Molecular Magnetic Properties. Workshop on Theoretical Chemistry. Mariapfarr, Salzburg, Austria February 14 17, Overview

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1 1 Molecular Magnetic Properties Workshop on Theoretical Chemistry Mariapfarr, Salzburg, Austria February 14 17, 2006 Trygve Helgaker Department of Chemistry, University of Oslo, Norway Overview The electronic Hamiltonian in an electromagnetic field The calculation of molecular magnetic properties

2 2 Nonrelativistic electronic Hamiltonian in an electromagnetic field classical mechanics of particles Newtonian mechanics (1687) Lagrangian mechanics (1788) Hamiltonian mechanics (1833) electromagnetic fields Maxwell s equations (1864) scalar and vector potentials gauge transformations spin-free electron in an electromagnetic field the Lagrangian and the Hamiltonian of a particle in an electromagnetic field the Schrödinger Hamiltonian (1925) spinning electron in an electromagnetic field the Dirac equation and electron spin (1928) the Schrödinger Pauli Hamiltonian (Pauli, 1927)

3 3 Classical mechanics Matter is described by Newton s equations: F (r,v,t) = ma the force defines the system and is obtained from experiment conservative forces (e.g., gravitational forces) are obtained from potentials: F (r) = V (r) Radiation is described by Maxwell s equations: E = ρ/ε 0 Coulomb s law B ε 0 µ 0 E/ t = µ 0 J Ampère s law B = 0 E + B/ t = 0 Faraday s law of induction The interaction between matter and radiation is described by the Lorentz force: F(r,v) = z (E + v B)

4 4 Lagrangian mechanics: the principle of least action For a system of n degrees of freedom, there are n generalized coordinates q i in configuration space n generalized velocities q i The principle of least action (Hamilton s principle): For each system, there exists a Lagrangian L(q i, q i,t) such that the action integral Z t2 S = L(q i, q i, t)dt t 1 assumes an extremum along the trajectory in configuration space taken by the system. q i t 2 q i t 1 t 1 t 2

5 5 t 1 Lagrange s equations From the principle of least action, we obtain Z t2 Z t2 L δs = δ L(q i, q i, t)dt = δq i + L «δ q i dt t 1 t 1 q i q i Z t2» L d L = δq i «δq i dt = 0 q i dt q i We conclude that the Lagrangian satisfies the following second-order differential equations (one for each degree of freedom): d dt L q i = L q i Lagrange s equations of motion the Lagrangian defines the system and is determined so that it reproduces the equations of motion consistent with experiment. Lagrange s equations preserve their form in any coordinate system. A more general formulation than the Newtonian one: unified description of matter and fields (Newton s and Maxwell s equations) springboard for quantum mechanics

6 6 Arbitrariness of the Lagrangian and gauge transformations The scalar Lagrangian is not uniquely defined. Assume that the Lagrangian L(q, q, t) satisfies the equations: d dt L q = L q. Consider now the following transformed Lagrangian where the arbitrary gauge function f (q, t) is independent of the velocity q: L (q, q,t) = L (q, q, t) + d f (q, t). dt The new Lagrangian satisfies the same equations of motion as the old one: d L dt q = d L dt q + d «f f q + dt q q t = L q + d f dt q = L + ddt «q f = L q. This is an example of a gauge transformation.

7 7 Conservative systems The Lagrangian of a particle in a conservative field may be written as L = T (q, q) {z } V (q) {z } kinetic energy potential energy The Lagrangian is thus easily set up for any conservative system, in any convenient coordinate system. Example: Assuming a Cartesian coordinate system L = 1 2 mv2 V (r), we find that Lagrange s equations immediately reduce to Newton s equations: d dt L v = L r d mv = V (r) ma = F. dt For particles in a (nonconservative) electromagnetic field, the Lagrangian can be cast in similar but slightly different form as discussed later.

8 8 The energy function In Lagrangian mechanics, the energy function is defined as: h(q, q,t) = L q q L (q, q, t). Assume a conservative Cartesian system with Lagrangian L = 1 2 mv2 V (r): h = L v v L = T v (T V ) = 2T T + V = T + V = total energy v More generally, h is equal to the total energy if T ( q) is quadratic in q and V (q) is independent of q. The energy function h is conserved if L does not depend explicitly on time: Proof: dh dt = d «L dt q q» «d L = dt q dh dt = L t dl dt q + L q q» L q q + L q L q + t = L t

9 9 Generalized momentum The generalized momentum p i conjugate to the generalized coordinate q i is defined as p i = L q i generalized momentum For a conservative system in Cartesian coordinates L = 1 2 mv2 V (r), the conjugate momentum corresponds to the linear momentum: p = L v = T v = 1 2 m v2 v = mv linear momentum The momentum conjugate to a coordinate that does not occur in the Lagrangian is conserved: dp i dt = d dt L q i = L q i = 0 if L is independent of q i such coordinates are said to be cyclic Compare: h is conserved if L does not depend explicitly on t, p i is conserved if L does not depend explicitly on q i

10 10 Hamiltonian mechanics For a system of n degrees of freedom, there are n second-order differential equations (Lagrange s equations): d dt L = L q i q i The motion is completely specified by the initial values of the n coordinates and the n velocities. In this sense, we may view q i and q i as 2n independent variables. Alternatively, let us treat q i and p i as 2n independent variables: Possible advantages of such a scheme: first-order equations better suited to cyclic coordinates {q i, q i } {q i, p i }

