Math Methods for Polymer Science Lecture 2: Fourier Transforms, Delta Functions and Gaussian Integrals In the first lecture, we reviewed the Taylor and Fourier series. These where both essentially ways of decomposing a given function into a differ- ent, more convenient, or more meaningful form. In this lecture, we review the generalization of the Fourier series to the Fourier transformation. In the context, it is also natural to review 2 special functions, Dirac delta functions and Gaussian functions, as these functions commonly arise in problems of Fourier analysis and are otherwise essential in polymer physics. For addi- tional reading on Fourier transforms, delta functions and Gaussian integrals see Chapters 15, 1 and 8 of Arken and Weber's text, Mathematical Methods for Physicists. 1 Fourier Transforms Conceptually, Fourier transforms are a straightforward generalizations of Fourier series which represent a function on finite domain of size L by an 2πn 2πn infinite sum over a discrete sets of functions, sin L x and cos L x . In the Fourier transform, that size of the domain is taken to 1 so that the domain becomes the all positive and negative values of x (see Fig. 1). The key differences between Taylor series, Fourier series and Fourier transforms are summarized as follows: Taylor Series - series representation in polynomials, \local" representation Fourier Series - series representation in sines and cosines, \global" repre- sentation (over finite or periodic domain) Fourier transform - integral representationin sines and cosines over infi- nite domain (L ! 1) How do we generalize a Fourier series to an infinite domain? For the Fourier transform it's more convenient to use complex representation of sine and cosine: eix = cos x + i sin x (1) 1 Figure 1: Schematic of Fourier transform of function f(x), where f(x) = 1 1 a0 X 2πn X 2πn + a cos x + b sin x . 2 n L n L n=1 n=1 Using this we can rewrite the Fourier series: 1 1 a0 X 2πn X 2πn f(x) = + a cos x + b sin x 2 n L n L n=1 n=1 1 X i2πnx = c exp (2) n L n=−∞ where ( an − ibn n > 0 cn = 2 (3) an + ibn 2 n < 0 a c = 0 : (4) 0 2 Notice also that complex functions ei2πnx=L, are also orthogonal. L L=2 Z 2 i2πm i2πn L 2π(m + n)i dx exp x exp x = exp x −L L L 2π(m + n)i L 2 −L=2 L π(m + n) = sin π(m + n) L 0 for m + n 6= 0 = (5) L for m + n = 0 Using this, we can extract Fourier coefficients by L 1 Z 2 2πn cn = dx exp − x f(x) (6) L −L L 2 Note that the complex notation takes care of factors of 2, etc. 2 We want to define Fourier transform as the L ! 1 limit of a Fourier series. This limit is unusual because we take L ! 1 while 2πn lim ≡ k (finite): (7) L!1 L Here, k is referred to as the wavenumber of the Fourier mode, eikx. The Fourier transform of a function, f(x), is defined as: Z 1 −ikx f~(k) = lim (cnL) = dx e f(x): (8) L!1 −∞ Often the Fourier transform is written as F [f(x)] = f~(k) (9) where F means Fourier transform. Notice that the Fourier transform takes f(x), function of \real space" variables, x, and outputs f~(k), a function of \Fourier space" variables, k. The Fourier transform can be inverted using the definition of the Fourier series 1 1 X X 1 f(x) = c e−ik(n)x = (c L)eik(n)x: (10) n L n n=−∞ n=−∞ Now as L ! 1 k takes on a continuum of values 2π 2π lim ∆k = k(n + 1) − k(n) = [(n + 1) − n] = ! 0 L!1 L L That is, δk becomes infinitely narrow in the L ! 1 limit. This means we can write the sum as 1 1 X 1 1 X 1 Z 1 lim = lim ∆k = dk (11) L!1 L 2π ∆k!0 2π n=−∞ n=−∞ −∞ The last step is the is Riemann's definition of an integral (see Fig. 2). And thus we get Z 1 dk f(x) = f~(k)eikx (12) −∞ 2π Eq. (12) is the inverse Fourier transform or h i F −1 f~(k) = f(x) (13) where F −1 means inverse Fourier transform, f~(x) is function of \Fourier space", and f(k) is function of \real space". 3 Figure 2: A figure showing that Fourier series becomes an integral of con- tinuous function f~(k) in limit that ∆k ! 