Menu Close

in-quantum-physics-is-this-a-true-statement-h-h-bar-d-dt-t-t-0-d-dt-t-t-t-t-dx-d-dt-dx-H-dx-1-ih-H-dx-1-ih-

Tinku Tara 0 Comments
Pin




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)⟩
$${in}\:{quantum}\:{physics}\:{is}\:{this}\:{a}\:{true}\: \\ $$$${statement}?\:{h}={h}\:{bar} \\ $$$$\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)\rangle=\mathrm{0} \\ $$$$\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)=\int\psi^{\ast} \left({t}\right)\psi\left({t}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int\frac{{d}\psi^{\ast} }{{dt}}\psi{dx}+\int\psi^{\ast} {H}\psi{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{{ih}}\int\left({H}\psi\right)^{\ast} \psi{dx}+\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx}−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx}=\mathrm{0} \\ $$$$\:\therefore−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {Hdx}−\frac{\mathrm{1}}{{ih}}\int\psi^{\ast} {H}\psi{dx} \\ $$$$\:\:\:\:\:\:=\frac{{d}}{{dt}}\langle\psi\left({t}\right)\mid\psi\left({t}\right)\rangle \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *