Suppose that the potential energy U of the particle is zero inside the box, but rises to infinity on the outside, that is

Particle in a box

( Infinite potential well )

Let us consider a particle restricted to move along the x-axis between x=0 and x=L by ideally reflecting, infinitely high walls of a box as shown in figure below.

Suppose that the potential energy U of the particle is zero inside the box, but rises to infinity on the outside, that is

U=0 for 0≤x≤L

U=∞ for x<0 and x>L

Because the particle cannot have an infinite amount of energy, it cannot exist outside the box, and so its wave function Ψ is 0 for x ≤ 0 and x ≥ L. our task is to find what Ψ is within the box namely between x = 0 and x = L.

In such a case, the particle is said to be

moving in an infinitely deep potential

well.

Within the box (U = 0) the Schrodinger’s steady state equation becomes

(1)

𝑑

𝜓

𝑑𝑥

+ 2𝑚

ℏ

𝐸𝜓 = 0

This equation has the solution

(2) Which we can verify by substitution back into Eq (1). A and B are constants to be evaluated.

𝜓 = 𝐴 sin 2𝑚𝐸

ℏ 𝑥 + 𝐵 cos 2𝑚𝐸

ℏ 𝑥

This solution is subject to a boundary conditions that Ψ = 0 for x = 0 and for x = L. Since cos0 = 1, the second term cannot describe the particle because it does not vanish at x = 0. Hence we conclude that B = 0. Since sin0 = 0, the sine terms always yields Ψ = 0, as required, but Ψ will be 0 at x = L only when

(3)

This result comes about because the sines of angles π, 2π, 3π,...

are all 0.

2𝑚𝐸

ℏ

𝐿 = 𝑛𝜋 n= 1, 2, 3,. . .

From equation (3) it is clear that the energy of the particle can have only certain values which are the eigenvalues. These eigenvalues, constituting the energy levels of the system, are found by solving equation (3) for E

, which gives

𝐸

=

2𝑚𝐿²

n=1, 2, 3, . . .

Wave Functions of a Particle in a Box

The wave functions of a particle in a box whose energies are E

are, from Eq. (2) with B=0,

𝜓

= 𝐴 sin 2𝑚𝐸

(5)

ℏ 𝑥

Substituting Eq. (4) for E

gives

(6)

for the eigen functions corresponding to the energy eigenvalues E

.

𝜓

= 𝐴 sin 𝑛𝜋𝑥

𝐿

With the help of the trigonometric identity sin

θ =1/2 (1-cos 2θ) we find that

න

−∞

∞

𝜓_𝑛 ²𝑑𝑥 = න

0 𝐿

𝜓_𝑛 ²𝑑𝑥 = 𝐴² න

0 𝐿

𝑆𝑖𝑛² 𝑛𝜋𝑥

𝐿 𝑑𝑥

= 𝐴²

2 න

0 𝐿

𝑑𝑥 − න

0 𝐿

cos 2𝑛𝜋𝑥

𝐿 𝑑𝑥

= 𝐴²

2 [𝑥 − 𝐿

2𝑛𝜋 sin 2𝑛𝜋𝑥 𝐿 ]₀^𝐿

=𝐴²(^𝐿

2) (7)

Normalization condition requires

න

−∞

∞

𝜓

𝑑𝑥 = 1

Comparing eqs. (7) and (8), we see that the wave functions of a particle in a box are normalized if

The normalized wave functions of a particle are therefore

(9)

𝐴 = 2 𝐿

𝜓_𝑛 = 2

𝐿 sin 𝑛𝜋𝑥 𝐿

The normalized wave functions Ψ

, Ψ

, and Ψ

together with the probability densities |Ψ

|

, |Ψ

|

,

and |Ψ

|

are plotted in figure. Although Ψ

may

be negative as well as positive, |Ψ

|

is always

positive and, since Ψ

is normalized, its value at a

given x is equal to the probability density of

finding the particle there. In every case |Ψ

|

=0 at

x=0 and x=L, the boundaries of the box.

Question 1

Qno.1): Find the probabilities of finding a particle trapped in a box of

length L in the region from 0.45L to 0.55L? For the ground state and the first excited state.

