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Given the energy functional $$E[\Psi] = \frac{\langle \Psi \vert H \vert \Psi \rangle}{\langle \Psi \vert \Psi \rangle},$$ its functional gradient is $$\frac{\delta E[\Psi]}{\delta \langle \Psi \vert}=\frac{H\vert \Psi \rangle -E[\Psi]\vert \Psi \rangle}{\langle \Psi \vert \Psi \rangle}.$$

I do not understand how to obtain this expression. What it the rule to evaluate a functional gradient of the function $E[\Psi]$?

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  • $\begingroup$ You might get better help if you detailed how you understand the derivative of a scalar involving vectors with respect to a vector. $\endgroup$ – Cosmas Zachos Jun 17 '18 at 20:35
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    $\begingroup$ Is this taken from a reference? $\endgroup$ – Qmechanic Jun 17 '18 at 20:48
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OP's formula seems to be the natural functional generalization of the partial derivative $$\frac{\partial E(z,z^{\dagger})}{\partial z^{\dagger}}~\stackrel{(2)}{=}~\frac{(H-E(z,z^{\dagger}))z}{z^{\dagger}z},\tag{1}$$ where $$E(z,z^{\dagger})~=~\frac{z^{\dagger}Hz}{z^{\dagger}z},\tag{2}$$ and where the variables $z$ and $z^{\dagger}$ are treated as independent.

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  • $\begingroup$ thank you a lot! You're right..I just got puzzled foolishly $\endgroup$ – Galuoises Jun 18 '18 at 21:06

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