Doubt regarding the completeness of ${\psi_n}$ in infinite potential well The wavefunctions (without the time factor) for an infinite potential well (width: $0$ to $a$):
$$\psi_n=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a} \right).$$
The set of $\psi_n$ is complete as any other wavefunction can be written as:
$$ \Psi(x \text{,} 0) = \sum_{n=1}^{\infty} c_n \psi_n (x).$$
I was thinking this is just a Fourier series, but Fourier series contain both sine and cosine terms. Since $\psi_n$ are all sine here, how can we justify this completeness?
 A: You may be interested in reading about the Fourier sine series.  The set of vectors
$$\psi_n(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$$
does indeed constitute an orthonormal basis for $L^2([0,a])$ in the sense that for each $\Phi\in L^2([0,a])$, there exists a sequence $\{c_n\}$ such that
$$\lim_{N\rightarrow \infty} \left\Vert \Phi - \sum_{n=1}^N c_n \psi_n\right\Vert = 0$$
This is what we mean by the existence of a complete orthonormal (Schauder) basis.

Note that this notion of convergence is not the same as pointwise convergence; since $\psi_n(0)=0$ for all $n$, $\sum_{n=1}^\infty c_n \psi_n(0) = 0$. However, in quantum mechanics we don't care about the value of wavefunctions at individual points; this is a mathematical subtlety buried in the theory of $L^2$ spaces.  If two wavefunctions $\alpha$ and $\beta$ differ at a countable number of points (or more generally, in such a way that $\Vert \alpha-\beta \Vert = 0$), they are to be regarded as exactly the same vector. If you are interested, see here for more.
