\Lambda$ cofuses me even more, since some textbooks call this function the “lagrangean”.

How does the variational method applies on situations like this?

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However, my understanding of variational techniques is that you are supposed to minimize (or extremize) an action integral of a lagrangean.

You can minimize whatever you want.

See, for example, Chapter 17 (Calculus of Variations) of Mathematical Methods For Physicists (4th Edition) by Arfken and Weber. In particular section 17.1 where is it proved that the integral: $$ J=\int f(y,y;x)dx $$ is stationary for: $$ \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'} = 0\;. $$

No mention of Lagrangians whatsoever in that entire Section!

How does the variational method applies on situations like this?

It applies because making the integral stationary results in the desired differential equation, in particular, since there's no $\psi'$ term, and since we can treat $\psi$ and $\psi^*$ independently: $$ \frac{\partial }{\partial \psi(x)} \left( \int\psi^*V\psi-\int\int\psi*VGV\psi \right)=0 \to $$ $$ \psi^*(x)V(x)-\int \psi^*(r)V(r)G(r,x)V(x)=0 \to $$ $$ \psi(\vec x)=\int d^3r \psi(\vec r)V^*(\vec r)G^*(r,x)=0 $$

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  • $\begingroup$ Hi, thanks a lot for your answer! May I ask you how this derivative is made? I mean, how the derivative of the wave function eliminates the integral, which is on the space coordinates? $\endgroup$ – Lucas Francisco Jun 27 '18 at 3:40
  • $\begingroup$ It's a variational derivative, I used the symbol $\partial$, but it should be understood it's a variational derivative in the calculus or variations sense... I can explain more later, but not at the moment. $\endgroup$ – hft Jun 27 '18 at 3:58

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