11 11 The Hamiltonian The differential of the Lagrangian L (q, q, t) is given by dl (q, q, t) = L q dq + L q d q + L t dt = ṗdq + p d q + L t dt We now introduce the Hamiltonian, whose differential should be given by: The Legendre transformation dh (q, p,t) = H q H = p q L dq + H p dp + H t the Hamiltonian function dt. (1) gives the required differential: dh = (pd q + q dp) (ṗ dq + pd q + L t A comparison of (1) and (2) yields dt) = ṗdq + q dp L t dt. (2) H q = ṗ, H p = q Hamilton s equations

12 12 Prescription for setting up the Hamiltonian 1. Choose n generalized coordinates q i. 2. Set up the Lagrangian L (q i, q i, t) such that reproduces the equations of motion. d dt L q i = L q i (1) 3. Introduce the conjugate momenta 4. Construct the Hamiltonian H = X i p i = L q i. (2) p i q i L (q i, q i, t) (3) and invert (2) to express the Hamiltonian H (q i,p i ) in terms of q i and p i. 5. Write down Hamilton s equations of motion: q i = H p i, ṗ i = H q i. (4)

13 13 Comparison of Lagrangian and Hamiltonian mechanics Lagrangian mechanics n second-order equations: d dt L q i = L q i q i are the primary variables in configuration space, q i secondary variables the state of the system is determined when the variables (q i, q i ) are known at a given time t Hamiltonian mechanics 2n first-order equations: q i = H p i, ṗ i = H q i q i and p i are independent variables in phase space, connected only by the equations of motion the state of the system is defined by a point (q i, p i ) in phase space, moving on a trajectory that satisfies Hamilton s equations of motion

14 14 Poisson brackets The Poisson bracket of two dynamical variables A (q, p,t) and B (q,p, t) is defined as [A, B] = X i A B B «A q i p i q i p i the fundamental Poisson brackets among conjugate coordinates and momenta: The total time derivatives are given by [q i, q j ] = 0, [p i, p j ] = 0, [q i, p j ] = δ ij da dt = A q A q + p ṗ + A t = A q H p A p H q + A t, and may therefore be expressed compactly as da dt = [A, H] + A t. important special cases: q = [q, H], ṗ = [p, H], dh dt = H t

15 15 Quantization of a particle in conservative force field The Hamiltonian formulation is more general than the Newtonian formulation: it is invariant to coordinate transformations it provides a uniform description of matter and radiation it constitutes the springboard to quantum mechanics The Hamiltonian function (the total energy) of a particle in a conservative force field: H(q, p) = p2 2m + V (q) Standard rule for quantization (in Cartesian coordinates): carry out the substitutions p i h, H i h t multiply the resulting expression by the wave function Ψ(q) from the right: i h Ψ(q) =» h2 t 2m 2 + V (q) Ψ(q) This approach is sufficient for a treatment of electrons in an electrostatic field. It is insufficient for nonconservative systems that is, for systems in a general electromagnetic field.

16 16 Review: Hamiltonian mechanics In classical Hamiltonian mechanics, a system of particles is described in terms their positions q i and conjugate momenta p i. For each such system, there exists a scalar Hamiltonian function H(q i, p i ) such that the classical equations of motion are given by: q i = H p i, ṗ i = H q i (Hamilton s equations of motion) Example: a single particle of mass m in a conservative force field F(q) the Hamiltonian function is constructed from a scalar potential: q = H(q,p) p ṗ = H(q,p) q = p m = H(q, p) = p2 2m + V (q), V (q) q (q) F(q) = V q Hamilton s equations are equivalent to Newton s equations: 9 = ; m q = F(q) (Newton s equations of motion) Whereas Newton s equations of motion are second-order differential equations, Hamilton s equations are first-order. We must now generalize this approach to particles in an electromagnetic field!

17 17 The Lorentz force and Maxwell s equations In the presence of an electric field E and a magnetic field (magnetic induction) B, a classical particle of charge z experiences the Lorentz force: F = z (E + v B) since this force depends on the velocity v of the particle, it is not conservative The electric and magnetic fields E and B satisfy Maxwell s equations ( ): E = ρ/ε 0 Coulomb s law B ε 0 µ 0 E/ t = µ 0 J Ampère s law with Maxwell s correction B = 0 E + B/ t = 0 Faraday s law of induction when the charge and current densities ρ(r, t) and J(r, t) are known, Maxwell s equations can be solved for E(r, t) and B(r, t) on the other hand, since the charges (particles) are driven by the Lorentz force, ρ(r, t) and J(r, t) are functions of E(r, t) and B(r, t) In the following, we shall consider the motion of particles in a given (fixed) electromagnetic field.

18 18 Scalar and vector potentials The second, homogeneous pair of Maxwell s equations involve only E and B: E + B t B = 0 (1) 1. Equation (1) is satisfied by introducing the vector potential A: = 0 (2) B = 0 B = A vector potential (3) 2. Inserting Eq. (3) in Eq. (2) and introducing a scalar potential φ, we obtain E + A «= 0 E + A t t = φ scalar potential The second pair of Maxwell s equations are thus automatically satisfied by writing E = φ A t B = A The potentials (φ,a) contain four rather than six components as in (E,B). They are obtained by solving the first, inhomogeneous pair of Maxwell s equations, which contains ρ and J.

