Chemistry 365: Normal Mode Analysis David Ronis McGill University 1. Quantum Mechanical Treatment Our starting point is the Schrodinger wave equation: Σ h 2 2 2m i N i=1 r 2 i + U( r 1,..., r N ) Ψ( r 1,..., r N ) = E Ψ( r 1,..., r N ), (1.1) where N is the number of atoms in the molecule, m i is the mass of the i th atom, and U( r 1,..., 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 Hamiltonian associated with Eq. (1.1) can be written as: H vib Σ N h 2 2 2m i i=1 2 i + U + 1 2 N Σ K i, j : i j, (1.2) i, j=1 R i is the equilibrium posi- where U is the minimum value of the potential energy, i r i R i, tion of the i th atom, and K i, j 2 U (1.3) r i r j r k = R k is the matrix of (harmonic) force constants. Henceforth, we will shift the zero of energy so as to make U =. 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 1, y 1, z 1,..., x N, y N, z N ]T, (1.4) where "T "denotes a matrix transpose. Similarly, we can encode the force constants or masses into 3N 3N matrices, and thereby rewrite Eq. (1.2) as H vib = h 2 M 1 2 + 1 2 K, (1.5) where all matrix quantities are emboldened, the superscript " " denotes the Hermitian conjugate (matrix transpose for real matrices), and where the mass matrix, M, is a diagonal matrix with the masses of the given atoms each appearing three times on the diagonal. The Hamiltonian given by Eq. (1.5) is the generalization of the usual harmonic oscillator Hamiltonian to include more particles and to allow for "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 now show, it can be separated. This is first shown using using the quantum mechanical framework we ve just set up.
Chemistry 365-2- Normal Mode Analysis Later, we will discuss the separation using classical mechanics. This is valid since, for harmonic oscillators, you get the same result for the vibration frequencies. We now make the transformation i m 1/2 i i,which allows Eq. (1.5) to be reexpressed in the new coordinates as where H vib = h 2 2 + 1 2 K, (1.6) K M 1/2 KM 1/2. (1.7) Since the matrix K is Hermitian (or symmetric for real matrices), it is possible to find a unitary (also referred to as an orthogonal matrix for real matrices) matrix which diagonalizes it; i.e., you can find a matrix P which satisfies where λ is diagonal (with real eigenvalues) and where P KP = λ or K = PλP, (1.8) PP = 1 (1.9) where 1 is the identity matrix. Note that P is the unitary matrix who s columns are the normalized eigenvectors. (See a good quantum mechanics book or any linear algebra text for proofs of these results). We now make the transformation in Eq. (1.5). This gives P (1.1) H vib = h 2 P P 2 + 1 2 λ, (1.11) Finally, Eq. (1.8) allows us to cancel the factors of P in this last equation and write H vib = 3N Σ 1 2 h 2 i=1 2 2 i + 1 2 λ i 2 i. (1.12) Thus the two transformations described above have 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 s equations 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.
