2 Fourier Series. 2.1 Complex Fourier Series The phases exp(inx)/ p 2, one for each integer n, are orthonormal on an interval of length 2 Z 2

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1 2 Fourier Series 2. Comlex Fourier Series The hases ex(inx)/, one for each integer n, are orthonormal on an interval of length e imx e inx e i(n m)x dx = dx = m,n (2.) where n,m =ifn = m, and n,m =ifn6= m. So if a function f(x) isa sum of these hases e inx f(x) = f n (2.2) then their orthonormality (2.) gives the nth coe cient f n as the integral e inx f(x) dx = e inx X m= f m e imx dx = m= n,mf m = f n (2.3) (Joseh Fourier ). The Fourier series (2.2) is eriodic with eriod because the hases are eriodic with eriod, ex(in(x + )) = ex(inx). Thus even if the function f(x) which we use in (2.3) to make the Fourier coe cients f n is not eriodic, its Fourier series (2.2) will nevertheless be strictly eriodic, as illustrated by Figs. 2.2 & 2.4. If the Fourier series (2.2) converges uniformly (section 2.7), then the termby-term integration imlicit in the formula (2.3) for f n is ermitted. How is the Fourier series for the comlex-conjugate function f (x) related

2 88 Fourier Series to the series for f(x)? The comlex conjugate of the Fourier series (2.2) is so the coe f (x) = f n e inx = f n e inx (2.4) cients f n (f ) for f (x) are related to those f n (f) for f(x) by Thus if the function f(x) is real, then f n (f )=f n(f). (2.5) f n (f) =f n (f )=f n(f). (2.6) Thus the Fourier coe cients f n for a real function f(x) satisfy f n = f n. (2.7) Examle 2. (Fourier Series by Insection). The doubly exonential function ex(ex(ix)) has the Fourier series ex e ix = n= in which n! = n(n )... is n-factorial with!. n! einx (2.8) Examle 2.2 (Beats). The sum of two sines f(x) =sin! x +sin! 2 x of similar frequencies!! 2 is the roduct (exercise 2.) f(x) = 2 cos 2 (!! 2 )x sin 2 (! +! 2 )x. (2.9) The first factor cos 2 (!! 2 )x is the beat; it modulates the second factor sin 2 (! +! 2 )x as illustrated by Fig The Interval In equations (2. 2.3), we singled out the interval [, ], but to reresent a eriodic function f(x) of eriod, we could have used any interval of length, such as the interval [, ] or [r, r +] f n = Z r+ r e inx f(x) dx. (2.) This integral is indeendent of its lower limit r when the function f(x) is eriodic with eriod. The choice r = often is convenient. With this

3 2.3 Where to ut the s 89 2 Beats.5 sin(ωx) +sin(ω2x) Figure 2. The curve sin! x +sin! 2 x for! = 3 and! 2 = 32. x choice of interval, the coe cient f n is the integral (2.3) shifted by f n = Z e inx f(x) dx. (2.) But if the function f(x) is not eriodic with eriod, then the Fourier coe cients (2.) do deend uon the choice r of interval. 2.3 Where to ut the s In Eqs.(2.2 & 2.3), we used the orthonormal functions ex(inx)/, and so we had factors of / in both equations. One can avoid having two square roots by setting d n = f n / and writing the Fourier series (2.2) and the orthonormality relation (2.3) as f(x) = d n e inx and d n = dx e inx f(x). (2.2)

4 9 Fourier Series Fourier Series for e 2 x.8 e 2 x Figure 2.2 The -term (dashes) Fourier series (2.6) for the function ex( 2 x ) on the interval (, ) is lotted from to. All Fourier series are eriodic, but the function ex( 2 x ) (solid) is not. x Alternatevely one set c n = f n and may use the rules f(x) = c n e inx and c n = Z f(x)e inx dx. (2.3) Examle 2.3 (Fourier Series for ex( m x )). et s comute the Fourier series for the real function f(x) =ex( m x ) on the interval (, ). Using Eq.(2.) for the shifted interval and the -lacement convention (2.2), we find that the coe cient d n is the integral d n = Z dx e inx e m x (2.4)

5 2.4 Real Fourier Series for Real Functions 9 which we may slit into the two ieces d n = = Z Z dx e(m in)x dx + e (m+in)x m ( ) n m 2 + n 2 e m (2.5) which shows that d n = d n.sincemisreal, the coe cients d n also are real, d n = d n. They therefore satisfy the condition (2.7) that holds for real functions, d n = d n, and give the Fourier series for ex( m x ) as e m x = d n e inx = = ( e m ) m + m m 2 + n 2 ( ) n e m e inx 2 m ( ) n m 2 + n 2 e m cos(nx). (2.6) In Fig. 2.2, the -term (dashes) Fourier series for m = 2 is lotted from x = to x =. The function ex( 2 x ) itself is reresented by a solid line. Although ex( 2 x ) is not eriodic, its Fourier series is eriodic with eriod. The -term Fourier series reresents the function ex( 2 x ) quite well within the interval [, ]. In what follows, we usually won t bother to use di erent letters to distinguish between the symmetric (2.2 & 2.3) and asymmetric (2.3) conventions on the lacement of the s. 2.4 Real Fourier Series for Real Functions The rules ( and ) for Fourier series are simle and aly to functions that are continuous and eriodic whether comlex or real. If a function f(x) is real, then its Fourier coe cients obey the rule (2.7) that holds for real functions, d n = d n.thusd is real, d = d, and we may

