Download Lab 2 - Computational Materials

Electronic Structure Calcula/ons: Density Func/onal Theory and Predic/ng Cataly/c Ac/vity Review •  Examples of experiments that do not agree with classical physics: –  Photoelectric effect •  Photons and the quan/za/on of light –  Double slit experiment •  Wave par/cle duality •  Introduc/on to /me-­‐independent Schrödinger Equa/on which formalizes these observa/ons in solving the following eigenvalue problem for the total energy of a system: # ! 2 ∂2
&
+V (r)( Ψ(r) = Etotal Ψ(r)
%−
2
$ 2m ∂r
'
Ĥ Ψ(r) = Etotal Ψ(r)
Review: 1D Par/cle in a box We solved the Schrödinger Equa/on for 1D par/cle in a box … Ĥ Ψ = EΨ
! 2 ∂2
−
Ψ(r) = Etotal Ψ(r)
2
2m ∂r
r By Benjamin D. Esham (bdesham) -­‐ Created by bdesham in Inkscape., Public Domain, hQps://commons.wikimedia.org/w/index.php?curid=3057717 And found the following set of solu/ons… " nπ r %
Ψ n (r) = Asin $
'
# L &
nπ !
En =
2
2mL
2
r By Papa November -­‐ Own work by uploader. Created using inkscape func/on ploQer extension, CC BY-­‐
SA 3.0, hQps://commons.wikimedia.org/w/index.php?curid=7724946 2
2
3D Par/cle in a box Next, let’s write down the Schrödinger Equa/on for a systems with more than one dimension. First we will consider the 3D par/cle in a box: # ! 2 ∂2
&
+ V (r)( Ψ(r) = Etotal Ψ(r)
%−
2
$ 2m ∂r
'
# !2 2
&
∇ + V (x, y, z)( Ψ(x, y, z) = Etotal Ψ(x, y, z)
%−
$ 2m
'
In the second equa/on we introduce the where
Laplacian, ∇2. This is needed to define the kine/c 2
2
2
∂
∂
∂
∇2 = 2 + 2 + 2
∂x
∂y ∂z
This is known as the Laplacian energy for a par/cle free to move in Cartesian space (3D, required for most real world physical systems). It sums the components of the kine/c energy from mo/on in the x, y, and z direc/ons. 3D Par/cle in a box This leads to the Schrödinger Equa/on for a 3D par/cle in a box highlighted in green in the lower right hand corner. # ! 2 ∂2
&
+ V (r)( Ψ(r) = Etotal Ψ(r)
%−
2
$ 2m ∂r
'
# !2 2
&
∇ + V (x, y, z)( Ψ(x, y, z) = Etotal Ψ(x, y, z)
%−
$ 2m
'
# !2 2
&
where
2
2
2
∂
∂
∂
∇ = 2+ 2+ 2
∂x
∂y ∂z
2
This is known as the Laplacian ∇ + 0 ( Ψ(r) = Etotal Ψ(r)
%−
$ 2m
'
!2 2
−
∇ Ψ(r) = Etotal Ψ(r)
2m
Hydrogen Atom Next we will consider our first real chemical system, the hydrogen atom. How would you write the total energy of a hydrogen atom with classical physics? Etotal = V + T
Poten/al Energy (V): Coulomb’s Law q1q2
V = ke
r
Kine/c Energy (T): 1
1
2
2
T = me ve + m p v p
2
2
Hydrogen Atom electron -­‐ q2=-­‐e + q1=+e proton Note that the charge on proton will be +e and the charge on an electron will be –e. Hydrogen Atom Now, we can write down the total energy using quantum mechanics: # ! 2 ∂2
&
+ V (r)( Ψ(r) = Etotal Ψ(r)
%−
2
$ 2m ∂r
'
2
2
# !2
&
!
