Download Introduction to Quantum Mechanics

Introduction to Quantum Mechanics
Classical Mechanics:
 Cause and effect
 Future history of particle is determined by its initial position and momentum
together with the force acts upon it
Quantum Mechanics
 Uncertainty principle
 Exploring the probability
The actors:
Erwin Schrodinger, Werner Heisenberg, Max Born, Paul Dirac,
and others
Wave function 
𝜓
2
is proportional to the probability to find the particle at certain time and place
Example
𝜓 = 𝐴 + 𝑖𝐵
𝜓 ∗ = 𝐴 − 𝑖𝐵
𝜓
Normalization
∞
𝜓 2 𝑑𝑉 = 1
−∞
∞
𝜓 2 𝑑𝑉 = 1
−∞
2
= 𝜓𝜓 ∗ = 𝐴2 + 𝐵2
Summary of the rules
𝜓 must be continous and single valued everywhere
𝑑𝜓 𝑑𝜓 𝑑𝜓
,
,
, must be continous and single valued everywhere
𝑑𝑥 𝑑y 𝑑z
𝜓 must normalizable, meaning 𝜓 must be 0 as x, y, z → ±∞,
so that
𝜓 2 𝑑𝑉 over all space be finite
Wave Equation
𝑑2𝑦
1 𝑑2 𝑦
=
𝑑𝑥 2 𝑣 2 𝑑𝑡 2
𝑥
Try to apply above formula to the case of 𝑦 𝑥 = 𝐴 exp − iω(𝑡 − 𝑣)
Formulating the wave function
𝑥
 𝑥 = 𝐴 exp − iω(𝑡 − 𝑣)
Remember:
𝜔 = 2𝜋𝑓 , 𝑣 = 𝜆𝑓, 𝐸 = ℎ𝑓 = 2𝜋ℏ𝑓, 𝜆 =
 𝑥 = 𝐴 exp − i/ℏ(𝐸𝑡 − 𝑝𝑥)
𝜕2𝜓
𝑝2
=− 2 𝜓
𝜕𝑥 2
ℏ
𝜕𝜓
𝑖𝐸
=− 𝜓
𝑑𝑡
ℏ
𝑝2
𝐸=
+ 𝑈(𝑥, 𝑡)
2𝑚
2𝜓
𝜕
𝑝2 𝜓 = −ℏ2 2
𝜕𝑥
E𝜓 = -
ℏ 𝜕𝜓
𝑖 𝑑𝑡
ℎ 2𝜋ℏ
=
𝑝
𝑝
Schrödinger’s equation: Time Dependent From
𝑝2
𝐸𝜓 =
𝜓 + 𝑈(𝑥, 𝑡)𝜓
2𝑚
𝜕𝜓
ℏ2 𝜕 2 𝜓
𝑖ℏ
=−
+ 𝑈(𝑥, 𝑡)𝜓
𝑑𝑡
2𝑚 𝑑𝑥 2
𝜕𝜓
ℏ2 𝜕 2 𝜓 𝜕 2 𝜓 𝜕 2 𝜓
𝑖ℏ
=−
+
+
+ 𝑈𝜓
𝑑𝑡
2𝑚 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
Wave Function Superposition
𝜓 = 𝑎1 𝜓1 + 𝑎2 𝜓2
Probability: (take a simple case of a1=a2=1
