Answer to Question #194200 in Quantum Mechanics for Aman Kumar

Question

3) (a) 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 (x, t)|²dx = 0]

(b) Show that

d(p)

dt

=

dx

i.e. expectation values follow Newton's law.

Answer 1

Suppose we have a normalized wave function at time t = 0

$\dfrac{d}{dt}\int_{-\infin}^{\infin}|\Psi(x,t)|^2dx=\int_{-\infin}^{\infin}\dfrac{\partial}{\partial t}|\Psi(x,t)|^2dx$

Writing Schrodinger equation as

$\dfrac{d\Psi}{dt}=\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}-\dfrac{i}{\hbar}V\Psi$

Similarly,

$\dfrac{\partial\Psi^*}{\partial t}=-\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}+\dfrac{i}{\hbar}V\Psi$

Substituting in the first equation and rearranging, we get,

$\Rightarrow \dfrac{\partial|\Psi|^2}{\partial t}=\dfrac{i\hbar}{2m}\bigg(\Psi^*\dfrac{\partial^2\Psi}{\partial x^2}-\Psi\dfrac{\partial^2\Psi^*}{\partial x^2}\bigg)$

$\Rightarrow \dfrac{\partial|\Psi|^2}{\partial t}=\dfrac{\partial}{\partial x}\bigg[\dfrac{i\hbar}{2m}\bigg(\Psi^*\dfrac{\partial^2\Psi}{\partial x^2}-\Psi\dfrac{\partial^2\Psi^*}{\partial x^2}\bigg)\bigg]$

$\Rightarrow \dfrac{d}{dt}\int_{-\infin}^{\infin}|\Psi(x,t)|^2dx=\dfrac{i\hbar}{2m}\bigg(\Psi^*\dfrac{\partial^2\Psi}{\partial x^2}-\Psi\dfrac{\partial^2\Psi^*}{\partial x^2}\bigg)_{-\infin}^{\infin}$

Since, $\Psi(x,t)\longrightarrow0$ as $x\longrightarrow\infin$ for $\Psi(x,t)$ to be non normalizable,

it follows

$\dfrac{d}{dt}\int_{-\infin}^{\infin}|\Psi(x,t)|^2dx=0$