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 =, (2.3) where remember that Y is a column vector and M, and K are matrices. We want to solve this homogeneous system of linear equations for Y. In general, the only way to get a nonzero solution is to make the matrix [Mω 2 K] singular; i.e., we must set det(mω 2 K) = det(ω 2 M 1/2 KM 1/2 )det(m) =. (2.4) Since, det(m) = (M 1 M 2...M N ) 3 wesee 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 given byeq. (2.7) above. The remaining steps are equivalent to what was done quantum mechanically. 3. Force Constant Calculations Here is an example of a force constant matrix calculation. We will consider a diatomic molecule, where the two atoms interact with a potential of the form: U(r 1, r 2 ) 1 2 2 K r 1 r 2 R ; (3.1) i.e., a simple Hookian spring. It is easy to take the various derivatives indicated in the preceding sections; here, however, we will explicitly expand the potential in terms of the atomic displacements, i. Bywriting, r i = R i + i (where R i is the equilibrium position of the i th nucleus), Eq., (3.1) can be rewritten as: U(r 1, r 2 ) = 1 1/2 2 2 K R2 12 + 2R 12 ( 1 2 ) + 1 2 2 R, (3.2) where R 12 R 1 R 2. Clearly, the equilibrium will have R 12 = R. Moreover, we expect that the vibrational amplitude will be small, and thus, the terms in the s in the square root in Eq. (3.2) will be small compared with the first term. The Taylor expansion of the square root implies that we can write which can be rewritten as A + B A + U(r 1, r 2 ) = 1 2 K R 12 + 2R 12 ( 1 2 ) + 1 2 2 B +..., (3.3) 2 A 2R 12 2 R, (3.4) U(r 1, r 2 ) = 1 2 K[ ˆR 12 ( 1 2 )] 2, (3.5)
Chemistry 365-4- Normal Mode Analysis where the ˆ denotes a unit vector and where all terms smaller than quadratic in the nuclear displacements have been dropped. If the square is expanded, 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 do so, define the equilibrium bond to point in along the x axis. Equation (3.5) shows that only x-components of the displacements cost energy, and hence, there will no force in thy y or z directions (thereby resulting in 4 zero eigen-frequencies). For the x components, Newton s equations become: m 1 m 2 1 x = K K x 2 K K 1 x 2 x, (3.6) where m i is the mass of the i th nucleus. This in turn leads to the following characteristic equation for the remaining frequencies: and = det m 1 m 2 ω 2 K K = (m 1 ω 2 K)(m 2 ω 2 K) K 2 = ω 2 (µω 2 K), where µ m 1 m 2 /(m 1 + m 2 )isthe reduced mass. Thus we pick up another zero frequency and a nonzero one with ω = K/µ, which is the usual result. (Note that we don t count ± 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 matrices are huge (3N 3N with N O(N A ))! Crystalline materials differ from small molecules in one important aspect; they are periodic structures made up of identical unit cells; specifically, each unit cell is placed at K K R = n 1 a 1 + n 2 a 2 + n 3 a 3, (4.1) where n i is an integer and a i is a primitive lattice vector. The atoms are positioned within each unit cell at positions labeled by an index, α. Thus, the position of any atom in entire crystal is determined by specifying R and α.hence, Eq. (2.5) can be rewritten as λu R,α = Σ R α Σ K R,α ;R,α u R,α. (4.2) The periodicity of the lattice implies that only the distance between unit cells can matter; i.e., K R,α ;R,α = K R 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 ũ k,α Σ e ik R u R,α. (4.3) R
Chemistry 365-5- Normal Mode Analysis When applied to both sides of Eq. (2.5), this gives * where λũ k,α K k;α,α Σ K k;α,α ũ k,α, (4.4) α Σ e ik R K R;α,α. (4.5) R The resulting eigenvalue problem has rank 3N cell,where N cell is the number of atoms in a unit cell, and is easily solved numerically. We haven t specified the values for the k s. Itturns out that a very convenient choice is to use k s expressed in terms of the so-called reciprocal lattice vectors, G,i.e., G m 1 b 1 + m 2 b 2 + m 3 b 3, (4.6) where the m i s are integers, and where the reciprocal lattice basis vectors are defined as a b 1 2 a 3 2π a 1 ( a 2 a 3 ), a b 2 3 a 1 2π a 2 ( a 3 a 1 ), and b a 3 1 a 2 2π a 3 ( a 1 a 2 ). (4.7) Note the following: a) The denominators appearing in the definitions of the b i are equal; i.e., a 1 ( a 2 a 3 ) = a 2 ( a 3 a 1 ) = a 3 ( a 1 a 2 ) = ν cell, (4.8) where ν cell is the volume of the unit cell. The volume of the reciprocal lattice unit cell is (2π ) 3 /ν cell. b) By construction, where δ i, j is a Kronecker-δ. a i b j = 2πδ i, j, (4.9) c) The real and reciprocal lattices need not be the same, even ignoring how the basis vectors are normalized; e.g., they are for the SCC, but the reciprocal lattice for the BCC lattice is a FCC lattice. d) For any lattice vector, R,and reciprocal lattice vector, G, e i R G = e 2π i(m 1n 1 +m 2 n 2 +n 3 m 3 ) = 1, (4.1) cf. Eqs. (4.1), (4.6), and (4.9). Hence, adding any reciprocal a lattice vector to the k in Eq. (4.4) changes nothing, and so, we restrict k to what is known as the First Brillouin Zone; specifically, k = k 1 b 1 + k 2 b 2 + k 3 b 3, (4.11) where k i m i /N i with m i =, 1, 2,..., N i 1. Note that N 1 N 2 N 3 = N, the total number of * Strictly speaking, in obtaining Eq. (4.5) we have assumed some special properties of the lattice. We don t use a finite lattice, rather one that obeys periodic boundary conditions (see below). Here we re using the definition of C. Kittel, Introduction to Solid State Physics (Wiley, 1966), p. 53. There are others, see, e.g., M. Born and K. Huang, Dyamical Theory of Crystal Lattices (Oxford, 1968), p. 69.
