Beyond the independent oscillator model

The Taylor series of the BO potential in nuclear coordinatesdoes not immediately yield independent oscillators.

Consider a diatomic molecule (e.g. H2), and only allowmotion in one dimension for simplicity.

Taylor series of Bo potential about equilibrium would yield:

V (x1, x2) =1

2

∂2V

∂x21

∣∣∣∣0

x21 +1

2

∂2V

∂x22

∣∣∣∣0

x22 +∂2V

∂x1∂x2

∣∣∣∣0

x1x2

=1

2γax

21 +

1

2γbx

22 + γcx1x2

where γa, γb, γc are derivatives that define the potential.

Cross term x1x2 gives rise to “coupled” oscillators.

In a crystal, there are many cross terms.

Need a general approach for coupled oscillators.

09/23/2021 at 13:24:09 slide 1

Coupled Harmonic Oscillator in 1d

m mxa→

xb→γγ γ

H =p2a2m

+p2b2m

+1

2γx2a +

1

2γx2b +

1

2γ(xb − xa)

2

Eqn of motion for each mass, mxi = Fi = −∂H∂xi

mxa = −γxa − γ(xa − xb) = −γ(2xa − xb)

mxb = −γxb − γ(xb − xa) = −γ(−xa + 2xb)

Get two new equations by first adding, then subtracting:

md2

dt2(xa + xb) = −γ(xa + xb) → xB +

γ

mxB = 0

md2

dt2(xa − xb) = −3γ(xa − xb) → xA +

3γ

mxA = 0

Where we have defined new variables:

xB = xa + xb xA = xa − xb

09/23/2021 at 13:24:09 slide 2

Coupled Harmonic Oscillator in 1d

We see that xB and xA are independent SHO with frequenciesω2B = γ

m and ω2A = 3γ

m .

What appeared to be a hard problem is in fact not: it is“uncoupled” or “non-interacting”.

The general solution is immediately known:

xB(t) = xoB cosωBt +xoBωB

sinωBt

xA(t) = xoA cosωAt +xoAωA

sinωAt

where xoB , xoB , x

oA, x

oA are the positions/velocities at t = 0.

We can then simply recover xa and xb as follows:

xa(t) =1

2(xB(t) + xA(t)) xb(t) =

1

2(xB(t)− xA(t))

What do modes A and B look like? Consider each alone.09/23/2021 at 13:24:09 slide 3

Picturing the uncoupled A and B modes

First a schematic of the undistorted system.

m m Undistorted System

Pure B mode xa(t) =12xB(t) and xb(t) =

12xB(t):

m→

m→

mω2B = γ

Pure A mode xa(t) =12xA(t) and xb(t) = −1

2xA(t):

m←

m→

mω2A = 3γ

These two mode do not disturb each other in any way: anamplitude on one generates no forces on the other.

09/23/2021 at 13:24:09 slide 4

Hamiltonian of uncoupled modes

Instructive to write Hamiltonian in terms of uncoupled modes.

One point to consider here: when adding/subtractingequations of motion, we can multiply by arbitrary constant.

A useful constant is 1√2, as this makes transformation

“unitary”; to be proven shortly.

So let us assume our decoupling transformation is:

xB =1√2(xa + xb) xA =

1√2(xa − xb)

Inverting this gives:

xa =1√2(xA + xB) xb =

1√2(xB − xA)

Same thing for momentum:

pa =1√2(pA + pB) pb =

1√2(pB − pA)

09/23/2021 at 13:24:09 slide 5

Hamiltonian of uncoupled modes - continued

Our orginal Hamiltonian:

H =p2a2m

+p2b2m

+1

2γx2a +

1

2γx2b +

1

2γ(xb − xa)

2

Substitute in our transformation:

xa =1√2(xA + xB) xb =

1√2(xB − xA)

Consider these two chunks:

x2a + x2b = x2A + x2B (xb − xa)2 = 2x2A

The transformed Hamiltonian is:

Hd =p2A2m

+p2B2m

+1

23γx2A +

1

2γx2B

A and B are called “normal modes”.09/23/2021 at 13:24:09 slide 6

Coupled Harmonic Oscillator in 1d - matrix approach

m mxa→

xb→γγ γ

We won’t be able to see the solution in a more complicatedcase (ie. more masses).

Let us formulate this as a linear algebra problem.