e
2
2
%% −
(( Ψ(xe , ye ...z p ) = Etotal Ψ(xe , ye ...z p )
∇e −
∇ p − ke
2m p
r(xe ,..z p ) '
$ 2me
Kine/c energy electron Kine/c Poten/al energy Energy: proton r can be wriQen in terms of the cartesian coordinates of the electron and proton Hydrogen Atom Then, to solve this problem we can convert to polar coordinates. If you are not familiar with polar coordinates, that is OK; you do not need grasp this to understand the solu/ons! Just know that the equa/on below is equivalent to the equa/on on the previous slide. And that the system is now described by distance between the proton and electron (r), and the angle between the vector poin/ng to the electron from the proton and two reference vectors (Φ and Θ). # !2 2
e2 &
∇ − ke ( Ψ(r, θ , φ ) = Etotal Ψ(r, θ , φ )
%−
r '
$ 2m
The laplacian in the equa/on above captures the kine/c energy of both the electron and proton in polar coordinates… Another Example: Hydrogen Atom Solu/ons Then, we can solve the Schrödinger Equa/on for the hydrogen atom and get the following solu/ons (Note: I have simplified these solu/ons and not included all terms): General Solu+ons First, the energy levels found matched perfectly with "1%
En = −C $ 2 '
experimental results of the hydrogen atom, an #n &
experiment performed by Niels Bohr. The solu<on for ψ n,l,m = Rn,l (r)ϒ l,m (θ , φ ) the energy is highlighted in blue. The variable, C, n
Rn,l (r) = e−αr ∑ ak r k
k=l+1
Where Υ represents the spherical harmonics– the equa<on is long so I have le? it out! hQp://www.feynmanlectures.caltech.edu/III_19.html represents all constants lumped into one, and n is the principle quantum number. Another Example: Hydrogen Atom Solu/ons General Solu+ons "1%
En = −C $ 2 '
#n &
ψ n,l,m
Next, the solu<ons for the corresponding wavefunc<ons are shown in blue. Do not worry = Rn,l (r)ϒ l,m (θ , φ ) about the equa<ons shown. No<ce quantum numbers; n, l, m! n
−α r
k
Rn,l (r) = e
∑a r
k
k=l+1
Where Υ represents the spherical harmonics– the equa<on is long so I have le? it out! hQp://www.feynmanlectures.caltech.edu/III_19.html Another Example: Hydrogen Atom Solu/ons General Solu+ons "1%
En = −C $ 2 '
#n &
The quantum numbers n, l, and m are the same quantum numbers you have seen in general chemistry ψ n,l,m = Rn,l (r)ϒ l,m (θ , φ ) n=principle quantum number =1,2,3… l=angular momentum quantum number =0,1,2..(n-­‐1) n
m=magne/c quantum number=-­‐l,-­‐(l-­‐1),…0…(l-­‐1),l Rn,l (r) = e−αr
ak r k
s=spin quantum number=±1/2 k=l+1
Where Υ represents the spherical Addi/onally there is a spin quantum number, s. This harmonics– the equa<on is long comes from the Pauli Exclusion principle states that so I have le? it out! no more than two fermions (e.g. electrons) can occupy the same quantum state. The par/cles have hQp://www.feynmanlectures.caltech.edu/III_19.html ½ opposite spin. ∑
Another Example: Review of quantum number rules from general chemistry n l m s/ħ Allowable states in subshell Allowable states in complete shell 1 0 0 ±1/2 2 2 2 0 0 ±1/2 2 1 -­‐1 ±1/2 0 ±1/2 1 ±1/2 0 0 ±1/2 1 -­‐1 ±1/2 0 ±1/2 1 ±1/2 -­‐2 ±1/2 -­‐1 ±1/2 0 ±1/2 1 ±1/2 2 ±1/2 3 2 6 8 2 6 18 10 Another Example: Hydrogen Atom Wavefunc/on Solu/ons Hydrogen Atom: Takeaways •  The agreement between the results from the Schrödinger Equa/on of the hydrogen atom with experimental results was a huge break through for quantum mechanics. •  These results confirmed that the Schrödinger Equa/on could be used to describe the physical world. •  The results found in this work are also influencing the way you think and learn about chemical systems today. •  Moving forward it makes sense to try and find similar results for larger chemical systems… What is the total energy for H2? How would you write the total energy of H2 with classical physics? Etotal = V + T
Kine/c Energy (T): 1
1
2
2
T = me (ve1 + ve2) + mn (vn21 + vn22)
2
2
Hydrogen Molecule e2 -­‐ e1 -­‐ n1 + n2 + Poten/al Energy (V): e1e2
n1n2
n1e1
n1e2
n2 e1
n2 e2
V = ke
+ ke
− ke
− ke
− ke
− ke
r
r
r
r
r
r
In quantum mechanics we can write the total energy as… 2
2 2
$ 2
2
2
2
2
2 '
"
"
!
