PROBLEM SET #2 SOLUTIONS by Robert A. DiStasio Jr.
Q1. 1-particle density matrices and idempotency.
(a) A matrix M is said to be idempotent if M 2 = M . Show from the basic definition that the HF density matrix is idempotent when expressed in an orthonormal basis.
An element of the HF density matrix is given as (neglecting the factor of two for the restricted closed-shell HF density matrix):
* μν = ∑ CCP νμ ii . (1) i
In matrix form, (1) is simply equivalent to
† = CCP occocc (2)
where Cocc is the occupied block of the MO coefficient matrix. To reformulate the nonorthogonal Roothaan-Hall equations
= SCFC ε (3) into the preferred eigenvalue problem ~ ~~ = CCF ε (4) the AO basis must be orthogonalized. There are several ways to orthogonalize the AO basis, but
I will only discuss symmetric orthogonalization, as this is arguably the most often used procedure in computational quantum chemistry. In the symmetric orthogonalization procedure, the MO coefficient matrix in (3) is recast as
~ + 1 − 1 ~ 2 =⇔= 2CSCCSC (5) which can be substituted into (3) yielding the eigenvalue form of the Roothaan-Hall equations in
(4) after the following algebraic manipulation: FC = SCε − 1 ~ − 1 ~ ( 2 = () 2CSSCSF )ε − 1 − 1 ~ 1 −− 1 ~ 2 ( 2 = 2 () 2CSSSCSFS )ε . (6) 1 −− 1 ~ 1 −− 1 ~ 2 2 )( = 2 2 )( CSSSCFSS ε ~ ~~ FC = Cε
~ ~ 1 −− 1 1 −− 1 In (6), the dressed Fock matrix, F , was clearly defined as ≡ 2 FSSF 2 and 2 2 SSSS 0 == 1 was used. In an orthonormal basis, therefore, the density matrix of (2) takes on the following form:
~ ~† = CCP occocc . (7)
In order to prove that the density matrix is idempotent in an orthonormal basis, we need to first prove the following proposition.
~† ~ Proposition 1: In an orthonormal basis, CC occocc =1.
~† ~ Proof. In order to prove that CC occocc =1, expand the occupied MO coefficient matrices as follows:
~† ~ ~* ~ ~ * ~ ~ ~ occocc = ∑ ~ CCCC μμ ~ji = ∑ μμ = ∑ μμ jijiji δ ij === 1 (8) μ~ μ~ μ~
~† ~ which proves the proposition that CC occocc =1.
QED
With Proposition 1, we are now in a position to directly prove that the density matrix is idempotent in an orthonormal basis.
Proposition 2: The density matrix is idempotent in an orthonormal basis with P 2 = P .
Proof. From the definition of the density matrix in an orthonormal basis, given by (7) above, we
~ ~ † 2 have = CCP occocc and want to prove that P = P , i.e., the idempotency condition in an orthonormal basis. In order to do so, expand P 2 as follows 2 ~ ~ 2† ~ ~ † ~ ~† ~ ~ † ~ ~ † ~ ~ † = ( occocc = () occocc )( occocc ) = ( ) occoccoccocc occocc == PCCCCCCCCCCCCP (9) in which the results of Proposition 1 were used in a fairly straightforward manner (c.f. (8)).
QED
(b) Derive the generalization of the idempotency condition for the case of a nonorthonormal basis, with overlap matrix S.
In a nonorthonormal basis, the results of Proposition 1 must be adjusted to derive the general idempotency condition for the 1-PDM. In analogy to Q1(a), we have two propositions which require proof.
† Proposition 3: In a nonorthonormal basis, C.SCoccocc =1
~† ~ Proof. Using the results of Proposition 1, namely that CC occocc =1in an orthonormal basis and
~ + 1 the fact that = 2CSC from (5), we have:
1 1 1 1 1 1 ~ † ~ + 2 † + 2 † 2 ++ 2 † 2 ++ 2 † occocc = occ occ = occ occ ))(()()( = occ occ = SCCCSSCCSSCCSCSCC occocc = 1 (10) which completes the proof.
QED
With Proposition 3 at our disposal, we are now in a position to show that = PPSP is the generalization of the idempotency condition for the case of a nonorthonormal basis.