Construct forces on each mass, mx = F = −∂V∂x

mxa = −γxa − γ(xa − xb) = −γ(2xa − xb)

mxb = −γxb − γ(xb − xa) = −γ(−xa + 2xb)

− m

γ

[xaxb

]=

[2 −1−1 2

] [xaxb

]Now we can try a solution xa = c1e

iωt and xb = c2eiωt

09/23/2021 at 13:24:09 slide 7

Coupled Harmonic Oscillator in 1d - matrix approach

Plugging this guess into matrix equation gives:

m

γω2

[c1c2

]=

[2 −1−1 2

] [c1c2

]We have converted a coupled differential equation to analgebraic equation: Eigenvalue Problem.[

2 −1−1 2

] [c1c2

]=

m

γω2

[c1c2

]↔ V |C ⟩ = λ|C ⟩

V is second derivative of potential (or H) wrt displacement.

Stiffness Matrix Vij =∂2H

∂xi∂xj=

∂2V

∂xi∂xj

First derivative comes from generating equations of motion.

The second derivative extracts the prefactor.

09/23/2021 at 13:24:09 slide 8

What are we solving for in Eigenvalue problem?

m and γ are known from experiment or first-principles.[2 −1−1 2

] [c1c2

]=

m

γω2

[c1c2

]↔ V |C ⟩ = λ|C ⟩

We are trying to solve for c1, c2, and ω.

This may seem strange given that we have three unknownsbut only two equations.

We will see that this problem can be solved.

Step back and take a general look at the Eigenvalue problem,but a few points first.

Dimension of V is number of displacements nd .

V is square, real, symmetric matrix (V = V⊤) given thatpotential is analytic (i.e. order of derivatives does not matter).

V =

∂2V∂x2a

∂2V∂xa∂xb

∂2V∂xb∂xa

∂2V∂x2b

09/23/2021 at 13:24:09 slide 9

The Eigenvalue problem V |C ⟩ = λ|C ⟩V is a matrix/operator

|C ⟩ is (column) vector/function (ie. Eigenvector)

λ is the Eigenvalue, which is a scalar.

If V is an nd × nd matrix, there are nd Eigenvalues andcorresponding Eigenvectors; need sub/superscript:

V |Cn⟩ = λn|Cn⟩ n = 1, . . . , nd

We will only consider Hermitian matrices:

V † = V , where V † ≡ (V⊤)∗

⟨C | ≡ (|C ⟩)† is a row matrix.

Note that ⟨Ci |Cj⟩ is a scalar and |Ci ⟩⟨Ci | is matrix.

Hermitian matrices are nice (prove later):

Eigenvalues of V are realEigenvectors of V form an orthonormal set: ⟨Ci |Cj⟩ = δij

Eigenvectors of V form complete set:∑

i |Ci ⟩⟨Ci | = 1

Symmetric matrices are Hermitian matrices.09/23/2021 at 13:24:09 slide 10

How to solve Eigenvalue problem

Move λ|C ⟩ to other side

V |C ⟩ = λ|C ⟩ → (V − λ1)|C ⟩ = 0

If (V − λ1)−1 exists, trivial solution:

(V − λ1)−1(V − λ1)|C ⟩ = |C ⟩ = 0

Demand (V − λ1)−1 not exist.

Inverse of a matrix B can be written as

B−1 =1

det(B)C⊤

C is cofactor matrix; not essential.

B−1 doesn’t exist if det(B) = 0.

Therefore, we must solve polynomial eqn

|V − λ1| = 0

Inversion with 2x2 matrix.

B =

[a bc d

]|B| = ad − cb

B−1 =1

|B|

[d −b−c a

]

BB−1 =

[1 00 1

]

09/23/2021 at 13:24:09 slide 11

How to solve Eigenvalue problem - continued

Need to solve polynomial equation for λ of order nd .

The following defines the polynomial equation.

|V − λ1| = 0

Let us return to the two mass problem.∣∣∣∣[ 2 −1−1 2

]− 1

m

γω2

∣∣∣∣ = 0 →∣∣∣∣ 2− λ −1−1 2− λ

∣∣∣∣ = 0

Determinant results in polynomial eqn.

Zeros of order 2 polynomial is exactly solvable.

