Deriving the Solution to $$\frac{\partial}{\partial \phi^{*}}$$

[Irishdoug]

Homework Statement

We would like to find the lowest energy tight binding wavefunction of the form $\ket{\Psi} = \sum_{n} \phi_{n} \ket{n}$

Relevant Equations

$H \phi = E \phi$

$\phi$ is the vector of N co-efficients and H is the N by N matrix $H_{n,m} = \bra{n} H \ket{m}$

We construct the energy $E = \frac{\bra{\psi} H \ket{psi}}{\bra{\psi}\ket{\psi}}$ and minimise it with respect to each $\phi_{n}$ to reproduce the eigenvalue equation $H \phi = E \phi$

The solution can be viewed here on page 41

https://usermanual.wiki/Document/St...ford20University20Press202015.1463186034/view

What I have is

$$\frac{\partial}{\partial \phi^{*}} (\frac{\sum_{n,m} \phi_{n}^{*} H_{n,m}\phi_{n}} {\sum_{n} \phi_{n}^{*} \phi_{n}}) = 0$$

I have access to the solution but I have no idea how this derivative is carried out. I have tried the product rule (where $\phi_{n}$ is considered constant) and quotient rule but neither work to give this solution:

$$\frac{\sum_{n} \phi_{m} H_{n,m}}{\sum_{p}\phi_{p}^{2}} - (\frac{\sum_{n,m} \phi_{n}^{*} H_{n,m}\phi_{n}} {\sum_{n} \phi_{n}^{*} \phi_{n}}) \frac{\phi_{n}}{\sum_{p}\phi_{p}^{*}\phi_{p}} = 0$$

Thankyou for your help.

 

[vela]

Please show your work. It isn't very helpful to say "I tried X, but it didn't work." Think about how helpful it would be to you if someone replied with only "Well, I tried X and it did work."

What did you get when you differentiated the numerator with respect to $\phi_n^*$?

 

[ergospherical]

The notation is messy and prone to mistakes. Write instead $\mathbf{x} = (\phi_1, \dots, \phi_n)^T$ as a column vector containing the numbers $\phi_n$ and write the energy as\begin{align*}
E(\mathbf{x}) = \frac{\mathbf{x}^{\dagger} H \mathbf{x}}{\mathbf{x}^{\dagger} \mathbf{x}}
\end{align*}Now you vary the vector $\mathbf{x} \mapsto \mathbf{x} + \delta \mathbf{x}$, where $\delta \mathbf{x} = (\delta \phi_1, \dots, \delta \phi_n)^T$, giving
\begin{align*}
\delta E \equiv E(\mathbf{x} + \delta \mathbf{x}) - E(\mathbf{x}) = \frac{(\mathbf{x}^{\dagger} + \delta \mathbf{x}^{\dagger})H(\mathbf{x} + \delta \mathbf{x})}{(\mathbf{x}^{\dagger} + \delta \mathbf{x}^{\dagger})(\mathbf{x} + \delta \mathbf{x})} - E(\mathbf{x})
\end{align*}Can you complete the proof by imposing that $\delta E$ vanishes to first order for all possible $\delta \mathbf{x}$?