Proposition 4: The density matrix is idempotent in a nonorthonormal basis with = PPSP .
Proof. From the definition of the density matrix in a nonorthonormal basis, given by (2) above,
† we have = CCP occocc and we wish to prove that = PPSP , i.e., the idempotency condition in a nonorthonormal basis. In order to do so, expand PSP as follows
† † † † † = ( occocc occocc = )()() occoccoccocc occocc == PCCCSCCCCCSCCPSP (11) which completes the proof.
QED (c) Derive the eigenvalues of an idempotent matrix (you may assume an orthonormal basis). Hence write an operator form for it, and describe the function of this operator.
If we write a general eigenvalue equation for the density operator, Pˆ , as Pˆ = ψλψ , and acknowledge that the density operator is idempotent, i.e., 2 = PP ˆˆ , we have the following two simultaneous conditions:
Pˆ = ψλψ (12) and
2 PP ˆˆ == ψλψψ . (13)
However, we could write (13) as follows:
2 ˆˆ ( ˆ ) PPPP ˆ ()( Pˆ )() ===== 2 ψλψλλψλψλψψ . (14)
Combining (13) and (14), we note that
= 2 ψλψλ (15) which implies that = λλ 2 , which can only be satisfied if λ = 0 or λ = 1. Therefore, the eigenvalues of the density operator are 0 and 1—this should make sense since the matrix form of the density matrix is diagonal with 1’s in the occupied block and 0’s in the virtual block
(neglecting the factor of two arising from the restricted closed-shell treatment).
To find the eigenvectors of the density operator, one must consider the consequences of having an eigenspectrum comprised of λ λ = 1,0|{ } . In essence, the action of the density operator is to either return an eigenvector unchanged ( λ = 1) or to annihilate an eigenvector ( λ = 0 ). This defines a projection operator, i.e., an operator of the form:
ˆ = ∑ ppX (16) p which is nothing more than the analog of the unit dyadic in function space. Therefore, the density
operator is a projection operator for the occupied space, ≡ SpanO ψ k }{ with )dim( = NO occ , with the following explicit form:
Nocc ˆ P = ∑ ψψ kk . (17) k
ˆ When acting on an occupied MO such as ψ i , an eigenfunction of P that is contained within
the occupied space, the density operator will return ψ i unchanged:
Nocc Nocc ˆ P i = ∑ ikk = ∑ kik = ψδψψψψψ i . (18) k k
ˆ When acting on a virtual (or unoccupied) MO such as ψ a , an eigenfunction of P that is NOT
contained within the occupied space, the density operator will annihilate ψ a :
Nocc Nocc ˆ P a = ∑ akk = ∑ δψψψψψ kak = 0 (19) k k
since , ≠∀ akk . In other words, the virtual space, ≡ SpanV ψ c }{ with dim( ) = NV virt and the occupied space are orthogonal spaces, ⊥ VO . Of course, one could define a projection operator corresponding to virtual space, namely,
Nvirt ˆ Q = ∑ ψψ aa (20) a from which one can show that ˆ QP =+ 1ˆˆ . Why?
(d) Suppose a matrix M (in an orthonormal basis) is approximately idempotent, with a deviation from idempotency that is D (i.e. = 0 + DMM with M 0 idempotent). Consider then a transformed matrix defined as M ′ 2 −= 23 MM 3 . Derive an expression for its deviation from idempotency in terms of D (to leading order), and hence discuss the action of this transformation.
2 Since 0 += DMM , with 0 = MM 0 and D the deviation from idempotency, we have:
2 2 2 0 )( 00 0 +++=+= DDMDMMDMM (21) which clearly illustrates that M deviates from idempotency to DO )( , i.e., linear in order of D. If we now consider the transformed matrix defined as ′ = 3MM 2 − 2M 3 , then the deviation from idempotency of M ′ can be derived by computing M ′2 − M ′ in terms of D. Lets first consider
M ′ , in which we will need to take successive powers of M. Equation (21) provides us with an expression for M 2 , so we need to explicitly consider M 3 now:
3 2 ( 00 0 0 ++++= DMDDMDMMM ))( 2 2 3 000 0 00 0 0 +++++++= DDDMDMDMMDDMDMMM (22) 2 2 000 0 00 0 ++++++≈ 0 DDMDMDMMDDMDMMM where, the term that was cubic in D was omitted before we proceed any further in the derivation.
