Application of Linear Algebra in (Polymer) Physics

Martin KrögerComputational Polymer Physics, D-MATL, ETH Zurich, http://www.complexfluids.ethz.ch

Example related to the lecture ’Polymere I’

I. LINEAR ALGEBRA: POWERS OF MATRICES

Here we recall a result obtained in the lecture ’Lineare Al-gebra’, using (polymer) physicists notation and wording. LetA be a square symmetric matrix, that is diagonalizable, anddenote with {λ1,2,..} the set of eigenvalues of A. With thetransformation matrix S that is constructed from the eigen-vectors of A, one has

A = S−1 ·D · S,(A)ij =

∑k,l

S−1ik DklSlj , , (1)

where D is a diagonal matrix with components λ1,2,.. on itsdiagonal, ie. Dkl = λkδkl. We are going to prove (i) that theN ’s power of A can be written as

AN = S−1 ·DN · S(AN

)ij

=∑k

S−1ik λ

Nk Skm (2)

and (ii) that the trace becomes

tr(AN ) =∑k

λNk (3)

Equations 2 and 3 allow us to calculate AN and its trace witharbitrary, integer-valued N with minor effort, while the directevaluation of AN requires summation overN2 terms for eachof the N2 components of AN .

Sketch of proof of eq 2

AN = (S−1 ·D · S) · (S−1 ·D · S) · ... · (S−1 ·D · S)

= S−1 ·D · (S · S−1) ·D · S · ... · S−1 ·D · S= S−1 ·DN · S (4)

Furthermore, because D is diagonal, D2 and hence DN arediagonal as well,

(DN )ij =∑

k,l,..,m

λiδikλkδkl...λmδmj = λNi δij (5)

Sketch of proof of eq 3

Using tr(B ·C) = tr(C ·B) and eqs 2 and 5 one has

tr(AN ) = tr(S−1 ·DN · S) = tr(S · S−1 ·DN ) = tr(DN )

=∑i

(DN )ii =∑i

λNi δii =∑i

λNi (6)

II. (POLYMER) PHYSICS: ONE-DIMENSIONAL SPINCHAIN

The one-dimensional spin chain has wide applications inphysics, and is known as 1D Ising model (cf. wikipedia).The quantum-mechanical analogue is known as Heisenbergmodel. Here, spin stands for a (binary) state that can be either’up’ or ’down’, ’on’ or ’off’, or take the values ’0’ or ’1’. Inpolymer physical chemistry it can also stand for a sequence ofmonomers within a polymer that form either a ’cis’ or ’trans’bond. Spin models in two and higher dimensions are of rele-vance as well, but too difficult to be treated with the methodswe mentioned above, while the 1D spin chain can be solvedanalytically. Modified spins that can take more than two val-ues can be solved as well, using the very same method, andare known as Potts models.

A spin with value σ ∈ {−1, 1} thus represents one of twopossible states. Consider a linear chain of N such spins. Sucha state we denote as Φ = (σ1, σ2, ..., σN ). The situation is de-picted in Fig. 1. The total energy of a spin chain is the sum ofcontributions from (i) interactions between neighboring spinsand (ii) an external orienting field. The sign and magnitude ofJ characterizes the interaction, andH the coupling strength tothe external field. If H = 0 this model is known as classical1D Heisenberg model. For J = 0 the spins do not interactand couple only to the external field. The full model had beensolved analytically in 3 dimensions by Lars Onsager (nobelprize, calculation too complicated for us). Here we want tosolve the 1D model analytically.

There are exactly 2N different spin chain configurations Φ.The energy of a single spin chain configuration (or state) isdenoted as hamiltonian H (cf. Fig. 1)

H(Φ) = −JN∑i=1

∑j

σiσj −BN∑i=1

σi (7)

where j assumes the values i − 1 and i + 1 only becauseonly neighboring spins interact. For a spin chain with peri-odic boundaries we define σ0 = σN and σN+1 = σ1, whilefor a spin chain without periodic boundaries the terms involv-ing spins σ0 and σN+1 are ignored. Let us consider the peri-odic case here because it is slightly simpler. The hamiltonian

2

-1 1 -1 -1 1 1 -1 1

-1 1 -1 -1 1 1 -1 1

spin chain

spin chain with periodic boundary conditions

interaction energy between a neighboring pair of spins

electrostatic energy of an individual spin with external �eld

-J -J +J +J

-B +B

FIG. 1: One-dimensional chain of N = 8 spins without and withperiodic boundary conditions.

defines the spin chain model and it comes with exactly twoparameters, J and B.

Now we have to make a connection between this well-defined model and physics. We assume that all possible 2N

different states Φ can potentially exist (in reality). Becauseeach state Φ has a different energy H(Φ), its probability ofoccurrence tends to be low if it has a high energy for the rea-son that the required energy cost has to be paid by ’somebody’.This ’somebody’ is the thermal energy related to temperature.At zero temperature, only the state with lowest energy wouldbe occupied. Depending on the signs of J and B, the statewith lowest possibleH(Φ) is either a spin-chain with all spinsup (σi = 1), all spins down (σi = −1), or the spin chain withalternating spins.

