Question

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 d dt int - infty ^ infty | psi(x,t)|^ 2 dx=0] (b) Show that d(p)/dt =(- partial V/partial x ) i.e.,expectation values follow Newton's law.

Answer 1

(a)  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 $|x| \longrightarrow∞$

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

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

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

(b) Expectation value of momentum p is given by

$<p>=\int_{-\infin}^{\infin}\Psi^*\hat p\Psi dx$

$\hat p$ is given by

$\hat p=-i\hbar\dfrac{\partial\Psi}{\partial x}$

Substituting $\hat p$ in above equation

$<p>=\int_{-\infin}^{\infin}\Psi^*\bigg(-i\hbar\dfrac{\partial\Psi}{\partial x}\bigg)dx$

$<p>=-i\hbar\int\Psi^*\bigg(\dfrac{\partial\Psi}{\partial x}\bigg)dx$

Differentiating above equation with respect to time

$\dfrac{d}{dt}<p>=(-i\hbar)\dfrac{d}{dt}\int\Psi^*\bigg(\dfrac{\partial\Psi}{\partial x}\bigg)dx$

$=(-i\hbar)\int\dfrac{\partial}{\partial t}\bigg(\Psi^*\dfrac{\partial\Psi}{\partial x}\bigg)dx$

$\dfrac{d}{dt}<p>=(-i\hbar)\int\bigg[\dfrac{\partial\Psi^*}{\partial t}\dfrac{\partial\Psi}{\partial x}+\Psi^*\dfrac{\partial}{\partial x}\dfrac{\partial\Psi}{\partial t}\bigg]dx\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$

Time dependent Schrodinger's equation is given as

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

$\dfrac{\partial\Psi}{\partial t}=\dfrac{1}{i\hbar}\bigg[\dfrac{-\hbar^2}{2m}\dfrac{\partial^2\Psi}{\partial x^2}+V\Psi\bigg]$

$=\dfrac{-i}{\hbar}\bigg[\dfrac{-\hbar^2}{2m}\dfrac{\partial^2\Psi}{\partial x^2}+V\Psi\bigg]$

$\dfrac{\partial\Psi}{\partial t}=\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}-\dfrac{i}{\hbar}V\Psi\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(2)$

Similarly,

$\dfrac{\partial\Psi^*}{\partial t}=\dfrac{-i\hbar}{2m}\dfrac{\partial^2\Psi^*}{\partial x^2}+\dfrac{i}{\hbar}V\Psi^*\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(3)$

Substituting the value of equation (2) and (3) in equation (1)

$\dfrac{d}{dt}<p>=i\hbar\int\bigg[\bigg(\dfrac{-i\hbar}{2m}\dfrac{\partial^2\Psi^*}{\partial x^2}+\dfrac{i}{\hbar}V\Psi^*\bigg)\dfrac{\partial\Psi}{\partial x}+\Psi^*\dfrac{\partial}{\partial x}\bigg(\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}-\dfrac{i}{\hbar}V\Psi\bigg)\bigg]dx$

$\dfrac{d}{dt}<p>=\dfrac{-\hbar^2}{2m}\int\bigg[\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial}{\partial x}\bigg(\dfrac{\partial^2\Psi}{\partial x^2}\bigg)\bigg]dx+\bigg<\dfrac{-\partial V}{\partial x}\bigg>$

Solving the middle term of the equation goes to zero as

$\Psi\longrightarrow0$ as $x\longrightarrow\pm\infin$

$\dfrac{d}{dt}<p>=\bigg<\dfrac{-\partial V}{\partial x}\bigg>$