Question Number 2234 by madscientist last updated on 10/Nov/15
$${in}\:{quantum}\:{physics}\:{is}\:{this}\:{a}\:{true}\: \\ $$$${statement}?\:{h}={h}\:{bar} \\ $$$$\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)\rangle=\mathrm{0} \\ $$$$\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)=\int\psi^{\ast} \left({t}\right)\psi\left({t}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int\frac{{d}\psi^{\ast} }{{dt}}\psi{dx}+\int\psi^{\ast} {H}\psi{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{{ih}}\int\left({H}\psi\right)^{\ast} \psi{dx}+\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx}−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx}=\mathrm{0} \\ $$$$\:\therefore−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {Hdx}−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx} \\ $$$$\:\:\:\:\:\:=\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)\rangle \\ $$