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‎Project#12/README.md

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@@ -6,7 +6,7 @@ and time-dependent Hartree-Fock (TDHF), which is also known as the random-phase
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approximation (RPA). The purpose of this project is to develop simple
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implementations of these two methods and to understand the basic differences
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between them. This project assumes you have completed at least the
9-
[CCSD programming project](https://github.com/CrawfordGroup/ProgrammingProjects/tree/master/Project%2305).
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[CCSD programming project](../Project%2305).
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You can read more about these methods in
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[J.B. Foresman, M. Head-Gordon, J.A. Pople, and M.J. Frisch, J. Phys. Chem. 96, 135-141 (1992)](http://pubs.acs.org/doi/pdf/10.1021/j100180a030) (CIS)
@@ -18,39 +18,27 @@ The fundamental idea behind CIS is the representation of the excited-state wave
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functions as linear combinations of singly excited determinants relative to the
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Hartree-Fock reference wave function, *viz.*
2020

21-
```
22-
EQUATION
23-
\[ |\Psi(m)\rangle = \sum_{jb} c_j^b(m) \{ a^+_b a_j \} |\Phi_0\rangle = \sum_{jb} c_j^b(m) |\phi_j^b\rangle, \]
24-
```
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<img src="./figures/singly-excited-determinant.png" height="60">
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where *m* identifies the various excited states, and we will use *i* and *j*
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(*a* and *b*) to denote occupied (unoccupied) spin-orbitals. Inserting this
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into the Schr<html>&ouml;</html>dinger equation and left-projecting onto a
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particular singly excited determinant gives
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31-
```
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EQUATION
33-
\sum_{jb} \langle \phi_i^a|H|\phi_j^b\rangle c_j^b(m) = E(m) c_i^a(m).
34-
```
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<img src="./figures/excited-det-schrod-eqn.png" height="60">
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If we recognize that we have one of these equations for every combination of
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*i* and *a* spin-orbitals, then this equation may be viewed as a matrix
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eigenvalue problem:
3933

