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 position 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 separated. This is ﬁrst shown using using the quantum mechanical framework we’ve just set up.

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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 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 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 normalized 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’sequations 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.

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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 solution 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 displacements, ∆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) KR12 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

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where the ˆ denotes a unit vector and where all terms smaller than quadratic in the nuclear displacements 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-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). 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 equation 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 matrices 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

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When applied to both sides of Eq. (2.5), this gives* λ ≡ ˜ u˜ k,α Σ Kk;α ,α ′u˜ k,α ′,(4.4) α where ˜ ≡ ik⋅R ˜ Kk;α ,α ′ Σ e KR;α ,α ′.(4.5) R

The resulting eigenvalue problem has rank 3Ncell,where Ncell is the number of atoms in a unit cell, and is easily solved numerically. We hav en’tspecified 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 m1b1 m2b2 m3b3,(4.6) † where the mi’s are integers, and where the reciprocal lattice basis vectors are defined as → → → → → → → a × a → a × a → a × a ≡ π 2 3 ≡ π 3 1 ≡ π 1 2 b1 2 → ⋅ → × → , b2 2 → ⋅ → × → ,and b3 2 → ⋅ → × → .(4.7) |a1 (a2 a3)| |a2 (a3 a1)| |a3 (a1 a2)| Note the following: → a) The denominators appearing in the definitions of the bi are equal; i.e., → ⋅ → × → = → ⋅ → × → = → ⋅ → × → = ν |a1 (a2 a3)| |a2 (a3 a1)| |a3 (a1 a2)| cell,(4.8)

where ν cell is the volume of the unit cell. The volume of the reciprocal lattice unit cell is 3 (2π ) /ν cell. b) By construction, → → ⋅ = πδ ai b j 2 i, j,(4.9) δ δ where i, j is a Kronecker- . c) The real and reciprocal lattices need not be the same, evenignoring howthe basis vectors are normalized; e.g., theyare for the SCC, but the reciprocal lattice for the BCC lattice is a FCC lattice. → → d) For anylattice vector, R,and reciprocal lattice vector, G,

→ → ⋅ π + + eiR G = e2 i(m1n1 m2n2 n3m3) = 1, (4.10) → cf. Eqs. (4.1), (4.6), and (4.9). Hence, adding anyreciprocal 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 k1b1 k2b2 k3b3,(4.11) ≡ = − = where ki mi/Ni with mi 0, 1, 2, . . . , Ni 1. Note that N1 N2 N3 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.

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→ * cells in the crystal. The Ni’s are the number of cells in the ai direction. With this choice , consider   → → → → → 3 Ni−1 1 −ik⋅R 1 ik⋅(R′−R) 1 2π im (n ′−n )/N → = → = → Π i i i i  Σ e ˜u k,α ΣΣ e uR′,α Σ uR′,α Σ e .(4.12) N → N → → → i=1 N = k k R′ R′  i mi 0  The sums in parenthesis are geometric series and give

π ′− N −1  2 i(ni ni)  1 i π ′− 1 e − 1 2 imi(ni ni)/Ni = = δ Σ e  π ′−  n ′,n ,(4.13) = 2 i(ni ni)/Ni − i i Ni mi 0 Ni  e 1 which when used in Eq. (4.12) shows that

→ → 1 −ik⋅R → = → Σ e ˜u α u α .(4.14) → k, R, N k → The sums over k are really sums overthe mi’s,and thus, ki hardly changes as we go from mi → mi + 1for large Ni,cf. Eq. (4.11). This allows the sums over mi to be replaced by integrals; i.e.,

3 N −1 → → Π i = V Σ ... ∫ dmi ... ∫∫∫ dk ..., (4.15) → i=1 0 (2π )3 k 1st Brillouin zone

where V = Nν cell is the volume of the system. The main goal of this discussion is to obtain the vibrational density of states. By using Eq. (4.15), it is easy to showthat

V → g(ω ) = ∫∫∫ dk δ (ω − ω (k)), (4.16) (2π )3 1st Brillouin zone where δ (x)isthe Dirac δ -function. Once the frequencies are known, it is relatively easy to numerically bin them by frequencyasafunction 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’sequations of motion for this atom become

→ ∆¨ = ∆x +∆x − ∆x + ∆y +∆y − ∆y m 0,0,0 K[ˆex( 1,0,0 −1,0,0 2 0,0,0) eˆy( 0,1,0 0,−1,0 2 0,0,0) + ∆x +∆x − ∆x eˆz( 0,0,1 0,0,1 2 0,0,0)], (5.1)

where eˆi 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 mi’s is mi =−Ni/2,..., 0,..., Ni/2,which, cf. Eq. (4.11), means that −1/2 ≤ ki ≤ 1/2.

