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Normal Mode Analysis ©David Ronis Mcgill University Chemistry 365: Normal Mode Analysis ©David Ronis McGill University 1. Quantum Mechanical Treatment Our starting point is the Schrodinger wav e equation: N −2 ∂2 − h + → → Ψ → → = Ψ → → Σ → U(r1,...,r N ) (r1,...,r N ) E (r1,...,r N ), (1.1) = ∂ 2 i 1 2mi ri where N is the number of atoms in the molecule, mi is the mass of the i’th atom, and → → U(r1,...,r N )isthe effective potential for the nuclear motion, e.g., as is obtained in the Born- Oppenheimer approximation. If the amplitude of the vibrational motion is small, then the vibrational part of the Hamil- tonian associated with Eq. (1.1) can be written as: N −2 ∂2 N ↔ → → ≈− h + + 1 ∆ ∆ Hvib Σ →2 U0 Σ Ki, j: i j,(1.2) i=1 2mi ∂∆ 2 i, j=1 i → ∆ ≡ → − → → where U0 is the minimum value of the potential energy, i ri Ri, Ri is the equilibrium posi- tion of the i’th atom, and 2 ↔ ∂ ≡ U Ki, j → → (1.3) ∂ ∂ → ri r j → r k = Rk is the matrix of (harmonic) force constants. Henceforth, we will shift the zero of energy so as to = make U0 0. Note that in obtaining Eq. (1.2), we have neglected anharmonic (i.e., cubic and higher order) corrections to the vibrational motion. The next and most confusing step is to change to matrix notation. We introduce a column vector containing the displacements as: ∆≡ ∆x ∆y ∆z ∆x ∆y ∆z T [ 1 , 1, 1,..., N , N , N ] ,(1.4) where "T"denotes a matrix transpose. Similarly,wecan encode the force constants or masses into 3N × 3N matrices, and thereby rewrite Eq. (1.2) as −2 † h ∂ − ∂ 1 H =− M 1 + ∆†K∆,(1.5) vib 2 ∂∆ ∂∆ 2 where all matrix quantities are emboldened, the superscript "†" denotes the Hermitian conjugate (matrix transpose for real matrices), and where the mass matrix, M,isadiagonal matrix with the masses of the givenatoms each appearing three times on the diagonal. The Hamiltonian givenbyEq. (1.5) is the generalization of the usual harmonic oscillator Hamiltonian to include more particles and to allowfor "springs" between arbitrary particles. Unless K is diagonal (and it usually isn’t) Eq. (1.5) would seem to suggest that the vibrational problem for a polyatomic is non-separable (why?); nonetheless, as we nowshow, itcan be sepa- rated. This is first shown using using the quantum mechanical framework we’ve just set up. Winter Term, 2019 Chemistry 365 -2- Normal Mode Analysis Later,wewill discuss the separation using classical mechanics. This is valid since, for harmonic oscillators, you get the same result for the vibration frequencies. → → ∆ → −1/2∆ We now makethe transformation i mi i,which allows Eq. (1.5) to be reexpressed in the newcoordinates as −h 2 ∂ † ∂ 1 H =− + ∆†K˜ ∆,(1.6) vib 2 ∂∆ ∂∆ 2 where − − K˜ ≡ M 1/2KM 1/2.(1.7) Since the matrix K˜ is Hermitian (or symmetric for real matrices), it is possible to find a uni- tary (also referred to as an orthogonal matrix for real matrices) matrix which diagonalizes it; i.e., you can find a matrix P which satisfies P†KP˜ = λ or K˜ = PλP†,(1.8) where λ is diagonal (with real eigenvalues) and where PP† = 1(1.9) where 1 is the identity matrix. Note that P is the unitary matrix who’scolumns are the normal- ized eigenvectors. (See agood quantum mechanics book or anylinear algebra text for proofs of these results). We now makethe transformation ∆→P∆ (1.10) in Eq. (1.5). This gives −h 2 ∂ † ∂ 1 H =− P†P + ∆†λ˜ ∆,(1.11) vib 2 ∂∆ ∂∆ 2 Finally,Eq. (1.8) allows us to cancel the factors of P in this last equation and write 3N 1 ∂2 1 H = Σ − −h 2 + λ ∆2 .(1.12) vib ∂∆2 i i i=1 2 i 2 Thus the twotransformations described above hav e separated the Hamiltonian into 3N uncoupled harmonic oscillator Hamiltonians, and are usually referred to as a normal mode transformation. Remember that in general the normal mode coordinates are not the original displacements from the equilibrium positions, but correspond to collective vibrations of the molecule. 