!
e
e
e
2
2
&∑−
)
∇
+
−
∇
+
k
+
k
−
k
Ψ(
r
)
=
E
Ψ(
r
)
∑
∑
∑
i
i
e
e
e
total
&
)
2mnuclei
re1e2
rn1n2 i=1 j=1 rnie j (
i=1
% i=1 2melectron
Increasing dimensionality •  The logically next step would be to solve this equa/ons for larger systems like He, H2+, or H2 •  Amazingly, there are no analy/cal solu/ons to the Schrödinger Equa/on for any of the chemical systems listed above!! •  In order to solve this equa/on for high dimensional systems, we will need to make some approxima/ons.. Increasing dimensionality It is easy to imagine how complicated Schrödinger's Equa/ons gets when we increase the dimensionality. First let’s write down the number degrees of freedom (DOF) for a chemical system with N nuclei with charge +Ze and E electrons with charge -­‐e: DOF = 3N + 3E
The DOF were determined by considering that x, y, and z coordinates are needed to describe the loca/on of each par/cle. This is the DOF needed to describe wave func/on. Increasing dimensionality Here is a general equa/on for a chemical system with N nuclei with charge +Ze and E total electrons with charge -­‐e: 2
N
E
E
N
N
N E
$E
2
2
2
2 2
2 '
"
"
!
!
e
Z
e
Ze
2
&∑−
)
∇
+
−
∇
+
k
+
k
−
k
Ψ(
r
)
=
E
Ψ(
r
)
∑
∑
∑
∑
∑
∑
∑
e
p
e
e
e
total
&
)
2m
2m
r
r
r
e
p
e1e2
n1n2
ne (
n=1
e1 =1 e2 =e1 +1
n1 =1 n2 =n1 +1
n=1 e=1
% e=1
Kine/c Kine/c Poten/al Poten/al Poten/al energy energy Energy: Energy: Energy: Electrons Protons electron-­‐
nuclei-­‐
Electron-­‐
nuclei nuclei electron interac/ons interac/ons interac/ons E + N + ~0.5(E2) + NE = total # of terms ~0.5(N2) + in Hamiltonian! Born Oppenheimer Approxima/on •  All atoms are composed of electrons and protons. How does the mass of an electron compare to the mass of a proton? mp
~ 1800
me
•  Should higher mass objects move faster or slower than lower mass objects? –  Higher mass objects more slower. –  The large difference in mass allows us to assume that electrons can respond almost instantaneously to the mo/on nuclei. Born Oppenheimer Approxima/on •  Instead of trying to solve Schrödinger's equa/on while relaxing all degrees of freedom (electrons and nuclei), fix the nuclei in place and solve for the loca/on of the electrons (or electronic wavefunc/on). •  With the electronic wavefunc/on, we can get the total energy for the fixed posi/ons of the nuclei and forces on the nuclei at those posi/ons. •  The set of solu/ons for the energy for various loca/ons of the nuclei will produce the poten<al energy surface Born Oppenheimer Approxima/on With the Born Oppenheimer approxima/on, here are the terms we can eliminate from the Hamiltonian... 2
N
E
E
N
N
N E
$E
2
2
2
2 2
2 '
"
"
!
!
e
Z
e
Ze
2
&∑−
)
∇
+
−
∇
+
k
+
k
−
k
Ψ(
r
)
=
E
Ψ(
r
)
∑
∑
∑
∑
∑
∑
∑
e
p
e
e
e
total
&
)
2m
2m
r
r
r
e
p
e1e2
n1n2
ne (
n=1
e1 =1 e2 =e1 +1
n1 =1 n2 =n1 +1
n=1 e=1
% e=1
Kine/c Kine/c Poten/al Poten/al Poten/al Constants! energy energy Energy: Energy: Energy: Electrons Protons Electron-­‐
Nuclei-­‐
electron-­‐
Nuclei Nuclei electron interac/ons Interac/ons Interac/ons E + N + 0.5(E2) + 0.5(N2) + NE = total # of terms in Hamiltonian! 0 0 Born Oppenheimer Approxima/on Which leads to the following simplified equa/on… 2
E
E
N E
$E
2
2
2 '
"
"
!
