Question

3. Consider a particle is in 1-D infinite potential well. (25%) (a) Find Wri(x), (7%) (b) Find the operator (t). (996) (c) Calculate 〈x(t). (9%)

Answer 1

Let us consider a particle or electron of mass ‘m’ moving along X-axis, enclosed in a one dimensional potential box as shown in figure.

Since the walls are of infinite potential,the particle does not penetrate out from the box.

i.e, potential energy of the particle V = ∞ at the walls.

The particle is free to move between the walls A&B at x=0 and x=L . The potential energy of the particle between the two walls is constant because no force is acting on the particle.

Therefore the potential energy is taken as zero for simplicity.

i.e, V = 0 between x=0 and x = L .

Boundary Conditions:-

The potential energy is,V(x) = 0 , when 0 < x < L

V(x) = ¥ , when 0³ x³ L

The Schrödinger one - dimensional time independent equation for the particle is given by,

d2ydx2+8π2mh2 ( E-V)Ψ = 0

For freely moving particle between the walls, V = 0

d2ydx2+8π2mh2 EΨ = 0----------(1)

Consider, 8π2mh2 E = K²

Then equation(1) becomes,

d2ydx2+K2 $sR8HOvFwzrbdIwAAAABJRU5ErkJggg==$ Ψ = 0    -------------------(2)

The general solution of abov equation is given by,

Y(x) = A sin kx + B cos kx   -----------(4) ,       where A, B are two constants and k is a propagation constant (or) wave vector.

Applying boundary conditions for equation(4), we get

(i). Y(x) = 0 at x = 0

Y(x) = A sin kx + B cos kx

Þ 0 = A sin k(0) + B cos k(0)

Þ 0 = 0 + B

ÞB = 0

Substitute ‘B’ value in equation(4), we get,Y(x)= A sin kx   -------------(5)

(ii). Y(x) = 0 at   x = L

Then equation(5) becomes, Y(x)= A sin kx  

Þ 0 = A sin kL

But A ¹ 0, and    sinKL = 0

ÞsinKL = sin np

Þ K L= np

Therefore,   K = nπL     ---------------(6)

Substitute equation (6) in (5), we get

Y(x)= A sin nπL x    --------------(6)  

To find the ‘A’ value, apply the normalization condition:-

-∞∞ôyô2 dx = 1

0LA2 sin2 ( n P x / L ) dx = 1

A20L [1 - cos (2 P n (x / L)) / 2] dx = 1

   Þ   A² . L2 =1

Þ    A² = 2L

Þ   A= 2L $cLKRrXDTd6mt8AAAAASUVORK5CYII=$    ----------(7)

   Substitute equation (7) in (6), we get

Y(x)=2L $cLKRrXDTd6mt8AAAAASUVORK5CYII=$ sin nπL x    (or) Ψ_n(x)= √ 2L Sin (nπx L )    --------(8)

                          This is the normalized wave function of electron.

Energy of the Electron:-

From equation (6), K = nπL $wpSSkw4ADf2ylF4xl5kAAAAASUVORK5CYII=$ Þ    K² = n2π2L2

We have, K² =8π2mEh2

Then, n2π2L2       = 8π2mEh2

Þ E_n = n2h28mL2     -------(9)

Therefore, the energy (or)energy eigen function of the electron, E_n = n2h28mL2 (or)Enx $kZc8PkPoB78O8OuWToWMAAAAASUVORK5CYII=$ = nx2h28mL2 $sswudNVSZQAAAABJRU5ErkJggg==$

Energy levelsof an Electron:-

               The energy of the electron, E_n = n2h28mL2

When n=1, E₁= h28mL2

When n=2, E₂= 4 .h28mL2 $RLXtSLOlUAAAAASUVORK5CYII=$ = 4.E₁

When n=3, E₁= 9. h28mL2 = 9.E₁

                        -----------

In general, E_n = n² . E₁

Therefore, the particle (electron) in the box has discrete values of energies. These values are quantized .

The normalized wave functions Y1 ,Y2,Y3, electron probability densities ôy1ô2 , ôy2ô2 , ôy3ô2 $4bhAThtygAAAABJRU5ErkJggg==$ ,ôy4ô2 ,,

energy eigen function En and their corresponding Eigen values E1, E2, E3, E4, are plotted, as shown in fig.

The wave function Y1, has two nodes at x = 0 & x = L

The wave function Y2, has three nodes at x = 0, x = L / 2 & x = L

The wave function Y3, has three nodes at x = 0, x = L / 3, x = 2 aL/ 3 & at x = L