 Solution

In the position representation the Time-dependent Schrödinger equation isFor a normalized state define probability density and probability current byMultiplying the equation by $\overline{\psi }$ and its complex conjugate by $\psi$, then subtracting, cancels the real potential. The resulting identity isThus $\boxed{{\partial }_t\rho +{\partial }_{x}j=0}$. Integrating over an interval gives $d{\int }_b^c\rho dx/dt=j(b)-j(c)$: probability changes only through flux across the endpoints. This is local conservation of quantum probability.

Continuity with the zero exterior imposes zero endpoint values on the infinite square well. Solving the stationary equation with these boundary conditions giveswith zero extension outside. The integral of ${\sin }^2(n\pi x/a)$ is $a/2$, which fixes the normalization. The initial coefficients are $3/5$ and $4/5$, whose squared moduli sum to one. Each energy component acquires its stationary phase:At the midpoint, ${\Psi }_1=\sqrt{2/a}$, ${\Psi }_1^{\prime }=0$, ${\Psi }_2=0$, ${\Psi }_2^{\prime }=-(2\pi /a)\sqrt{2/a}$. HenceTaking its imaginary part proves