Physics Quantum mechanics One dimensional box power point presentation

1. BY:-BHAGWAN SAHAY JANGID
B.TECH FIRST YEAR (CS)

2. A QUANTUM MECHANICAL FREE PARTICLE
IN ONE DIMENSIONAL BOX
Region-I Region-IIRegion-III
X=0 X=L X
V=∞ V=∞V=0
V=∞ V=∞
Let as consider a particle of mass m , confined in a one dimensional box or
square well potential of length L and of perfectly rigid walls. The potential
function of particle is given as
V(x)=0 for 0 < x < L ; inside the box (Ist region)
V(x)=∞ for x ≤ 0 𝑎𝑛𝑑 x ≥ 𝐿 ; out side the box (II and III region)
Since inside the box potential is zero the particle is free to move within the
dimensions of the box but as it reaches to the boundaries it faces infinite
potential and gets reflected back and forth within the box –dimensions.
The Schrödinger's wave equation
𝑑2Ψ
𝑑𝑥2 +
2𝑚
ℏ2 (E-V) Ψ=0
𝑑2Ψ
𝑑𝑥2 +
2𝑚
ℏ2 EΨ=0

3. 𝑑2Ψ
𝑑𝑥2 + 𝑘2
Ψ=0OR (1)
Where 𝑘 =
2𝑚𝐸
ℏ2
(2)
The general solution of this equation can be written as
Ψ = 𝐴 sin 𝑘𝑥 + 𝐵 cos 𝑘𝑥 (3)
Applying boundary conditions
i. At x=0, wave function should vanish⇒ Ψ = 0
Then from equation (3)Ψ = 0 = 𝐴 sin 0 + 𝐵 cos 0
⇒ 𝐵 = 0
Which provides Ψ = 𝐴 sin 𝑘𝑥
Ψ
(4)
(5)

4. ii. At x=L, again wave function Ψ should vanish ⇒ Ψ=0
Then from equation (4), Ψ=0=𝐴 sin 𝑘𝐿
Here A cannot be equal to zero i.e., A≠0 because in that condition wave function vanishes therefore
sin 𝑘𝐿 = 0
sin 𝑘𝐿=sin 𝑛𝜋 ⇒ 𝑘 =
𝑛𝜋
𝐿
(6)
Hence , from equations (5) and (6) the wave function becomes
Ψ = 𝐴 sin
𝑛𝜋
𝐿
Now applying normalization condition to determine normalization constant
𝐴
−∞
∞
ΨΨ∗ 𝑑𝑥 = 1
⇒ 0
𝐿
Ψ 2
𝑑𝑥 = 1
⇒ 𝐴2
0
𝐿
sin2 𝑛𝜋𝑥
𝐿
𝑑𝑥 = 1
Or 𝐴2 𝐿
2
= 1 ⇒ 𝐴 =
2
𝐿
(8)
(7)

5. The Eigen wave function Ψ =
2
𝐿
sin
𝑛𝜋𝑥
𝐿
From equation (7)and (8)
Again from equations (2)and(6)
𝑘 =
2𝑚𝐸
ℏ2 =
𝑛𝜋
𝐿
On solving 𝐸 =
𝑛2 𝜋2ℏ2
2𝑚𝐿2
Or 𝐸 𝑛 =
𝑛2 𝜋2ℏ2
2𝑚𝐿2
Hence Ψ 𝑛 =
2
𝐿
sin
𝑛𝜋𝑥
𝐿
(9)
(10)
(11)
(13)
(12)

6. And
Wave function correspond to different energy states Discrete energy level in one dimensional box
For n=1, Ψ1 =
2
𝐿
sin
𝜋𝑥
𝐿
; 𝐸1 =
n=2, Ψ2 =
2
𝐿
sin
2𝜋𝑥
𝐿
; 𝐸2 = 4𝐸1
n=3, Ψ3 =
2
𝐿
sin
3𝜋𝑥
𝐿
; 𝐸3 = 9𝐸1
n=m, Ψ 𝑚 =
2
𝐿
sin
𝑚𝜋𝑥
𝐿
; 𝐸 𝑚 = 𝑚2 𝐸1
𝜋2ℏ2
2𝑚𝐿2

7. Problem:-
Wave function is Ψ 𝑥 = 𝑁𝑒 𝑖𝑐𝑥. Determine the normalization constant over the
region −𝑎 ≤ 𝑥 ≤ 𝑎.
Solution :- Given wave functionΨ 𝑥 = 𝑁𝑒 𝑖𝑐𝑥
Now from normalization condition
−𝑎
𝑎
Ψ 𝑥 Ψ∗
𝑥 𝑑𝑥 = 1
⇒ 𝑁2
−𝑎
𝑎
𝑒 𝑖𝑐𝑥 𝑒−𝑖𝑐𝑥 𝑑𝑥 = 1
⇒ 𝑁2 𝑥 −𝑎
𝑎 =1
𝑁2
2𝑎 = 1
𝑁 =
1
2𝑎
Hence ,the normalized wave function Ψ 𝑥 =
1
2𝑎
𝑒 𝑖𝑐𝑥