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If I have $$E(C)\langle C|C\rangle=\langle C|H|C\rangle \tag{1}$$ where $H$ is an operator and $$|C\rangle=\sum_{i=1}^m C_i \;|i\rangle \tag{2}$$ is a function as and $\langle C|C\rangle$ simply means $\int_0^{\infty} r^2 C^\ast C dr$. By differentiating (1) How can I reach $$E_i^{(1)}(C)\langle C|C\rangle+2E(C)\langle i|C\rangle=2\langle i|H|C\rangle \tag{3}$$

J.G.
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Wisdom
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1 Answers1

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Since $\langle C|C\rangle=\sum_{ij}C_i^\ast C_j\langle i|j\rangle$,$$\partial_{C_i^\ast}\langle C|C\rangle=\sum_jC_j\langle i|j\rangle=\langle i|C\rangle.$$Differentiating $(1)$ with respect to $C_i^\ast$,$$\frac{\partial E(C)}{\partial C_i^\ast}\langle C|C\rangle+E(C)\langle i|C\rangle=\frac{\partial}{\partial C_i^\ast}\langle C|H|C\rangle.$$A similar treatment of the right-hand side shows it to be$$\frac{\partial}{\partial C_i^\ast}\sum_{hj}C_h^\ast C_j\langle h|H|j\rangle=\sum_jC_j\langle i|H|j\rangle=\langle i|H|C\rangle.$$So (3) holds with$$E_i^{(1)}(C)=\frac12\frac{\partial E(C)}{\partial C_i^\ast}.$$

J.G.
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  • Thanks a lot. If I replace $|C>$ by $e^{-ar-br^2}$ what happens? I mean a form similar to equation (3) – Wisdom Dec 21 '19 at 14:47
  • @Wisdom I think you mean to take $\langle r|C\rangle$ as proportional to $e^{-ar-br^2}$. – J.G. Dec 21 '19 at 14:49
  • No, In fact in the original problem, function is a linear combination of $|C_i>$ but for my case. the function is $e^{-ar-br^2}$ where two parameters are $a$ and $b$ – Wisdom Dec 21 '19 at 14:52
  • @Wisdom You seem to get conflating a lot of scalars with similarly named kets. – J.G. Dec 21 '19 at 14:53
  • I have to calculate the energy gradients using a general formula which led to equation (3) but my wave function is not similar to $|C>$, so how can I calculate the energy gradients? – Wisdom Dec 21 '19 at 14:56
  • @Wisdom You could try differentiating $\langle\psi|H|\psi\rangle/\langle\psi|\psi\rangle$ with respect to some parameter(s) in your wavefunction $\langle x|\psi\rangle$. – J.G. Dec 21 '19 at 15:20