From (21) and (22), we have M ′ as:
2 3 2 2 2 ′ 23 00 0 −+++≈−= 23 00 0 0 −−− 222 0 DDMDMMDDMMDDMDMMMMM . (23)
From (23), we can compute the deviation from idempotency of M ′ , by considering the quantity
M ′2 − M ′ directly (keeping in mind that only the leading terms in D must be considered)
2 2 2 2 2 ′ − ′ (3 0 −≈ 0 − 000 − 00 0 0 −++ DDDMMDDMDMDDMMMDMDMMM ) . (24) if the leading terms in D, i.e., (DO 2 ) are retained. Hence, the transformation matrix is now idempotent to first order in the deviation D from idempotency. In this sense, the transformation matrix has purified M ′ and is this procedure is referred to as a “purification of an almost idempotent density matrix” (For an extremely readable detailed review of the central role the density matrix plays in quantum chemistry, see R. McWeeny, Phys. Rev. 32, 335 (1960)).
2. Koopman’s theorem for ionization energies and electron affinities.
(a) Use the Slater-Condon matrix element expressions to derive the Koopman’s expression for the energy difference between an optimized n-electron HF determinant, and the (n-1) electron determinant obtained by removing 1 electron from the ith spin-orbital.
N Let 0 21 ⋅⋅⋅⋅⋅⋅=Ψ χχχχ Ni be the optimized N-electron ground state HF determinant and
N −1 i 21 ii +− 11 ⋅⋅⋅⋅⋅⋅=Ψ χχχχχ N be the N-1 electron determinant obtained from removing an
electron from spin orbital χi in the optimized N-electron ground state HF determinant. Note:
The N-1 electron determinant has NOT been reoptimized in the variational sense, as it was simply obtained by removing an electron from the HF N-electron reference. The ionization potential (IP) for removing an electron from the ith spin orbital is given as the difference between the energy of the N-1 electron determinant obtained from removing an electron from spin orbital
χi in the optimized N-electron ground state HF determinant and the energy of the optimized N- electron ground state HF determinant, i.e.,
N −1 N i −= EEIP 0 . (25)
Using the Slater-Condon rules for matrix elements, we need to determine the energies corresponding to both of these wavefunctions. Lets start with the optimized N-electron ground state HF determinant, in which:
N N ˆ N E0 0 H ΨΨ= 0 . (26)
As was discussed in both lecture and discussion, the Hamiltonian operator is the sum of one- and
ˆ ˆ two-electron operators ( & OO 21 ). From the Slater-Condon rules, we have
N ˆ O 010 =ΨΨ ∑ jhj (27) j and N ˆ 1 O 020 =ΨΨ ∑ jkjk (28) 2 jk where the sums are over all occupied spin orbitals. Using (27) and (28), we can immediately write down the energy of the optimized N-electron HF ground state determinant as
N N N 1 0 = + ∑∑ jkjkjhjE (29) j 2 jk
and the energy of the N-1 electron determinant obtained by removing an electron from χi as
N N N N −1 1 i = ∑ jhjE + ∑∑ jkjk (30) ≠ij 2 ≠≠ij ik since the ith spin orbital is now unoccupied in the N-1 electron determinant. Substitution of (29) and (30) into (25) yields the desired expression for the IP:
N −1 N i −= EEIP 0 N 1 N N N 1 N = ∑ jhj + ∑∑ − − ∑∑ jkjkjhjjkjk ≠ij 2 ≠≠ij ik j 2 jk 1 N 1 N . (31) −−= − ∑∑ ikikjijiihi 2 j 2 k N −−= ∑ jijiihi j
Of course, the energy difference in (31) is nothing more than the negative of the orbital energy
ε i , thereby yielding a very simple, but interesting physical interpretation of the canonical HF orbital energy eigenvalue, namely, that
N −1 N i EEIP 0 −=−= ε i . (32)
This result is known as Koopman’s Theorem for ionization potentials. In the next part, we will investigate Koopman’s Theorem for electron affinities.
