Find Fourier coefficients - M. Chester text

[GreyNoise]

Homework Statement

I am self studying an introductory quantum physics text by Marvin Chester Primer of Quantum Mechanics. I am stumped at a problem (1.10) on page 11. We are given

f(x) = \sqrt{ \frac{8}{3L} } cos^2 \left ( \frac {\pi}{L} x \right )

and asked to find its Fourier coefficients C_0 and C_1 from

C_k = \frac{1}{L} \displaystyle {\int_0^L} e^{-ikx} f( x ) dx

I found C_0 = \sqrt{ \frac{2}{3L} }; it was easy as k=0 greatly simplified the integral. But I am completely baffled at an answer for the C_1 coefficient. The text has the answer as \frac{1}{ \sqrt{6L} }, and I do not get anything remotely close that; I keep getting answers with nonzero imaginary parts and \pi in the terms.

Homework Equations

f(x) = \sqrt{ \frac{8}{3L} } cos^2 \left (\frac{\pi}{L} x \right)

C_k = \frac{1}{L} \displaystyle {\int_0^L} e^{-ikx} f( x ) dx

C_1 = \frac{1}{ \sqrt{6L} }

The Attempt at a Solution

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To begin, I substituted f(x) into the equation for C_k for k = 1

C_1 = \frac{1}{L} \displaystyle {\int_0^L e^{-ix} \sqrt{ \frac {8}{3L} }} cos^2 \left ( \frac {\pi}{L} x \right ) dx

Then I factored the constants out of the integral

C_1 = \frac{1}{L} \sqrt{ \frac {8}{3L} } \displaystyle {\int_0^L e^{-ix}} cos^2 \left ( \frac {\pi}{L} x \right ) dx

Then I attempted to solve the integral

\displaystyle {\int_0^L e^{-ix}} cos^2 \left ( \frac {\pi}{L} x \right ) dx

I tried doing it by parts, then I tried by substituting e^{-ix} = cos (x) - i sin (x) and distributing cos^2 \left ( \frac {\pi}{L} x \right ) into that and integrating the result, but I only generated 4 pages of blundering chicken scratch that gave me things like \textstyle {\frac {2i}{3L} \sqrt{ \frac{8}{3L}} }; another attempt led to \textstyle {\frac {1}{\pi} \frac{2i}{3} }. I even went to my TI-89 and the Wolfram website and got an analytical solution to the integral, but the definite integrals they returned gave complex numbers that did NOT look anything like \frac{1}{ \sqrt{6L} }. Can anyone tell me; am I even setting up the integral correctly?

 

[Dick]

Can anyone tell me; am I even setting up the integral correctly?

Double check your form defining the Fourier coefficents. The $e^{-ikx}$ doesn't have period $L$ for $k=1$. Shouldn't it be something more like $e^{-i\frac{2k\pi}{L}x}$? $e^{-ikx}$ is used in the transform over the whole line.

 

[GreyNoise]

Double check your form defining the Fourier coefficents. The $e^{-ikx}$ doesn't have period $L$ for $k=1$. Shouldn't it be something more like $e^{-i\frac{2k\pi}{L}x}$? $e^{-ikx}$ is used in the transform over the whole line.

Thnx much for that Dick! That was the fix I needed. Now it brings to my mind a question about the form of the exponential argument. The author
wrote
$$ C_k = \frac{1}{L} \int_0^L e^{-ikx} f(x) dx $$
and I used the form literally, but now I am concerned that I missed a basic Fourier series concept. The author had written (before the above equation) that
$$ kL = 2 \pi n $$
So the form I should have used would appear as
$$ C_n = \frac{1}{L} \int_0^L e^{-i \frac{2 \pi n}{L} x} f(x) dx $$
where I substituted $k = \textstyle \frac{2 \pi n}{L}$ in the exponent's argument. Am I correct on this point?

 

[Ray Vickson]

Thnx much for that Dick! That was the fix I needed. Now it brings to my mind a question about the form of the exponential argument. The author
wrote
$$ C_k = \frac{1}{L} \int_0^L e^{-ikx} f(x) dx $$
and I used the form literally, but now I am concerned that I missed a basic Fourier series concept. The author had written (before the above equation) that
$$ kL = 2 \pi n $$
So the form I should have used would appear as
$$ C_n = \frac{1}{L} \int_0^L e^{-i \frac{2 \pi n}{L} x} f(x) dx $$
where I substituted $k = \textstyle \frac{2 \pi n}{L}$ in the exponent's argument. Am I correct on this point?

Not quite, but it depends in part on how you plan to use the $C_n$.

Basically, it is always easiest to start with an orthonormal set of functions $\{ u_k(x) \}$, which have the inner-product properties that
$\langle u_j, u_k \rangle = 0$ if $k \neq j$ and $\langle u_k,u_k \rangle = 1$ for all $k$. Here, $\langle \cdot , \cdot \rangle$ denotes the inner product for functions on $[0,L]$:
$$ \langle f,g \rangle = \int_0^L f^*(x) g(x) \, dx, $$
where $f^*$ is the complex congugate of $f$.

You can easily verify that the functions $u_k(x) =(1/\sqrt{L}) \: e^{i 2 \pi k x/L}, \; k = 0, \pm 1, \pm 2, \ldots$ constitute an orthonormal set. So, if we expand a function $f(x)$ as
$$f(x) = \sum_{k=-\infty}^{\infty} c_k u_k(x), $$
we get (with almost no work) that the coefficients are
$$c_k = \langle u_k , f \rangle = \int_0^L u^*(x) f(x) \, dx = \frac{1}{\sqrt{L}} \int_0^L e^{-i 2 \pi k x/L} f(x) \; dx.$$

In other words, we can either use the expansion $\sum c_k u_k$ or use the expansion $\sum_k \sqrt{L} C_k e^{i 2 \pi k x/L}$ with your $C_k$ (because $\text{my} \: c_k = \sqrt{L} \times \text{your} \: C_k$). The factor $\sqrt{L}$ needs to go somewhere, and IMHO it is easier to put it into the definition of the basis functions $\{ u_k \}$.

Note added in edit: it turns out that what YOU wrote is also correct! You are using expansion functions $f_k(x) = e^{i 2 \pi x k/L}, \; k = 0, \pm 1, \pm 2, \ldots$ and coefficients $C_k = (1/L) \int_0^L f^*_k(x) f(x) \, dx,$, so your $k$th term of the expansion has the form
$$\text{term}\; k = \frac{1}{L} \int_0^L f^*_k(t) f(t) \, dt \times f_k(x),$$
and this is the same as the $k$th term $c_k u_k(x)$ given above.

 

[GreyNoise]

Sorry for delayed response; thnx are in order; I was away on vacation. Thank you Dick and Ray. The posts got me back on top of the curve (my Fourier forays are ten years ago now), so I committed them to my notes. I might be back for more regarding the M. Chester text I am reading. Thnx again.