6 92 Fourier Series write the Fourier series (2.2) for a real function f(x) as f(x) =d + d n e inx + X d n e inx = d + dn e inx + d n e inx = d + dn e inx + d n e inx = d + = d + d n (cos nx + i sin nx)+d n (cos nx i sin nx) (d n + d n) cos nx + i(d n d n)sinnx. (2.7) In terms of the real coe cients a n = d n + d n and b n = i(d n d n), (2.8) the Fourier series (2.7) of a real function f(x) is f(x) = a 2 + X a n cos nx + b n sin nx. (2.9) What are the formulas for a n and b n? By (2.8 & 2.2), the coe is cient a n a n = e inx f(x)+e inx f (x) Z dx = e inx + e inx f(x) dx 2 (2.2) since the function f(x) is real. So the coe cient a n of cos nx in (2.9) is the cosine integral of f(x) a n = cos nx f(x) dx. (2.2) Similarly, equations (2.8 & 2.2) and the reality of f(x) imly that the coe cient b n is the sine integral of f(x) b n = i e inx 2 e inx f(x) dx = sin nx f(x) dx (2.22) The real Fourier series (2.9) and the cosine (2.2) and sine (2.22) integrals

7 for the coe 2.4 Real Fourier Series for Real Functions 93 cients a n and b n also follow from the orthogonality relations if n = m 6= sin mx sin nx dx = (2.23) otherwise, 8 < if n = m 6= cos mx cos nx dx = if n = m = (2.24) : otherwise, and sin mx cos nx dx =, (2.25) which hold for integer values of n and m. If the function f(x) iseriodicwitheriod, then instead of the interval [, ], one may choose any interval of length such as [, ]. What if a function f(x) is not eriodic? The Fourier series for an aeriodic function is itself strictly eriodic, is sensitive to its interval (r, r + ) of definition, may di er somewhat from the function near the ends of the interval, and usually di ers markedly from it outside the interval. Examle 2.4 (The Fourier Series for x 2 ). The function x 2 is even and so the integrals (2.22) for its sine Fourier coe cients b n all vanish. Its cosine coe cients a n are given by (2.2) a n = Z Integrating twice by arts, we find for n 6= and a n = 2 n Z nx f(x) cos dx Z = cos nx x 2 dx. (2.26) x sin nx dx = ale 2x cos nx n 2 a = Z =( ) n 4 n 2 (2.27) x 2 dx = 2 3. (2.28) Equation (2.9) now gives for x 2 the cosine Fourier series x 2 = a 2 + X a n cos nx = X n cos nx ( ) n 2. (2.29) This series raidly converges within the interval [, ] as shown in Fig. 2.3, but not near the endoints ±. Examle 2.5 (The Gibbs overshoot). The function f(x) =x on the interval [, ] is not eriodic. So we exect trouble if we reresent it as a Fourier

8 94 Fourier Series Fourier Series for x x Figure 2.3 The function x 2 (solid) and its Fourier series of 7 terms (dot dash) and 2 terms (dashes). The Fourier series (2.29) for x 2 quickly converges well inside the interval (, ). x series. Since x is an odd function, equation (2.2) tells us that the coe a n all vanish. By (2.22), the b n s are b n = Z cients dx x sin nx =2( )n+ n. (2.3) As shown in Fig. 2.4, the series 2( ) n+ sin nx n (2.3) di ers by about from the function f(x) =x for 3 <x< and for <x<3 because the series is eriodic while the function x isn t. Within the interval (, ), the series with terms is very accurate excet for x & and x., where it overshoots by about 9% of the discontinuity, a defect called a Gibbs overshoot (J. Willard Gibbs Incidentally, Gibbs s father heled defend the Africans of the schooner

9 2.5 Stretched Intervals 95 6 Fourier Series for the Aeriodic Function x x Gibbs Overshoot for the Function x Overshoot Figure 2.4 (to) The Fourier series (2.3) for the function x (solid line) with terms (... ) and terms (solid curve) for <x<. The Fourier series is eriodic, but the function x is not. (bottom) The di erences between x and the -term (... ) and the -term (solid curve) on (, ) exhibit a Gibbs overshoot of about 9% at x & and at x.. x Amistad). Any time we use a Fourier series to reresent an aeriodic function, a Gibbs overshoot will occur near the endoints of the interval. 2.5 Stretched Intervals If the interval of eriodicity is of length instead of, thenwemayuse the hases ex(inx/ ) which are orthonormal on the interval [,] Z dx! e inx/ e imx/ = nm. (2.32)

10 96 Fourier Series The Fourier series f(x) = f n e inx/ (2.33) is eriodic with eriod. Thecoe cientf n is the integral f n = Z e inx/ f(x) dx. (2.34) These relations ( ) generalize to the interval [,] our earlier formulas (2. 2.3) for the interval [, ]. If the function f(x) iseriodicf(x ± ) =f(x) witheriod, thenwe may shift the domain of integration by any real number r f n = Z +r r e inx/ f(x) dx (2.35) without changing the coe cients f n. An obvious choice is r = /2 for which (2.33) and (2.34) give f(x) = e inx/ Z /2 e inx/ f n and f n = f(x) dx. (2.36) /2 If the function f(x) is real, then on the interval [,] in lace of Eqs.(2.9), (2.2), & (2.22), one has f(x) = a 2 + X nx a n cos + b n sin nx, (2.37) and a n = 2 b n = 2 Z Z nx dx cos f(x), (2.38) nx dx sin f(x). (2.39) The corresonding orthogonality relations, which follow from Eqs.(2.23),