0 (area). 1.1 Delta Function Related to the Fourier transform is a special function called the Dirac delta function, δ(x). It's essential properties can be deduced by the Fourier trans- form and inverse Fourier transform. Here, we simply insert the definition of the Fourier transform, eq. (8), into equation for the inverse transform, eq. (12), 1 1 1 Z dk Z dk Z 0 f(x) = f~(k)eikx = eikx dx0 eikx f(x0) −∞ 2π −∞ 2π −∞ 1 1 Z Z dk 0 = dx0 eik(x−x ) f(x0) −∞ −∞ 2π Z 1 = dx0 δ(x − x0)f(x0) = f(x) (14) −∞ This last line defines the properties of delta function, which is defined im- plicitly by the integral in the parentheses on the second line. When you integrate the product of the Dirac delta function with another function, it returns the value of that function at the point where the argument of δ(x) vanishes. Geometrically, you can think of it as an infinitely tall and narrowly peeked function, with area 1 under the curve (see Fig. ??). Using this definition of δ(x) we can derive the Fourier transform of os- cillatory functions. 4 Figure 3: Sketch of a Dirac delta function. Figure 4: Plot of f(x), which is only non-zero near x = 0. Example 1: Compute Fourier transform of A cos(qx). Z 1 f~(k) = dx A cos(qx)eikx −∞ A Z 1 = dx eiqx + e−iqx eikx 2 −∞ A Z 1 h i = dx ei(q+k)x + e−i(q−k)x 2 −∞ = πA [δ(q + k) + δ(q − k)] which is only non-zero for k = ±q. The Fourier transform is particularly useful for studying the properties of a function which is non-zero only over a finite region of space. For example, density or probability distribution for polymer chain. For this case, we can use Taylor series expansion to cast light on what 5 Figure 5: Plot of f(x). the Fourier transform tells us. Z 1 f~(k) = dx e−ikxf(x) −∞ Z 1 1 i 1 = dx 1 − ikx − k2x2 + k3x3 + k4x4 + ::: f(x) (15) −∞ 2! 3! 4! Clearly, Z 1 f~(k = 0) = dx f(x) (16) −∞ which is the total area under curve, named N. But we see from Taylor series that various powers of k represent certain averages. That is, 1 i 1 f~(k) = N 1 − ikhxi − k2hx2i + k3hx3i + k4hx4i + ::: (17) 2! 3! 4! where Z 1 dx xnf(x) n −∞ hx i = Z 1 (18) dx f(x) −∞ is an average of xn weighted by f(x). That is what I mean when we say that f~(k) is something of a global representation. f~(k) seems to encodes properties of the function over its entire range not just locally. Let's try an example. Compute Fourier transform of A jxj ≤ a f(x) = 0 jxj > a 6 Figure 6: Plot of Gaussian function. Z 1 Z a f~(k) = dx f(x)e−ikx = A dx e−ikx −∞ −a A h i 2A sin(ka) = e−ika − eika = (19) ik k Now that we have full expression, let's examine small k behavior. sin(ka) ka − 1 (ka)3 + ::: lim f~(k) = 2A = 2A 3! k!0 k k 1 = 2Aa 1 − (ka)2 + ::: (20) 3! This is just the form we derived above. Note that 2Aa = N(area), and you can check that Z a a 1 3 dx x2 x 3 a2 hx2i = −a = −a = (21) Z a 2a 3 dx −a 1.2 Gaussian Integral Let's use the Fourier transform to study an important function, the Gaussian bump 2 − x f(x) = Ae 2a2 (22) 7 This function is very important in random systems, especially in polymer physics. Aside: Integrating a Gaussian function (A trick!). Z 1 x2 I = dx exp − 2 (23) −∞ 2a Z 1 2 2 Z 1 Z 1 2 2 2 x x + y I = dx exp − 2 = dx dy exp − 2 (24) −∞ 2a −∞ −∞ 2a This double integral is carried out over whole x − y plane. Let's do same integral in polar coordinates: r = px2 + y2, x = r cos φ and y = r sin φ. Z 2π Z 1 2 2 r I = dφ dr r exp − 2 0 0 2a Z 1 2 d 2 r = 2π dr −a exp − 2 0 dr 2a 1 r2 = 2π −a2 exp − = 2πa2 (25) 2a2 0 p thus, I = 2πa. Let's go back to Fourier transform of a Gaussian bump. 1 Z 2 2 f~(k) = A dx e−x =2a e−ikx (26) −∞ This can be done by "completing the square" of argument in exponential x2 1 k2a2 + ikx = (x + ika2)2 + (27) 2a2 2a2 2 then Z 1 2 2 2 2 ~ (x + ika ) k a f(k) = A dx exp − 2 exp − −∞ 2a 2 k2a2 Z 1 u2 = A exp − du exp − 2 2 −∞ 2a p k2a2 = A 2πa exp − (28) 2 Note that this was done by changing variables u = x + ika2 and du = dx.