- #1

GreyNoise

Gold Member

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## 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

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

and asked to find its Fourier coefficients [itex] C_0 [/itex] and [itex] C_1 [/itex] from

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

I found [itex] C_0 = \sqrt{ \frac{2}{3L} } [/itex]; it was easy as

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

## Homework Equations

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

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

[itex] C_1 = \frac{1}{ \sqrt{6L} } [/itex]

## The Attempt at a Solution

[/B]

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

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

Then I factored the constants out of the integral

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

Then I attempted to solve the integral

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

I tried doing it by parts, then I tried by substituting [itex] e^{-ix} = cos (x) - i sin (x) [/itex] and distributing [itex] cos^2 \left ( \frac {\pi}{L} x \right ) [/itex] into that and integrating the result, but I only generated 4 pages of blundering chicken scratch that gave me things like [itex] \textstyle {\frac {2i}{3L} \sqrt{ \frac{8}{3L}} } [/itex]; another attempt led to [itex] \textstyle {\frac {1}{\pi} \frac{2i}{3} } [/itex]. 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 [itex] \frac{1}{ \sqrt{6L} } [/itex]. Can anyone tell me; am I even setting up the integral correctly?