11 2.6 Fourier Series in Several Variables 97 (2.24), & (2.25), are: Z mx nx /2 if n = m 6= dx sin sin = (2.4) otherwise, 8 Z mx nx < /2 if n = m 6= dx cos cos = if n = m = (2.4) : otherwise, and Z mx nx dx sin cos =. (2.42) They hold for integer values of n and m, and they imly Eqs.(2.37) 2.39). On the interval [ 2.6 Fourier Series in Several Variables, ], the Fourier-series formulas (2.33 & 2.34) are f(x) = f n = Z f n e inx/ 2 (2.43) e inx/ 2 f(x) dx. (2.44) We may generalize these equations from a single variable to N variables x =(x,...,x N )withn x = n x n N x N f(x) = f n = n = Z... dx... n N = Z f n e in x/ (2) N/2 (2.45) e in x/ dx N f(x). (2.46) (2) N/2 2.7 How Fourier Series Converge A Fourier series reresents a function f(x) as the limit of a sequence of functions f N (x) given by f N (x) = NX k= N f k e ikx/ in which f k = Z f(x) e ikx/ dx. (2.47) Since the exonentials are eriodic with eriod, a Fourier series always is eriodic. So if the function f(x) is not eriodic, then its Fourier series will

12 98 Fourier Series reresent the eriodic extension f of f defined by f (x + n) =f(x) for all integers n and for ale x ale. A sequence of functions f N (x) converges to a function f(x) on a (closed) interval [a, b] if for every > and each oint a ale x ale b, there exists an integer N(, x) such that f(x) f N (x) < for all N>N(, x). (2.48) If this holds for an N(, x) =N( ) that is indeendent of x 2 [a, b], then the sequence of functions f N (x) converges uniformly to f(x) on the interval [a, b]. A function f(x) iscontinuous on an oen interval (a, b) if for every oint a<x<bthe two limits f(x ) lim f(x ) and f(x + ) lim f(x + ) (2.49) <! <! agree; it also is continuous on the closed interval [a, b] iff(a + ) = f(a) and f(b ) = f(b). A function continuous on [a, b] is bounded and integrable on that interval. If a sequence of continuous functions f N (x) converges uniformly to a function f(x) on a closed interval a ale x ale b, then we know that f N (x) f(x) < for N>N( ), and so Z b a f N (x) dx Z b a f(x) dx ale Z b a f N (x) f(x) dx < (b a). (2.5) Thus one may integrate a uniformly convergent sequence of continuous functions on a closed interval [a, b] termbyterm Z b lim N! a f N (x) dx = Z b a lim f N(x) dx = N! Z b a f(x) dx. (2.5) A function is iecewise continuous on [a, b] if it is continuous there excet for finite jums from f(x ) to f(x + ) at a finite number of oints x. Atsuchjums,wedefine the eriodically extended function f to be the mean f (x) =[f(x ) + f(x + )]/2. Fourier s convergence theorem (Courant, 937,. 439): The Fourier series of a function f(x) that is iecewise continuous with a iecewise continuous first derivative converges to its eriodic extension f (x). This convergence is uniform on every closed interval on which the function f(x) is continuous (and absolute if the function f(x) has no discontinuities). Examles 2. and 2.2 illustrate this result.

13 2.7 How Fourier Series Converge 99 A function whose kth derivative is continuous is in class C k.onthe interval [, ], its Fourier coe cients (2.3) are Z f n = f(x) e inx dx. (2.52) If f is both eriodic and in C k, then one integration by arts gives f n = Z d alef(x) e inx dx in and k integrations by arts give f (x) e inx in dx = Z f (x) e inx in dx f n = Z f (k) (x) e inx dx (2.53) (in) k since the derivatives f (`) (x) of a C k eriodic function also are eriodic. Moreover if f (k+) is iecewise continuous, then Z ale d f n = f (k) e inx (x) dx (in) k+ f (k+) e inx (x) (in) k+ dx Z = f (k+) (x) e inx (2.54) dx. (in) k+ Since f (k+) (x) is iecewise continuous on the closed interval [, ], it is bounded there in absolute value by, let us say, M. So the Fourier coe cients of a C k eriodic function with f (k+) iecewise continuous are bounded by f n ale n k+ Z f (k+) (x) dx ale M. (2.55) nk+ We often can carry this derivation one ste further. In most simle examles, the iecewise continuous eriodic function f (k+) (x) actually is iecewise continuously di erentiable between its successive jums at x j.inthis case, the derivative f (k+2) (x) is a iecewise continuous function lus a sum of a finite number of delta functions with finite coe cients. Thus we can integrate once more by arts. If for instance the function f (k+) (x) jumsj times between and by f (k+) j, then its Fourier coe cients are f n = = Z f (k+2) (x) e inx (in) JX Z xj+ f s (k+2) j= x j k+2 dx (x) e inx JX (in) k+2 dx + j= f (k+2) j e inx j (in) k+2 (2.56)

