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Chemical potential-fictitious atoms

  1. May 10, 2009 #1
    Hello,

    I am sorry to ask a relatively unrelated question. I need to know if the chemical potential of fictitious atoms like hydrogen used to saturate the dangling bonds is different from the chemical potential of hydrogen in common chemical potential tables.

    Regards,
    Sarah
     
  2. jcsd
  3. May 10, 2009 #2

    alxm

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    It's definitely different. Stealing some text from http://en.wikipedia.org/wiki/Chemical_potential#Electronic_chemical_potential"
    [tex]\mu(\mathbf{r})=\left[ \frac{\delta E[\rho]}{\delta \rho(\mathbf{r})}\right]_{\rho=\rho_{\mathrm{ref}}}[/tex]

    Formally, a functional derivative yields many functions, but is a particular function when evaluated about a reference electron density - just as a derivate yields a function, but is a particular number when evaluated about a reference point. The density functional is written as

    [tex]E[\rho] = \int \rho(\mathbf{r})\nu(\mathbf{r})d^3r + F[\rho][/tex]

    where [tex]\nu(\mathbf{r})[/tex] is the external potential, e.g., the electrostatic potential of the nuclei and applied fields, and F is the Universal functional, which describes the electron–electron interactions, e.g., electron Coulomb repulsion, kinetic energy, and the non-classical effects of exchange and correlation. With this general definition of the density functional, the chemical potential is written as

    [tex]\mu(\mathbf{r}) = \nu(\mathbf{r})+\left[\frac{\delta F[\rho]}{\delta\rho(\mathbf{r})}\right]_{\rho=\rho_{\mathrm{ref}}}[/tex]

    And [tex]\nu(\mathbf{r}) = \frac{Z}{|\mathbf{R}-\mathbf{r}|}[/tex]. So since Z=1 for ordinary hydrogen but some different number for your fictitious hydrogens, they must differ by at least that term.
     
    Last edited by a moderator: Apr 24, 2017
  4. May 11, 2009 #3
    Thanks for the very useful reply.

    Now, according to the formula of chemical potential you have kindly mentioned and the information we have from the pseudopotential of the fictitious "H", is there a way to calculate the derivative of "F" and then calculate the chmical potential of the "H"

    Also, if we calculated the chemical potential, could it be used for calculating the formation energy? Or in the formation energy real atoms are considered?
     
  5. May 11, 2009 #4

    alxm

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    Well, that'd be the usual methods for determining a http://en.wikipedia.org/wiki/Functional_derivative" [Broken] (the WP article conveniently includes some exact values of [tex]\frac{\delta E[\rho]}{\delta\rho}[/tex] for some of the simpler approximate density-functionals.) The exact density functional is not known, though.

    Well, no, not really. And using the formula given requires that you already know the energy (E[rho]). But no quantum-chemical method I know of requires Z to be an integer.
     
    Last edited by a moderator: May 4, 2017
  6. May 11, 2009 #5
    Thanks so much again.

    In the enclosed paper, Solid State Communications 148 (2008) 101–104, they have saturated the dangling bonds with real hydrogen and they have calculated the formation energy.

    If, we have the fictitious hydrogen with its pseudopotential, as you kindly said, we do not know its exact density functional. In this case, there's no way to calculate the chemical potential or I am wrong?
     
  7. May 11, 2009 #6

    alxm

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    Well, the fact that the exact density functional is unknown doesn't mean there aren't approximations. It's what DFT theory is all about. The paper you cited, for instance, used the PW91 functional. I can't say whether their software had the feature of calculating chemical potential directly using the equation above.

    But there's a much simpler way (with the added bonus of being just as easily used with wave-based methods) Using the Mulliken definition of chemical potential, it's [tex]\mu_{Mulliken} = -1/2(IP+EA)[/tex], where IP is the ionization potential (relative energy of the system with an electron removed) and EA is the electron affinity (relative energy of the system with an electron added).
     
  8. May 11, 2009 #7
    Your answers have really helped me.

    All the questions I asked have arised from the contrast between this papare:

    http://link.aps.org/doi/10.1103/PhysRevB.77.115349

    And the paper I mentioned in the previous post !!!!

    As you see, the formation energy has been presented by putting the real H molecule energy ?!?! (while the saturation's done with fictitious ones) !!!

    Also, they have considered the energy of a Ga atom obtained from bulk Ga metal. What does it mean? do we have to divide the energy of the bulk to the number of atoms to get one atom energy?
     
  9. May 11, 2009 #8

    alxm

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    Unless I missed something, I believe the heat of formation was calculated with fractionally-charged hydrogens in the calculations where they used them (Dmol3). So it's a 'fictitious' heat of formation. That's not a problem though, because what they're interested in is the bulk property, not what's going on at the edges. Secondly because they're counting on the effect to more-or-less cancel out, because they chose pairs with charges 0.75 and 1.25, so the overall charge remains the same.

    Yes, more or less like that. Usually you calculate a single unit cell or two, with periodic boundary conditions at the 'edges' - giving you the effect of calculating on an infinitely large crystal, not a tiny model.
     
  10. May 11, 2009 #9
    This is like what we do for the cohesive energy, is that right? increasing the lattice parameteres and then calculating the energy difference. But as they have also mentioned the energy of N2 molecule it appears that they calculate the energy of the bulk for Ga metal and N2 molecule and then they divide by the number of the atoms. Mixed up!!! because they have already divided the energy of N2 by 2 but
     
  11. May 11, 2009 #10
    continuing the last message, for Ga, the number of atoms in its orthorhombic cell has not been divided?!?!?! It mixes me up.
     
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