Problem Set #1 Solutions
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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): * PCCμν = ∑ μi ν i . (1) i In matrix form, (1) is simply equivalent to † PCC= occ occ (2) where Cocc is the occupied block of the MO coefficient matrix. To reformulate the nonorthogonal Roothaan-Hall equations FC= SCε (3) into the preferred eigenvalue problem ~~ ~ FC= Cε (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 ~ CSCCSC=2 ⇔ = 2 (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 ~ FSCSSC( 2 )(= 2 )ε − 1 − 1 ~ −1 − 1 ~ SFSCSSSC2 ( 2 )(= 2 2 )ε . (6) −1 − 1 ~ −1 − 1 ~ ()S2 FS2 C= () S2 SS2 Cε ~~ ~ FC = Cε ~ ~ −1 − 1 −1 − 1 In (6), the dressed Fock matrix, F , was clearly defined as F≡ S2 FS 2 and S2 SS2 = S 0 =1 was used. In an orthonormal basis, therefore, the density matrix of (2) takes on the following form: ~ ~† PCC= occ occ . (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, CCocc occ =1. ~† ~ Proof. In order to prove that CCocc occ =1, expand the occupied MO coefficient matrices as follows: ~† ~ ~* ~ ~ * ~ ~ ~ CCCCocc occ = ∑ μ~i μ~ j = ∑ μi μ j= ∑ iμ μ j= i j =δ ij = 1 (8) μ~ μ~ μ~ ~† ~ which proves the proposition that CCocc occ =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 PCC= occ occ 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 ~ ~ † ~ ~† ~ ~ † ~ ~ † ~ ~ † PCCCCCCCCCCCCP= ( occ occ )(= occ occ )( occ occ ) = occ( occ occ) occ =occ occ = (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.occSC occ =1 ~† ~ Proof. Using the results of Proposition 1, namely that CCocc occ =1in an orthonormal basis and ~ + 1 the fact that CSC= 2 from (5), we have: 1 1 1 1 1 1 ~ † ~ + 2 † + 2 † +2 + 2 † +2 + 2 † Cocc C occ = ( S Cocc ) ( S Cocc )= ( Cocc S )( S Cocc ) = Cocc S S Cocc = Cocc SC occ = 1 (10) which completes the proof. QED With Proposition 3 at our disposal, we are now in a position to show that PSP= P 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 PSP= P . Proof. From the definition of the density matrix in a nonorthonormal basis, given by (2) above, † we have PCC= occ occ and we wish to prove that PSP= P , i.e., the idempotency condition in a nonorthonormal basis. In order to do so, expand PSP as follows † † † † † PSP= ( Cocc C occ )()() S Cocc C occ = Cocc C occ SC occ C occ = Cocc C occ = P (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 ψ= λ ψ , and acknowledge that the density operator is idempotent, i.e., PPˆ2 = ˆ , we have the following two simultaneous conditions: Pˆ ψ= λ ψ (12) and PPˆ2 ψ= ˆ ψ = λ ψ . (13) However, we could write (13) as follows: PPPPˆ2 ψ= ˆ( ˆ ψ) =ˆ()( λψ = λψPˆ )() = λλψ = λψ2 . (14) Combining (13) and (14), we note that λ ψ= λ2 ψ (15) which implies that λ= λ2 , 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 = 0,1} . 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: Xˆ = ∑ p p (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, O≡ Span{}ψ k with dim(ON ) = occ , with the following explicit form: Nocc ˆ P = ∑ ψk ψ k . (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 ψi = ∑ ψk ψ k ψ i = ∑ ψk δ ki = ψ 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 ψa = ∑ ψk ψ k ψ a = ∑ ψk δ ka = 0 (19) k k since ∀k, k ≠ a . In other words, the virtual space, V≡ Span{}ψ c with dim(VN) = virt and the occupied space are orthogonal spaces, OV⊥ . Of course, one could define a projection operator corresponding to virtual space, namely, Nvirt ˆ Q = ∑ ψa ψ a (20) a from which one can show that PQˆ +ˆ = 1 ˆ . Why? (d) Suppose a matrix M (in an orthonormal basis) is approximately idempotent, with a deviation from idempotency that is D (i.e. MMD= 0 + with M 0 idempotent). Consider then a transformed matrix defined as M ′ =3MM2 − 2 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 MMD=0 + , with MM0 = 0 and D the deviation from idempotency, we have: 2 2 2 M=() M0 + D = M0 + M 0 D + DM0 + D (21) which clearly illustrates that M deviates from idempotency to OD() , i.e., linear in order of D. If we now consider the transformed matrix defined as MM′ = 3 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 M=( M0 + M 0 D + DM0 + D)( M0 + D ) 2 2 3 =M0 + M 0 DM 0 + DM0 + D M0 + M 0 D + M0 D + DM0 D + D (22) 2 2 ≈M0 + M 0 DM 0 + DM0 + D M0 + M 0 D + M0 D + DM0 D 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 M′ =3 M − 2 M ≈ M0 + M 0 D + DM0 +3 D − 2 M0 DM 0 −2 D M0 − 2 M0 D − 2 DM0 D . (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 M′ − M′ ≈3( M0 D − M0 D M0− M 0 DM 0 D− DM0 DM 0 + D M0 + DM0 D − D ) . (24) if the leading terms in D, i.e., OD( 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.