14 Fourier Series in which the subscrit s means that we ve searated out the delta functions. The Fourier coe cients then are bounded by f n ale M n k+2 (2.57) in which M is related to the maximum absolute values of f s (k+2) (x) and of the. The Fourier series of eriodic C k functions converge very f (k+) j raidly if k is big. Examle 2.6 (Fourier Series of a C Function). The function defined by 8 < ale x< f(x) = x ale x</2 (2.58) : x /2 ale x ale is continuous on the interval [, ] and its first derivative is iecewise continuous on that interval. By (2.55), its Fourier coe cients f n should be bounded by M/n. In fact they are (exercise 2.8) f n = Z f(x)e inx dx = ( )n+ (i n ) 2 n 2 (2.59) bounded by 2 2//n 2 in agreement with the stronger inequality (2.57). Examle 2.7 (Fourier Series for a C Function). The function defined by f(x) = + cos 2x for x ale/2 and f(x) = for x /2 has a eriodic extension f that is continuous with a continuous first derivative and a iecewise continuous second derivative. Its Fourier coe cients (2.52) f n = Z /2 /2 ( + cos 2x) e inx dx = satisfy the inequalities (2.55) and (2.57) for k =. 8sinn/2 (4n n 3 ) Examle 2.8 (The Fourier Series for cos µx). The Fourier series for the even function f(x) = cos µx has only cosines with coe cients (2.2) Z a n = cos nx cos µx dx Z = [cos(µ + n)x + cos(µ n)x] dx = ale sin(µ + n) sin(µ n) + = 2 µ( ) n µ + n µ n µ 2 sin µ. n2 Thus whether or not µ is an integer, the series (2.9) gives us 2µ sin µ cos x cos 2x cos 3x cos µx = 2µ 2 µ µ µ (2.6) (2.6)

15 2.7 How Fourier Series Converge which is continuous at x = ± (Courant, 937, cha. IX). Examle 2.9 (The Sine as an Infinite Product). In our series (2.6) for cos µx, wesetx =, dividebysinµ, relace µ with x, and so find for the cotangent the exansion cot x = 2x 2x 2 + x x x (2.62) or equivalently cot x x = 2x 2 x x x (2.63) For ale x ale q<, the absolute value of the nth term on the right is less than 2q/((n 2 q 2 )). Thus this series converges uniformly on [,x], and so we may integrate it term by term. We find (exercise 2.) Z x sin x cot t dt =ln t x = X Z x 2tdt n 2 t 2 = X ale x 2 ln n 2. (2.64) Exonentiating, we get the infinite-roduct formula " sin x x =ex X x 2 # Y x 2 ln n 2 = n 2 (2.65) for the sine from which one can derive the infinite roduct (exercise 2.2)! Y x 2 cos x = (2.66) (n 2 )2 for the cosine (Courant, 937, cha. IX). Fourier series can reresent a much wider class of functions than those that are continuous. If a function f(x) is square integrable on an interval [a, b], then its N-term Fourier series f N (x) will converge to f(x) in the mean, that is lim N! Z b a dx f(x) f N (x) 2 =. (2.67) What haens to the convergence of a Fourier series if we integrate or di erentiate term by term? If we integrate the series e inx/ f(x) = f n (2.68)

16 2 Fourier Series then we get a series Z x F (x) = dx f(x )= f (x a) i a f n n e inx/ e ina/ (2.69) that converges better because of the extra factor of /n. An integrated function f(x) is smoother, and so its Fourier series converges better. But if we di erentiate the same series, then we get a series f (x) =i 3/2 nf n e inx/ (2.7) that converges less well because of the extra factor of n. A di erentiated function is rougher, and so its Fourier series converges less well. 2.8 Quantum-Mechanical Examles Suose a article of mass m is traed in an infinitely dee one-dimensional square well of otential energy if <x< V (x) = (2.7) otherwise. The hamiltonian oerator is d 2 H = ~2 + V (x), (2.72) 2m dx2 in which ~ is Planck s constant divided by. This tiny bit of action, ~ = J s, sets the scale at which quantum mechanics becomes imortant. Quantum-mechanical corrections to classical redictions can be big in rocesses whose action is less than ~. An eigenfunction (x) of the hamiltonian H with energy E satisfies the equation H (x) =E (x) which breaks into two simle equations: and ~ 2 ~ 2 d 2 (x) 2m dx 2 = E (x) for <x< (2.73) d 2 (x) 2m dx 2 + (x) =E (x) for x< and for x>. (2.74) Every solution of these equations with finite energy E must vanish outside the interval <x<. So we must find solutions of the first equation (2.73) that satisfy the boundary conditions (x) = for x ale and x. (2.75)

17 2.8 Quantum-Mechanical Examles 3 For any integer n 6=, the function r 2 nx n(x) = sin for x 2 [,] (2.76) and n (x) = for x/2 (,) satisfies the boundary conditions (2.75). When inserted into equation (2.73) ~ 2 d 2 2m dx 2 n(x) = ~2 n 2 n(x) =E n n (x) (2.77) 2m it reveals its energy to be E n =(n~/) 2 /2m. These eigenfunctions n (x) are comlete in the sense that they san the sace of all functions f(x) that are square-integrable on the interval (,) and vanish at its end oints. They rovide for such functions the sine Fourier series r 2 nx f(x) = f n sin (2.78) which is eriodic with eriod 2 and is the Fourier series for a function that is odd f( x) = f(x) on the interval (, ) and zero at both ends. Examle 2. (Time Evolution of an Initially Piecewise Continuous Wave Function). Suose now that at time t = the article is confined to the middle half of the well with the square-wave wave function (x, ) = r 2 for 4 <x<3 4 (2.79) and zero otherwise. This iecewise continuous C wave function is discontinuous at x = /4 and at x =3/4. Since the functions hx ni = n (x) are orthonormal on [,] Z r 2 nx r dx sin 2 mx sin = nm (2.8) the coe cients f n in the Fourier series are the inner roducts (x, ) = f n = hn, i = Z dx r 2 nx f n sin r 2 nx sin (2.8) (x, ). (2.82)