(b) Repeat for the electron affinity, adding an extra electron to the ath virtual orbital.
N As above in 2(a), let 0 21 ⋅⋅⋅⋅⋅⋅=Ψ χχχχ Ni be the optimized N-electron ground state HF
N +1 determinant and a a 21 ⋅⋅⋅⋅⋅⋅=Ψ χχχχχ Ni be the N+1 electron determinant obtained from
adding an electron to spin orbital χ a in the optimized N-electron ground state HF determinant.
Note: The N+1 electron determinant has NOT been reoptimized in the variational sense, as it was simply obtained by adding an electron to the HF N-electron reference. The electron affinity
(EA) for adding an electron to the ath spin orbital is given as the difference between the energy of the optimized N-electron ground state HF determinant and the energy of the N+1 electron
determinant obtained from adding an electron to spin orbital χ a in the optimized N-electron ground state HF determinant and the energy of the optimized N-electron ground state HF determinant, i.e.,
NN +1 0 −= EEEA a . (33)
Using the Slater-Condon rules for matrix elements given by (27) and (28) above, we can write
the energy for the N+1 electron determinant obtained from adding an electron to spin orbital χ a in the optimized N-electron ground state HF determinant as:
N N N N +1 1 a = ∑ ahajhjE ++ + ∑∑ jajajkjk . (34) j 2 jk j
Using the energy expression for the optimized N-electron HF ground state determinant given by
(29) and the energy just computed (34) in the EA expression given by (33) yields:
NN +1 0 −= EEEA a N 1 N N 1 N N = + − ∑∑∑ ahajhjjkjkjhj −− − ∑∑ jajajkjk . (35) j 2 jk j 2 jk j N −−= ∑ jajaaha j As seen above in the derivation of Koopman’s Theorem for ionization potentials, the energy
expression in (35) is nothing more than the negative of the orbital energy ε a . Therefore,
NN +1 0 EEEA −=−= ε aa (36) which is Koopman’s Theorem for electron affinities.
(c) Discuss how ionization potentials and electron affinities computed as the difference of separate HF calculations on the neutral and ion might differ from the Koopman’s estimate. On this basis, how do you think the quality of the Koopman’s IP’s and EA’s would compare?
In (a) and (b), we are using the so-called “frozen orbital” approximation, i.e., assuming that the spin orbitals in the N±1 electron wavefunctions are the same as the spin orbitals in the N electron wavefunction. By performing a separate reoptimization of the N±1 electron wavefunctions, the variational SCF procedure would yield lower energies corresponding to a set of MO coefficients uniquely tailored to the respective ionic species. Therefore, neglect of relaxation often overestimates IPs and underestimates EAs computed using Koopman’s theorem. There are two distinct reasons why estimates of IPs are considerably better than EAs obtained in this manner.
First, correlation effects (neglected at the HF level) tend to increase with the number of electrons, thereby tending to cancel the relaxation error for IPs while adding to the relaxation error for EAs. Second, the eigenvalues of HF virtual orbitals on neutral molecules are almost always positive, despite the fact that most neutral molecules will form a stable anion, thereby adding to the error in predicting EAs.
Q3. Toy SCF calculations on HeH+. Use either paper/calculator, or write a small program, or use a matrix-aware math environment like MatLab (or SciLab). The main information (hopefully the only information) you’ll need is the following: the overlap matrix, S, the core Hamiltonian matrix, H (derived from kinetic energy, and nuclear attraction to He and H), and the symmetry distinct non-zero 2-electron integrals in the minimal basis at the equilibrium geometry (R=1.4632 a.u.). The first function is on He and the 2nd is on H.
(a) While a simple initial guess is P=0, form a better initial guess density matrix by assuming that both electrons are in the He (1s) orbital. Write the P matrix and verify that it is properly idempotent.
