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Math Methods for Polymer Science

Lecture 2: Fourier Transforms, Delta Functions and

Gaussian Integrals

In the rst lecture, we reviewed the Taylor and Fourier series. These where both essentially ways of decomposing a given function into a dier- 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 nite domain of sizeLby an innite sum over adiscretesets of functions, sin2nL xand cos2nL x. In the Fourier transform, that size of the domain is taken to1so that the domain becomes the all positive and negative values ofx(see Fig. 1). The key dierences 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 nite or periodic domain) Fourier transform- integral representationin sines and cosines over in- nite domain (L! 1) How do we generalize a Fourier series to an innite domain? For the Fourier transform it's more convenient to use complex representation of sine and cosine: e ix= cosx+isinx(1) 1 Figure 1: Schematic of Fourier transform of functionf(x), wheref(x) = a 02 +1X n=1a ncos2nL x +1X n=1b nsin2nL x

Using this we can rewrite the Fourier series:

f(x) =a02 +1X n=1a ncos2nL x +1X n=1b nsin2nL x 1X n=1c nexpi2nxL (2) where c n=( anibn2 n >0 a n+ibn2 n <0(3) c 0=a02 :(4) Notice also that complex functionsei2nx=L, are also orthogonal. Z L2 L2 dxexpi2mL x expi2nL x =L2(m+n)iexp2(m+n)iL x L=2 L=2

L(m+n)sin(m+n)L

=0 form+n6= 0

Lform+n= 0(5)

Using this, we can extract Fourier coecients by

c n=1L Z L2 L2 dxexp 2nL x f(x) (6) Note that the complex notation takes care of factors of 2, etc. 2 We want to dene Fourier transform as theL! 1limit of a Fourier series. This limit is unusual because we takeL! 1while lim

L!12nL

k(nite):(7) Here,kis referred to as thewavenumberof the Fourier mode,eikx. The Fourier transform of a function,f(x), is dened as: f(k) = limL!1(cnL) =Z 1 1 dx eikxf(x):(8)

Often the Fourier transform is written as

F[f(x)] =~f(k) (9)

whereFmeans 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 denition of the Fourier series f(x) =1X n=1c neik(n)x=1X n=11L (cnL)eik(n)x:(10)

Now asL! 1ktakes on a continuum of values

lim

L!1k=k(n+ 1)k(n) =2L

[(n+ 1)n] =2L !0 That is,kbecomes innitely narrow in theL! 1limit. This means we can write the sum as lim L!11 X n=11L =12limk!01 X n=1k=12Z 1 1 dk(11) The last step is the is Riemann's denition of an integral (see Fig. 2). And thus we get f(x) =Z 1

1dk2~f(k)eikx(12)

Eq. (12) is theinverseFourier transform or

F

1h~f(k)i

=f(x) (13) whereF1means inverse Fourier transform,~f(x) is function of \Fourier space", andf(k) is function of \real space". 3 Figure 2: A gure 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,(x). It's essential properties can be deduced by the Fourier trans- form and inverse Fourier transform. Here, we simply insert the denition of the Fourier transform, eq. (8), into equation for the inverse transform, eq. (12), f(x) =Z 1

1dk2~f(k)eikx=Z

1

1dk2eikxZ1

1 dx0eikx0f(x0) Z 1 1 dx0Z1

1dk2eik(xx0)

f(x0) Z 1 1 dx0(xx0)f(x0) =f(x) (14) This last line denes the properties of delta function, which is dened 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(x) vanishes. Geometrically, you can think of it as an innitely tall and narrowly peeked function, with area 1 under the curve (see Fig.??). Using this denition of(x) we can derive the Fourier transform of os- cillatory functions. 4

Figure 3: Sketch of a Dirac delta function.

Figure 4: Plot off(x), which is only non-zero nearx= 0.

Example 1:Compute Fourier transform ofAcos(qx).

f(k) =Z 1 1 dx Acos(qx)eikx A2 Z 1 1 dxeiqx+eiqxeikx A2 Z 1 1 dxh ei(q+k)x+ei(qk)xi =A[(q+k) +(qk)] which is only non-zero fork=q. The Fourier transform is particularly useful for studying the properties of a function which is non-zero only over a nite 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 off(x).

the Fourier transform tells us. f(k) =Z 1 1 dx eikxf(x) Z 1 1 dx

1ikx12!

k2x2+i3! k3x3+14! k4x4+::: f(x) (15)

Clearly,

f(k= 0) =Z 1 1 dx f(x) (16) which is the total area under curve, named N. But we see from Taylor series that various powers ofkrepresent certain averages. That is, f(k) =N

1ikhxi 12!

k2hx2i+i3! k3hx3i+14! k4hx4i+::: (17) where hxni=Z 1 1 dx xnf(x)Z 1 1 dx f(x)(18) is an average ofxnweighted byf(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 f(x) =Ajxj a

0jxj> a

6

Figure 6: Plot of Gaussian function.

f(k) =Z 1 1 dx f(x)eikx=AZ a adx eikx Aik h eikaeikai =2Asin(ka)k (19) Now that we have full expression, let's examine smallkbehavior. lim k!0~f(k) = 2Asin(ka)k = 2Aka13! (ka)3+:::k = 2Aa 113!
(ka)2+::: (20) This is just the form we derived above. Note that 2Aa=N(area), and you can check that hx2i=Z a adx x2Z a adx=13 x3a a2a=a23 (21)

1.2 Gaussian Integral

Let's use the Fourier transform to study an important function, the Gaussian bump f(x) =Aex22a2(22) 7 This function is very important in random systems, especially in polymer physics. Aside: Integrating a Gaussian function (A trick!). I=Z 1 1 dxexp x22a2 (23) I 2= Z1 1 dxexp x22a2 2 =Z 1 1 dxZ 1 1 dyexp x2+y22a2 (24) This double integral is carried out over wholexyplane. Let's do same integral in polar coordinates:r=px

2+y2,x=rcosandy=rsin.

I 2=Z 2 0 dZ 1 0 dr rexp r22a2 = 2Z 1 0 drddr a2exp r22a2 = 2 a2exp r22a2 1 0 = 2a2(25) thus,I=p2a. Let's go back to Fourier transform of a Gaussian bump. f(k) =AZ 1 1 dx ex2=2a2eikx(26) This can be done by "completing the square" of argument in exponential x

22a2+ikx=12a2(x+ika2)2+k2a22

(27) then f(k) =AZ 1 1 dxexp (x+ika2)22a2 exp k2a22 =Aexp k2a22 Z1 1 duexp u22a2 =Ap2aexp k2a22 (28) Note that this was done by changing variablesu=x+ika2and du= dx. Once again we can learn something off(x) by examining smallkprop- erties of~f(k). f(k) =Aap2 1k22! a2+k44! (3a2) +::: (29) 8 Figure 7: Schematic of an electron density distribution. There is no odd terms sincehxni= 0 for n odd sincef(x) =f(x). By taking Fourier transform of Gaussian function, we have automatically cal- culated all moments, or averages, of the distribution. The most important is the second moment hx2i=Z 1 1 dxexp x22a2 x 2Z 1 1 dxexp x22a2quotesdbs_dbs20.pdfusesText_26