2. Normal Modes in Classical Mechanics The starting point for the normal mode analysis in classical mechanics are Newton’sequa- tions for a system of coupled harmonic oscillators. Still, using the notation that led to Eq. (1.12), we can write the classical equations of motion as: M∆¨ (t) =−K∆(t), (2.1) where components of the left-hand-side of the equation are the rates of change of momentum of the nuclei, while the right-hand-side contains the harmonic forces. Winter Term, 2019 Chemistry 365 -3- Normal Mode Analysis Since we expect harmonic motion, we’ll look for a solution of the form: ∆(t) = cos(ω t + φ)Y,(2.2) where φ is an arbitrary phase shift. If Eq. (2.2) is used in Eq. (2.1) it follows that: (Mω 2 − K)Y = 0, (2.3) where remember that Y is a column vector and M,and K are matrices. We want to solvethis homogeneous system of linear equations for Y.Ingeneral, the only way to get a nonzero solu- tion is to makethe matrix [Mω 2 − K]singular; i.e., we must set − − det(Mω 2 − K) = det(ω 2 − M 1/2KM 1/2)det(M) = 0. (2.4) = 3 ≠ Since, det(M) (M1 M2...M N ) 0wesee that the second equality is simply the characteristic equation associated with the eigenvalue problem: Ku˜ = λu,(2.5) with λ ≡ ω 2 and with K˜ givenbyEq. (2.7) above.The remaining steps are equivalent to what wasdone quantum mechanically. 3. Force Constant Calculations Here is an example of a force constant matrix calculation. We will consider a diatomic molecule, where the twoatoms interact with a potential of the form: 1 2 U(r , r ) ≡ K |r − r | − R ;(3.1) 1 2 2 1 2 0 i.e., a simple Hookian spring. It is easy to takethe various derivativesindicated in the preceding sections; here, however, wewill explicitly expand the potential in terms of the atomic displace- ments, ∆i.Bywriting, ri = Ri +∆i (where Ri is the equilibrium position of the i’th nucleus), Eq., (3.1) can be rewritten as: 2 1 1/2 = 2 + ⋅ ∆ −∆ + ∆ −∆ 2 − U(r1, r2) K R12 2R12 ( 1 2) | 1 2| R0 ,(3.2) 2 ≡ − = where R12 R1 R2.Clearly,the equilibrium will have |R12| R0.Moreover, weexpect that the vibrational amplitude will be small, and thus, the terms in the ∆’s inthe square root in Eq. (3.2) will be small compared with the first term. The Taylor expansion of the square root implies that B √ √ A+ B ≈ √ A + +..., (3.3) 2 √ A we can write ⋅ ∆ −∆ + ∆ −∆ 2 2 = 1 + 2R12 ( 1 2) | 1 2| − U(r1, r2) KR12 R0 ,(3.4) 2 2R12 which can be rewritten as 1 U(r , r ) = K[Rˆ ⋅ (∆ −∆ )]2,(3.5) 1 2 2 12 1 2 Winter Term, 2019 Chemistry 365 -4- Normal Mode Analysis where the ˆ denotes a unit vector and where all terms smaller than quadratic in the nuclear dis- placements have been dropped. If the squareisexpanded, notice the appearance of cross terms in the displacements of 1 and 2. It is actually quite simple to finish the normal mode calculation in this case. To doso, define the equilibrium bond to point in along the x axis. Equation (3.5) shows that only x-com- ponents of the displacements cost energy,and hence, there will no force in thy y or z directions (thereby resulting in 4 zero eigen-frequencies). Forthe x components, Newton’sequations become: ∆¨x − ∆x m1 0 ⋅ 1 =− K K ⋅ 1 ∆x − ∆x,(3.6) 0 m2 2 K K 2 where mi is the mass of the i’th nucleus. This in turn leads to the following characteristic equa- tion for the remaining frequencies: m 0 K −K = 1 ω 2 − 0 det − 0 m2 K K = ω 2 − ω 2 − − 2 (m1 K)(m2 K) K and = ω 2(µω 2 − K), ≡ + where µ m1m2/(m1 m2)isthe reduced mass. Thus we pick up another zero frequencyand a nonzero one with ω = √ K/µ,which is the usual result. (Note that we don’tcount ± roots twice--WHY?). 4. Normal Modes in Crystals The main result of the preceding sections is that the characteristic vibrational frequencies are obtained by solving for eigenvalues, cf., Eq. (2.5). For small to mid-size molecules this can be done numerically,onthe other hand, this is not practical for crystalline solids where the matri- ces are huge (3N × 3N with N∼O(N A))! Crystalline materials differ from small molecules in one important aspect; theyare periodic structures made up of identical unit cells; specifically,each unit cell is placed at → = → + → + → R n1a1 n2a2 n3a3,(4.1) → where ni is an integer and ai is a primitive lattice vector.The atoms are positioned within each unit cell at positions labeled by an index, α .Thus, the position of anyatom in entire crystal is determined by specifying R and α .Hence, Eq. (2.5) can be rewritten as λ = ˜ uR,α ΣΣ KR,α ;R′,α ′uR′,α ′.(4.2) R′ α ′ The periodicity of the lattice implies that only the distance between unit cells can matter; i.e., ˜ = ˜ KR,α ;R′,α ′ KR−R′;α ,α ′,and this turns Eq. (4.2) into a discrete convolution that can be simplified by introducing a discrete Fourier transform, i.e., we let ≡ ik⋅R u˜ k,α Σ e uR,α .(4.3) R Winter Term, 2019 Chemistry 365 -5- Normal Mode Analysis When applied to both sides of Eq.
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