If the initial guess assumes that both electrons are in the He (1s) orbital, then the occupied block of the MO coefficient matrix takes on the following form:
⎛1⎞ Cocc = ⎜ ⎟ (37) ⎝0⎠ from which we can compute the restricted closed-shell HF density matrix as
† ⎛1⎞ ⎛ 01 ⎞ ⎛ 02 ⎞ = 2 CCP occocc 2 ⋅= ⎜ ⎟ ()201 ⋅=× ⎜ ⎟ = ⎜ ⎟ (38) ⎝0⎠ ⎝ 00 ⎠ ⎝ 00 ⎠ which clearly corresponds to placing 2 electrons in the He (1s) orbital. The density matrix in (38) is clearly written in a nonorthonormal basis, so to demonstrate idempotency, we need to use the results of Proposition 4 above, i.e., that = PPSP . However, this form must be further modified for the restricted closed-shell case density matrix as follows. Since the restricted closed-shell HF
† density matrix is defined as = 2CP occCocc in (38), instead of = PPSP for the idempotency condition, we have:
† † † † † = occocc occocc = occoccoccocc = 4)(4)2()2( occocc = 2PCCCSCCCCCSCCPSP . (39)
1 Therefore, redefining ′ == CCPP † yields 2 occocc
† † † † † ′′ = ( occocc occocc = )()() occoccoccocc occocc == PCCCSCCCCCSCCPSP ′ (40) which defines the generalization of the idempotency condition for the restricted closed-shell density matrix in a nonorthonormal basis. So, using the overlap matrix given in the problem
⎛ 4508.00.1 ⎞ S = ⎜ ⎟ (41) ⎝ 0.14508.0 ⎠ we can demonstrate idempotency as: 1 ⎛ 02 ⎞ ⎛ 4508.00.1 ⎞ 1 ⎛ 02 ⎞ ⎛ 4508.01 ⎞ ⎛ 01 ⎞ ⎛ 01 ⎞ PSP ′′ ⋅= ⎜ ⎟ × ⎜ ⎟ ⋅× ⎜ ⎟ = ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟ = P′ . (42) 2 ⎝ 00 ⎠ ⎝ 0.14508.0 ⎠ 2 ⎝ 00 ⎠ ⎝ 00 ⎠ ⎝ 00 ⎠ ⎝ 00 ⎠
(b) Form the Fock matrix associated with this initial guess, and evaluate the total energy associated with this initial guess (don’t forget nuclear repulsion!). Use the fact that this calculation is restricted to do the minimum calculations (i.e. Cα = Cβ so you only need Fα, etc.).
The Fock matrix associated with this initial guess is a restricted closed-shell Fock matrix with the following form:
N core core μν μν += ∑ λσμνλσ μν += GHPIIHF μν (43) λσ
core where H μν and Gμν are the one- and two-electron parts of the Fock matrix, i.e.,
N N core nuclear 1 H μν = Tμν + Vμν and μν = PIIG λσμνλσ = ∑∑ − 2 λνμσλσμν )]|()|[( Pλσ . To compute the λσ λσ
Fock matrix from the guess density given in (38), one needs the kinetic energy matrix (Tμν ) given as
⎛ 1670.01643.2 ⎞ T = ⎜ ⎟ (44) ⎝ 7600.01670.0 ⎠ the nuclear-electron attraction potential energy matrices (V He and V H ) given as
He ⎛ −− 1029.11398.4 ⎞ V = ⎜ ⎟ (45) ⎝ −− 2652.11029.1 ⎠ and
H ⎛ −− 4113.06772.0 ⎞ V = ⎜ ⎟ , (46) ⎝ −− 2266.14113.0 ⎠ respectively, as well as the set of two-centered integrals given below (in Chemist’s notation) = = 6057.0)11|22(3072.1)11|11( = = 3118.0)21|22(4373.0)11|21( . (47) = = 7746.0)22|22(1773.0)21|21(
Using (43), (44), (45), (46), and (47), the Fock matrix can be computed as:
⎛ −− 9099.03455.1 ⎞ Fμν = ⎜ ⎟ (48) ⎝ −− 6977.09099.0 ⎠
To evaluate the electronic RHF energy, we need the following formula:
N RHF 1 core E = ∑ [ μνμν + FHP μν ] (49) 2 μν from which the total RHF energy follows as
MM TOT RHF RHF ZZ βα nn EVEE +=+= ∑∑ . (50) ααβ > Rαβ
From (38), (44), (45), (46), (48), and (49), E RHF is computed as -3.9982 hartrees. When the nuclear-nuclear repulsion energy, which is equal to 2/1.4632 = +1.3669 hartrees, is included, the total energy ETOT = -3.9982 + 1.3669 = -2.6313 hartrees.
