PHYS212 : Lectures

Subsections

• 2.2.1 An alternative Fourier Series
• 2.2.2 Fourier Transformations
• 2.2.3 A Square Pulse
• 2.2.4 A Square Wave Packet
• 2.2.5 A Gaussian Pulse

2.2 Localized Packets

2.2.1 An alternative Fourier Series

• The sine/cosine form of the Fourier series we have considered thus far can be recast in terms of complex exponentials.

• Eq. 2.14 can be written
  

  $$f(x) = a_o + \sum_{n=1}^{\infty} \left[ a_n \left( \frac{e^{i k_n x} + e^{-i k_n x}}{2} \right) \right.$$

  $$\left. + b_n \left( \frac{e^{i k_n x} - e^{-i k_n x}}{2i} \right) \right]$$ (2.25)

  
  
  where
  

  $$k_n = \frac{2 \pi n}{\lambda}$$ (2.26)

  
  
  which can be rearranged to give
  

  $$f(x) = a_o + \sum_{n=1}^{\infty} \left[ \left( \frac{a_n -... ...x} + \left( \frac{a_n + ib_n}{2} \right) e^{-i k_n x} \right]$$ (2.27)

  
  

• We can now see a way to write the Fourier series in a more compact form
  

  $$f(x) = \sum_{n=-\infty}^{\infty} c_n e^{i k_n x}$$ (2.28)

  
  
  with
  

  $$c_o = a_o$$

  $$c_n = \frac{a_n - ib_n}{2}$$

  $$c_{-n} = \frac{a_n + ib_n}{2} = c_n^*$$ (2.29)

  
  

• An improved inner product
  • The functions $e^{ik_n x}$ ought to be orthogonal to each other. Investigating the issue,
    
    $$\int_0^\lambda e^{ik_n x} e^{ik_m x} dx =$$
    
    
    
    
    $$\int_0^\lambda \left[\cos(k_n x) + i\sin(k_n x)\right] \left[\cos(k_m x) + i\sin(k_m x)\right] dx$$
    
    
    
    

    $$= \int_0^\lambda \cos(k_n x)\cos(k_m x) dx$$

    $$- \int_0^\lambda \sin(k_n x)\sin(k_m x) dx$$

    $$+ \int_0^\lambda i\cos(k_n x)\sin(k_m x) dx$$

    $$+ \int_0^\lambda i\sin(k_n x)\cos(k_m x) dx$$ (2.30)

    
    
    which we can evaluate using the orthogonality relations of Eq.'s 2.12 to find
    

    $$\int_0^\lambda e^{ik_n x} e^{ik_m x} dx = 0$$ (2.31)

    
    
    for $n \neq m$, as expected.

  • For $n = m$, the first two terms are not zero, but they cancel,
    

    $$\int_0^\lambda e^{2ik_n x} dx = \frac{\lambda}{2} - \frac{\lambda}{2} = 0$$ (2.32)

    
    
    This is troublesome, because we need the inner product of a function with itself to be nonzero.

  • We have a problem with the inner product we have been using - a problem we did not discover working with the sine/cosine form of the Fourier series, because sine and cosine are real functions.

  • We extend the inner product of Eq. 2.11 to apply to complex functions as follows,
    

    $$\vec{f}(x) \cdot \vec{g}(x) = \int_{x_o}^{x_o + \lambda} f(x) g^*(x) dx$$ (2.33)

    
    
    which changes the relative sign of the first two terms of Eq. 2.30 so that they do not cancel, while preserving orthogonality.

  • With this inner product, we find the compact orthogonality relation
    

    $$\int_{x_o}^{x_o + \lambda} e^{ik_n x} e^{-ik_m x} dx = \lambda \delta_{nm}$$ (2.34)

    
    

  • Applying Eq. 2.34 to Eq. 2.28, the alternative form of the Fourier series, we find
    

    $$c_n = \frac{1}{\lambda} \int_{x_o}^{x_o + \lambda} f(x) e^{-ik_n x} dx$$ (2.35)

    
    
    which applies to positive and negative $n$ as well as to $n = 0$.

2.2.2 Fourier Transformations

• We can use a kind of Fourier series to describe localized wave packets instead of periodic ones by taking the limit as $\lambda$, the interval of periodicity, goes to infinity.

• The wavenumber spacing between subsequent terms in the Fourier series is
  

  $$\Delta k = \frac{2 \pi}{\lambda} \Delta n = \frac{2 \pi}{\lambda}$$ (2.36)

  
  
  Hence, $\Delta k$ becomes an infinitesimal interval $dk$ in the limit.

