# General solution of a particle in a box

The general solution of particle in a 1D infinite potential well(width L) is given by: $$\psi(x,t)=\sum_n a_{n}.\sqrt{\frac{2}{L}}.\sin\bigg(\frac{n\pi x}{L}\bigg).\exp\bigg(-\frac{ iE_n.t}{\hbar}\bigg)$$ Here each of the term is an eigenfunction of the Schrodinger equation. If $$\phi_n(x)=\sqrt{\frac{2}{L}}\sin\bigg(\frac{n\pi x}{L}\bigg)$$ is a solution of schrodinger equation, then so is $$\phi_n(x)=\sqrt{\frac{2}{L}}\sin\bigg(\frac{n\pi x}{L}\bigg).\exp(i\delta_n)$$ . So the general solution should be - $$\psi(x,t)={\sum_n} a_{n}.\sqrt{\frac{2}{L}}\sin\bigg(\frac{n\pi x}{L}\bigg).\exp\bigg(-\frac{ iE_n.t}{\hbar}+i\delta_n\bigg),$$ where $$\delta_n$$ is real. Shouldn't this be the correct form? Why everyone neglects the phase?

• Do you intend $\delta_n$ to be a specific expression? Because right now, you have placed no restrictions on it, which means that the wavefunction is in general not normalized if $\delta_n$ has any nonzero real part. – probably_someone Sep 5 '19 at 9:40

The coefficient $$a_n$$ is usually not presumed to be real. If we go with this convention, the phase $$\delta_n$$ can be absorbed into $$a_n$$ without a loss of generality.
The point is that an overall phase is physically irrelevant. So, when one talks about a specific energy eigenfunction such as a particular $$\phi_n$$, one doesn't care about an overall phase. You are right in noticing that when we put a bunch of individual eigenfunctions together in a linear combination, the relative phases among them matter. But, as I said, this relative phase is absorbed into the complex coefficients $$a_n$$ in a natural manner.
By the way, I am assuming that you meant $$\exp{(i\delta_n)}$$ with $$\delta_n\in\mathbb{R}$$ and not $$\exp{(\delta_n)}$$ because $$\delta_n$$ is not a phase if you meant $$\exp{(\delta_n)}$$. As pointed out in a comment, you cannot really say that if $$\phi_n$$ is a solution then $$\phi_n\exp{(\delta_n)}$$ is also a solution unless $$\delta_n$$ is purely imaginary (in which case, $$-i\delta_n$$ would be the phase and not $$\delta_n$$).