e
Ze
&∑−
)
∇ e + ∑ ∑ ke
− ∑∑ ke
Ψ(r ) = Etotal Ψ(r )
&
)
re1e2 n=1 e=1 rne (
e1 =1 e2 =e1 +1
% e=1 2me
Kine/c energy Electrons E Poten/al Poten/al Energy: Energy: Electron-­‐
electron-­‐
electron Nuclei interac/ons Interac/ons NE + 0.5(E2) + = total # of terms in Hamiltonian! Now the wavefunc/on, ψ(r), only describes the posi/ons of the !
electrons. So r is now the posi/ons of all E electrons. Born Oppenheimer Approxima/on With the Born-­‐Oppenheimer approxima/on, we can remove the nuclei as degrees of freedom from the wave func/on. DOF = 3N + 3E
DOF = 3E
This is s<ll a lot of degrees of freedom! If we wanted to solve this for a 100 atom pla<num nanopar<cle, there would be 23,000 degrees of freedom! Hartree Product •  The dimensionality of the electronic wave func/on is s/ll very high! •  First we can write the wave func/on for all electrons as a product of individual wave func/ons: ψ (r1, r2 ...rN ) = φ1 (r1 )φ2 (r2 )...φ N (rN )
•  Where φ is the wave func/on for an individual electron •  This is known as the Hartree Product and is used in many electronic structure methods. Electron Density •  Next, recall the following property of the wave func/on: –  The probability that E electrons are at a par/cular set of coordinates r1,r2,…rE is propor/onal to |Ψ(r1,r2,..rE)|2 •  Another way of describing this is the density of electrons at a par/cular posi/ons in space, notated as n(r): E
n(r) = n(x, y, z) = ψ * (x1, y1,...zE )ψ (x1, y1,...zE ) = 2∑φi* (x, y, z)φi (x, y, z)
i=1
•  In the above equa/on, we include the individual electron wave func/ons, φ, from the Hartree Product and the total electronic wave func/on, Ψ. •  We can write the electron density with only 3 degrees of freedom. Density Func/onal Theory •  Hohenberg and Kohn no/ced that using the electron density could reduce the degrees of freedom dras/cally from 3E to 3 and proved the following two theorems. •  Theorem 1: There exists a one-­‐to-­‐one mapping between the ground state electron density and the ground state electronic wave func<on n(r) = ψ * (r1, r2 ,...rN )ψ (r1, r2 ,...rN )
Θ[n(r)] = ψ (r1, r2 ,...rN )
•  Θ[n(r)] is known as a func/onal Density Func/onal Theory •  Hohenberg and Kohn Theorem: “The electron density that minimizes the energy of the overall func/onal is the true electron density corresponding to the full Schrödinger Equa/on.” •  The Varia/onal principle can be used: E[Ψ trial ] ≥ E[Ψ 0 ]
where Ψ0 is the ground state wave func/on •  With these theorems, we can rewrite Schrödinger's Equa/on in terms of the electron density. This means we have reduced the 3N dimensions to only 3! •  And if we minimize the overall energy from the electron density, we can find the ground state wave func/on. Density Func/onal Theory Hohenberg and Kohn rewrote the Schrödinger Equa/on in terms of the energy and wavefunc/on of one electron: # !2 2
&
∇ + V (r) + VH (r) + VXC (r)(φi (r) = Eiφi (r)
%−
$ 2m
'
Electron Electron Addi/onal Kine/c Electron correc/onal nuclei energy interac/on– term– Electron dependent beyond the on electron scope of density, n(r)! lecture = 3 + 1 + 1(read above) + few total # of terms in one electron Hamiltonian! Increasing dimensionality With the Born-­‐Oppenheimer approxima/on, we can remove the nuclei as degrees of freedom.. Wri/ng the Schrödinger Equa/on in terms of individual electrons and the electron density instead of the wave func/on… DOF = 3E
DOF = 3
Going back to the 100 atom pla<num nanopar<cle, we have reduced 23,000 degrees of freedom to 3. General procedure for calcula+ng the electronic wave func+on with fixed nuclei posi+ons with DFT 1.  Define an ini/al trial electron density, n(r). 2.  Solve the following equa/ons defined using the trial electron density to find the single-­‐par/cle wave func/ons, φi(r): # !2 2