• In terms of $\Delta k$, a Fourier coefficient on the interval $-\frac{\lambda}{2} \leq x \leq \frac{\lambda}{2}$ is given by
  

  $$c_n = \frac{\Delta k}{2 \pi} \int_{-\frac{\pi}{\Delta k}}^{\frac{\pi}{\Delta k}} f(x) e^{-i k_n x} dx$$ (2.37)

  
  

• Inserting Eq. 2.37 into the Fourier series (Eq. 2.28), we have
  

  $$f(x) = \sum_{n=-\infty}^{\infty} \left[ \frac{\Delta k}{2 \p... ...\frac{\pi}{\Delta k}} f(x') e^{-i k_n x'} dx' \right] e^{ikx}$$ (2.38)

  
  
  It is important here to distinguish between the variable of integration $x'$ in the $c_n$ calculation from $x$ in the series.

• Taking the limit $\lambda \rightarrow \infty$ which is equivalent to $\Delta k \rightarrow 0$, the sum becomes an integral, and we obtain
  

  $$f(x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \left... ...t_{-\infty}^{\infty} f(x') e^{-i k x'} dx' \right] e^{ikx} dk$$ (2.39)

  
  
  where we distribute the factor of $\frac{1}{2\pi}$ following one of the several conventions for defining Fourier transformations.^2.2

• Our discrete spectrum of Fourier coefficients $\{..., c_{-2}, c_{-1}, c_o, c_1, c_2, ...\}$ has become the expression in square brackets in Eq. 2.39, a continuous function of $k$,
  

  $$F(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) e^{-i k x} dx$$ (2.40)

  
  
  called the Fourier transformation of $f(x)$.

• Eq. 2.39 implies that the function $f(x)$ can be obtained from $F(k)$ via
  

  $$f(x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} F(k) e^{i k x} dk$$ (2.41)

  
  
  This is called the inverse Fourier transformation.

• We can describe the evolution of a wave packet with time by including the plane wave time dependence in its inverse Fourier transformation
  

  $$f(x, t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} F(k) e^{i (k x - \omega t)} dk$$ (2.42)

  
  

• Orthogonality
  • Using the inverse Fourier transformation (Eq. 2.41) to replace $f(x)$ in the Fourier transformation (Eq. 2.40), we have
    

    $$F(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \lef... ...{-\infty}^{\infty} F(k') e^{i k' x} dk' \right] e^{-i k x} dx$$ (2.43)

    
    

  • We can reverse the order of integration in Eq. 2.43 to give
    

    $$F(k) = \int_{-\infty}^{\infty} F(k') \left[ \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{i (k'- k) x} dx \right] dk'$$ (2.44)

    
    

  • The expression in square brackets in Eq. 2.44 is one form of the Dirac delta function^2.3
    

    $$\delta(k-k') = \frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i (k-k') x} dx$$ (2.45)

    
    
    which has properties in integration analogous to the Kronecker delta in discrete summation, namely,
    

    $$F(k) = \int_{-\infty}^{\infty} F(k') \delta(k-k') dx'$$ (2.46)

    
    

  • Hence, the orthogonality relation of the complex exponentials on an infinite interval of periodicity is given by
    

    $$\int_{-\infty}^{\infty} e^{-i (k-k') x} dx = 2 \pi \delta(k-k')$$ (2.47)

    
    

  • The symmetry of Eq.'s 2.39 and 2.43 suggests that a similar argument can be used to find a Dirac delta function in $x$. The result is
    

    $$\delta(x-x') = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i k (x-x')} dk$$ (2.48)

    
    

2.2.3 A Square Pulse

• Consider a single square pulse of length $L$ centered on the origin, described by
  

  $$f(x) = \left\{ \begin{array}{lc} A & (\vert x\vert < \frac{L}{2})\\ 0 & (\vert x\vert > \frac{L}{2}) \end{array}\right.$$ (2.49)

  
  

• The Fourier transformation of the pulse is
  

  $$F(k) = \frac{A}{\sqrt{2 \pi}} \int_{-\frac{L}{2}}^{\frac{L}{2}} e^{-ikx} dx$$

  $$= \frac{A}{\sqrt{2 \pi}} \left. \frac{1}{-ik} e^{-ikx} \right\vert _{-\frac{L}{2}}^{\frac{L}{2}}$$

  $$= \frac{A}{\sqrt{2 \pi}} \frac{e^{-i\frac{kL}{2}} - e^{i\frac{kL}{2}}}{-ik}$$

  $$= A \sqrt{\frac{2}{\pi}} \frac{\sin \left(\frac{kL}{2}\right)}{k}$$ (2.50)

  
  
  A plot of this ``wavenumber spectrum'' is shown in Fig. 2.3.

  Figure 2.3: The Fourier transformation $F(k)$ of the square pulse described by Eq. 2.49 with $A = 1$. The solid curve has $L=1$, and the dashed curve has $L=2$.

  

• The pulse is then described in space and time by the inverse transformation of Eq. 2.42,
  

  $$y(x,t) = \frac{A}{\pi} \int_{-\infty}^{\infty} \frac{\sin \left(\frac{kL}{2}\right)}{k} e^{i k(x-v_p t)} dk$$ (2.51)

  
  
  where we have included the plane wave time dependence in the complex exponential factor.