Chemistry 365-6- Normal Mode Analysis cells in the crystal. The N i s are the number of cells in the a i direction. With this choice *, consider 1 N k Σ e i k Rũ k,α = 1 N k Σ Σ e i k ( R R) 3 u = R,α Σ u Π 1 R,α The sums in parenthesis are geometric series and give 1 N i N i 1 m i = R Σ e 2π im i(n i n i )/N i which when used in Eq. (4.12) shows that 1 N k R i=1 N i N i 1 Σ e 2π im i(n i n i )/N i. (4.12) m i = = 1 e 2π i(n i n ) i 1 N i e 2π i(n i n i )/N i = δ 1 ni,n i, (4.13) Σ e i k Rũ k,α = u R,α. (4.14) The sums over k are really sums over the m i s, and thus, k i hardly changes as we go from m i m i + 1for large N i,cf. Eq. (4.11). This allows the sums over m i to be replaced by integrals; i.e., Σ... k 3 Π N i 1 i=1 dm i...= where V = Nν cell is the volume of the system. V (2π ) d k..., (4.15) 3 1st Brillouin The main goal of this discussion is to obtain the vibrational density of states. By using Eq. (4.15), it is easy to show that g(ω ) = zone V (2π ) d k 3 δ (ω ω (k)), (4.16) 1st Brillouin zone where δ (x) is the Dirac δ -function. Once the frequencies are known, it is relatively easy to numerically bin them by frequency asafunction of k. 5. Normal Modes in Crystals: An Example Consider a crystal with one atom of mass m per unit cell and nearest neighbor interactions of the type considered in Sec. 3. Newton s equations of motion for this atom become m,, = K[ê x ( x 1,, + x 1,, 2 x,,) + ê y ( y,1, + y, 1, 2 y,, ) +ê z ( x,,1 + x,,1 2 x,,)], (5.1) where ê i is a unit vector in the i direction. Note that this model implies that vibrations in the x, y, z, directions are separable. By using this in Eq. (4.5) we see that * The more common choice for the range of the m i s is m i = N i /2,...,,..., N i /2, which, cf. Eq. (4.11), means that 1/2 k i 1/2.
