# Is there any physical significance to the orthonormality of states in a wave function?

In the infinite square well with bounds $0$ and $a$, the solutions to the time independent Schrodinger equation are: $$\psi_n(x)=\sqrt{\dfrac{2}{a}}\sin{\left(\dfrac{n\pi}{a}x\right)}$$ One of the properties of these wave functions is that they are "mutually orthogonal", meaning that $$\int_0^a \psi_m(x)^* \psi_n(x)dx=\delta_{mn}$$ Where $\delta_{mn}$ is the Kronecker delta. This fact is useful for finding the coefficient $c_n$ in the linear combination $$f(x)=\sum_{n=1}^{\infty}c_n\psi_n(x)$$ However, does orthonormality mean anything in terms of the particle? Do particles with mutually orthogonal stationary states differ from particles without them from a physical point of view?

These orthogonal states are energy eigenstates. Every measurable quantity provides an orthogonal basis of eigenstates. The physical meaning of their orthogonality is that, when energy (in this example) is measured while the system is in one such state, it has no chance of instead being found to be in another. Thus a general state's probability of being observed in state $n$ upon making such a measurement is $c_n^\ast c_n$.
A similar analysis for two consecutive measurements, be they of the sme observable or different observables, can be used to derive the probability distribution for the second measurement's result. This requires understanding the state's time-dependence between measurements. The energy eigenstates' probability distribution doesn't change over time, as the $c_n$ are simply multiplied by the unit complex number $\exp -\frac{iE_n t}{\hbar}$ over a time $t$.