342 Chapter 9. Cooperative transitions in macromolecules[[Student version, January 17, 2003]]
replace Θ^2 bythe more convenient function 2(1−cos Θ). Note that we have written every force
term twice and divided by two; the reason for this choice will become clear in a moment.
Exactly as in Your Turn 9g, we can reformulate Equation 9.36 as a matrix raised to the power
N−1, sandwiched between two vectors. We need the largest eigenvalue of the matrixT,just as
before. Remembering that our objects now have continuous indices, a “vector”Vin this context
is specified by afunction,V(ˆt), and the matrix product is the “convolution integral”:
(TV)(ˆt)=∫
d^2 nˆT(ˆt,ˆn)V(ˆn), (9.37)where
T(ˆt,nˆ)=exp[
f
2 kBT(ˆt·ˆz+ˆn·ˆz)+A
(
nˆ·ˆt−1)]
. (9.38)
The reason for our apparently perverse duplication of the force terms is that nowTis a symmetric
matrix, so the mathematical facts quoted in Section 9.3.2′on page 341 apply.
Wewill use a simple technique—theRitz variational approximation—to estimate the maximal
eigenvalue (see Marko & Siggia, 1995).^12 The matrixTis real and symmetric. As such, it must
have a basis set of mutually perpendicular eigenvectorseisatisfyingTei=λieiwith real, positive
eigenvaluesλi.AnyvectorVmay be expanded in this basis:V=
∑
iciei.Wenext consider the
“estimated eigenvalue”
λmax,est≡V·(TV)
V·V
=
∑
iλi(ci)(^2) ei·ei
∑
i(ci)^2 ei·ei
. (9.39)
The last expression above is no larger than the largest eigenvalue,λmax.Itequalsλmaxwhen we
chooseVto be equal to the corresponding eigenvectore 0 ,that is, ifc 0 =1and the otherci=0.
Your Turn 9p
Suppose that there is one maximal eigenvalueλ 0. Show thatλmax,estis maximized precisely
whenVis a constant timese 0 .[Hint: Trythe 2× 2 case first; here you can see the result
explicitly. For the general case, letxi=(ci/c 0 )^2 ,Ai=(ei·ei)/(e 0 ·e 0 ), andLi=(λi/λ 0 )for
i>1. Then show that the estimated eigenvalue has no maximum other than at the point where
all thexi=0.]Toestimateλmax,then, we need to find the functionV 0 (ˆt)that saturates the boundλmax,est≤
λmax,orinother words maximizes Equation 9.39. This may sound as hard as finding a needle in
the infinite-dimensional haystack of functionsV(ˆt). The trick is to use physical reasoning to select
apromisingfamilyof trial functions,Vw(ˆt), depending on a parameterw(or several). We substitute
Vw(ˆt)intoEquation 9.39 and choose the valuew∗that maximizes our estimated eigenvalue. The
correspondingVw∗(ˆt)isthen our best proposal for the true eigenvector,V 0 (ˆt). Our estimate for
the true maximal eigenvalueλmaxis thenλ∗≡λmax,est(w∗).
Tomake a good choice for the family of trial functionsVw(ˆt), we need to think a bit about the
physical meaning ofV.You found in Your Turn 9g(c) that the average value of a link variable can
beobtained by sandwiching
( 10
0 − 1)
between two copies of the dominant eigenvectore 0 .Atzero
force, each link of our chain should be equally likely to point in all directions, while at high force
it should be most likely to point in the +ˆzdirection. In either case the link should not prefer any
(^12) Amore general approach to maximal-eigenvalue problems of this sort is to find a basis of mutually orthogonal
functions ofˆt,expandV(ˆt)inthis basis, truncate the expansion after a large but finite number of terms, evaluateT
on this truncated subspace, and use numerical software to diagonalize it.