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 2 π  sin (k1 ) 0 0 ˜ =− ω 2 2 π  Kk 4 0 0 sin (k2 ) 0 ,(5.2) 2 π  0 0 sin (k3 )  where ω ≡ (K/m)1/2 and where the identity,1− cos(x) = 2sin2(x/2)was used. Since K˜ is 0 → k ω = ω π = already diagonal, we see that i(k) 2 0 sin(ki ), i 1, 2, 3, which is also the result for a one dimensional chain (as was expected, giventhe separability of the vibrations in the x, y,and z directions). The normal mode is just eî,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 kz.When k x = 0wedon’tget a single zero fre- 2/3 quency, asexpected, but one for anyofthe N y Nz values of (k y, kz). Thus, with 3N zero frequencies, not 6, the crystal is unstable! The problem is easily fixed by modifying the interaction potential, i.e., we replace Eq. (3.5) by K → → U ≡ (∆ − ∆ )2,(5.3) 2 1 2 which is invariant under rigid translations and rotations, and allows us to rewrite Eq. (5.1) as

→ → → → → → → → → → ∆¨ = ∆ + ∆ − ∆ + ∆ + ∆ − ∆ + ∆ + ∆ − ∆ m 0,0,0 K( 1,0,0 −1,0,0 2 0,0,0 0,1,0 0,−1,0 2 0,0,0 0,0,1 0,0,1 2 0,0,0). (5.4) With this, it follows that ↔ ˜ =− ω 2 2 π + 2 π + 2 π Kk 4 0[sin (k1 ) sin (k2 ) sin (k3 )]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 = ω 2 π + 2 π + 2 π 1/2 i(k) 2 0[sin (k1 ) sin (k2 ) sin (k3 )] .(5.6) → → ω ∼ ω π ≡ 2 + 2 + 2 1/2 If all the ki 0, Eq. (5.6) becomes (k) 2 0k ,where k (k1 k2 k3) which is the expected linear dispersion lawfor long wevelength, acoustic vibrations. Some constant fre- quencysurfaces in the First Brillouin Zone are shown in Figs. 5.1-5.3.

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ω ω = Fig. 5.1. Aconstant normal mode frequencysurface Fig. 5.2. As in Fig. 5.1 but with / 0 2. Notice ω ω = for / 0 1. The reciprocal lattice basis was used the C4 axises through the centers of anyofthe unit which need not be orthogonal (although it is for the cell faces. SCC lattice). Also note that we’ve switched to the other deﬁnition of the ﬁrst Brillouin zone, with −1/2 ≤ ki ≤ 1/2.The surface is roughly spherical, as expected for long wav elength acoustic phonons.

Fig. 5.3. As in Fig. 5.1 but with ω ω = / 0 3.

Normal mode frequencies for this model were computed numerically for a uniform sample of ki’s inthe First Brillouin Zone and binned in order to get the vibrational density of of states as shown in Fig. 5.4

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Fig. 5.4. Vibrational density of states for the isotropic model deﬁned by Eq. (5.3). Anysingle atom per unit cell lattice will give the same result. The data wasobtained by numerically binning the frequencies in the First Brillouin √ ω Zone. Note that the maximum frequencyfor this model is 2 3 0.

By noting that δ (x − x ) δ ( f (x)) = Σ i ,(5.7) i  df (x )  i   dxi 

where f (x)has zeros at x = xi, i = 1, . . ., and that ∞ ds δ (x) = ∫ eixs,(5.8) −∞ 2π we can rewrite Eq. (4.16) as g(ω ) 1 1 1 = ω δ ω 2 − ω 2 2 ∫ dk1 ∫ dk2 ∫ dk3 ( (k)) 3N 0 0 0 ∞ ω 1 1 1 = is(ω 2−ω 2(k)) ∫ ds ∫ dk1 ∫ dk2 ∫ dk3 e π −∞ 0 0 0 ∞ ω 2 = ds eisω Φ3(s), π ∫−∞ where the extra factor of 3 is due to the triple degeneracyofeach mode and where 1 − iω 2s 2 π k Φ(s) ≡ ∫ dk e 4 0 sin ( ) 0 When we use Eq. (5.7) for the frequencies, it follows that the wav evector integrations separate and

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∞ ds 2 g(ω ) = 2Vω ∫ eisω Φ3(s), (5.9) −∞ 2π where − ω 2 2 0is π dk1 −is ω 2 2 π k e ω 2is z Φ(s) ≡ ∫ e 4 0 sin ( 1) = ∫ dz e2 0 cos( ) 2π 2π 2 0

− ω 2is ω 2isz − ω 2is e 2 0 1 e2 0 e 2 0 = = ω 2 ∫ dz J0(2 0 s), (5.10) 2π 2 −1 (1 − z2)1/2 2π * where J0(x)isaBessel function of the ﬁrst kind. When this is used in Eq. (5.10), we see that ∞ 2Vω ω 2− ω 2 ω = is( 6 0) 3 ω 2 g( ) ∫ ds e J0 (2 0 s){5.11) (2π )4 −∞

*The integral leading to the last equality can be found in I.S. Gradsheyn and I.M. Ryzhik, Table of Inte- grals, Series, and Products,A.Jeffreyeditor,(Academic Press, 1980), Eq. (3.387.2) on p. 321.

Winter Term, 2019