18 4 Fourier Series Fourier Series for a Piecewise Continuous Wave Function.2.8 ψ (x, ) Figure 2.5 The iecewise continuous wave function (x, ) for = 2 (2.79, straight solid lines) and its Fourier series (2.8) with terms (solid curve) and terms (dashes). Gibbs overshoots occur near the discontinuities at x =/2 and x =3/2. x They are roortional to /n in accord with (2.57) f n = 2 Z 3/4 /4 dx sin nx = 2 n ale n cos 4 3n cos 4. (2.83) Figure 2.5 lots the square wave function (x, ) (2.79, straight solid lines) and its -term (solid curve) and -term (dashes) Fourier series (2.8) for an interval of length = 2. Gibbs s overshoot reaches.93 at x =.52 for terms and.898 at x =.52 for terms (not shown), amounting to about 9% of the unit discontinuity at x =/2. A similar overshoot occurs at x =3/2. How does (x, ) evolve with time? Since n (x), the Fourier comonent (2.76), is an eigenfunction of H with energy E n, the time-evolution oerator

19 2.8 Quantum-Mechanical Examles 5.8 ψ (x, t) 2 for a Piecewise Continuous Wave Function ψ (x, t) Figure 2.6 For an interval of length = 2, the robability distributions P (x, t) = (x, t) 2 of the -term Fourier series (2.85) for the wave function (x, t) at t =(thickcurve),t = 3 (medium curve), and =2m 2 /~ (thin curve). The jaggedness of P (x, t) arises from the two discontinuities in the initial wave function (x, ) (2.86) at x = /4 and x =3/4. x U(t) =ex( iht/~) takes (x, ) into (x, t) =e iht/~ (x, ) = r 2 nx f n sin e ient/~. (2.84) Because E n =(n~/) 2 /2m, the wave function at time t is (x, t) = r 2 nx f n sin e i~(n)2 t/(2m 2). (2.85) It is awkward to lot comlex functions, so Fig. 2.6 dislays the robability distributions P (x, t) = (x, t) 2 of the -term Fourier series (2.85) for the wave function (x, t) at t =(thickcurve),t = 3 (medium curve), and

20 6 Fourier Series.6 Fourier Series of a Continuous Function.4.2 ψ (x, ) Figure 2.7 The continuous wave function (x, ) (2.86, solid) and its - term Fourier series ( , dashes) are lotted for the interval [, 2]. x =2m 2 /~ (thin curve). The discontinuities in the initial wave function (x, ) cause both the Gibbs overshoots at x =/2 and x =3/2 seeninthe series for (x, ) lotted in Fig. 2.5 and the choiness of the robability distribution P (x, t) exhibited in Fig.(2.6). Examle 2. (Time Evolution of a Continuous Function). What does the Fourier series of a continuous function look like? How does it evolve with time? et us take as the wave function at t =thec function (x, ) = 2 sin (x /4) for 4 <x<3 4 (2.86) and zero otherwise. This initial wave function is a continuous function with a iecewise continuous first derivative on the interval [,], and it satisfies the eriodic boundary condition (, ) = (,). It therefore satisfies the conditions of Fourier s convergence theorem (Courant, 937,. 439), and so its Fourier series converges uniformly (and absolutely) to (x, ) on [,].

21 Quantum-Mechanical Examles 7 Probability Distribution of a Continuous Wave Function 2 ψ (x, t) Figure 2.8 For the interval [, 2], the robability distributions P (x, t) = (x, t) 2 of the -term Fourier series (2.92) for the wave function (x, t) at t =, 2,, =2m 2 /~,, and are lotted as successively thinner curves. x As in Eq.(2.82), the Fourier coe cients f n are given by the integrals Z r 2 nx f n = dx sin (x, ), (2.87) which now take the form f n = 2 2 Z 3/4 /4 dx sin nx (x sin /4). (2.88) Doing the integral, one finds for f n that for n 6= f n = n 2 [sin(3n/4) + sin(n/4)] (2.89) 4 while c 2 =. These Fourier coe cients satisfy the inequalities (2.55) and (2.57) for k =. The factor of /n 2 in f n guarantees the absolute convergence

22 8 Fourier Series of the series (x, ) = r 2 nx f n sin (2.9) because asymtotically the coe cient f n is bounded by f n ale A/n 2 where A is a constant (A = 44/(5 ) will do) and the sum of /n 2 converges to the Riemann zeta function (4.99) 2 = (2) = n2 6. (2.9) Figure 2.7 lots the -term Fourier series (2.9) for (x, ) for = 2. Because this series converges absolutely and uniformly on [, 2], the - term and -term series were too close to (x, ) to be seen clearly in the figure and so were omitted. As time goes by, the wave function (x, t) evolves from (x, ) to (x, t) = r 2 nx f n sin e i~(n)2 t/(2m 2). (2.92) in which the Fourier coe cients are given by (2.89). Because (x, ) is continuous and eriodic with a iecewise continuous first derivative, its evolution in time is much calmer than that of the iecewise continuous square wave (2.79). Figure 2.8 shows this evolution in successively thinner curves at times t =, 2,, =2m 2 /~,, and. The curves at t = and t = 2 are smooth, but some wobbles aear at t = and at t = due to the discontinuities in the first derivative of (x, ) at x =.5 and at x =.5. Examle 2.2 (Time Evolution of a Smooth Wave Function). Finally, let s try a wave function (x, ) that is eriodic and infinitely di erentiable on [,]. An infinitely di erentiable function is said to be smooth or C.The infinite square-well otential V (x) of equation (2.7) imoses the eriodic boundary conditions (, ) = (, ) =, so we try Its Fourier series (x, ) = (x, ) = r 2 3 ale cos x. (2.93) r 2 e ix/ e ix/ (2.94) 6 has coe cients that satisfy the uer bounds (2.55) by vanishing for n >.

