Question #196614

Suppose, I have normalized the wave function at some point of time. The wave

function evolves with time according to time dependent Schrodinger equation. How do

I know that the wave function remains normalized after some time?

[Hint: Show that 𝑑

𝑑𝑡

∫−∞

∞

|𝜓(𝑥,𝑡)|

2𝑑𝑥 = 0]


Expert's answer

 Schrodinger’s equation says that the dynamical evolution of Ψ is given by

i ∂Ψ∂t=− ℏ22m∂2Ψ∂x2+V(x)Ψi ~\dfrac{∂Ψ}{∂t} = −~\dfrac{\hbar^2}{2 m}\dfrac{∂^2Ψ}{∂x^2}+ V (x) Ψ

for a potential V(x)


Our statistical interpretation is provided by viewing Ψ(x,t)∗Ψ(x,t)Ψ(x, t) ∗ Ψ(x, t) as a density: ρ(x,t)ρ(x, t) =∣Ψ(x,t)∣2= |Ψ(x, t)|^ 2 so that

P(a,b)=∫abρ(x,t)dx=∫ab∣Ψ(x,t)∣2dxP(a, b) = \int ^b_aρ(x, t) dx =\int ^b_a|Ψ(x, t)|^2dx


ddt∫−∞∞∣Ψ(x,t)∣2dx=∫−∞∞∂∂t(Ψ∗(x,t)Ψ(x,t))dx\dfrac{d}{dt} \int^ ∞_{−∞}|Ψ(x, t)|^2dx =\int^ ∞_{−∞}\dfrac{∂}{∂t} (Ψ^∗(x, t) Ψ(x, t)) dx


then we just need the temporal derivative of the norm (squared) of Ψ(x, t):

∂∂t(Ψ∗(x,t)Ψ(x,t))=∂Ψ∗∂tΨ+Ψ∗∂Ψ∂t\dfrac{∂}{∂t} (Ψ^∗(x, t) Ψ(x, t)) = \dfrac{∂Ψ^∗}{∂t} Ψ + Ψ^∗ \dfrac{∂Ψ}{∂t}


Schrodinger’s equation, and its complex conjugate provide precisely the desired connection:

∂Ψ∂t=iℏ2m∂2Ψ∂x2−iℏ V(x)Ψ\dfrac{∂Ψ}{∂t} =\dfrac{i\hbar}{ 2 m}\dfrac{∂^2Ψ}{∂x^2}−\dfrac{i}{\hbar}~V (x) Ψ  


∂Ψ∗∂t=−iℏ 2m∂2Ψ∗∂x2+iℏ V(x)Ψ∗\dfrac{∂Ψ^∗}{∂t} = −\dfrac{i\hbar}{ ~2 m}\dfrac{∂^2Ψ^∗}{∂x^2}+\dfrac{i}{\hbar}~V (x) Ψ^∗


and we can write the above as:

∂∂t(Ψ∗(x,t)Ψ(x,t))=iℏ 2m(−∂2Ψ∗∂x2Ψ+Ψ∗∂2Ψ∂x2).\dfrac{\partial}{∂t} (Ψ^∗(x, t) Ψ(x, t)) = \dfrac{i\hbar}{ ~2 m}\bigg(−\dfrac{∂^2Ψ^∗}{∂x^2}Ψ + Ψ^∗\dfrac{ ∂^2Ψ}{∂x^2}\bigg).


Now, under the integral, we can use integration by parts1 to simplify the above – consider the first term :

∫−∞∞∂2Ψ∗∂x2Ψdx=∣(∂Ψ∗∂xΨ)∣−∞∞−∫−∞∞∂Ψ∗∂x∂Ψ∂xdx\int^∞_{−∞}\dfrac{∂^2Ψ^∗}{∂x^2}Ψ dx =\bigg|\bigg(\dfrac{∂Ψ^∗}{∂x} Ψ\bigg)\bigg|^∞_{−∞}−\int^ ∞_{−∞}\dfrac{∂Ψ^∗}{∂x}\dfrac{∂Ψ}{∂x} dx


We can get rid of the boundary term by assuming (an additional requirement) that Ψ(x,t)⟶0Ψ(x, t) \longrightarrow 0 as ∣x∣⟶∞|x| \longrightarrow∞

ddt∫−∞∞(Ψ∗(x,t)Ψ(x,t))=iℏ 2m∫−∞∞[(−∂2Ψ∗∂x2Ψ+Ψ∗∂2Ψ∂x2)].\dfrac{d}{dt}\int_{-\infin}^{\infin} (Ψ^∗(x, t) Ψ(x, t)) = \dfrac{i\hbar}{ ~2 m}\int^\infin_{-\infin}\bigg[\bigg(−\dfrac{∂^2Ψ^∗}{∂x^2}Ψ + Ψ^∗\dfrac{ ∂^2Ψ}{∂x^2}\bigg)\bigg].

ddt∫−∞∞∣Ψ(x,t)∣2dx=iℏ 2m∫−∞∞[∂Ψ∗∂x∂Ψ∂x−∂Ψ∗∂x∂Ψ∂x]dx\dfrac{d}{dt} \int^ ∞_{−∞}|Ψ(x, t)|^2dx=\dfrac{i\hbar}{ ~2 m}\int^ ∞_{−∞}\bigg[\dfrac{∂Ψ^∗}{∂x}\dfrac{∂Ψ}{∂x}-\dfrac{∂Ψ^*}{∂x}\dfrac{∂Ψ}{∂x}\bigg] dx

ddt∫−∞∞∣Ψ(x,t)∣2dx=0\dfrac{d}{dt} \int^ ∞_{−∞}|Ψ(x, t)|^2dx=0




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