&
−
∇
+
V
(r)
+
V
(r)
+
V
(r)
%
(φi (r) = Eiφi (r)
H
XC
$ 2m
'
Note: VH depends on n(r) which we guessed! 3.  Calculate the new electron density defined by the Kohn–Sham single par/cle wave func/ons from step 2 E
nnew (r) = 2∑φi* (r)φi (r)
i=1
4.  Compare the calculated electron density, n(r), from steps 1 and 3. If the two densi/es are the same, then this is the ground-­‐
state electron density, and it can be used to compute the total energy. If the two densi/es are different, then the trial electron density must be updated in some way. General Procedure for Op/miza/on The general procedure for numerical geometry op/miza/on with an empirical poten/al is as follows (this is what you did for lab 1-­‐partA): 1.  Calculate the force (or nega/ve of the gradient) on all atoms for some configura/on of an atomic system. 2.  If the force is less than threshold, you have found a cri/cal point! STOP. 3.  If not, move the atoms such that they go towards a cri/cal points 4.  Repeat. General Procedure for Op/miza/on with Density Func/onal Theory The general procedure for numerical geometry op/miza/on with DFT is as follows: 1.  Solve for op/mal posi/ons of electrons with the general procedure for electronic wave func/on op/miza/on. Note: see a few slides back for details and recall that the nuclei are fixed in this calcula<on. 2.  Calculate the force (or nega/ve of the gradient) on all nuclei for some configura/on of an atomic system. 3.  If the force is less than threshold, you have found a cri/cal point! STOP. 4.  If not, move the atoms such that they go towards a cri/cal points 5.  Repeat. This concludes the introduc/on to electronic structure calcula/ons: Ques/ons? Now, let’s briefly talk about catalyst Review of Fuel Cells Anode (oxida/on—loss of electrons): 2H2à4H++4e-­‐ Cathode (reduc/on—gain of electrons) O2+4H++4e-­‐ à2H2O Overall reac/on (redox): 2H2 + O2 à 2H2O We will par/cularly interested in the oxygen reduc/on reac/on (ORR) in this class O2+4H++4e-­‐ à2H2O Poten/al Energy ORR example mechanism Overall Fuel Cell Reac/on 2H2 + O2 à 2H2O We can use the energe<cs of one of the intermediate states in this mechanism and predict cataly<c ac<vity from this! See the next slide for data. Reac/on Mechanism Volcano Plots By calcula/ng the binding energy of oxygen (x-­‐axis below, blue box in the previous slide, and equa/on on the next slide), we can predict the cataly/c ac/vity of a material. The idea behind this plot is that we do not want the catalyst to interact too strongly or too weekly with substrate, but find the op/mal trade off between the two. optimal
reactive
metals
strong
binding
noble
metals
high
barrier
Binding Energy •  The x axis in the volcano plot is the binding energy of oxygen •  This is the energy required to separate a whole into parts •  For example, if we consider oxygen binding to a 32 atom nanopar/cle –  E_binding = E(Pt32 + O) – E(Pt32)-­‐1/2 E(O2) Cohesive Energy •  Energy gained from individual atoms forming nanopar/cle •  Gives quan/ty measuring rela/ve stabili/es of nanopar/cles •  For example if you had a 32 atom pla/num par/cle: –  E_cohesive = E(Pt32) – 32*E(Pt-­‐atom) References •  Introduc<on to Quantum Mechanics –  By Sy. M. Binder –  Publisher: Elsevier Science –  ISBN: 9780080489285 (Ebook) –  This book can be viewed online through UT library •  Density Func<onal Theory: A Prac<cal Introduc<on –  By David S. Sholl and Janice A. Steckel –  Publisher: Wiley –  ISBN 978 0 470 37317 0 (cloth) –  This book can be viewed online through UT library Suggested Reading •  Sec/ons 1.3 and 1.4 in Density Func<onal Theory: A Prac<cal Introduc<on