40-
```
41-
EQUATION
42-
\[ {\mathbf {\underline{\underline H}}}\ {\mathbf {\underline c}}(m) = E(m) {\mathbf {\underline c}}(m). \]
43-
```
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<img src="./figures/matrix-eigenvalue-problem.png" height="60">
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To solve this equation, we need an expression for the matrix elements in terms
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of things we already know, i.e. Fock matrix elements and two-electron
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integrals. This can be done using either algebraic or diagrammatic techniques
4839
to obtain (in the spin-orbital notation of previous projects):
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50-
```
51-
EQUATION
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\langle \phi_i^a | H | \phi_j^b \rangle = H_{ia,jb} = f_{ab} \delta_{ij} - f_{ij} \delta_{ab} + \langle aj || ib \rangle.
53-
```
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<img src="./figures/matrix-elements.png" height="60">
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Our task is then relatively simple: Build the Hamiltonian matrix (expressed in
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the basis of all singly excited determinants) using the above expression and
@@ -60,7 +48,7 @@ occupied orbitals times the number of unoccupied orbitals. For our
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STO-3G/H<sub>2</sub>O test case, with its ten occupied and four unoccupied
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spin-orbitals, the matrix will be 40 x 40.
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63-
* Hint: CIS Hamiltonian for STO-3G H<sub>2</sub>O
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* [Hint](./hints/hint1.md): CIS Hamiltonian for STO-3G H<sub>2</sub>O
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Make sure you can compute the correct CIS excitation energies for each of the
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four test cases provided at the end of the page. Before you move on to the
@@ -84,99 +72,11 @@ from a simple two-electron/two-orbital example (such as the *1s 2s* excited
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state configuration of the He atom). One can easily show that the four
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possible determinants arising from this configuration,
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87-
```
88-
EQUATION
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\[ \left|
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\begin{array}{cc}
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\uparrow & \\
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\hline
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\uparrow & \\
94-
\hline
95-
\end{array}
96-
\right\rangle,\ \
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\left|
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\begin{array}{cc}
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& \downarrow \\
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\hline
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& \downarrow \\
102-
\hline
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\end{array}
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\right\rangle,\ \
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\left|
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\begin{array}{cc}
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\uparrow & \\
108-
\hline
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& \downarrow\\
110-
\hline
111-
\end{array}
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\right\rangle,\ \
113-
\left|
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\begin{array}{cc}
115-
& \downarrow \\
116-
\hline
117-
\uparrow & \\
118-
\hline
119-
\end{array}
120-
\right\rangle,\ \ \]
121-
```
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<img src="./figures/four-possible-determinants.png" height="60">
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are components of one singlet and one triplet in the following combinations:
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125-
```
126-
EQUATION
127-
\[ |\Phi^3_1\rangle = \left|
128-
\begin{array}{cc}
129-
\uparrow & \\
130-
\hline
131-
\uparrow & \\
132-
\hline
133-
\end{array}
134-
\right\rangle,\ \
135-
|\Phi^3_{-1}\rangle = \left|
136-
\begin{array}{cc}
137-
& \downarrow \\
138-
\hline
139-
& \downarrow\\
140-
\hline
141-
\end{array}
142-
\right\rangle,\ \
143-
|\Phi^3_{0}\rangle =
144-
\frac{1}{\sqrt{2}}\left\{
145-
\left|
146-
\begin{array}{cc}
147-
\uparrow & \\
148-
\hline
149-
& \downarrow\\
150-
\hline
151-
\end{array}
152-
\right\rangle -
153-
\left|
154-
\begin{array}{cc}
155-
& \downarrow \\
156-
\hline
157-
\uparrow & \\
158-
\hline
159-
\end{array}
160-
\right\rangle\right\},\ \ {\rm and}\ \
161-
|\Phi^1_{0}\rangle =
162-
\frac{1}{\sqrt{2}}\left\{
163-
\left|
164-
\begin{array}{cc}
165-
\uparrow & \\
166-
\hline
167-
& \downarrow\\
168-
\hline
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\end{array}
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\right\rangle +
171-
\left|
172-
\begin{array}{cc}
173-
& \downarrow \\
174-
\hline
175-
\uparrow & \\
176-
\hline
177-
\end{array}
178-
\right\rangle\right\}, \]
179-
```
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<img src="./figures/singlet-triplet-combinations.png" height="60">
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where the superscript is the spin multiplicity (*2S+1*) and the subscript is
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the *M<sub>S</sub>* value of the wave function. So, if we wanted to compute
@@ -192,13 +92,7 @@ expression and the equation for the CIS Hamiltonian matrix elements in the
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previous section, we may write a spin-factored equation for the
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<html>&alpha;</html> coefficients as
