Quantum Mechanics (1) · Problem 7

A few things first

Physics — Quantum Mechanics · fourier-series-hilbert-space

Problem

Let f∈L2(−π,π)f \in L^2(-\pi, \pi) be a summable complex function on the real interval [−π,π][-\pi, \pi] (with Lebesgue measure). (a) Define the scalar product by: ⟨f∣g⟩=∫−ππf∗(x)g(x) dx,(1.128)\langle f|g\rangle = \int_{-\pi}^{\pi} f^*(x)g(x)\,dx, \tag{1.128} for f,g∈L2(−π,π)f, g \in L^2(-\pi, \pi). Starting with the intuitive, but non-trivial, assumption that there is no vector in L2(−π,π)L^2(-\pi, \pi) other than the trivial vector (f∼0f \sim 0) which is orthogonal to all of the functions sin⁡(nx)\sin(nx), cos⁡(nx)\cos(nx), n=0,1,2,…n = 0, 1, 2, \ldots, show that any vector ff may be expanded as: f(x)=∑n=0∞(ancos⁡nx+bnsin⁡nx),(1.129)f(x) = \sum_{n=0}^{\infty}(a_n \cos nx + b_n \sin nx), \tag{1.129} where a0=12π∫−ππf(x) dx(1.130)a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\,dx \tag{1.130} an=1π∫−ππf(x)cos⁡nx dx(n>0)(1.131)a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\,dx \quad (n > 0) \tag{1.131} bn=1π∫−ππf(x)sin⁡nx dx.(1.132)b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx\,dx. \tag{1.132} [You may consult a text such as Fano's Mathematical Methods of Quantum Mechanics for a full proof of the completeness of such functions.] (b) Consider the function: f(x)={−1x<0,0x=0,+1x>0.(1.133)f(x) = \begin{cases} -1 & x < 0, \\ 0 & x = 0, \\ +1 & x > 0. \end{cases} \tag{1.133} Determine the coefficients an,bna_n, b_n, n=0,1,2,…n = 0, 1, 2, \ldots for this function for the expansion of part (a). (c) We wish to investigate the partial sums in this expansion: fN(x)=∑n=0N(ancos⁡nx+bnsin⁡nx).(1.134)f_N(x) = \sum_{n=0}^{N}(a_n \cos nx + b_n \sin nx). \tag{1.134} Find the position, xNx_N of the first maximum of fNf_N (for x>0x > 0). Evaluate the limit of fN(xN)f_N(x_N) as N→∞N \to \infty. Give a numerical answer. In so doing, you are finding the maximum value of the series expansion in the limit of an infinite number of terms. [You may find the following identity useful: ∑n=1Ncos⁡(2n−1)x=12sin⁡2Nxsin⁡x.(1.135)\sum_{n=1}^{N} \cos(2n-1)x = \frac{1}{2}\frac{\sin 2Nx}{\sin x}. \tag{1.135} (d) Obviously, the maximum value of f(x)f(x), defined in part (b), is 1. If the value you found for the series expansion is different from 1, comment on the possible reconciliation of this difference with the theorem you demonstrated in part (a).

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