23 2.8 Quantum-Mechanical Examles 9.4 Fourier Series of a Smooth Function.2 ψ (x, ) Figure 2.9 The wave function (x, ) (2.93) is infinitely di erentiable, and so the first terms of its uniformly convergent Fourier series (2.96) o er a very good aroximation to it. x The coe cients of the Fourier sine series for the wave function (x, ) are given by the integrals (2.82) Z r 2 nx f n = dx sin (x, ) = 2 Z nx ale x dx sin cos 3 = 8[( )n ] 3 n(n 2 4) (2.95) with all the even coe cients zero, c 2n =. The f n s are roortional to /n 3 which is more than enough to ensure the absolute and uniform convergence

24 Fourier Series.4 Probability Distribution of a Smooth Wave Function.2 ψ (x, t) Figure 2. The robability distributions P (x, t) = (x, t) 2 of the - term Fourier series (2.97) for the wave function (x, t) at t =, 2,, =2m 2 /~,, and are lotted as successively thinner curves. The time evolution is calm because the wave function (x, ) is smooth. x of its Fourier sine series (x, ) = r 2 nx f n sin. (2.96) As time goes by, it evolves to r 2 nx (x, t) = f n sin e i~(n)2 t/(2m 2 ) (2.97) and remains absolutely convergent for all times t. The e ects of the absolute and uniform convergence with f n / /n 3 are obvious in the grahs. Figure 2.9 shows (for = 2) that only terms are required to nearly overla the initial wave function (x, ). Figure 2. shows that the evolution of the robability distribution (x, t) 2 with time is

25 2.9 Dirac Notation smooth, with no sign of the jaggedness of Fig. 2.6 or the wobbles of Fig Because (x, ) is smooth and eriodic, it evolves calmly as time asses. 2.9 Dirac Notation Vectors ji that are orthonormal hk ji = k,j san a vector sace and exress the identity oerator I of the sace as (.4) I = NX jihj. (2.98) Multilying from the right by any vector gi in the sace, we get j= gi = I gi = NX jihj gi (2.99) which says that every vector gi in the sace has an exansion (.42) in terms of the N orthonormal basis vectors ji. Thecoe cientshj gi of the exansion are inner roducts of the vector gi with the basis vectors ji. These roerties of finite-dimensional vector saces also are true of infinitedimensional vector saces of functions. We may use as basis vectors the hases ex(inx)/. They are orthonormal with inner roduct (2.) e imx e inx e i(n m)x (m, n) = dx = dx = m,n (2.) which in Dirac notation with hx ni =ex(inx)/ and hm xi = hx mi is hm ni = hm xihx ni dx = j= e i(n m)x dx = m,n. (2.) The identity oerator for Fourier s sace of functions is I = nihn. (2.2) So we have and hx fi = hx I fi = fi = I fi = hx nihn fi = nihn f i (2.3) e inx hn fi (2.4)

26 2 Fourier Series which with hn fi = f n is the Fourier series (2.2). The coe are the inner roducts (2.3) cients hn fi = f n hn fi = hn xihx fi dx = e inx hx fi dx = e inx f(x) dx. (2.5) 2. Dirac s Delta Function A Dirac delta function is a (continuous, linear) ma from a sace of suitably well-behaved functions into the real or comlex numbers. It is a functional that associates a number with each function in the function sace. Thus (x y) associates the number f(y) with the function f(x). We may write this association as Z f(y) = f(x) (x y) dx. (2.6) Delta functions o u all over hysics. The inner roduct of two of the kets xi that aear in the Fourier-series formulas (2.4) and (2.5) is a delta function, hx yi = (x y). The formula (2.5) for the coe cient hn f i becomes obvious if we write the identity oerator for functions defined on the interval [, ] as for then hn fi = hn I fi = I = hn xihx fi dx = xihx dx (2.7) e inx hx fi dx. (2.8) The equation yi = I yi with the identity oerator (2.7) gives yi = I yi = xihx yi dx. (2.9) Multilying (2.7) from the right by fi and from the left by hy, we get f(y) =hy I fi = hy xihx fi dx = hy xif(x) dx. (2.) These relations (2.9 & 2.) say that the inner roduct hy xi is a delta function, hy xi = hx yi = (x y). The Fourier-series formulas (2.4) and (2.5) lead to a statement about

27 the comleteness of the hases ex(inx)/ f(x) = 2. Dirac s Delta Function 3 f n e inx = Interchanging and rearranging, we have f(x) = e e iny f(y) einx dy. (2.) in(x y)! f(y) dy. (2.2) But f(x) and the hases e inx are eriodic with eriod, so we also have Z! ein(x y) f(x +`) = f(y) dy. (2.3) Thus we arrive at the Dirac comb ein(x y) = or more simly e inx = " +2 `= # cos(nx) = (x y `) (2.4) `= (x `). (2.5) Examle 2.3 (Dirac s Comb). The sum of the first, terms of this cosine series (2.5) for the Dirac comb is lotted for the interval ( 5, 5) in Fig. 2.. Gibbs overshoots aear at the discontinuities. The integral of the first, terms from -5 to 5 is 5.. The stretched Dirac comb is e in(x y)/ = `= (x y `). (2.6) Examle 2.4 (Parseval s Identity). Using our formula (2.34) for the Fourier coe cients of a stretched interval, we can relate a sum of roducts fn g n of the Fourier coe cients of the functions f(x) and g(x) to an integral of the roduct f (x) g(x) f n g n = Z dx einx/ f (x) Z dy e iny/ g(y). (2.7)