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```
196-
EQUATION
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c_{i_\alpha}^{a_\alpha}(m) E(m) = \sum_{j_\alpha b_\alpha}
198-
\left[ f_{a_\alpha b_\alpha} \delta_{i_\alpha j_\alpha} - f_{i_\alpha j_\alpha} \delta_{a_\alpha b_\alpha} + \langle a_\alpha j_\alpha || i_\alpha b_\alpha \rangle \right]
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c_{j_\alpha}^{b_\alpha}(m) + \sum_{j_\beta b_\beta} \langle a_\alpha j_\beta ||
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i_\alpha b_\beta \rangle c_{j_\beta}^{b_\beta}(m).
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```
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<img src="./figures/pin-factored-eqn.png" height="60">
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Note that the mix-spin cases (where *j=*<html>&alpha;</html> and
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*b=*<html>&beta;</html> or *vice versa*) do not contribute since the Fock
@@ -207,39 +101,27 @@ carry out spin integration on the integrals in the above expression and assume
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that the <html>&alpha;</html> and <html>&beta;</html> CI coefficients are
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identical for the same spatial orbitals, i.e.,
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```
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EQUATION
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c_{i_\alpha}^{a_\alpha}(m) = c_{i_\beta}^{a_\beta}(m).
213-
```
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<img src="./figures/identical-ci-coeff.png" height="60">
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we obtain the ***spatial orbital*** expression
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we obtain the <b><i>spatial orbital</i></b> expression
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217-
```
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EQUATION
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c_i^a(m) E(m) = \sum_{jb} \left[ f_{ab} \delta_{ij} - f_{ij} \delta_{ab} + 2 <aj|ib> - <aj|bi> \right] c_j^b(m).
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```
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<img src="./figures/spatial-orbital-expression.png" height="60">
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The part in brackets above is an expression for the spatial-orbital CIS
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Hamiltonian, spin-adapted for singlet excited states, and diagonalization of
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this matrix will yield only the singlet eigenvalues you obtained from your
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spin-orbital matrix earlier.
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* Hint: Spin-adapted CIS singlet Hamiltonian for STO-3G H<sub>2</sub>O
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* [Hint](./hints/hint2.md): Spin-adapted CIS singlet Hamiltonian for STO-3G H<sub>2</sub>O
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How about the triplets? We use exactly the same spin-factorization, but
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instead require
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232-
```
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EQUATION
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c_{i_\alpha}^{a_\alpha}(m) = -c_{i_\beta}^{a_\beta}(m).
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```
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<img src="./figures/inverse-ci-coeff.png" height="60">
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This yields a slightly simpler Hamiltonian:
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239-
```
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EQUATION
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c_i^a(m) E(m) = \sum_{jb} \left[ f_{ab} \delta_{ij} - f_{ij} \delta_{ab} - <aj|bi> \right] c_j^b(m),
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```
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<img src="./figures/simpler-hamiltonian.png" height="60">
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which, upon diagonalization, will yield only the triplet eigenvalues (but each only occurring once) from your earlier diagonalziation.
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@@ -266,47 +148,23 @@ TDHF/RPA wave function expansion in terms of orbital rotations instead of
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Slater determinants, but that's a discussion for another day.) The TDHF/RPA
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eigenvalue equations take the form
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269-
```
270-
EQUATION
271-
\left(
272-
\begin{array}{cc}
273-
{\mathbf {\underline{\underline A}}} & {\mathbf {\underline{\underline B}}} \\
274-
-{\mathbf {\underline{\underline B}}} & - {\mathbf {\underline{\underline A}}} \\
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\end{array}
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\right) \left(
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\begin{array}{c}
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{\mathbf {\underline X}} \\
279-
{\mathbf {\underline Y}} \\
280-
\end{array}
281-
\right) = E \left(
282-
\begin{array}{c}
283-
{\mathbf {\underline X}} \\
284-
{\mathbf {\underline Y}} \\
285-
\end{array}
286-
\right).
287-
```
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<img src="./figures/tdhf-eqn.png" height="60">
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289153
The definition of the ***A*** matrix is just the CIS matrix itself, *viz.*
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291-
```
292-
EQUATION
293-
A_{ia,jb} \equiv \langle \phi_i^a | H | \phi_j^b \rangle = f_{ab} \delta_{ij} - f_{ij} \delta_{ab} + \langle aj || ib \rangle,
294-
```
155+
