Lecture 3
MSE 515
Solutions of Schrodinger equation for special cases
Last time:
−ħ2
𝜕Ψ
2
∇
𝜓 + 𝑉𝜓 = 𝐸𝜓 = 𝑖ℏ 𝜕𝑡
2𝑚
Ψ 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎 𝑤𝑎𝑣𝑒𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
+∞
∫
|Ψ(𝑥)|2 𝑑𝑥 = 1
−∞
The probability of finding it must be unity. This is a critical point
and will come up over and over.
1. Free electrons: (1-D) (Simple but very important one)
- We consider electrons that propagate freely in potential
free space in the positive x-direction
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- ∇ 𝜓 →
𝑑2 𝜓
𝑑𝑥 2
-
−ħ2 𝑑2 𝜓
2𝑚 𝑑𝑥 2
= 𝐸𝜓 _____ Eq. 1
- 𝜓(𝑥 ) = 𝑒 𝑖𝛼𝑥
1
-
𝑑𝜓
𝑑𝑥
= 𝑖𝛼𝑒 𝑖𝛼𝑥
-
𝑑2 𝜓
𝑑𝑥 2
= 𝑖 2 𝛼 2 𝑒 𝑖𝛼𝑥
- In Eq. 1
ħ2
2 𝑖𝛼𝑥
𝛼
𝑒
= 𝐸𝜓
2𝑚
-
ħ2 𝛼 2 𝑖𝛼𝑥
𝑒
= 𝐸𝜓 𝑟𝑒𝑚𝑒𝑚𝑏𝑒𝑟 𝜓 = 𝑒 𝑖𝛼𝑥
2𝑚
ħ2 𝛼 2
∴𝐸=
2𝑚
-Since no boundary conditions, all values of energy are allowed.
α =√
2𝑚𝐸
ħ2
______Eq. 2
𝑝2
= 𝐸(𝐾. 𝐸. ) 𝑝 (𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚)
2𝑚
𝑝2 = 2𝑚𝐸 𝑖𝑛 𝐸𝑞. 1.
α=√
𝑝2
ħ2
2
𝑝
=ħ =
α=
2𝜋
𝜆
𝑝 ∙ 2𝜋
ℎ
′
→
𝜆
𝑑𝑒
𝐵𝑟𝑜𝑔𝑙𝑖𝑒
𝑠 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠
𝑝
ℎ
=𝑘
= k (wave number)
K is a vector → 𝑘𝑥 𝑘𝑦 𝑘𝑧
λ→ wave vector
Note: k – vector describes the wave properties of an electron.
Quantum Mechanics (k)
Classical Mechanism (p)
𝑝
And remember k = ħ
Back to Energy E, we can find
ħ2
E = 2𝑚 𝑘 2
ψ(x) = 𝑒 𝑖𝑘𝑥
Separation of variables
ψ(x,t) = 𝑒 𝑖𝑘𝑥 ∙ 𝑒 𝑖𝜔𝑡
This is an equation of traveling wave. It represents a free
particle.
2. Bound electrons [ potential wall]
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- We consider that the electron can move freely between
infinitely high potential
barriers
Electrons cannot escape
because of the potential
wall
Note: The potential inside the wall is zero. The
difference between this case and last case is the
boundary conditions.
−ħ2 𝑑 2 𝜓
= 𝐸𝜓
2𝑚 𝑑𝑥 2
Assume the solution is ψ(x) = AeiαL + Be-iαL
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𝛼= √
2𝑚𝐸
ħ2
Applying Boundary Conditions
at x = zero ψ = 0
at x = L ψ = 0
since ψ = 0 , x ≤ 0 , x ≥ L
0 = A𝑒 𝑧𝑒𝑟𝑜 + 𝐵𝑒 𝑧𝑒𝑟𝑜
A = -B or B = -A
A𝑒 𝑖𝛼𝑎 − 𝐴𝑒 −𝑖𝛼𝑎 = 0
A (𝑒 𝑖𝛼𝑎 − 𝑒 −𝑖𝛼𝑎 ) = 0  A
EULER'S FORMULA IS THE KEY TO UNLOCKING THE SECRETS OF
QUANTUM PHYSICS
From Euler’s Equation
1 𝑖𝛿
sin 𝛿 = (𝑒 − 𝑒 −𝑖𝛿 )
2𝑖
2𝑖𝐴 𝑖𝛼𝐿
(𝑒 − 𝑒 −𝑖𝛼𝐿 ) = 0
2𝑖
2A* 𝑖 sin (α L) = 0
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∴ sin α L = 0 → αL = n𝜋
α=
𝑛𝜋
n= 0, 1, 2, 3... (integer)
𝐿
from α = √
2𝑚𝐸
ħ2
ħ2
E = 2𝑚 𝛼 2
ħ2 𝑛 2 𝜋 2
= 2𝑚
𝑎𝐿2
ħ2 𝜋 2
E = 2𝑚𝑎𝐿2 𝑛2 n = 1,
2, 3…
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- The second case is similar to an electron bound to its
atomic nucleus
- Only certain energy levels are allowed for the electron
- This is “Energy quantization”
- Probability of finding the electron at inside the well:
𝜓𝑛 = A (𝑒 𝑖𝛼𝑥 − 𝑒 −𝑖𝛼𝑥 ) na
= 2𝐴𝑖 ∙ sin 𝛼𝑥
𝜓𝜓 ∗ = 4𝐴2 𝑠𝑖𝑛2 𝛼𝑥
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3. Finite potential barrier: (Tunnel effect)
Assume a free electron propagating in the positive xdirection meets a potential barrier V0 (higher than the
total energy of electron).
