TPT results

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Contents

VASP computational details

2 TPT molecules, 5 gold layers (bottom 2 frozen bulk), 5x5x1 M.-P. k mesh. PBE functional. xxx eV cutoff. XxYxZ unit cell.

Definition of angles

I define the ring angle as the angle between the normal of the (least square fitted) plane through the six carbon atoms in the ring, and the normal of the frozen gold planes. I define the peptide angle as the angle between the surface normal and the NCO plane.


Optimized 2x1 TPT VASP structure

Download .xyz file here[1] (supercell!). Two inequivalent molecules in this unit cell.

"Geometrical" angles:

Peptide plane normal/surface normal: 65.852, 65.568 degrees.

Ring plane normal/surface normal: 81.769, 77.278 degrees. Average using formula below: 79.296 (almost the same as arithmetic average).

Optimized VASP structure with waters

.xyz file of a supercell can be downloaded[2] "Geometrical" angles:

Peptide plane normal/surface normal: 64.458, 65.778 degrees.

Ring plane normal/surface normal: 78.172, 75.234 degrees. Average using formula below: 76.628.

Water binding energies

E(H20 optimized) = -14.208769 eV E(2 TPT optimized on Au) = -448.617830 eV E(2 TPT+ 2 H2O optimized on Au) = -478.387588 eV

Ebinding = (E(2 TPT+ 2 H2O on Au) - E(2 TPT+ 2 H2O on Au) - 2*E(H20 optimized))/2 = -0.67 eV/water (negative sign means that the water wants to be bound to the molecules).

I did use the same unit cell for all calculations, so the number should be comparable.

X-ray absorption

Take a number of absorption sites absorbing at the same energy, but with different dipole transition moments \bar{m}_i. Then we get a total absorption intensity proportional to

 I(\hat{\epsilon}) = \sum_i (\bar{m}_i\cdot \hat{\epsilon})^2,

where \hat{\epsilon} is the electric field polarization vector. This is a sum of (up to) quadrupoles, so the result is again a quadrupole, although possibly with three nonzero eigenvalues instead of one. If we demand cylindrical symmetry around the z axis we get two independent moments. Calculate these for the actual dipole moments to get the angle we predict from calculation. We should have

 I(\hat{\epsilon}) =  \epsilon_z^2 I_{zz} + (\epsilon_x^2 + \epsilon_y^2)I_{xx}

(note that this can include a monopole component). From a cylindrically symmetrized absorber I get

I_{xx} = \frac{1}{2}(m_x^2 + m_y^2)
I_{zz} = m_z^2

From this it follows how to add absorbers with different transition moment vectors. The procedure by Stöhr is described in [3]

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