Let Probability Density ρ\rho and wave function ψ=ψ(x,t)\psi=\psi(x,t) be

ρ=ψ∗ψ=∣ψ∣2\rho=\psi^*\psi=|\psi|^2

Differentiate w.r.t time via product rule

∂∂t(ψ∗ψ)=∂ψ∗∂tψ+ψ∗∂ψ∂t\frac{\partial}{\partial t}(\psi^*\psi) = \frac{\partial\psi^*}{\partial t}\psi + \psi^*\frac{\partial\psi}{\partial t}

via Schrödinger equation, Hamiltonian II we know

iℏ∂ψ∂t=−ℏ22m∂2ψ∂x2+Vψi\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V\psi ∂ψ∂t=1iℏ[−ℏ22m∂2ψ∂x2+Vψ]\frac{\partial\psi}{\partial t} = \frac{1}{i\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} + V\psi\right] ∂ψ∗∂t=−1iℏ[−ℏ22m∂2ψ∗∂x2+Vψ∗]\frac{\partial\psi^*}{\partial t} = -\frac{1}{i\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2\psi^*}{\partial x^2} + V\psi^*\right]

where V(x)V(x) is the potential energy and xx is the position along one dimension.

so the V(x)V(x) terms cancel leaving kinetic pieces

∂ρ∂t=−iℏ2m[∂2ψ∗∂x2ψ−ψ∗∂2ψ∂x2]\frac{\partial\rho}{\partial t} = -\frac{i\hbar}{2m}\left[\frac{\partial^2\psi^*}{\partial x^2}\psi - \psi^*\frac{\partial^2\psi}{\partial x^2}\right]

Recall the Probability Current

This causes

∂ρ∂t=−∂J∂x\frac{\partial \rho}{\partial t}=-\frac{\partial J}{\partial x} ⇒∂∂t∣ψ(x,t)∣2=−∂J∂x\Rightarrow\quad \boxed{\frac{\partial}{\partial t}|\psi (x,t)|^2=-\frac{\partial J}{\partial x}}

This means that probability is locally conserved. It can't appear or vanish, only flow.