At finite temperature, the probability for each state is (ac-cording to statistical physics) proportional to its Boltzmannweight p(Φ)

p(Φ) = e−H/kBT = e−βH (8)

where β = 1/kBT and T denotes temperature, and kB Boltz-mann’s constant. The task is to calculate, as an example, themeasurable mean orientation of a spin, 〈σ〉 (known as magne-tization), as function of β, J , andB. The mean spin is definedby, and subsequently rewritten as

〈σ〉 =1

N

⟨∑i

σi

⟩=

1

Z

∑Φ

(1

N

N∑i=1

σi

)p(Φ) (9)

with normalization constant (so-called partition sum)

Z =∑Φ

p(Φ) (10)

Notice that p(Φ)/Z is a probability while p(Φ) is a weight. Ifwe were able to calculate Z, we know 〈σ〉. Why?

d lnZ

dB=

1

Z

d

dB

∑Φ

p(Φ)

=1

Z

d

dB

∑Φ

exp

βJ∑i,j

σiσj + βB∑i

σi

=

β

Z

∑Φ

p(Φ)∑i

σi = βN〈σ〉 (11)

where we used the above relationships only. We thus need toonly calculate Z to know about the mean spin. This problemcan be mapped onto a problem of Linear Algebra as follows.Consider a 2× 2 matrix

A =

(A−− A−+

A+− A++

)(12)

where A−− is going to represent a pair of neighboring spinsthat both point downwards, while A−+ stands for pair whoseleft spin is down, and whose right spin is up etc. In any caseone has

tr(A1) = A−− +A++

tr(A2) = A−−A−− +A−+A+− +A+−A−+ +A++A++,

tr(AN ) = 2N terms (13)

Notice that due to the trace only terms appear that have thestructure ...A∗+A+∗... or ...A∗−A−∗.... For a spin-chain witha single spin N = 1 and periodic boundary conditions, thereare exactly two possible states, σ1 = −1 or σ1 = +1. In theformer case, considering periodic boundary conditions, spin−1 interacts with spin −1, represented by A−−, while in thelatter case spin +1 interacts with itself, represented by A++.Similarly for all N . In other words, each of the 2N termsof tr(AN ) represents a contribution to Z, because p(Φ) is aproduct of pairwise contributions. To be specific

Z =∑Φ

p(Φ)

=∑Φ

∏i,j

eβJσiσj+βBσi

=∑

σ1∈{−1,1}

∑σ2∈{−1,1}

...∑

σN∈{−1,1}

∏i,j

eβJσiσj+βBσi

=∑

σ1∈{−1,1}

∑σ2∈{−1,1}

...∑

σN∈{−1,1}

∏i,j

Aσiσj

= tr(AN ) (14)

To summarize, to solve the problem of calculating the meanspin, we have to evaluate

Z = tr(AN ) (15)

3

with components of A given by

Aσ1,σ2= exp [βJσ1σ2 + βBσ1] (16)

We moreover have the freedom to define A in a symmetricfashion because it does not matter in eq 14

Aσ1,σ2= exp

[βJσ1σ2 +

βB

2(σ1 + σ2)

](17)

Inserting spin values ±1 into eq 17 we find

A =

(eβ(J−B) e−βJ

e−βJ eβ(B+J)

)(18)

Lets now calculate eq 15 with eq 16 using eq 3 from Lin-ear Algebra. For the two eigenvalues λ± of A (physicistsproblem-adapted notation, λ1 and λ2 can be used instead) weobtain

λ± =1

2eβ(J−B)

[1 + e2βB

± 2eβ(B−2J)√

1 + [e2βJ sinh(βB)]2]

(19)

The partition sum is thus

Z = λN+ + λN− (20)

If we are interested in large systems (so-called thermody-namic limit), only the absolutely larger of the two eigenvaluesis relevant.

There are several cases one may consider including positiveand negative J and B (antiferro, ferro etc.). If spins prefer tobe parallel, J > 0. If spins prefer to align in the direction ofthe applied field, B > 0. The larger of the two eigenvalues isλ+ in either case. We have thus obtained, for large N ,

lnZ ≈ N lnλ+ (21)

and using eq 11 we can proceed and finally calculate the mag-netization

〈σ〉 =1

βN

d lnZ

dB(22)

N�1=

1

β

d lnλ+

dB=

(e2βB − 1)e−β(B−2J)

2√

1− [e2βJ sinh(βB)]2(23)

This is our final result for large N , while we also know theexact result for arbitrary N from the exact Z given by eq 20.With a tiny modification we could have also calculated the 〈σ〉for a spin chain without periodic boundary conditions.

Let us now discuss this result eq 23. For noninteractingspins J = 0 and eq 23 simplifies to

〈σ〉 = tanh (βB) (J = 0) (24)

This is a well-known result, that applies also to a single spin ina magnetic field. It represents the magnetization of a magneticmoment subjected to an external magnetic field, at finite tem-perature. At huge temperature, corresponding to β → 0, the

β Jβ B

magnetization < σ >

huge N

FIG. 2: Magnetization 〈σ〉 (or mean value of the two-state system)according to eq 23. J charaterizes the strength and sign of inter-action, B characterizes the strength and orientation of the externalfield, β = 1/kBT is inverse thermal energy.

β Jβ B

magnetization < σ >

N = 2

FIG. 3: Same as previous figure for a spin chain with only N = 2spins, and periodic boundary conditions. Calculated using Z from eq20 in eq 22. While the result is qualitatively similar to the behaviorin the thermodynamic limit, the transitions are less sharp.

magnetization vanishes. At small temperature, correspondingto large β, the mean spin is 〈σ〉 = ±1, depending on the di-rection of the applied magnetic field (sign of B), as long as|B| > |J |. When B = 0,

〈σ〉 = 0 (B = 0) (25)

is also expected for symmetry reasons, while each particularrealization of the spin chain may prefer to be uniformly ori-ented for J > 0 (or anti-oriented for J < 0). The full resultfor an infintely long spin chain (N � 1) is visualized in Fig.2. For comparison, the exact result for a spin chain with onlyN = 2 spins is shown in Fig. 3.