19 19 Particle in an electromagnetic field For a particle in an electromagnetic field, we must set up a Lagrangian such that d dt L q i = L q i reduces to Newton s equations with the Lorentz force F = z (E + v B), This is not a conservative system, for which L = T V, E = φ A t, F i = V q i Rather, it belongs to a broader class of systems, for which L = T U, F i = U + d «U q i dt q i B = A For a particle subject to the Lorentz force, the generalized potential is given by U = z (φ v A) velocity-dependent potential and the Lagrangian becomes L = T z (φ v A) particle in an electromagnetic field

20 20 Conjugate momentum in an electromagnetic field We recall that, for a conservative system described by Lagrangian of the form L (q, q) = T ( q) V (q), the conjugate momentum in Cartesian coordinates p = L v = T v = mv π is equal to the linear (kinetic) momentum π: p = π particle in a conservative field By contrast, for a nonconservative system described by Lagrangian L (q, q) = T ( q) U (q, q), the conjugate and kinetic momenta are no longer the same. In particular, for a particle in an electromagnetic field L = T z (φ v A) we obtain p = L v = T v + za p = π + za particle in an electromagnetic field

21 21 The Hamiltonian in an electromagnetic field We recall that, for a conservative system with Lagrangian L = T ( q) V (q), where T ( q) is quadratic in q, the Hamiltonian is given by H = T ( q) + V (q). Let us now consider the nonconservative system consisting of a particle in a field From the conjugate momentum we obtain the Hamiltonian H = p v L = `mv 2 + zv A L = T ( q) U (q, q) = 1 2 mv2 z (φ v A). p = mv + za, «1 2 mv2 zφ + zv A Expressed in canonical coordinates, the Hamiltonian now becomes: = 1 2 mv2 + zφ = T + zφ. H = T + zφ = (p za) (p za) 2m + zφ note: H = T + U + zv A T + U

22 22 Gauge transformations The scalar and vector potentials φ and A are not unique. Consider the following transformation of the potentials: 9 φ = φ f = t f = f(q, t) gauge function of position and time A = A + f ; This gauge transformation of the potentials does not affect the physical fields: E = φ A = φ + f t t A t f t B = (A + f) = B + f = B We are free to choose f(q, t) so as to make φ and A satisfy additional conditions. In the Coulomb gauge, the gauge function is chosen such that the vector potential becomes divergenceless: = E A = 0 Coulomb gauge Note: Gauge transformations induce the following transformations: L = L + z df dt, p = p + z f, H = H z f t, However, the equations of motion are unaffected! T = T

23 23 Quantization of a particle in an electromagnetic field We have now constructed a Hamiltonian function such that Hamilton s equations are equivalent to Newton s equation with the Lorentz force: q i = H p i & ṗ i = H q i ma = z (E + v B) To this end, we introduced scalar and vector potentials φ and A such that E = φ A t, B = A In terms of these potentials, the classical Hamiltonian function takes the form H = π2 2m + zφ, π = p za kinetic momentum Quantization is now accomplished in the usual manner, by the substitutions p i h, H i h t This results in the following time-dependent Schrödinger equation for a particle in an electromagnetic field: i h Ψ t = 1 2m ( i h za) ( i h za)ψ + zφψ

24 24 Electron spin According to our previous discussion, the nonrelativistic Hamiltonian for an electron in an electromagnetic field is given by: H = π2 2m eφ, π = i h + ea However, this description ignores a fundamental property of the electron: spin. Spin was introduced by Pauli in 1927, to fit experimental observations: H = (σ π)2 2m π2 eφ = 2m + e h 2m B σ eφ where σ contains three operators, represented by the two-by-two Pauli spin matrices σ x 0 1 A, σ y 0 i A, σ z 1 0 A 1 0 i The Schrödinger equation now becomes a two-component equation π2 e h eφ + 2m 2m B e h z 2m (B x ib y ) Ψ α A e h 2m (B π x + ib y ) 2 e h eφ 2m 2m B = Ψ α A z Ψ β Ψ β the two components are only coupled in the presence of an external magnetic field

25 25 Spin and relativity The introduction of spin by Pauli in 1927 may appear somewhat ad hoc. By contrast, spin arises naturally from Dirac s relativistic treatment in is spin is a relativistic effect? However, reduction of Dirac s equation to nonrelativistic form yields the Hamiltonian (σ π)2 π2 H = eφ 2m 2m eφ spin is therefore not a relativistic property of the electron Indeed, it is possible to take the factorized form (σ π)2 H = 2m eφ as the starting point for a nonrelativistic treatment, with unspecified operators σ. All algebraic properties of σ then follow from the requirement (σ p) 2 = p 2 : [σ i, σ j ] + = δ ij, [σ i, σ j ] = 2ǫ ijk σ k these operators are represented by the two-by-two Pauli spin matrices We interpret σ by associating an intrinsic angular momentum (spin) with the electron: s = hσ/2

26 26 Classical relativistic Hamiltonian Hamiltonian for an electron in an electromagnetic field q H = m 2 c 4 + c 2 (p + ea) 2 eφ Hamilton s equations give us ṙ = H p ṗ = H r where the relativistic kinetic momentum is given by π = mv p 1 v 2 /c 2 p = π ea conjugate momentum π = e(e + v B) Lorentz force Lorentz factor nonrelativistic momentum Relationship to nonrelativistic mechanics p m 2 c 4 + c 2 π 2 = mc 2 + π2 + O ˆ(v/c) 2 2m π = mv + O ˆ(v/c) 2 the nonrelativistic limit is obtained as (v/c) 2 0