λ2 − 4λ+ 3 = 0 →λB = 1

λA = 3→

ωB =√

γm

ωA =√

3γm

Now let us find the eigenvectors by plugging in eigenvalues...09/23/2021 at 13:24:09 slide 12

Finding the Eigenvectors

Begin by plugging in λB = 1[2 −1−1 2

] [c1c2

]= 1

[c1c2

]→ c1 = c2 → |CB⟩ =

1√2

[11

]Now plug in λA = 3[

2 −1−1 2

] [c1c2

]= 3

[c1c2

]→ c1 = −c2 → |CA⟩ =

1√2

[1−1

]Check that vectors are orthonormal → ⟨Ci |Cj⟩ = δij

Completeness given by∑

i |Ci ⟩⟨Ci | = 1. Let us check:

|CA⟩⟨CA|+ |CB⟩⟨CB | =1

2

[1 11 1

]+

1

2

[1 −1−1 1

]=

[1 00 1

]

09/23/2021 at 13:24:09 slide 13

Picturing the uncoupled modes

We show the undistorted configuration, the asymmetric mode,and the symmetric mode.

We can see the effective spring constant of each mode bytaking the matrix element.

m m

m→

m→

⟨CB |V |CB⟩ = γ2

[1 1

] [ 2 −1−1 2

] [11

]= γ

m←

m→

⟨CA|V |CA⟩ = γ2

[−1 1

] [ 2 −1−1 2

] [−11

]= 3γ

m m→

γ[0 1

] [ 2 −1−1 2

] [01

]= 2γ

09/23/2021 at 13:24:09 slide 14

Proving properties of Hermitian matrix

Hermitian matrices have three key properties: Real Evals,orthonormal Evecs, and complete Evecs.

Let us first recall the transpose of a matrix product:

(AB)⊤ = B⊤A⊤

This follows from definitions of transpose and multiplication.

A⊤ij = Aji (AB)ij =

∑k

AikBkj

We can then prove:

(AB)⊤ij =∑k

AjkBki =∑k

A⊤kjB

⊤ik =

∑k

B⊤ikA

⊤kj = (B⊤A⊤)ij

Dagger is conjugate transpose; same property holds.

(AB)† = B†A†

09/23/2021 at 13:24:09 slide 15

Eigenvalues are real

Let us start with Eval probem and take dagger:

V |C ⟩ = λ|C ⟩ ⟨C |V † = λ∗⟨C |

For Hermitian matrix V = V †

V |C ⟩ = λ|C ⟩ ⟨C |V = λ∗⟨C |

Multiply 1st eq. from left by ⟨C |; 2nd from right by |C ⟩:

λ =⟨C |V |C ⟩⟨C |C ⟩ = λ∗

This proves Eigenvalue must be real.

09/23/2021 at 13:24:09 slide 16

Proving orthonormality of Eigenvectors - nondegenerate

We can always choose to normalize a given vector to 1, so wehave ⟨Ci |Ci ⟩ = 1

Prove that ⟨Ci |Cj⟩ = 0 when i = j .

First treat case where Evals are all distinct, which are called“nondegenerate” Evals; consider two Evals/Evecs:

V |Ci ⟩ = λi |Ci ⟩ V |Cj⟩ = λj |Cj⟩

Eq. i : dagger and mutliply by |Cj⟩; Eq. j : mutliply by ⟨Ci |

⟨Ci |V |Cj⟩ = λi ⟨Ci |Cj⟩ ⟨Ci |V |Cj⟩ = λj⟨Ci |Cj⟩

Subtract two equations.

(λi − λj)⟨Ci |Cj⟩ = 0→ ⟨Ci |Cj⟩ = 0

09/23/2021 at 13:24:09 slide 17

Proving orthonormality of Eigenvectors - degenerate

Consider degenerate Evecs.

V |Ci ⟩ = λ|Ci ⟩ V |Cj⟩ = λ|Cj⟩

For degenerate Evecs, any linear combination is also an Evec.

|Ck⟩ = ai |Ci ⟩+ aj |Cj⟩

V |Ck⟩ = aiλ|Ci ⟩+ ajλ|Cj⟩ = λ|Ck⟩

The two Evecs must be linearly independent given that theyare not the same vector.

Can always use Gram-Schmidt to orthoganilize linearlyindependent set.

|Ci ⟩ and1√

⟨Cj |(1− |Ci ⟩⟨Ci |)(1− |Ci ⟩⟨Ci |)|Cj⟩(1− |Ci ⟩⟨Ci |)|Cj⟩

09/23/2021 at 13:24:09 slide 18

Proving Completeness

Completeness follows from orthogonality and dimensioncounting.

