Infinite square well potential

The particle of mass m is in one dimensional infinite square well potential characterized by

V(x) = 0 ; -L< x< L

= ∞ ; l xl ≥ L

The time independent Schrodinger equation for -L< x< L, V(x) = 0 can be written as

d²u(x)/ dx² + 2mE u(x)/ ℏ² = 0....i

where u(x) represents the wave functio.

We have

k² = 2mE/ℏ² ......ii

Eqn i becomes

d²u(x)/ dx² + k²u(x) = 0 ...iii

The general solution of this will be

u(x) = A cos kx + B sin kx ...iv

The condition l x l ≥ L ; v(x)-> ∞ satisfies the condition of bound state. For bound state, the boundary condition is

l u(x) l² -> 0 for lxl -> L

That means wave functions vanishes at x = ± L

Thus for x = -L, u(x) = 0

Applying this in the equation iv,

0 = A cos k(-L) ∓ B sin k(-L)

A cos kL - B sin kL = 0 ....v

Similarly for x = + L , u(x) = 0

Eqn iv becomes

0 = A cos kL + B sin kL

Adding v and vi we get

2 A cos kL = 0

cos kL = 0

cos kL = cos n𝜋/ 2 for n = 1, 3, 5....

Similarly, subtracting vi and v

2B sin kL = 0

sin kL = sin n 𝜋/ 2 for n= 2,4,6....

k = n 𝜋/ 2L for even n ......vii

Thus for both odd and even n

k = n 𝜋/ 2L

Substituting this in equation i

( n 𝜋/ 2L)² = 2mEn/ ℏ²

En = n²π²ℏ²/ 8mL² .........ix

Equation ix gives the energy spectrum of the particle inside square well potential. It is found that the energy spectrum is discrete as n =1,2,3,... are as follows

E_1 = E

E_2 = 4E

E_3 = 9E

For odd n, the wavefunction can be expressed using iv

U_odd (x) = A cos (nπ/ 2L) x .....x

Similarly for even n,

U_even (x) = B sin (nπ/ 2L) x .....xi

Equation x and xi can be normalized using the normalized condition

Let us normalized equation xi

Similarly normalizing equation x yield

A = 1 / \sqrt{L}

Thus, the normalized eigen functions of the particle are

u_even (x) = 1 / \sqrt{L} sin (nπ/ 2L) x ..... xii

u_odd (x) = 1 / \sqrt{L} cos (nπ/ 2L) x ..... xiii

The parity of eigen functions are as follows

u_even (x) = 1 / \sqrt{L} sin (nπ/ 2L) x = - u_even (x) ....xiv

Even eigen function has an odd parity


u_even (x) = 1 / \sqrt{L} cos (nπ/ 2L) x = u_odd (x) ....xv

Odd eigen function has even parity

The energy interval between two successive levels

E_n+1 - E_n = π²ℏ²/ 8mL² [ (n+1)² - n² ]

when L-> ∞, E_n+1 - E_n becomes zero, this means the discrete spectrum becomes continuous spectrum.

This note is a part of the Physics Repository.