Chemistry 365-7- Normal Mode Analysis sin 2 (k 1 π ) K k = 4ω 2 sin 2 (k 2 π ) sin 2 (k 3 π ), (5.2) where ω (K/m) 1/2 and where the identity, 1 cos(x) = 2sin 2 (x/2) was used. Since K k is already diagonal, we see that ω i ( k) = 2ω sin(k i π ), i = 1, 2, 3, which is also the result for a one dimensional chain (as was expected, given the separability of the vibrations in the x, y, and z directions). The normal mode is just ê i,which shows that our model potential is too simple. Consider the first eigen-vector. Itcorresponds to an arbitrary displacement in the x direction, with an eigenvalue that is independent of k y and k z. When k x = wedon t get a single zero frequency, asexpected, but one for any ofthe N y N z values of (k y, k z ). Thus, with 3N 2/3 zero frequencies, not 6, the crystal is unstable! by The problem is easily fixed by modifying the interaction potential, i.e., we replace Eq. (3.5) U K 2 ( 1 2 ) 2, (5.3) which is invariant under rigid translations and rotations, and allows us to rewrite Eq. (5.1) as m,, = K( 1,, + 1,, 2,, +,1, +, 1, 2,, +,,1 +,,1 2,, ). (5.4) With this, it follows that K k = 4ω 2 [sin 2 (k 1 π ) + sin 2 (k 2 π ) + sin 2 (k 3 π )] 1, (5.5) where 1 is a 3 3identity matrix and where we have expressed k in terms of the reciprocal lattice basis vectors, cf. Eq (5.11). The triply degenerate vibrational frequencies are simply ω i ( k) = 2ω [sin 2 (k 1 π ) + sin 2 (k 2 π ) + sin 2 (k 3 π )] 1/2. (5.6) If all the k i, Eq. (5.6) becomes ω ( k) 2ω kπ,where k (k 2 1 + k 2 2 + k 2 3) 1/2 which is the expected linear dispersion law for long wevelength, acoustic vibrations. Some constant frequency surfaces in the First Brillouin Zone are shown in Figs. 5.1-5.3.
Chemistry 365-8- Normal Mode Analysis Fig. 5.1. A constant normal mode frequency surface for ω /ω = 1. The reciprocal lattice basis was used which need not be orthogonal (although it is for the SCC lattice). Also note that we ve switched to the other definition of the first Brillouin zone, with 1/2 k i 1/2. The surface is roughly spherical, as expected for long wavelength acoustic phonons. Fig. 5.2. As in Fig. 5.1 but with ω /ω = 2. Notice the C 4 axises through the centers of any ofthe unit cell faces. Fig. 5.3. ω /ω = 3. As in Fig. 5.1 but with Normal mode frequencies for this model were computed numerically for a uniform sample of k i s in the First Brillouin Zone and binned in order to get the vibrational density of of states as shown in Fig. 5.4
Chemistry 365-9- Normal Mode Analysis Fig. 5.4. Vibrational density of states for the isotropic model defined by Eq. (5.3). Any single atom per unit cell lattice will give the same result. The data was obtained by numerically binning the frequencies in the First Brillouin Zone. Note that the maximum frequency for this model is 2 3ω. By noting that where f (x) has zeros at x = x i, i = 1,..., and that we can rewrite Eq. (4.16) as g(ω ) 3N δ ( f (x)) = Σ δ (x x i), (5.7) i df (x i ) dx i δ (x) = ds 2π eixs, (5.8) = 2ω 1 dk 1 1 dk 2 1 dk 3 δ (ω 2 ω 2 (k)) = ω π = ω π ds 1 dk 1 1 dk 2 1 dk 3 e is(ω 2 ω 2 (k)) ds eisω 2 Φ 3 (s), where the extra factor of 3 is due to the triple degeneracy ofeach mode and where Φ(s) 1 dk e 4iω 2 s sin2 (π k) When we use Eq. (5.7) for the frequencies, it follows that the wavevector integrations separate and
Chemistry 365-1- Normal Mode Analysis where g(ω ) = 2Vω Φ(s) dk 2 1 2 2π e is4ω sin 2 (π k ) 1 = e 2ω is 2π 2 ds 2π eisω 2 Φ 3 (s), (5.9) π dz e2ω 2 is cos(z) 1 2π dz e 2ω 2 isz 2 2 1 (1 z 2 ) = e 2ω 1/2 2π = e 2ω 2 is is J (2ω 2 s), (5.1) where J (x) isabessel function of the first kind. * When this is used in Eq. (5.1), we see that g(ω ) = 2Vω (2π ) 4 ds eis(ω 2 6ω 2 ) J 3 (2ω 2 s) {5.11) * The integral leading to the last equality can be found in I.S. Gradsheyn and I.M. Ryzhik, Table of Integrals, Series, and Products, A.Jeffrey editor, (Academic Press, 198), Eq. (3.387.2) on p. 321.