Completeness of energy eigenfunctions of the infinite potential well vs Fourier series

Question

The energy eigenfunctions $\varphi_n(x)=\sqrt{\frac{2}{a}} \sin(\frac {n\pi x}{a})$ are said to be complete (and form a basis) within the interval $(0,a)$ meaning any square integrable function can be expanded as

\begin{equation}\tag{*}\label{*} f(x)=\sum_{n=1}^{\infty} c_n \sqrt{\frac{2}{a}} \sin(\frac {n\pi x}{a}). \end{equation}

Note that I have excluded the boundary points because all the basis functions are zero there by construction. However, here is my confusion: the Fourier series theorem when applied to a function with period $2a$ gives us the expansion

\begin{equation}\tag{**}\label{**} f(x)=a_0+\sum_{n=1}^{\infty}a_n \sin(\frac {n\pi x}{a})+ b_n \cos(\frac {n\pi x}{a}). \end{equation} where the set of functions $\{1,\sin(\frac {n\pi x}{a}),\cos(\frac {n\pi x}{a})\}$ form the basis. So, my question is why do coefficients $a_0$ and all the $b_n$'s go to zero in the Fourier series (see \ref{**}) for the functions under consideration in the first expression (see \ref{*})?

Edit: Quoting from Wikipedia,

Sines and cosines form an (orthonormal) Schauder basis for square-integrable functions on a finite domain. As a particular example, the collection: $$\{\sqrt{2} \sin (2 \pi n x) \mid n \in \mathbb{N}\} \cup\{\sqrt{2} \cos (2 \pi n x) \mid n \in \mathbb{N}\} \cup\{1\}$$ forms a basis for $L^2[0,1]$.

Again, this basis set is larger than the supposed basis set just consisting of $\sin$ functions.

Answer 1

In fact, these are just two different bases for the same space. Any function on a compact interval $[0,a]$ can be written as a sum of sines, cosines, or both.

In the language of Fourier series, what we’re doing is taking the function defined on $[0,a]$ and extending it to a periodic function on the entire real line in three different ways.

1. We can extend the original function to an odd function on $[-a,a]$ and then copy and paste that function across the rest of the real line. Because this function is odd the Fourier expansion includes only sines.
2. If we first extend the original function to an even function on $[-a,a]$, then the corresponding expansion will include only cosines.
3. If we don’t bother with the initial extension and just copy and paste our original function as-is, then the expansion will include both sines and cosines.

Note that in the first two cases, the periodic function being expanded has period $2a$, and so the arguments of the trig functions will look like $\frac{2n\pi x}{2a}=\frac{n\pi x}{a}$. In the third, the function will be periodic with period $a$ and so the arguments will take the form $\frac{2n\pi x}{a}$.

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From a physical perspective, this can be understood by noting that $$b_1=\left\{\sin\left(\frac{n\pi x}{a}\right)\right\}_{n=1}^\infty \qquad b_2=\left\{\cos\left(\frac{n\pi x}{a}\right)\right\}_{n=0}^\infty$$ $$b_3=\left\{1,\sin\left(\frac{2n\pi x}{a}\right),\cos\left(\frac{2n\pi x}{a}\right)\right\}_{n=1}^\infty$$ are sets of eigenfunctions corresponding to three different self-adjoint operators:

1. $b_1$ corresponds to the operator $-\frac{d^2}{dx^2}$ which may act on twice (weakly) differentiable functions with Dirichlet boundary conditions (i.e. $f(0)=f(a)=0$). This is often taken to be the Hamiltonian of a free particle confined to a box because it is the set of bound states for the finite potential well when the height of the walls goes to infinity.
2. $b_2$ corresponds to the operator $-\frac{d^2}{dx^2}$ which may act on twice (weakly) differentiable functions with Neumann boundary conditions (i.e. $f'(0)=f'(a)=0$). This is an alternative Hamiltonian which also confines the particle to the box without demanding that energy eigenstates vanish at the edges, and is in some sense more natural; note that the uniform wavefunction $\psi(x)=1/\sqrt{a}$ is an energy eigenstate with zero energy and therefore evolves trivially with time, whereas if we choose the Hamiltonian with Dirichlet boundary conditions $\psi$ would distort very rapidly with time.
3. $b_3$ corresponds to the operator $-i\frac{d}{dx}$ which may act on (weakly) differentiable functions with periodic boundary conditions (i.e. $f(0)=f(2\pi)$). This is the (angular) momentum operator for a particle on a ring of circumference $a$.