27 27 The Hamiltonian is given by Linearization of Hamiltonian H = c p π 2 + m 2 c 2 eφ but we would like time and space coordinates to appear symmetrically in the equation Following Dirac, we write To determine the α i, we note that i h Ψ t = HΨ π 2 + m 2 c 2 = (α x π x + α y π y + α z π z + α 0 mc) 2 (α x π x + α y π y + ) 2 = α 2 x π2 x + α2 y π2 y + (α xα y + α y α x ) π x π y + = π 2 x + π2 y if the α i operators anticommute α 2 x = α 2 y = 1 α x α y + α y α x = 0 9 = ; [α i, α j ] + = 2δ ij The Hamiltonian may now be written as H D = cα π + βmc 2 eφ, α = [α x, α y, α z ], β = α 0

28 28 The Dirac equation In matrix representation, the anticommuting operators α i are represented by four 4 4 matrices α i 0 σ i A, β I 0 A, σ i 0 0 I where I is the 2 2 unit matrix and the σ i are the usual Pauli spin matrices: σ x 0 1 A, σ y 0 i A, σ z 1 0 A 1 0 i In this representation, the Dirac equation i h Ψ t = `cα π + βmc 2 eφ Ψ therefore has a four-component solution: 0 i h Ψ 1 t Ψ 2 t Ψ 3 t Ψ 4 t 1 0 = C B mc 2 eφ 0 cπ z c(π x iπ y ) 0 mc 2 eφ c(π x + iπ y ) cπ z cπ z c(π x iπ y ) mc 2 eφ 0 c(π x + iπ y ) cπ z 0 mc 2 eφ 1 0 C B Ψ 1 Ψ 2 Ψ 3 Ψ 4 1 C A Positive solutions are associated with electrons (α and β spin), negative with positrons.

29 29 The Lévy-Leblond equation The time-independent Dirac equation may be written in the form: eφ cσ ϕ A = `E mc ϕ cσ π 2mc 2 eφ χ χ 1 A. Introducing the scaled energy and scaled small component E = E mc 2, χ = cχ, and rearranging, we obtain an equation where c occurs only as c 2 : eφ σ π ϕ A = ϕ σ π 2m c 2 eφ χ c 2 χ 1 A. Letting c, we obtain the Lévy-Leblond equation: eφ σ π ϕ A = ϕ σ π 2m χ 0 1 A. This is the nonrelativistic limit of the Dirac equation a useful zero-order equation for relativistic perturbation theory.

30 30 The Schrödinger equation The Lévy-Leblond equation is given by (dropping primes): eφ σ π ϕ A = ϕ σ π 2m χ 0 1 A. Solving the second equation for the small component χ χ = 1 2m σ πϕ and substituting the result into the first equation, we obtain» 1 (σ π) (σ π) eφ ϕ = Eϕ 2m Finally, invoking the identity (σ u)(σ v) = u v + iσ u v we arrive at the two-component Schrödinger equation:» 1 2m π2 + 1 iσ π π eφ 2m ϕ = Eϕ In the absence of a vector potential, the second term vanishes: A = 0 π π = p p = 0

31 31 Expansion of the kinetic momentum Assuming the Coulomb gauge A = 0, we obtain π 2 Ψ = (p + ea) (p + ea)ψ = p 2 Ψ + ep AΨ + ea pψ + e 2 A 2 Ψ = p 2 Ψ + e(p A)Ψ + 2eA pψ + e 2 A 2 Ψ = `p 2 + 2eA p + e 2 A 2 Ψ Recalling the relation A = B, we obtain (π π) Ψ = (p + ea) (p + ea) Ψ = ep AΨ + ea pψ = e(p A) Ψ + e(pψ) A + ea pψ = i he ( A)Ψ = i hebψ In the Coulomb gauge, the kinetic energy operator is therefore given by: T = 1 2m π m iσ π π = 1 2m p2 + e m A p + e m B s + e2 2m A2 where we have used hσ = 2s.

32 32 Electron spin The Dirac Hamiltonian does not commute with the orbital angular momentum operator but rather with the operator l + h 2 σ We therefore assign to the electron an intrinsic spin angular momentum s = h 2 σ Likewise, we interpret the Zeeman term by assigning to the electron a magnetic moment µ B σ B = m B, µ B = e h 2m where we have introduced is the anomalous spin magnetic moment: m = 2µ B s From quantum electrodynamics, one finds that the true spin magnetic moment differs slightly from that given by Dirac s theory: m = gµ B s, g 2.002

33 33 The nonrelativistic electronic Hamiltonian The general form of the nonrelativistic electronic Hamiltonian is H = π2 2m + e h 2m B σ eφ, π = i h + ea Assuming the Coulomb gauge A = 0, we expand the squared kinetic momentum: H = p2 2m + e m e2 (A p + B s) + 2m A2 eφ, s = h σ spin operator 2 For a many-electron system, we add the instantaneous Coulomb interactions and obtain: H mol = H 0 X φ i + X A i p i + X B i s i + 1 X A 2 i 2 i i i i where (in atomic units) H 0 is the spin- and field-free molecular electronic Hamiltonian and where the perturbation operators are given by: 1. φ i real singlet 2. A i p i paramagnetic imaginary singlet 3. B i s i paramagnetic real triplet A2 i diamagnetic real singlet We shall in the following consider nuclear magnetic fields and uniform external magnetic fields.

