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Self-Consistent Field Theory and Its Applications by M. W. Matsen

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1.5 Mathematical Techniques <strong>and</strong> Approximations 21<br />

for the entropic free energy. The logarithmic term can be dropped in the large-N limit. An<br />

integration <strong>by</strong> parts is then performed to reexpress the equation as<br />

∫<br />

f e<br />

k B T = a2 N<br />

dr |∇f 0 (r)| 2 (1.77)<br />

6V<br />

where the boundary term vanishes as a result of the assumption, ∇φ α (r) =0, along the edge<br />

of V. For the final step of the derivation, the eigenfunction, f 0 (r), is replaced <strong>by</strong> the segment<br />

concentration, φ α (r), using Eqs. (1.73) <strong>and</strong> (1.74). This produces the L<strong>and</strong>au-Ginzburg free<br />

energy expression,<br />

∫<br />

f e<br />

k B T = a2 ρ 0<br />

24<br />

dr |∇φ α(r)| 2<br />

φ α (r)<br />

(1.78)<br />

which gives the entropic energy without the need to evaluate the underlying field, w(r). The<br />

derivation assumes only that the molecular weight (i.e., N) of the polymer is large.<br />

1.5.3 Fourier Representation<br />

Quantum mechanical problems are often made more tractable <strong>by</strong> reexpressing them in socalled<br />

momentum space. The principle is very similar to that of the spectral method, where<br />

quantities are exp<strong>and</strong>ed in terms of energy eigenstates. The main difference is that the expansions<br />

are now in terms of eigenfunctions, exp(ik · r), of the momentum operator, ˆp ≡−i∇.<br />

(Here, i ≡ √ −1.) The other difference is that the momentum eigenvalue, k, is continuous,<br />

which implies that sums are now replaced <strong>by</strong> integrals <strong>and</strong> Kronecker delta functions are<br />

substituted <strong>by</strong> Dirac delta functions. For example, the orthogonality condition in Eq. (1.60)<br />

becomes<br />

∫<br />

1<br />

(2π) 3 dr exp(i(k 1 − k 2 ) · r) =δ(k 1 − k 2 ) (1.79)<br />

Furthermore, the expansion of an arbitrary function, g(r), becomes the integral,<br />

g(r) = 1 ∫<br />

(2π) 3 dk g(k)exp(ik · r) (1.80)<br />

<strong>and</strong> the coefficient of the expansion,<br />

∫<br />

g(k) ≡ dr g(r)exp(−ik · r) (1.81)<br />

is now a continuous function, referred to as the Fourier transform of g(r). Although g(k)<br />

is complex, its complex conjugate satisfies g ∗ (k) =g(−k), which implies that |g(k)| 2 =<br />

g(−k)g(k). Before moving on, we quote a useful identity,<br />

∫<br />

dr g 1 (r)g 2 (r) = 1 ∫<br />

(2π) 3 dk g 1 (−k)g 2 (k) (1.82)

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