(c) Solve the generalized eigenvalue problem to obtain a new occupied orbital and a new density matrix…. How different are they from your initial guess?
To solve the generalized eigenvalue problem, we need to first symmetrically orthogonalize the
− 1 − 1 Fock matrix using S 2 . To compute S 2 , first diagonalize the overlap matrix
† ⎛ 05492.0 ⎞ sSUU == ⎜ ⎟ (51) ⎝ 4508.10 ⎠ take the inverse square root of the eigenvalues
1 ⎛ 03493.1 ⎞ − 2 s = ⎜ ⎟ (52) ⎝ 8302.00 ⎠
− 1 and then undiagonalize to form the desired S 2 matrix 1 1 ⎛ − 2596.00898.1 ⎞ − 2 − 2 † UUsS == ⎜ ⎟ . (53) ⎝− 0898.12596.0 ⎠
We can now form the transformed Fock matrix as:
1 1 ⎛ −− 5639.01302.1 ⎞ ~ 2 −− 2 = FSSF = ⎜ ⎟ . (54) ⎝ −− 4045.05639.0 ⎠
The transformed Fock matrix has 2 eigenvalues, namely, -0.0967 and -1.4379, and to proceed forward toward the ground state energy, we must choose the lower eigenvalue, -1.4379. What are we converging to if we continuously choose the higher eigenvalue? The eigenvector corresponding to the lower eigenvalue is the orthogonal MO coefficient vector for the occupied orbital (using the normalization condition on the occupied MO coefficient):
~ ⎛ 8778.0 ⎞ Cocc = ⎜ ⎟ . (55) ⎝ 4790.0 ⎠
~ + 1 − 1 ~ Since 2 =⇔= 2CSCCSC , we have the nonorthonormal occupied MO coefficient vector as
1 ⎛ − 2596.00898.1 ⎞ ⎛ 8778.0 ⎞ ⎛ 8323.0 ⎞ − 2 ~ CSC occ == ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟ (56) ⎝− 0898.12596.0 ⎠ ⎝ 4790.0 ⎠ ⎝ 2941.0 ⎠ which gives rise to the following new (and improved) density matrix:
† ⎛ 8323.0 ⎞ ⎛ 4896.03854.1 ⎞ = 2 CCP occocc 2 ⋅= ⎜ ⎟ × ()2941.08323.0 = ⎜ ⎟. (57) ⎝ 2941.0 ⎠ ⎝ 1730.04896.0 ⎠
Clearly, the new occupied MO coefficients (56) and new density matrix (57) are quite different from our initial guess, in which both electrons were placed in the He (1s) orbital. There are several ways to quantify this difference (maximum or RMS deviations among elements of the density matrix for instance), but the main point here is that the electrons are now shared among the H and He atoms, as expected at equilibrium separation.
(d) Evaluate the total energy with the improved density matrix. If you have written a code or are using an environment, feel free to iterate the SCF procedure to convergence!
With the new density matrix in (57), the new Fock matrix is computed as:
⎛ −− 0354.14436.1 ⎞ F = ⎜ ⎟ (58) ⎝ −− 7958.00354.1 ⎠ from which the electronic and total RHF energies are given by -4.2228 and -2.8559 hartrees, respectively. At convergence, the total RHF energy is given by -2.8605 hartrees. Note that since the integrals were given to you with 4 significant figures, you should only expect to converge the energy to 4 significant figures (at best).
4. The minimum basis dissociation problem.:
(a) While I summarized the H2 case with restricted and unrestricted orbitals in lecture, it is still a useful exercise to carefully derive the expressions yourself. So, for the UHF treatment of H2 at dissociation, express the wavefunction in terms of spin-eigenstates, and discuss.