How many Evecs are there? Given by number of Evals.

How many Evals are there?

Evals obtained from finding roots of polynomial equationobtained from determinant |V − λ1|

a0 + a1λ+ · · ·+ anλn = 0

where n is numer of rows/cols of V .

Fundamental Theorem of Algebra dictates that there are nroots for an order n polynomial.

Given that the Evecs are orthonormal, and that there are n ofthem, they span the n-dimensional over which V is defined.

09/23/2021 at 13:24:09 slide 19

Generic considerations for 2×2 Eigenvalue problem

Let us consider the generic case (with real t):

V =

[ϵ1 tt ϵ2

]→

[ϵ1 tt ϵ2

] [c1c2

]= λ

[c1c2

]∣∣∣∣ϵ1 − λ t

t ϵ2 − λ

∣∣∣∣ = 0 → λ2 − (ϵ1 + ϵ2)λ+ ϵ1ϵ2 − t2

λ =ϵ1 + ϵ2

2± 1

2

√(ϵ2 − ϵ1)2 + 4t2 =

ϵ1 + ϵ22

± |t|√(

ϵ2 − ϵ12t

)2

+ 1

Rescaling V by a constant rescales λ by same constant.

First term does not affect splitting; shifts both levels. Forsplitting, examine ϵ2 − ϵ1 and t, or perhaps ϵ2−ϵ1

2t and t

For clarity, we can study the operator shifted by the average:

V − ϵ1 + ϵ22

1 → λ = ±|t|√(

ϵ2 − ϵ12t

)2

+ 1

09/23/2021 at 13:24:09 slide 20

Evals vs. t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t

λ±

ϵ2 − ϵ1 =12

ϵ2 = ϵ1

Evals are degenerate for t = 0

Evals split linearly when ϵ1 = ϵ2.

Evals split quadratically for ϵ1 = ϵ2.

09/23/2021 at 13:24:09 slide 21

Generic considerations for 2×2 Eigenvalue problem

What about the Eigenvectors?

1 = c21 + c22

There is only one degree of freedom.

Convenient to use polar coordinates.

c1 = cos θ c2 = sin θc2c1

= tan θ

1 = c21 + c22 = cos2 θ + sin2 θ

We will show θ can be written in terms of ϵ2−ϵ12t , and only this

parameter will change the Eigenvectors.

09/23/2021 at 13:24:09 slide 22

Generic considerations for 2×2 Eigenvalue problem

Solve for c2 in terms of c1 and λ, first eqn gives:[ϵ1 tt ϵ2

] [c1c2

]= λ

[c1c2

]→ ϵ1c1 + tc2 = λc1 → c2

c1=

1

t(λ− ϵ1)

c2c1

= tan θ =

ϵ2 − ϵ12t

± |t|t

√(ϵ2 − ϵ12t

)2

+ 1

Use normalization condition, solve for c1.

c1 =1√

1 + tan2 θc2 =

tan θ√1 + tan2 θ

Sanity check with two mass problem: ϵ1 = ϵ2 = 2, t = −1.

λ = 2± 1c2c1

= tan θ = ∓1

Stiff mode out of phase, soft mode in phase ✓.09/23/2021 at 13:24:09 slide 23

Plotting ci

0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ϵ2−ϵ12t

ci

√22

c1

c2

09/23/2021 at 13:24:09 slide 24

Plotting c2i

0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ϵ2−ϵ12t

c2i

c21

c22

09/23/2021 at 13:24:09 slide 25

Limitations to solving coupled oscillators

To solve the coupled oscillator, must diagonalize matrix.

This is equivalent to finding the roots of a polynomial.→ Can only solve polynomials using radicals up to order 4.→ Numerical methods scale like N3.

N = 104 → 14 min, N = 105 → 236 hrs, N = 106 → 26 yrs.

1000 1500 2000 2500 3000

matrix dimension

0

5

10

15

20

25

tim

e (s

econds)

8.5*10^-10*x^3data

intel i7-3770k - np.linalg.eigh(np.ones((dd,dd)))

Symmetry may allow for analytical solution.

09/23/2021 at 13:24:09 slide 26