28 4 Fourier Series 3.5 x 4 Dirac Comb Sum of Series Figure 2. The sum of the first, terms of the series (2.5) for the Dirac comb is lotted for 5 ale x ale 5. Both Dirac sikes and Gibbs overshoots are visible. x This sum contains Dirac s comb (2.6) and so fn g n = = Z Z dx dx Z Z dy f (x) g(y) dy f (x) g(y) `= e in(x y)/ (x y `). (2.8) But because only the ` = tooth of the comb lies in the interval [,], we have more simly fn g n = Z dx Z dy f (x) g(y) (x y) = Z dx f (x) g(x). (2.9)

29 2. The Harmonic Oscillator 5 In articular, if the two functions are the same, then Z f n 2 = dx f(x) 2 (2.2) which is Parseval s identity. Thus if a function is square integrable on an interval, then the sum of the squares of the absolute values of its Fourier coe cients is the integral of the square of its absolute value. Examle 2.5 (Derivatives of Delta Functions). Delta functions and other generalized functions or distributions ma smooth functions that vanish at infinity into numbers in ways that are linear and continuous. Derivatives of delta functions are defined so as to allow integrations by arts. Thus the nth derivative of the delta function (n) (x y) mas the function f(x) to ( ) n times its nth derivative f (n) (y) at y Z Z (n) (x y) f(x) dx = (x y)( ) n f (n) (x) dx =( ) n f (n) (y) (2.2) with no surface term. Examle 2.6 (The Equation xf(x) = a). Dirac s delta function sometimes aears unexectedly. For instance, the general solution to the equation xf(x) =a is f(x) =a/x+b (x)whereb is an arbitrary constant (Dirac, 967, sec. 5), (Waxman and Peck, 998). Similarly, the general solution to the equation x 2 f(x) =a is f(x) =a/x 2 +b (x)/x+c (x)+d (x) inwhich (x) is the derivative of the delta function, and b, c, and d are arbitrary constants. 2. The Harmonic Oscillator The hamiltonian for the harmonic oscillator is H = 2 2m + 2 m!2 q 2. (2.22) The commutation relation [q, ] q q = i~ imlies that the lowering and raising oerators r m! a = q + i r m! and a i = q (2.23) 2~ m! 2~ m! obey the commutation relation [a, a ] =. In terms of a and a, which also are called the annihilation and creation oerators, the hamiltonian H has

30 6 Fourier Series the simle form H = ~! a a + 2. (2.24) There is a unique state i that is annihilated by the oerator a, as may be seen by solving the di erential equation r m! hq a i = 2~ hq q + i i =. (2.25) m! Since hq q = q hq and hq i = ~ i the resulting di erential equation is dhq i dq (2.26) dhq i dq = m! ~ q hq i. (2.27) Its suitably normalized solution is the wave function for the ground state of the harmonic oscillator m! /4 m!q hq 2 i = ex. (2.28) ~ 2~ For n =,, 2,...,thenth eigenstate of the hamiltonian H is ni = n! a n i (2.29) where n! n(n )...isn-factorial and! =. Its energy is H ni = ~! n + 2 ni. (2.3) The identity oerator is I = nihn. (2.3) n= An arbitrary state i has an exansion in terms of the eigenstates ni i = I i = nihn i (2.32) n= and evolves in time like a Fourier series X, ti = e iht/~ i = e iht/~ nihn i = e i!t/2 X n= n= e in!t nihn i (2.33)

31 with wave function 2.2 Nonrelativistic Strings 7 (q, t) =hq, ti = e i!t/2 X n= e in!t hq nihn i. (2.34) The wave functions hq ni of the energy eigenstates are related to the Hermite olynomials (examle 8.6) H n (x) =( ) n x2 dn e dx n e x2 (2.35) by a change of variables x = m!/~ q sq and a normalization factor se (sq) 2 /2 m! /4 e m!q 2 /2~ m! /2q hq ni = 2 n n! H n(sq) = H n. ~ 2 n n! ~ (2.36) The coherent state i i = e 2 /2 e a i = e 2 /2 n= n n! ni (2.37) is an eigenstate a i = i of the lowering (or annihilation) oerator a with eigenvalue. Its time evolution is simly, ti = e i!t/2 e 2 /2 e i!t n ni = e i!t/2 e i!t i. (2.38) n! n= 2.2 Nonrelativistic Strings If we clam the ends of a nonrelativistic string at x = and x =, then the amlitude y(x, t) will obey the boundary conditions and the wave equation as long as y(x, t) remains small. The functions y n (x, t) =sin nx y(,t)=y(, t) = (2.39) v (2.4) f n sin nvt + d n cos nvt (2.4) satisfy this wave equation (2.4) and the boundary conditions (2.39). They reresent waves traveling along the x-axis with seed v.