At dissociation, the UHF wavefunction is given by (after some algebra):
UHF 1 =Ψ 3 BA + AB × − αββαϕϕϕϕ )]2()1()2()1([)]2()1()2()1([ 2 2 (59) 1 + 3 BA − AB × + αββαϕϕϕϕ )]2()1()2()1([)]2()1()2()1([ 2 2 in which there are equivalent amounts of the singlet (first term) and triplet (second term, Ms = 0) eigenfunctions. At the dissociation limit, the wavefunction is 50% singlet and 50% triplet in spin character. Of course, this is not a desirable feature of the UHF wavefunction and is commonly referred to as spin contamination. Since the Hamiltonian does not depend on spin, it will commute with the Sˆ 2 operator. Therefore, we should be able to choose a set of eigenfunctions that is common to both of these operators. It should be stressed that just because two operators commute does not mean that an eigenfunction of one operator is necessarily an eigenfunction of the other operator. If two operators commute, it can be shown that the intersection of their corresponding eigenspaces is not null. (b) For the restricted dissociation of H2 in a minimum basis, establish the expression for the wavefunction at dissociation in terms of ionic and covalent configurations.
At dissociation, the RHF wavefunction is given by:
RHF 1 =Ψ 3 AA × − αββαϕϕ )]2()1()2()1([)]2()1([ 2 2 1 3 BA + AB × − αββαϕϕϕϕ )]2()1()2()1([)]2()1()2()1([ . (60) 2 2 1 + 3 BB × + αββαϕϕ )]2()1()2()1([)]2()1([ 2 2
In the RHF case, the wavefunction is comprised of 2 ionic terms (terms 1 and 3) and 2 covalent terms (contained within the composite term given by term 3). Therefore, the RHF wavefunction is 50% covalent and 50% ionic at dissociation. Since the wavefunction should be 100% covalent at dissociation, it is the presence of the ionic terms that raise the energy and lead to deleterious behavior at infinite separation. However, unlike the UHF wavefunction, the RHF wavefunction is a pure singlet. Hence we have reached a compromise between accurate energetics and accurate wavefunctions. At the UHF level, the energetics at dissociation are superb (relative to the RHF level) due to the larger degree of variational flexibility, but the wavefunction is spin contaminated and is therefore not a pure spin eigenfunction. On the other hand, the RHF wavefunction, with less variational flexibility, provides worse energetics but is indeed a pure spin eigenfunction.
(c) For the HeH+ case that you worked on in q. 3, carefully repeat your calculation at infinite separation (the dissociation limit). • Write down the exact expressions for S, T, VHe, VH and the symmetry-unique 2-electron integrals at dissociation.
At dissociation, the integrals are given by:
⎛ 00.1 ⎞ RS )( =∞→ ⎜ ⎟ (61) ⎝ 0.10 ⎠ ⎛ 01643.2 ⎞ RT )( =∞→ ⎜ ⎟ (62) ⎝ 7600.00 ⎠
He ⎛− 01398.4 ⎞ RV )( =∞→ ⎜ ⎟ (63) ⎝ 00 ⎠
H ⎛ 00 ⎞ RV )( =∞→ ⎜ ⎟ (64) ⎝ − 2266.10 ⎠ and
= 3072.1)11|11( . (65) = 7746.0)22|22(
• Evaluate the Fock matrix and energy for the initial guess with the 2 electrons on He, and perform one SCF step again. Interpret the changes that you see.
Given the same initial guess density matrix as in Q3(a), the Fock matrix is now given as:
⎛− 06683.0 ⎞ F = ⎜ ⎟ (66) ⎝ − 4666.00 ⎠ with a corresponding electronic RHF energy of -2.6438 hartrees—which of course is completely
equivalent to the total RHF energy since Rαβ = ∞ and therefore Vnn = 0 . The total RHF energy at dissociation is clearly higher than the total RHF energy at equilibrium (-2.8605 hartrees).
Since the Fock matrix is already diagonal, we are done, i.e., we have found a basis in which the
Fock matrix is diagonal and no further iterations will change the density matrix. Therefore, our final density matrix at dissociation is equal to the guess density in which both electrons are placed in the He (1s) orbital:
⎛ 02 ⎞ RP )( =∞→ ⎜ ⎟ . (67) ⎝ 00 ⎠ This should make sense to you, since the off-diagonal, or coupling elements, of the integral matrices vanish at the dissociation limit. Therefore, the electrons on He really have no way of feeling the presence of the H (1s) orbital—it is in fact infinitely far away…
• Discuss your result in terms of the nature of the products, the ionization potentials of H and He, and the question of whether RHF is appropriate for this case.
