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

  1. Apr 11, 2017 #1

    HAYAO

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    I have a very fundamental question about the linear variational method (Huckel theory).

    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 of ε by Ci 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.
     
    Last edited: Apr 11, 2017
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  3. Apr 11, 2017 #2

    PeterDonis

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    Correct. You have to check the second derivative to see whether it's a minimum, maximum, or saddle point.
     
  4. Apr 11, 2017 #3

    HAYAO

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    Oh, okay. So then why, in Huckel method, do they say "minimum can be obtained from derivative = 0"? Is this empirical? Or is there a generalized proof?
     
  5. Apr 11, 2017 #4

    PeterDonis

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    Can you give a specific reference where this claim is made?
     
  6. Apr 11, 2017 #5

    HAYAO

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    Page 84 of "Quantum Chemistry: Fundamentals to Applications" by Tamás Veszprémi and Miklós Fehér.

    or from equation (10.6.9) from Chemwiki
     
  7. Apr 11, 2017 #6

    PeterDonis

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    I don't see anything in either of these references that proves that the first derivative being zero guarantees that it is a minimum. I suspect they are simply leaving out the part where they check the second derivative and verify that they have found a minimum.
     
  8. Apr 11, 2017 #7

    HAYAO

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    Yes indeed, which is why I was wondering why they are saying this without any proof. In fact, I want to know if it is possible to check the second derivative. It seems to me like a tedious work.
     
    Last edited: Apr 11, 2017
  9. Apr 12, 2017 #8

    DrClaude

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    In practice, because of the way the trial wave functions are constructed, the probability of ending up anywhere else than close to the ground state. In addition, the solution is found by minimization, not by finding the zeros. (The minimization method will surely involve calculating the Jacobian, so it is easy to check that the solution found corresponds to a minimum.)
     
  10. Apr 12, 2017 #9

    HAYAO

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    I'll be honest I don't know much about mathematical methods for minimization. However, I do know that a solution is found by minimization. That is pretty much the same for many methods, e.g. Hartree-Fock method.

    The problem I am having is the explanation used in above links. They do not explicitly do the actual procedure of using trial functions but instead says that they can provide secular determinant from the condition that a minimum can supposedly be met when [itex]\frac{d\varepsilon }{dC_{i}} = 0[/itex], which can be used for the Huckel Method. I feel like there is a leap in logic because [itex]\frac{d\varepsilon }{dC_{i}} = 0[/itex] is not always a minimum, provided that we don't know exactly what [itex]\varepsilon = \frac{\sum_{i,j}^{n}C_{i}C_{j}H_{ij} }{\sum_{i,j}^{n}C_{i}C_{j}S_{ij}}[/itex] is. I wanted to know the mathematical explanation why this is always true for LCAO variational method.
     
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