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

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