Postulates Of Quantum Mechanics MCQ Level - 2 - IIT JAM MCQ

For the states to be orthogonal

and for them to be normalised,

As wave number,

We know that group velocity given by

The correct answer is: 

V = 0 for a free particle
 ∴  

Schrödinger equation.

  (General solution)
∴    Ae^ikx   and  Be^–ikx  are solution to the free particle

The correct answer is: Energy and momentum eigen functions with both positive and negative momentum eigenvalues.

The correct answer is:  13/16

From the continuity equation

i.e. p is constant in time

 Probability current density
The correct answer is: If     then probability density is constant in time.

Given, v = 3 × 10⁷ m/s
Let    Δx_min be the minimum uncertainty in position of the electron and Δp the maximum uncertainty in the momentum of the electron.

Thus, we have,

= 0.03867 × 0.9949 × 10^–10 m
= 3.8 × 10^–12 m
The correct answer is: 3.8 × 10^–12 m

The correct answer is: All the energy eigenvalues are always real.

The particle cannot exist outside the box, as it cannot have infinite amount of energy. Thus, it’s wave function is between 0 and L, where L is the length of the side of the box.

The energy of the linear harmonic oscillator is

This is a constant of motion. We can represent the constant value of E by means of averages of the kinetic and potential over a cycle of motion by writing.

The average value of x and p should vanish for an oscillating particle. So, we can identifywith the square of the corresponding uncertainties.

Since from the uncertainty principle  To determine the minimum energy of the oscillator, we put

The minimum energy is

V = 0 for a free particle
 ∴  
Schrödinger equation.

∴  Ae^ikx and Be^–ikx     (General solution)
∴  are solution to the free particle

The correct answer is: Energy eigenvalue is   and momentum is