<img src="./figures/A-matrix.png" height="60">
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296157
while **X** and **Y** are the parameters of single excitations and
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de-excitations, respectively, and the ***B*** matrix is simply
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299-
```
300-
EQUATION
301-
B_{ia,jb} \equiv \langle ab || ij \rangle.
302-
```
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<img src="./figures/B-matrix.png" height="60">
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Thus, the row/column dimension of the TDHF/RPA Hamiltonian is twice that of the
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CIS Hamiltonian, and the matrix is non-symmetric (so you must be careful about
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the diagonalization function you choose). Do you obtain twice as many
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excitation energies?
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309-
* Hint: TDHF/RPA Hamiltonian for STO-3G H<sub>2</sub>O
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* [Hint](./hints/hint3.md): TDHF/RPA Hamiltonian for STO-3G H<sub>2</sub>O
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## A Better Approach to Solving the TDHF/RPA Eigenvalue Equations
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@@ -316,57 +174,28 @@ Hamiltonian storage cost), one can rearrange the eigenvalue equations. First
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write eigenvalue equation two separate equations, each in terms of the
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submatrices **A** and **B**:
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319-
```
320-
EQUATION
321-
\[ {\mathbf {\underline{\underline A}}} {\mathbf {\underline X}} + {\mathbf {\underline{\underline B}}} {\mathbf {\underline Y}} = E {\mathbf {\underline X}} \]
322-
```
177+
<img src="./figures/smarter-tdhf-1.png" height="60">
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and
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326-
```
327-
EQUATION
328-
\[
329-
-{\mathbf {\underline{\underline B}}} {\mathbf {\underline X}} - {\mathbf {\underline{\underline A}}} {\mathbf {\underline Y}} = E {\mathbf {\underline Y}}. \]
330-
```
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<img src="./figures/smarter-tdhf-2.png" height="60">
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Now take +/- combinations of these equations to obtain
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334-
```
335-
EQUATION
336-
\[
337-
({\mathbf {\underline{\underline A}}} - {\mathbf {\underline{\underline B}}})
338-
({\mathbf {\underline X}} - {\mathbf {\underline Y}})
339-
= E ({\mathbf {\underline X}} + {\mathbf {\underline Y}}) \]
340-
```
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<img src="./figures/smarter-tdhf-3.png" height="60">
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and
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344-
```
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EQUATION
346-
\[ ({\mathbf {\underline{\underline A}}} + {\mathbf {\underline{\underline B}}})
347-
({\mathbf {\underline X}} + {\mathbf {\underline Y}})
348-
= E ({\mathbf {\underline X}} - {\mathbf {\underline Y}}). \]
349-
```
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<img src="./figures/smarter-tdhf-4.png" height="60">
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Solve for ***(X+Y)*** in the second equation:
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353-
```
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EQUATION
355-
\[ ({\mathbf {\underline X}} + {\mathbf {\underline Y}})
356-
= E ({\mathbf {\underline{\underline A}}} + {\mathbf {\underline{\underline B}}})^{-1}
357-
({\mathbf {\underline X}} - {\mathbf {\underline Y}}). \]
358-
```
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<img src="./figures/smarter-tdhf-5.png" height="60">
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Insert this result into the first equation, rearrange a bit, and finally
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obtain:
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363-
```
364-
EQUATION
365-
\[ ({\mathbf {\underline{\underline A}}} + {\mathbf {\underline{\underline B}}})
366-
({\mathbf {\underline{\underline A}}} - {\mathbf {\underline{\underline B}}})
367-
({\mathbf {\underline X}} - {\mathbf {\underline Y}})
368-
= E^2 ({\mathbf {\underline X}} - {\mathbf {\underline Y}}). \]
369-
```
198+
<img src="./figures/smarter-tdhf-6.png" height="60">
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This is an eigenvalue equation of the same dimension as the CIS eigenvalue
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equation (number of occupied orbitals times number of unoccupied orbitals),

‎Project#12/figures/A-matrix.png

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‎Project#12/figures/B-matrix.png

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‎Project#12/figures/excited-det-schrod-eqn.png

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‎Project#12/figures/four-possible-determinants.png

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‎Project#12/figures/identical-ci-coeff.png

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‎Project#12/figures/inverse-ci-coeff.png

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‎Project#12/figures/matrix-eigenvalue-problem.png

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‎Project#12/figures/matrix-elements.png

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‎Project#12/figures/simpler-hamiltonian.png

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‎Project#12/figures/singlet-triplet-combinations.png

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‎Project#12/figures/singly-excited-determinant.png

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‎Project#12/figures/smarter-tdhf-1.png

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