I have a very fundamental question about the linear variational method (Huckel theory).(adsbygoogle = window.adsbygoogle || []).push({});

It says in any textbook that the variational method provides energy upper bound to the actual energy of a wavefunction by using test wavefunction.

[tex]\varepsilon = \frac{\sum_{i,j}^{n}C_{i}C_{j}H_{ij} }{\sum_{i,j}^{n}C_{i}C_{j}S_{ij}}[/tex]

This means that the derivative ofεby C_{i}is 0, which would provide the minimum.

However, the right side of the equation above is a quadratic/quadratic rational function. If I understand correctly, depending on the numerator and denominator, it can also have a maximum (or might not even have any). So [itex]\frac{d\varepsilon }{dC_{i}} = 0[/itex] does not necessarily provide the minimum.

I am guessing that I am referring to a general case in mathematics, rather than special case with several constraints like in quantum mechanics. So then how do we know in this "special case" that [itex]\frac{d\varepsilon }{dC_{i}} = 0[/itex] will always give the minimum?

This might be a very stupid noob question, but it was something I took for granted and never really thought about.

EDIT: In fact, actual calculation does indeed reach to an adequately right result. This is why I took it for granted. But, I wish to know why obtaining the minimum is a necessary condition.

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# I Derivative = 0 is always minima? (Linear variational method)

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