34 34 Uniform static electric and magnetic fields The scalar and vector potentials of the uniform (static) fields E and B are given by: 9 8 E = φ(r, t) A(r,t) = const = < t B = A(r, t) = const ; φ(r) = E r : A(r) = 1 2 B r Interaction with the electrostatic field: X φ(r i ) = E X r i = E d e, i i d e = X i r i electric dipole operator Orbital paramagnetic interaction with the magnetostatic field: X X A p i = 1 B r 2 i p i = 1 2 B L, L = X r i p i orbital ang. mom. op. i i i Spin paramagnetic interaction with the magnetostatic field: X B s i = B S, S = X s i spin ang. mom. op. i i Total paramagnetic interaction with a uniform magnetic field: H z = B d m, d m = 1 (L + 2S) Zeeman interaction 2

35 35 Nuclear magnetic fields and hyperfine interactions The nuclear moments set up a magnetic vector potential ( 10 8 a.u.): A(r) = α 2 X K M K r K r 3 K, α 2 = c a.u., M K = γ K hi K 10 4 a.u. This vector potential gives rise to the following paramagnetic hyperfine interaction A p = X K M T K hpso K, hpso K = α2 r K p r 3 K Taking the curl of this vector potential, we obtain: B(r) = A(r) = 8πα2 3 = α 2 L K r 3 K X δ(r K )M K + α X 2 K K paramagnetic SO (PSO) 3r K (r K M K ) r 2 K M K the first term contributes only when the electron is at the position of the nuclei the second term is a classical dipole field and contributes at a distance This magnetic field B(r) then gives rise to two distinct first-order triplet operators: 8 < B s = X K M T K (hfc K + hsd K ), : h FC K r 5 K = 8πα2 3 δ(r K )s Fermi contact (FC) h SD K = α2 3r Kr T K r2 K I 3 r K 5 s spin dipole (SD)

36 36 Review The nonrelativistic electronic Hamiltonian: H = H 0 + H (1) + H (2) = H 0 + A(r) p + B(r) s A(r)2 Rayleigh Schrödinger perturbation theory to second order: E (1) = 0 A p + B s 0 E (2) = 1 0 A 2 0 X 0 A p + B s n n A p + B s 0 2 E n n E 0 Vector potentials of the uniform external field and the nuclear magnetic moments: A (r) = 1 2 B r O, A K (r) = α 2 M K r K r 3 K, A(r) = B(r), A(r) = 0 Orbital and spin Zeeman interactions with the external magnetic field: H (1) Zeeman = 1 2 B L O + B s Orbital and spin hyperfine interactions with the nuclear magnetic moments: H (1) hyperfine = α2 M K L K r 3 K {z } PSO + 8πα2 δ (r K )M K s 3 {z } FC + α 2 3(s r K)(r K M K ) (M K s)r 2 K r 5 K {z } SD

37 37 Gauge transformation of the Schrödinger equation Consider a general gauge transformation: A = A + f, φ = φ f t It can be shown that the Hamiltonian then transforms in the following manner H = H + f t, which constitutes a unitary transformation: H i «= exp ( if) H i «exp (if) t t In order that the Schrödinger equation is still satisfied H i «Ψ H i «Ψ, t t the new wave function must be related to the old one by a compensating unitary transformation: Ψ = exp ( if) Ψ No observable properties such as the electron density are then affected: ρ = (Ψ ) Ψ = [Ψexp( if)] [exp( if)ψ] = Ψ Ψ = ρ

38 38 Gauge-origin transformations Different choices of gauge origin in the external vector potential A O (r) = 1 B (r O) 2 are related by a divergenceless gauge transformation: A K (r) = A O (r) A O (K) = A O (r) + f, f (r) = A O (K) r the exact wave function transforms accordingly and gives gauge-invariant results: Ψ exact K = exp [ia O (K) r]ψ exact O approximate wave functions are in general not able to carry out this transformation: Ψ approx K exp [ia O (K) r]ψ approx O different gauge origins therefore give different results We might contemplate attaching an explicit phase factor to the wave function: Ψ approx K def = exp [ia O (K) r]ψ approx O for any K, this approach produces the same result as with the gauge origin at O however, no natural, best gauge origin can usually be identified (except for atoms) in any case, we might as well carry out the calculation with the origin at O! applied to individual AOs, however, this approach makes much more sense!

39 39 Natural gauge origin for AOs Assume AOs positioned at K with the following properties: H 0 χ lm = E 0 χ lm, L K z χ lm = m l χ lm, L K = i(r K) We first choose the gauge origin to be at K: A K (r) = 1 B (r K) 2 The AOs χ lm are then correct to first order in B: H K (B) χ lm =»H BLKz + O `B 2 χ lm =»E m lb + O `B 2 χ lm Next, we put the gauge origin at O K: A O (r) = 1 B (r O) 2 The AOs χ lm are now correct only to zero order in B: 12 H O (B) χ lm =»H 0 + BLOz + O `B 2 χ lm»e m lb + O `B 2 Standard AOs are biased towards K! χ lm

40 40 London orbitals A traditional AO gives best description with the gauge origin at its position K. Attach to each AO a phase factor that represents the gauge-origin transformation to its position K from the global origin O: ω lm = exp [ia K (O) r] χ lm = exp ˆi 1 2 B (O K) r χ lm Each AO now behaves as if the global gauge origin were at its position! In particular, all AOs are now correct to first order in B, for any global origin O. The calculations become gauge-origin independent and uniform (good) quality is guaranteed. These are the London orbitals (1937), also known as GIAOs (gauge-origin independent AOs).