From the RHF total energetics at equilibrium (-2.8605) and at dissociation (-2.6438 hartrees), we can compute the binding energy for HeH+ as approximately 0.216 hartrees = 5.90 eV. For reference, the correct value is only 2.04 eV, with the error mainly due to the minimal basis set used and the HF approximation.
From the integrals given, we can also determine the energies of H, He, and He+ within the minimal STO-3G basis.
The energy for the H atom is:
H H VTE 2222 −=−=+= .02266.17600.0 4666 hartrees. (68)
The energy for the He atom is:
He He VTE 1111 =++= −=+− .23072.1)1398.41643.2(*2)11|11()(2 6438 hartrees. (69)
Finally, the energy for the He+ cation is:
He + VTE −=−=++= 9755.11398.41643.2)11|11( hartrees. (70) He 1111
With this set of energetics, we can now compare dissociation energies for the following two competing processes:
HeH+ → He + H+ ΔE = (-2.6438 + 0) – (-2.8605) hartrees = 0.2167 hartrees
HeH+ → He+ + H ΔE = (-1.9755 + -0.4666) – (-2.8605) hartrees = 0.4184 hartrees from which we find that HF calculations within a minimal basis correctly predict that HeH+ will
+ dissociate into He + H (which are both closed-shell species). Unlike the dissociation for H2, the products of the dissociation of HeH+ are closed shell species. This in fact was confirmed analytically above in which we found that the total RHF energy at dissociation was -2.6438 hartrees—in case you have not noticed, this is just the energy of the He atom (69). From an ionization potential point of view, the IP for H → H+ + e- is much smaller than for He → He+ + e-, thereby supporting our findings.
So, is RHF appropriate for this dissociation problem? Let’s revisit this question after we consider the UHF method below…
(d) To further clarify, evaluate the energy of a single determinant UHF wavefunction for HeH+ at dissociation with one electron on each atom, and compare against (c).
In the part, we will consider the UHF treatment of HeH+ dissociation with the products being open-shells: He+ and H. In this case, the integrals at R → ∞ , will be the same as given in (61),
(62), (63), (64), and (65). In general, the UHF energy is given by
UHF 1 α β core α α β β E = ∑ μν + μν )[( μν μν μν ++ μν FPFPHPP μν ] (71) 2 μν where the alpha and beta Fock matrix elements are
α core α β α μν HF μν ∑[( λσ ++= PP λσ λσμν − Pλσ λνμσ )]|()|)( (72) λσ and
β core α β β μν HF μν ∑[( λσ ++= PP λσ λσμν − Pλσ λνμσ )]|()|)( . (73) λσ
Assuming (WLOG) that the electron in the He (1s) orbital is of alpha spin and the electron in the
H (1s) orbital is of beta spin, the alpha and beta density matrices are given as:
α ⎛ 01 ⎞ P = ⎜ ⎟ (74) ⎝ 00 ⎠ and β ⎛ 00 ⎞ P = ⎜ ⎟ . (75) ⎝ 10 ⎠
From (62), (63), and (64), we have:
core ⎛ 01643.2 ⎞ ⎛− 01398.4 ⎞ ⎛ 00 ⎞ ⎛− 09755.1 ⎞ H μν = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ . (76) ⎝ 7600.00 ⎠ ⎝ 00 ⎠ ⎝ − 2266.10 ⎠ ⎝ − 4666.00 ⎠
Therefore,
α ⎛− 09755.1 ⎞ F = ⎜ ⎟ (77) ⎝ − 4666.00 ⎠ and
β ⎛− 09755.1 ⎞ F = ⎜ ⎟ . (78) ⎝ − 4666.00 ⎠
Using (71), (74), (75), (77), and (78), we can compute the UHF energy at dissociation as -2.4421 hartrees. This supports our earlier findings in that HeH+ would prefer to dissociate into closed- shell fragments He and H+ rather than open-shell fragments He+ and H. Therefore, the RHF method is completely applicable in this case. In the case of H2 dissociation, however, in which a bond is cleaved homolytically, extra care must be taken when comparing the restricted and unrestricted solutions to the HF equations.