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Chemistry LibreTexts

7.3: Orthogonal Expansions

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    106843
  • Note

    As stated in Section 7.2, the coefficients of \ref{eq:fourier} are defined as so:

    \[\label{ao} a_0=\dfrac{1}{L}\int_{-L}^{L}f(x)dx\]

    \[\label{an} a_n=\dfrac{1}{L}\int_{-L}^{L}f(x)\cos{\left(\dfrac{n\pi x}{L} \right)}dx\]

    \[\label{bn} b_n=\dfrac{1}{L}\int_{-L}^{L}f(x)\sin{\left(\dfrac{n\pi x}{L} \right)}dx\]

    The idea of expressing functions as a linear combination of the functions of a given basis set is more general than what we just saw. The sines and cosines are not the only functions we can use, although they are a particular good choice for periodic functions. There is a fundamental theorem in function theory that states that we can construct any function using a complete set of orthonormal functions.

    The term orthonormal means that each function in the set is normalized, and that all functions of the set are mutually orthogonal. For a function in one dimension, the normalization condition is:

    \[\label{eq:fourier_normalization} \int_{-\infty }^{\infty }{\left | f (x) \right |}^2\; dx=1\]

    Two functions \(f(x)\) and \(g(x)\) are said to be orthogonal if:

    \[\label{eq:fourier_orthogonal} \int_{-\infty }^{\infty }{f (x) g^*(x) }\; dx=0\]

    The idea that you can construct a function with a linear combination of orthonormal functions is analogous to the idea of constructing a vector in three dimensions by combining the vectors \(\vec{v_1}=\{(1,0,0)\}, \vec{v_2}=\{(0,1,0)\},\vec{v_3}=\{(0,0,1)\},\) which as we all know are mutually orthogonal, and have unit length.

    The basis set we use to construct a Fourier series is

    \[\{1, \sin{(\frac{\pi}{L} x)}, \cos{(\frac{\pi}{L} x), \sin{(2\frac{\pi}{L} x)}, \cos{(2\frac{\pi}{L} x)}, \sin{(3\frac{\pi}{L} x)}}, \cos{(3\frac{\pi}{L} x)}...\} \nonumber\]

    We will prove that these functions are mutually orthogonal in the interval \([0,2L]\) (one period).

    For example, let’s prove that \(\sin{(n\frac{\pi}{L} x)}\) and \(1\) are orthogonal:

    \[\int sin\left (\frac{n\pi x}{L} \right )dx=-\frac{L}{n\pi}cos\left (\frac{n\pi x}{L} \right ) \nonumber\]

    \[\int_{0}^{2L} sin\left (\frac{n\pi x}{L} \right )dx=-\frac{L}{n\pi}cos\left (2n\pi \right )+\frac{L}{n\pi}cos(0)=\frac{L}{n\pi}\left ( 1-cos(2n\pi) \right )=0 \nonumber\]

    We can also prove that any \(\sin{(nx)}\) is orthogonal to any \(\cos{(nx)}\):

    \[\int sin\left (\frac{n\pi x}{L} \right ) \cos\left (\frac{n\pi x}{L} \right )dx=-\frac{L}{4n\pi}\cos\left (\frac{2n\pi x}{L} \right ) \nonumber\]

    \[\int_{0}^{2L} \sin\left (\frac{n\pi x}{L} \right ) cos\left (\frac{n\pi x}{L} \right )dx=-\frac{L}{4n\pi}\cos\left (4n\pi\right )+\frac{L}{4n\pi}\cos (0)=0 \nonumber\]

    Following the same procedure, we can also prove that

    \[\int \sin\left (\frac{n\pi x}{L} \right ) \sin\left (\frac{m\pi x}{L} \right )dx=0\;n\neq m \nonumber\]

    \[\int \cos\left (\frac{n\pi x}{L} \right ) \cos\left (\frac{m\pi x}{L} \right )dx=0\;n\neq m \nonumber\]

    The functions used in a Fourier series are mutually orthogonal. Are they normalized?

    \[\int_{0}^{2L} \sin^2\left (\frac{n\pi x}{L} \right )dx=L \nonumber\]

    \[\int_{0}^{2L} \cos^2\left (\frac{n\pi x}{L} \right )dx=L \nonumber\]

    \[\int_{0}^{2L} 1^2\;dx=2L \nonumber\]

    They are not! The functions \(1/2L, \frac{1}{L}\sin{(\frac{\pi}{L} x)}\) and \(\frac{1}{L}\cos{(\frac{\pi}{L} x)}\) are normalized, so we may argue that our orthonormal set should be:

    \[\{\frac{1}{2L},\frac{1}{L} \sin{(\frac{\pi}{L} x)},\frac{1}{L} \cos{(\frac{\pi}{L} x),\frac{1}{L} \sin{(2\frac{\pi}{L} x)}, \frac{1}{L}\cos{(2\frac{\pi}{L} x)}}, ...\} \nonumber\]

    and the series should be written as:

    \[\label{eq:fourier2} f(x)=c_0\frac{1}{2L}+\frac{1}{L}\sum_{n=1}^{\infty}c_n \cos\left ( \frac{n\pi x}{L} \right )+\frac{1}{L}\sum_{n=1}^{\infty}d_n \sin\left ( \frac{n\pi x}{L} \right )\]

    where we used the letters \(c\) and \(d\) to distinguish these coefficients from the ones defined in Equations \ref{ao}, \ref{an} and \ref{bn}.

    However if we compare this expression to Equation \ref{eq:fourier}:

    \[\label{eq:fourier} f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n cos\left ( \frac{n\pi x}{L} \right )+\sum_{n=1}^{\infty}b_n sin\left ( \frac{n\pi x}{L} \right )\]

    we see that it is just a matter of how we define the coefficients. The coefficients in Equation \ref{eq:fourier} equal the coefficients in Equation \ref{eq:fourier2} divided by \(L\). In other words, the coefficients in Equation \ref{eq:fourier} already contain the constant \(L\) (look at Equations \ref{ao}, \ref{an} and \ref{bn}), so we can write the sines and cosines without writing the factor \(1/L\) every single time.

    In conclusion, the set

    \[\{1, \sin{\left(\frac{\pi}{L} x\right)}, \cos{\left(\frac{\pi}{L} x\right), \sin{\left(2\frac{\pi}{L} x\right)}, \cos{\left(2\frac{\pi}{L} x\right)}, \sin{\left(3\frac{\pi}{L} x\right)}}, \cos{\left(3\frac{\pi}{L} x\right)}...\} \nonumber\]

    is not strictly orthonormal the way it is written, but it is once we include the constant \(L\) in the coefficients. Therefore, the cosines and sines form a complete set that allows us to express any other function using a linear combination of its members.

    There are other orthonormal sets that are used in quantum mechanics to express a variety of functions. Just remember that we can construct any function using a complete set of orthonormal functions.

    We can construct any function using a complete set of orthonormal functions.

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