41 41 Review The nonrelativistic electronic Hamiltonian: H = H 0 + H (1) + H (2) = H 0 + A(r) p + B(r) s A(r)2 Rayleigh Schrödinger perturbation theory to second order: E (1) = 0 A p + B s 0 E (2) = 1 0 A 2 0 X 0 A p + B s n n A p + B s 0 2 E n n E 0 Vector potentials of the uniform external field and the nuclear magnetic moments: A (r) = 1 2 B r O, A K (r) = α 2 M K r K r 3 K, A(r) = B(r), A(r) = 0 Orbital and spin Zeeman interactions with the external magnetic field: H (1) Zeeman = 1 2 B L O + B s Orbital and spin hyperfine interactions with the nuclear magnetic moments: H (1) hyperfine = α2 M K L K r 3 K {z } PSO + 8πα2 δ (r K )M K s 3 {z } FC + α 2 3(s r K)(r K M K ) (M K s)r 2 K r 5 K {z } SD

42 42 Taylor expansion of the energy Expand the energy in the presence of an external magnetic field B and nuclear magnetic moments M K around zero field and zero moments: E (B,M) = E perm. magnetic moments z } { B T E (10) BT E (20) B {z } magnetizability X K hyperfine coupling z } { X K B T E (11) K M K {z } shieldings + 1 M T K E(01) K X KL M T K E(02) KL M L {z } spin spin couplings The first-order terms vanish for closed-shell systems because of symmetry: D c.c. ˆΩ E D imaginary c.c. c.c. ˆΩ E triplet c.c. 0 Higher-order terms are negligible since the perturbations are tiny: 1) the magnetic induction B is weak ( 10 5 a.u.) 2) the nuclear magnetic moments M K are small (µ 0 µ N 10 8 a.u.) We shall therefore consider only the second-order terms: the magnetizability, the shieldings, and the spin spin couplings +

43 43 The magnetizability Assume zero nuclear magnetic moments and expand the molecular electronic energy in the external magnetic induction B: E (B) = E 0 + B T E (10) BT E (20) B + The molecular magnetic moment at B is now given by M mol (B) def de (B) = db = E(10) E (20) B + = M perm + ξb +, where we have introduced the permanent magnetic moment and the magnetizability: M perm = E (10) = de permanent magnetic moment db B=0 describes the first-order change in the energy but vanishes for closed-shell systems ξ = E (20) = d2 E db 2 molecular magnetizability B=0 describes the second-order energy and the first-order induced magnetic moment The magnetizability is responsible for molecular diamagnetism, important for molecules without a permanent magnetic moment.

44 44 The calculation of magnetizabilities The molecular magnetizability of a closed-shell system: D 0 fi ξ = d2 E db 2 = 0 2 fl H B X n = 1 D E 0 r O r T O r 4 T O r O I X 2 {z } n diamagnetic term H B E D n n H B E 0 E n E 0 0 L O n n L T 0 O E n E 0 {z } paramagnetic term The (often) dominant diamagnetic term arises from differentiation of the operator: 1 ˆB2 r 2 O (B r O)(B r O ) 2 A2 (B) = 1 8 (B r O) (B r O ) = 1 8 the isotropic part of the diamagnetic contribution is given by: ξ dia = 1 3 Tr ξ dia = 1 0 x2 6 O + yo O z2 = 0 r2 6 O 0 Only the orbital Zeeman interaction contributions to the paramagnetic term: system surface S 0 0 singlet state for 1 S systems (closed-shell atoms), the paramagnetic term vanishes altogether: 1 S 0 gauge origin at nucleus 1 2 L O

45 45 Hartree Fock magnetizabilities Basis-set requirements for magnetizatibilities are modest if London orbitals are used: basis cc-pvdz cc-pvtz cc-pvqz HF basis-set error 2.8% 1.0% 0.4% The HF model overestimates the magnitude of magnetizabilities by 5% 10%: JT 2 HF exp. diff. H 2 O % NH % CH % CO % PH % H 2 S % C 3 H % CSO % CS % compare with polarizabilities, which require large basis sets and are underestimated

46 46 High-resolution NMR spin Hamiltonian Consider a molecule in the presence of an external field B along the z axis and with nuclear spins I K related to the nuclear magnetic moments M K as: M K = γ K hi K 10 4 a.u. where γ K is the magnetogyric ratio of the nucleus. Assuming free molecular rotation, the nuclear magnetic energy levels can be reproduced by the following high-resolution NMR spin Hamiltonian: H NMR = X K γ K h(1 σ K )BI K z + X K>L γ K γ L h 2 K KL I K I L where we have introduced {z } nuclear Zeeman interaction {z } nuclear spin spin interaction the nuclear shielding constants σ K the (reduced) indirect nuclear spin spin coupling constants K KL This is an effective nuclear spin Hamiltonian: it reproduces NMR spectra without considering the electrons explicitly the spin parameters σ K and K KL are adjusted to fit the observed spectra we shall consider their evaluation from molecular electronic-structure theory

47 47 Simulated 200 MHz NMR spectra of vinyllithium experiment RHF MCSCF B3LYP

48 48 Nuclear shielding constants Recall the energy expansion for a closed-shell molecule in the presence of an external field B and nuclear magnetic moments M K : E (B,M) = E BT E (20) B In this expansion, E (11) K magnetic moments: X K B T E (11) K M K X KL M T K E(02) KL M L + describes the coupling between the applied field and the nuclear in the absence of electrons (i.e., in vacuum), this coupling is identical to I 3 : HZeeman nuc = B X M K the purely nuclear Zeeman interaction K in the presence of electrons (i.e., in a molecule), the coupling is modified slightly: E (11) K = I 3 + σ K the nuclear shielding tensor Since the nuclear shielding constants arise from a hyperfine interaction between the electrons and the nuclei, it is proportional to α and is measured in ppm. The nuclear Zeeman interaction, which does not enter the electronic problem, has here been introduced in a purely ad hoc fashion. Its status is otherwise similar to that of the Coulomb nuclear nuclear repulsion operator.

