1
$\begingroup$

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 $E[\Psi]$?

$\endgroup$
2
  • 1
    $\begingroup$ You might get better help if you detailed how you understand the derivative of a scalar involving vectors with respect to a vector. $\endgroup$ Commented Jun 17, 2018 at 20:35
  • 1
    $\begingroup$ Is this taken from a reference? $\endgroup$
    – Qmechanic
    Commented Jun 17, 2018 at 20:48

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ thank you a lot! You're right..I just got puzzled foolishly $\endgroup$
    – Galuoises
    Commented Jun 18, 2018 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.