Introduction

The Application of Schrodinger equation contains many important phenomena which use to explain various theories. In this particular article Application of Schrodinger equation, we are going to discuss applications of Schrodinger equation i.e. particle in the one-dimensional potential box, eigenfunction, and eigenvalues, discreteness of energy levels, the difference between classical mechanics and Quantum mechanics result etc.

Application of Schrodinger equation: particle in the one-dimensional potential box

particle in one dimensional potential box

Let us consider a particle having mass, and moving along the x-direction. The region of motion is x=0 to x=a. This motion of the particle is restricted by opposite sides of the wall. In other words, the particle experiences a repulsive force at the walls of the box. The total energy E remains constant. Within the region of the box, (0<x<a) the potential energy is constant and for simplicity, we take it to be zero. i.e, V(x)=0 in the region and finds rigid walls at x=0 and x=a such that the potential appears to be infinite to the particle. The particle cannot exist outside the box, as it doesn’t have infinite energy. thus the box can be regarded as a square well potential of infinite depth and width ‘a’.

Now we know that the Schrodinger equation in general form-

δ²ψ /δx²+ 2m (E-V)ψ /h²=0

Inside the box,

δ²ψ/δx² + 2m (E)ψ /h²=0      (1)

Put 2mE/h² = K²

δ²ψ/δx² +K²ψ = 0                 (2)

This is the second order differential equation.

Solutions of the Schrodinger equation

Solution fo equation (2) is-

ψ(x) = Asin Kx + Bcos Kx      (3)

Now applying boundary conditions,

At x=0 ,the wave function must vanishes ψ(x=a)=0, requires,

0= Asin (0) + Bcos(0)

Or B=0

Using this in equation (3)

ψ(x) = Asin (Kx )             (4)

Where  constant A is determined by the normaliंzation condition.

The boundary condition at x=a,that is ψ(x=a)=0,

Using this in  equation (4) we get,

0= Asin (Ka)           (5)

Since the constant A can not be zero because we should then have no wave function, hence we can say that

Sin(Ka) = 0

This means

Ka =π or Ka=2π

In general Ka=nπ

K =nπ/ a                 (6)

Where n integer ,n=1,2,3,… solving for K

We get

( 2mE/h²)½ =  nπ/a

E= n²π²h²/2ma²         (7)

Now applying normalisation condition, we determined the constant A appearing in the wave function. This condition requires that,

∫₋∞⁺∞ ψ*(x)ψ(x)dx =1

Using equation (4)we get

,A²∫₀ª sin²(nπx/a)dx =1

Using identity,

Sin²θ = 1/2(1 – cos2θ)

And integrating we get

Or ,A²/2[(x+ (a/2nπ)sin(2nπx/a)]₀ª = 1

Or, A²/2(a) = 1

Now, A =√2/a             (8)

Therefore, the normalized wave function for the particle in one dimensional box are,

ψ(x) = √2/a sin (nπx/a)             (9)

Eigenfunction and energy Eigen values for n= 1,2,3..

Using equation (9) and (7)for energy values, we find that

For n =0,  ψ₁(x) = √2/a( sin (πx/a) ),   E= π²h²/2ma²

Again For n=2 , ψ₂(x) = √2/a (sin (2πx/a)) ,   E= 2²π²h²/2ma²

Now, For n=3 , ψ₃(x) = √2/a (sin (3πx/a) )     E= 3²π²h²/2ma²

For n=4 , ψ₃(x)= √2/a (sin (4πx/a))        E=4²π²h²/2ma²

And so on

In general, for the nth term, we introduce a subscript n,

Hence equation Schrodinger for one dimension box particle is given by-

Hψn = Enψn(x)

Here ψn is Eigen function and En is energy  Eigenvalue

ψn(x) = √2/a (sin nπx/a))      (10)

And En =n²π²h²/2ma²         (11)

 DiscreteNess of Energy Levels

Quantum mechanically, a free particle can have any possible energy value. But when the free particle is  constrained to move in a potential well (within certain boundries only) , then it can not have any possible enegy value. Only some specific energy values are allowed, or simply we can say energy is quantised. this is called  DISCRETENESS OF ENERGY LEVELS.

The energy spectrum for a particle of mass m, in the infinite well

When the particle is confined in a potential box, the energy it can have is one of the discrete values. the ground state energy term is called “zero point energy term, but its not actually zero it have some finite value .it is

E₁ = π²h²/2ma² = h²/8ma²       (12)

Plot of wave function

Difference between classical and quantum results
  1. In classical mechanics, if a particle is confined in a box, its ground state energy can be zero (particles are at rest) but in quantum mechanics, the ground state energy of the particle is not zero. It is E₁ = h²/8ma²
  2. In classical mechanics, if a particle is confined in a box its energy can vary continuously but in quantum mechanics, the energies are quantized. The energy can take one of the dicreate values E₁,4E₁, 9E₁,16E₁,   ..
  3. The expectation value of the position of the particle,(x), is found to be <x>= a/2 .in classical mechanics we  can say exactly where the particle was if we knew the initial condition.

So from all the above phenomena, the concept of the particle in one dimensional potential box is clear.

 

 

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