49 49 The calculation of nuclear shielding tensors Nuclear shielding tensors of a closed-shell system: σ K = = α2 d 2 fi E el = 0 2 fl H dbdm K B M 0 2 X K n * r T O 0 r + KI 3 r O r T K 2 rk 3 α 0 X 2 n {z } diamagnetic term D 0 H B E D n n E n E 0 0 L O n D n r 3 E 0 H M K K LT K E 0 E n E 0 {z } paramagnetic term The (usually) dominant diamagnetic term arises from differentiation of the operator: A (B) A(M K ) = 1 2 α2 r 3 K (B r O) (M K r K ) As for the magnetizability, there is no spin contribution for singlet states: S 0 0 singlet state For 1 S systems (closed-shell atoms), the paramagnetic term vanishes completely and the the shielding is given by (assuming gauge origin at the nucleus): σ Lamb = 1 D1 3 α2 S r 1 1S E Lamb formula K

50 50 Benchmark calculations of BH shieldings HF MP2 CCSD CCSD(T) FCI σ( 11 B) σ( 1 H) σ( 11 B) σ( 1 H) TZP+ basis, R BH = pm J. Gauss and K. Ruud, Int. J. Quantum Chem. S29 (1995) 437 J. Gauss and J. F. Stanton, J. Chem. Phys. 104 (1996) 2574

51 51 Calculated and experimental equilibrium shielding constants HF CAS MP2 CCSD CCSD(T) exp. HF F ± 6 H ± 0.2 H 2 O O ± 17 H ± 0.02 NH 3 N H ± 1.0 CH 4 C H F 2 F N 2 N ± 0.2 CO C ± 0.9 O ± 17 For references and details, see Helgaker, Jaszuński, and Ruud, Chem. Rev. 99 (1999) 293.

52 52 DFT shielding constants HF LDA BLYP B3LYP KT2 CCSD(T) exp. HF F ± 6 H 2 O O ± 17 NH 3 N CH 4 C F 2 F N 2 N ± 0.2 CO C ± 0.9 O ± 17

53 53 Coupled-cluster convergence of shielding constants CCSD CCSD(T) CCSDT CCSDTQ CCSDTQ5 FCI σ( 13 C) σ( 13 C) σ( 17 O) σ( 17 O) All calculations in the cc-pvdz basis and with a frozen core. Kállay and Gauss, J. Chem. Phys. 120 (2004) 6841.

54 54 Nuclear spin spin couplings The last term in the expansion of the molecular electronic energy in B and M K E (B,M) = E BT E (20) B PK BT E (11) K M K PKL MT K E(02) KL M L + describes the coupling of the nuclear magnetic moments in the presence of electrons. There are two distinct contributions to the coupling: the direct and indirect contributions E (02) KL = D KL + K KL The direct coupling occurs by a classical dipole mechanism: D KL = α 2 R 5 KL `R2 KL I 3 3R KL R T KL a.u. it is anisotropic and vanishes in isotropic media such as gases and liquids The indirect coupling arises from hyperfine interactions with the surrounding electrons: it is exceedingly small: K KL a.u. 1 Hz it does not vanish in isotropic media it gives the fine structure of high-resolution NMR spectra Experimentalists usually work in terms of the (nonreduced) spin spin couplings J KL = h γ K 2π γ L 2π K KL isotope dependent

55 55 K KL = The calculation of indirect nuclear spin spin coupling tensors The indirect nuclear spin spin coupling tensor of a closed-shell system is given by: D E E 0 H M n Dn 0 H K M L d 2 E el dm K dm L = fi 0 2 fl H M K M 0 2 X L n Carrying out the differentiation, we obtain: K KL = α *0 4 r T K r + LI 3 r K r T L rk 3 2α 0 X 4 r3 L n {z } diamagnetic spin orbit (DSO) D 0 r 3 K L K E n E 0 E D n n r 3 L LT L E 0 E n E 0 {z } paramagnetic spin orbit (PSO) fi fl fi fl 2α X 0 8π 4 3 δ(r K)s + 3r Kr T K r2 K I 3 r K 5 s n n 8π 3 δ(r L)s T + 3r Lr T L r2 L I 3 r L 5 s T 0 E n n E 0 {z } Fermi contact (FC) and spin dipole (SD) the isotropic FC/FC term often dominates short-range coupling constants the FC/SD and SD/FC terms often dominate the anisotropic part of K KL the orbital contributions (especially DSO) are usually but not invariably small for large internuclear separations, the DSO and PSO contributions cancel

56 56 Calculations of indirect nuclear spin spin coupling constants The calculation of spin spin coupling constants is a challenging task: triplet as well as singlet perturbations are involved electron correlation important the Hartree Fock model fails abysmally the dominant FC contribution requires an accurate description of the electron density at the nuclei (large decontracted s sets) We must solve a large number of response equations: 3 singlet equations and 7 triplet equations for each nucleus for shieldings, only 3 equations are required, for molecules of all sizes Spin spin couplings are very sensitive to the molecular geometry: equilibrium structures must be chosen carefully large vibrational corrections (often 5% 10%) However, unlike in shielding calculations, there is no need for London orbitals since no external magnetic field is involved. For heavy elements, a relativistic treatment may be necessary.