• Given the piecewise nature of the square pulse, perhaps it is not surprising that the integral is not straightforward to evaluate. However, Maple is equipped with conditional functions that allow it to express the result and plot it.

• Animation : squarePulse.mws [20 kB] (see Appendix A on running animations)

2.2.4 A Square Wave Packet

• Consider a segment of plane wave of length $L$ centered on the origin, described by
  

  $$f(x) = \left\{ \begin{array}{lc} Ae^{i k_o x} & (\vert x\ve... ...{L}{2})\\ 0 & (\vert x\vert > \frac{L}{2}) \end{array}\right.$$ (2.52)

  
  

• The Fourier transformation of the packet is
  

  $$F(k) = \frac{A}{\sqrt{2 \pi}} \int_{-\frac{L}{2}}^{\frac{L}{2}} e^{-i (k-k_o) x} dx$$

  $$= \frac{A}{\sqrt{2 \pi}} \left. \frac{1}{-i(k-k_o)} e^{-i(k-k_o)x} \right\vert _{-\frac{L}{2}}^{\frac{L}{2}}$$

  $$= \frac{A}{\sqrt{2 \pi}} \frac{e^{-i\frac{(k-k_o)L}{2}} - e^{i\frac{(k-k_o)L}{2}}}{-i(k-k_o)}$$

  $$= A \sqrt{\frac{2}{\pi}} \frac{\sin \left(\frac{(k-k_o)L}{2}\right)}{(k-k_o)}$$ (2.53)

  
  

• Eq. 2.53 gives the wavenumber spectrum identical in form to that of the square pulse shown in Fig. 2.3 but centered on $k=k_o$ instead of on $k=0$.

• The inverse transformation (Eq. 2.41) including the plane wave time dependence is
  

  $$y(x,t) = \frac{A}{\pi} \int_{-\infty}^{\infty} \frac{\sin \left(\frac{(k-k_o)L}{2}\right)}{(k-k_o)} e^{i k(x-v_p t)} dk$$ (2.54)

  
  

• Animation : squarePacket.mws [54 kB] (see Appendix A on running animations)

2.2.5 A Gaussian Pulse

Figure 2.4: The Gaussian pulse described by Eq. 2.55 with $A = 1$. The solid curve has $\sigma = 2$, and the dashed curve has $\sigma = 4$.

• Consider a Gaussian pulse (shaped like a ``bell curve''),
  

  $$f(x) = Ae^{-\frac{x^2}{2\sigma^2}}$$ (2.55)

  
  
  shown in Fig. 2.4.

• The constant $\sigma$ is a rough measure of the width of the pulse. In fact, Eq. 2.55 describes a normal distribution with standard deviation $\sigma$.

• In taking the Fourier transformation and its inverse, we will make use of the integral
  

  $$\int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi}$$ (2.56)

  
  

• Performing the Fourier transformation of $f(x)$, we have
  

  $$F(k) = \frac{A}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2} -ikx} dx$$ (2.57)

  
  

• In order to express Eq. 2.57 in the form of Eq. 2.56, we ``complete the square'' in the exponent,
  

  $$-\left( \frac{x^2}{2\sigma^2} + ikx \right)^2 = -\left( \fra... ... \frac{i\sigma k}{\sqrt{2}} \right)^2 - \frac{\sigma^2 k^2}{2}$$ (2.58)

  
  
  yielding
  

  $$F(k) = \frac{A}{\sqrt{2\pi}} e^{- \frac{\sigma^2 k^2}{2}} \... ...sqrt{2} \sigma} + \frac{i\sigma k}{\sqrt{2}} \right)^2} dx$$ (2.59)

  
  

• With the substitution
  

  $$u = \frac{x}{\sqrt{2} \sigma} + \frac{i\sigma k}{\sqrt{2}},  du = \frac{dx}{\sqrt{2} \sigma}$$ (2.60)

  
  
  we have
  

  $$F(k) = \frac{A}{\sqrt{2\pi}} \sqrt{2} \sigma e^{- \frac{\sigma^2 k^2}{2}} \int_{-\infty}^{\infty} e^{-u^2} du$$ (2.61)

  
  
  which, with the help of Eq. 2.56 gives the final result
  

  $$F(k) = \sigma A e^{- \frac{\sigma^2 k^2}{2}}$$ (2.62)

  
  
  shown in Fig. 2.5.

  Figure 2.5: The Fourier transformation of the Gaussian pulse described by Eq. 2.62 with $A = 1$.The solid curve has $\sigma = 2$, and the dashed curve has $\sigma = 4$.

  

• The inverse transformation (Eq. 2.41) including the plane wave time dependence is
  

  $$y(x,t) = \frac{\sigma A}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-\frac{\sigma^2 k^2}{2} + i k(x-v_p t)} dk$$ (2.63)

  
  

• You will show for homework that this integral can be evaluated to give
  

  $$y(x,t) = A e^{-\frac{(x - v_p t)^2}{2\sigma^2}}$$ (2.64)

  
  
  which is simply our original pulse (Eq. 2.55) including time dependence.

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006