Answer to Question #196614 in Quantum Mechanics for LUCKY

Question

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]

Answer 1

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

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

for a potential V(x)

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

$P(a, b) = \int ^b_aρ(x, t) dx =\int ^b_a|Ψ(x, t)|^2dx$

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

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

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

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

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

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

and we can write the above as:

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

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

$\int^∞_{−∞}\dfrac{∂^2Ψ^∗}{∂x^2}Ψ dx =\bigg|\bigg(\dfrac{∂Ψ^∗}{∂x} Ψ\bigg)\bigg|^∞_{−∞}−\int^ ∞_{−∞}\dfrac{∂Ψ^∗}{∂x}\dfrac{∂Ψ}{∂x} dx$

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

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

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

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