57 57 Relative importance of the contributions to spin spin coupling constants The isotropic indirect spin spin coupling constants can be uniquely decomposed as: J KL = J DSO KL + JPSO KL + JFC KL + JSD KL The spin spin coupling constants are often dominated by the FC term. Since the FC term is relatively easy to calculate, it is tempting to ignore the other terms. However, none of the contributions can be a priori neglected (N 2 and CO)! 200 PSO FC PSO FC FC FC FC FC SD FC FC PSO FC SD FC -100 SD H2 HF H2O O H NH3 N H CH4 C H C2H4 C C HCN N C N2 CO C2H2 C C

58 58 RHF and the triplet instability problem RHF does not in general work for spin spin calculations: the RHF wave function often becomes triplet unstable at or close to such instabilities, the RHF description of spin interactions becomes unphysical the spin spin coupling constants of C 2 H 4 : Hz 1 J CC 1 J CH 2 J CH 2 J HH 3 J cis 3 J trans exp RHF CAS Indeed, any method based on the RHF reference state may have problems: 1 J CN in HCN [Auer and Gauss, JCP 115 (2001) 1619] Hz RHF CCSD CCSD(T) CC3 CCSDT relaxed unrelaxed in CC theory, orbital relaxation should be treated through singles amplitudes noniterative CCSD(T) should not be used; use iterative CC3 instead all electrons should be correlated in unrelaxed CC calculations

59 59 Reduced spin spin coupling constants (10 19 kg m 2 s 2 Å 2 ) RHF CAS RAS SOPPA CCSD CC3 exp vib HF 1 J HF CO 1 J CO N 2 1 J NN H 2 O 1 J OH J HH NH 3 1 J NH J HH C 2 H 4 1 J CC J CH J CH J HH J cis J tns abs at R e %

60 60 Reduced spin spin coupling constants (10 19 kg m 2 s 2 Å 2 ) RHF LDA BLYP B3LYP RAS exp vib HF 1 J HF CO 1 J CO N 2 1 J NN H 2 O 1 J OH J HH NH 3 1 J NH J HH C 2 H 4 1 J CC J CH J CH J HH J cis J tns abs at R e %

61 61 Comparison of density-functional and wave-function theory normal distributions of errors for these molecules and some other systems for which vibrational corrections have been made: HF LDA BLYP B3LYP CAS RAS SOPPA CCSD some observations: HF has a very broad distribution and overestimates strongly LDA underestimates only slightly, but has a large standard deviation BLYP reduces the LDA errors by a factor of two B3LYP improves upon GGA (but not as dramatically as for other properties) B3LYP errors are similar to those of CASSCF and about twice those of the dynamically correlated methods RASSCF, SOPPA, and CCSD the most accurate method appears to be CCSD the situation is much less satisfactory than for geometries and atomization energies

62 62 Trends in spin spin coupling constants comparison of calculated (red) and B3LYP (blue) spin spin coupling constants plotted in order of decreasing experimental value trends are quite well reproduced by B3LYP, in particular for large couplings

63 63 Basis-set requirements Accurate spin spin calculations require large basis sets, augmented with steep s functions Huz-III-su0: [11s6p2d/6s2p] Huz-III-su3: [14s6p2d/9s2p] B3LYP calculations on benzene with and without added s functions: Hz MCSCF B3LYP-su3 B3LYP-su0 emp 1 J CC J CC J CC J CH J CH J CH J CH J HH J HH J HH Cancellation of errors gives excellent agreement without added steep functions :)

64 64 The Karplus curve Vicinal couplings depend critically on the dihedral angle: 3 J HH in ethane as a function of the dihedral angle: 14 empirical 12 DFT The agreement with the empirical Karplus curve is good.

65 65 Valinomycin C 54 H 90 N 8 O 18 DFT can be applied to large molecular systems such as valinomycin (168 atoms) there are a total of 7587 spin spin couplings to the carbon atoms in valinomycin below, we have plotted the magnitude of the reduced LDA/6-31G coupling constants on a logarithmic scale, as a function of the internuclear distance: the coupling constants decay in characteristic fashion, which we shall examine most of the indirect couplings beyond 500 pm are small and cannot be detected

66 66 Valinomycin LDA/6-31G spin spin couplings to CH, CO, CN, CC greater than 0.01 Hz

67 67 Valinomycin LDA/6-31G spin spin couplings to CH, CO, CN, CC greater than 0.01 Hz

68 68 Valinomycin LDA/6-31G spin spin couplings to CH, CO, CN, CC greater than 0.01 Hz

69 69 Valinomycin LDA/6-31G spin spin couplings to CH, CO, CN, CC greater than 0.01 Hz

70 70 Valinomycin LDA/6-31G spin spin couplings to CH, CO, CN, CC greater than 0.01 Hz

71 71 The different long-range decays of the Ramsey terms Letting M be the center of the product Gaussian G a G b, we obtain D E D E FC G a h K G b exp µr 2 SD KM, G a h K G b D E D E DSO PSO G a h KL G b G a h K G b R 2 KM R 2 LM, Insertion in Ramsey s expression gives (red positive, blue negative) R 3 KM, R 2 KM FC decays exponentially mixed signs SD decays as R 3 mixed signs DSO decays as R 2 negative PSO decays as R 2 positive

72 72 The long-range orbital contributions At large separations, the couplings are dominated by the orbital contributions: J DSO KL R 2 KL, JPSO KL R 2 KL large separations Moreover, in this limit, the DSO contributions all become negative: D 0 K DSO KL = 2α4 3 r 3 K r 3 L r K r L 0 E < 0 large separations Also, the PSO contributions become positive, nearly cancelling the DSO contributions. use of Taylor expansion, the virial theorem, and the resolution of identity give: J DSO KL + JPSO KL R 3 KL large separations, large basis However, the PSO contributions converge very slowly to the basis-set limit: LDA 6 31G 10 7 LDA HII