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Write sch. Eq. For each region:
Region I. V=0
𝑑2 𝜓
𝑑𝑥 2
+
2𝑚
ħ2
𝐸𝜓
Region II.
𝑑2 𝜓
𝑑𝑥 2
+
2𝑚
ħ2
(𝐸 − 𝑉𝑜 )𝜓 = 0
Region I solutions:
𝜓𝐼 = 𝐴𝑒 𝑖𝑎𝑥 + 𝐵𝑒 −𝑖𝑎𝑥
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𝑎= √
2𝑚𝐸
ħ2
Region II solution:
𝜓𝐼𝐼 = 𝐶𝑒 𝑖𝛽𝑥 + 𝐷𝑒 −𝑖𝛽𝑥
Only certain solutions exist (for which n is integer)
Let’s plot the energy solutions for the two cases:
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The second case is similar to an electron bond to a nucleus.
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First three levels of the stationary state and probability of those
states.
Region I solutions:
𝜓𝐼 = 𝐴𝑒 𝑖𝑎𝑥 + 𝐵𝑒 −𝑖𝑎𝑥
𝑎= √
2𝑚𝐸
ħ2
Region II solution:
𝜓𝐼𝐼 = 𝐶𝑒 𝑖𝛽𝑥 + 𝐷𝑒 −𝑖𝛽𝑥
2𝑚
β = √ ħ2 (𝐸 − 𝑉0 )
(E – V0) is less than zero
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β → imaginary
= if
2𝑚
∴ 𝛾 = √ ℎ2 (𝑉0 − 𝐸 )
𝜓𝐼𝐼 = 𝐶𝑒 𝛾𝑥 + 𝐷𝑒 −𝛾𝑥
Using boundary conditions
X→ ∞
𝜓𝐼𝐼 = 𝐶 ∙ ∞ + 𝐷 ∙ 0
C must be zero
∴ 𝜓𝐼𝐼 = 𝐷𝑒 −𝛾𝑥
Ψ decreases in region II exponentially
The decrease is higher for larger , for larger potential barrier
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The electron wave propagates in the finite potential barrier.
Tunneling effect: penetration of a potential barrier
- This is only quantum mechanical effect.
- In classical mechanics: If the electron kinetic energy is
smaller than V, the electron will be entirely reflected
and “cannot overcome the barrier”
Examples of tunneling :
- Tunneling of electrons from one metal to another
through an oxide film.
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- Emission of alpha particles from nuclei by tunneling
through the binding potential barrier.
4. Another Case
- We can find that electron can penetrate region II and
propagates in region III.
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−ħ2 2
𝜕Ψ
∇ 𝜓 + 𝑉𝜓 = 𝐸𝜓 = 𝑖ℏ
2𝑚
𝜕𝑡
Partial differ eq. in
2
-
∇ 𝜓 →
-
= h / 2p
Y
𝑑2 𝜓
𝑑𝑥 2
𝑑2 𝜓
+ 𝑑𝑦 2 +
𝑑2 𝜓
𝑑𝑧 2
mass = m
V = potential
|𝜓(𝑥, 𝑦, 𝑧, 𝑡)|2 dx dy dz
gives the probably of finding the electron in volume dx dy dz
−ħ2 ∇2 𝜓
2𝑚 𝜓
1 𝛿𝜔
+ 𝑉 = 𝑖ħ 𝜔
function r
𝛿𝑡
Eq. 1
function time
𝜓(𝑟, 𝑡) = 𝜓(𝑟)𝜔(𝑡)
Eq 2
Separation of variables substitute Eq 2 in Eq. 1
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For each Eq. to be correct it must be equal to a constant
- For each equation to be right, it must equal to constant
−ħ2 ∇2 𝜓
2𝑚 𝜓
𝜕𝜔
+𝑉 =𝐸
iħ 𝜕𝑡 = 𝐸𝜔
E is a constant → Eq. 3
E is the same const.
solution: 𝜔 = exp[
−𝑖𝐸𝑡
ħ
] from eq. 3
−ħ2
( 2𝑚 ∇2 + 𝑉) 𝜓 = 𝐸𝜓
time independent Schrödinger Equation.
It will be applied to be calculations of stationery conditions.
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