32 8 Fourier Series The sace S of functions f(x) that satisfy the boundary condition (2.39) is sanned by the functions sin(nx/). One may use the integral formula Z sin nx to derive for any function f 2 S the Fourier series f(x) = f n sin nx with coe cients f n = 2 mx sin dx = 2 nm (2.42) Z and the reresentation (x z 2m) = 2 m= for the Dirac comb on S. (2.43) sin nx f(x)dx (2.44) sin nx nz sin (2.45) 2.3 Periodic Boundary Conditions Periodic boundary conditions often are convenient. For instance, rather than study an infinitely long one-dimensional system, we might study the same system, but of length. The ends cause e ects not resent in the infinite system. To avoid them, we imagine that the system forms a circle and imose the eriodic boundary condition In three dimensions, the analogous conditions are (x ±, t) = (x, t). (2.46) (x ±,y,z,t)= (x,y,z,t) (x, y ±,z,t)= (x, y,z,t) (2.47) (x, y, z ±,t)= (x, y, z,t). The eigenstates i of the free hamiltonian H = 2 /2m have wave functions (x) =hx i = e ix /~ /(~) 3/2. (2.48) The eriodic boundary conditions (2.47) require that each comonent i of momentum satisfy i /~ =n i or = ~n = hn (2.49)

33 2.3 Periodic Boundary Conditions 9 where n is a vector of integers, which may be ositive or negative or zero. Periodic boundary conditions naturally arise in the study of solids. The atoms of a erfect crystal are at the vertices of a Bravais lattice x i = x + 3X n i a i (2.5) in which the three vectors a i are the rimitive vectors of the lattice and the n i are three integers. The hamiltonian of such an infinite crystal is invariant under translations in sace by i= 3X n i a i. (2.5) i= To kee the notation simle, let s restrict ourselves to a cubic lattice with lattice sacing a. Then since the momentum oerator generates translations in sace, the invariance of H under translations by a n ex(ian ) H ex( ian ) =H (2.52) imlies that e ian and H are comatible normal oerators [e ian,h] =. As exlained in section.3, it follows that we may choose the eigenstates of H also to be eigenstates of e ian which imlies that e ia n/~ i = e iak n i (2.53) (x + an) =hx + an i = hx e ia n/~ i = hx e iak n/~ i = e iak n (x). (2.54) Setting (x) =e ik x u(x) (2.55) we see that condition (2.54) imlies that u(x) is eriodic u(x + an) =u(x). (2.56) For a general Bravais lattice, this Born von Karman eriodic boundary condition is! 3X u x + n i a i,t = u(x,t). (2.57) i= Equations (2.54) and (2.56) are known as Bloch s theorem.

34 2 Fourier Series Exercises 2. Show that sin! x +sin! 2 x is the same as (2.9). 2.2 Find the Fourier series for the function ex(ax) on the interval < x ale. 2.3 Find the Fourier series for the function (x 2 2 ) 2 on the same interval (, ]. 2.4 Find the Fourier series for the function (+cos x)sinax on the interval (, ]. 2.5 Show that the Fourier series for the function x cos x on the interval [, ] is x cos x = 2 sin x +2 X n=2 ( ) n n n 2 sin nx. (2.58) 2.6 (a) Show that the Fourier series for the function x on the interval [, ] is x = 4 cos(2n + )x 2 (2n + ) 2. (2.59) n= (b) Use this result to find a neat formula for 2 /8. Hint: set x =. 2.7 Show that the Fourier series for the function sin x on the interval [, ] is sin x = 2 4 cos 2nx 4n 2. (2.6) 2.8 Show that the Fourier coe cients of the C function (2.58) on the interval [, ] are given by (2.59). 2.9 Find by insection the Fourier series for the function ex[ex( ix)]. 2. Fill in the stes in the comutation (2.27) of the Fourier series for x Do the first integral in equation (2.64). Hint: di erentiate ln sin x x. 2.2 Use the infinite-roduct formula (2.65) for the sine and the relation cos x =sinx/(2 sin x) to derive the infinite-roduct formula (2.66) for the cosine. Hint: " Y x 2 4 n2 # = " Y x 2 4 (2n )2 #" x 2 4 (2n)2 2.3 What s the general solution to the equation x 3 f(x) =a? #. (2.6)

35 Exercises Suose we wish to aroximate the real square-integrable function f(x) by the Fourier series with N terms Then the error f N (x) = a 2 + N X E N = (a n cos nx + b n sin nx). (2.62) [f(x) f N (x)] 2 dx (2.63) will deend uon the 2N + coe cients a n and b n.thebestcoe cients minimize this error and satisfy the n By using these conditions, find them. 2.5 Find the Fourier series for the function n =. (2.64) f(x) = (a 2 x 2 ) (2.65) on the interval [, ] for the case a 2 < 2.TheHeaviside ste function (x) is zero for x<, one-half for x =, and unity for x> (Oliver Heaviside, ). The value assigned to () seldom matters, and you need not worry about it in this roblem. 2.6 Derive or infer the formula (2.6) for the stretched Dirac comb. 2.7 Use the commutation relation [q, ] = i~ to show that the annihilation and creation oerators (2.23) satisfy the commutation relation [a, a ] =. 2.8 Show that the state ni =(a ) n i/ n! is an eigenstate of the hamiltonian (2.24) with energy ~!(n + /2). 2.9 Show that the coherent state i (2.37) is an eigenstate of the annihilation oerator a with eigenvalue. 2.2 Derive equations (2.44 & 2.45) from the exansion (2.43) and the integral formula (2.42). 2.2 Consider a string like the one described in section 2.2, which satisfies the boundary conditions (2.39) and the wave equation (2.4). The string is at rest at time t = and is struck recisely at t = and x = a so that y(x, ) = t) = v (x a). t=

36 22 Fourier Series Find y(x, t) and ẏ(x, t), where the dot means time derivative Same as exercise (2.2), but now the initial conditions are u(x, ) = f(x) and u(x, ) = g(x) (2.68) in which f() = f() = and g() = g() =. Find the motion of the amlitude u(x, t) of the string (a) Find the Fourier series for the function f(x) =x 2 on the interval [, ]. (b) Use your result at x = to show that n 2 = 2 (2.69) 6 which is the value of Riemann s zeta function (4.99) (x) at x = 2.