Solution:

The eigen function (wave function) of a particle trapped in a box of length L are:

ψ_n= ²

𝐿 sin ^𝑛π𝑥

𝐿

The probability of finding the particle between x₁ and x_2, when it is in the n^th state is ;

P =׬_𝑥

1

𝑥₂

𝜓_𝑛 ²=²

𝐿 ׬_𝑥

1

𝑥₂

sin(^𝑛π𝑥

𝐿 ) 𝑑𝑥

P = ²

𝐿 ׬_𝑥

1

𝑥₂ 1

2 (1 − cos ^2π𝑛𝑥

𝐿 ) 𝑑𝑥 P = ¹

𝐿 𝑥 − ^𝐿

2π𝑛 sin ^2π𝑛𝑥

𝐿

Here x

= 0.45L x

= 0.55L and for the ground state n = 1

P =

𝐿

{[0.55𝐿 −

2π

sin(1.10π)] − (0.45𝐿 −

2π

sin 0.90π)}

[(0.55-0.45) -

2π

(sin 198

− sin 162

) 0.10 -

2π

(sin 198

− sin 162

)

[sin 𝐴 − sin 𝐵 = 2 cos

2

sin

2

] 0.10 -

π

(cos 180

sin 18

) 0.10 +

3.14

= 19.8%

Similarly, for the first excited state (n = 2), repeat and get

P = 0.65%

Question 2

An electron is bound in one dimensional potential box which has width 2.5×10^-10m. Assuming the height of the box to be infinite, calculate the lowest two permitted energy values of the electron?

Solution

The energy of an electron in a potential box of infinite height is given by;

E_n = ^𝑛

2ℎ² 8𝑚𝐿²

Given L = 2.5×10^-10m

Hence E

=

8×9.1×10⁻³¹×(2.5×10⁻¹⁰)²

J

=

1.6×10⁻¹⁹

𝑛

= 6.038𝑛

eV

Two lowest permitted energy values of the electron are E

= 6.038 ×(𝟏)

= 𝟔. 𝟎𝟑𝟖 𝒆𝑽

E

= 6.038 × (𝟐)

= 𝟐𝟒. 𝟏𝟓𝟐 𝒆𝑽

Question 3

Find the energy of an electron moving in one dimension in an infinitely high potential box of width 1 Å

(mass of the electron is 9.11×10⁻³¹𝑘𝑔 𝑎𝑛𝑑 ℎ = 6.63 × 10⁻³⁴𝐽𝑠) Solution

The energy of an electron in a potential box of infinite height is given by;

E_n = ^𝑛2ℎ2

8𝑚𝐿²

Given L = 1 Å = 1×10⁻¹⁰𝑚

• For minimum possible energy n = 1 Hence E

=

−34)²

8×9.11×10⁻³¹×(10−10)²

= 6.03×10

𝐽

=

−18

1.6×10⁻¹⁹

= 𝟑𝟕. 𝟔𝟗 𝒆𝑽

Tunnel Effect

When a particle of energy E is less than U approaches a potential barrier, according to classical mechanics the particle must be reflected. In quantum mechanics, the de Broglie waves that correspond to the particle are partly reflected and partly transmitted, which means that the particle has a finite chance of penetrating the barrier.

• Tunneling also occurs in the operation of certain

semiconductor diodes in which electrons pass

through potential barriers even though their kinetic

energies are smaller than the barrier heights

• Let us consider a beam of identical particles all of which have the kinetic energy E. The beam is incident from the left on a potential barrier of height U and width L as shown in figure. On both sides of the barrier U=0, which means that no forces act on the particle there. The wavefunction ψ_І+ represents the incoming particles moving to the right and ψ_І- represents the reflected particles moving to the left; ψ_{І І І} represents the transmitted particles moving to the right. The wavefunction ψ_{І І} represents the particles inside the barrier, some of which end up in region III while the others return to region I. the transmission probability T for a particle to pass through the barrier is equal to the fraction of the incident beam that gets through the barrier. Its approximate value is given by;