𝑃 = 𝜓 2 = 𝜓1 + 𝜓2 2 = 𝜓1∗ + 𝜓2∗ 𝜓1 + 𝜓2
= 𝜓1∗ 𝜓1 + 𝜓2∗ 𝜓2 + 𝜓1∗ 𝜓2 + 𝜓2∗ 𝜓1
= 𝑃1 + 𝑃2 + 𝜓1∗ 𝜓2 + 𝜓2∗ 𝜓1
Expectation value
∞
𝑥 𝜓 2 𝑑𝑥
𝑥 =
−∞
Operators
 𝑥 = 𝐴 exp − i/ℏ(𝐸𝑡 − 𝑝𝑥)
𝜕𝜓 𝑖
= 𝑝𝜓
𝜕𝑥 ℏ
𝑝𝜓 =
𝜕𝜓
𝑖
= − 𝐸𝜓
𝜕𝑡
ℏ
𝐸𝜓 = 𝑖ℏ
Operator momentum
ℏ 𝜕
𝑝=
𝑖 𝜕𝑥
Operator Energi
𝐸 = 𝑖ℏ
𝜕
𝜕𝑡
ℏ 𝜕𝜓
𝑖 𝜕𝑥
𝜕𝜓
𝜕𝑡
Another way to obtain Schrodinger’s Equation
𝐸 = 𝐾𝐸 + 𝑈
Kinetic energy operator
𝑝2
1 ℏ 𝜕
𝐸𝐾 =
=
2𝑚 2𝑚 𝑖 𝑑𝑥
2
ℏ2 𝜕 2
=−
2𝑚 𝜕𝑥 2
Energy operator
𝜕
𝐸 = 𝑖ℏ
𝜕𝑡
Total energy operator in Hamiltonian form
𝜕
ℏ2 𝜕 2
𝑖ℏ
=−
+ 𝑈(𝑥, 𝑡)
𝑑𝑡
2𝑚 𝑑𝑥 2
Operator and Expectation Values
∞
∞
𝜓 ∗ 𝑝𝜓𝑑𝑥 =
𝑝 =
𝐸 =
𝜓∗
−∞
−∞
∞
∞
𝜓 ∗ 𝐸𝜓𝑑𝑥
−∞
𝜓∗
=
−∞
ℏ 𝜕
ℏ
𝜓𝑑𝑥 =
𝑖 𝜕𝑥
𝑖
𝜕
𝑖ℏ
𝜓𝑑𝑥 = 𝑖ℏ
𝜕𝑡
∞
𝜓∗
−∞
∞
𝜓∗
−∞
𝜕
𝜓𝑑𝑥
𝜕𝑥
𝜕𝜓
𝑑𝑥
𝜕𝑡
Steady State Condition
 𝑥 = 𝐴 exp −
i
ℏ
i
𝐸𝑡 − 𝑝𝑥 = 𝐴 𝑒
−ℏ𝐸𝑡
𝑒
i
𝑝𝑥
ℏ
i
= 𝜓𝑠 𝑒
− ℏ𝐸𝑡
Subtitute in to Hamiltonian form
ℏ2 𝜕 2
𝐸=−
+ 𝑈(𝑥, 𝑡)
2𝑚 𝑑𝑥 2
Time dependent Schrodinger ‘s equation
𝐸𝜓𝑠
𝑒 −𝑖𝐸𝑡/ℏ
ℏ2 𝜕 2
−𝑖𝐸𝑡/ℏ + 𝑈(𝑥, 𝑡)𝜓 𝑒 −𝑖𝐸𝑡/ℏ
=−
𝜓
𝑒
𝑠
𝑠
2𝑚 𝑑𝑥 2
Stready State for one dimension
𝜕2
2𝑚
𝜓 +
𝐸 − 𝑈 𝜓𝑠 = 0
𝑑𝑥 2 𝑠 ℏ2
Stready State for three dimension
𝜕2
𝜕2
𝜕2
2𝑚
𝜓 +
𝜓 +
𝜓 +
𝐸 − 𝑈 𝜓𝑠 = 0
𝑑𝑥 2 𝑠 𝑑𝑦 2 𝑠 𝑑𝑧 2 𝑠 ℏ2
Particle in a box with infinite potential wall
U=∞
𝜓=0
U=∞
𝜓≠0
Time independet, the particle is always in the box
for any time.
𝜓=0
𝜕2
2𝑚
𝜓
+
𝐸 − 𝑈 𝜓𝑠 = 0
𝑑𝑥 2 𝑠 ℏ2
U=0
X=0
X=L
Solution:
𝜓 𝑥 = 𝐴𝑒𝑥𝑝(𝑖𝛼𝑥)
Question: A?,  ?
𝜕2
2𝑚
𝜓 + 2 𝐸𝜓 = 0
𝑑𝑥 2
ℏ
Determine 
𝜕2
2𝑚
𝜓 = − 2 𝐸𝜓
𝑑𝑥 2
ℏ
Subtitute the solution 𝜓 𝑥 = 𝐴𝑒𝑥𝑝 𝑖𝛼𝑥 to the left side of equation
Remeber: 𝑒 𝑖𝛼𝑥
2𝑚𝐸
𝛼=
ℏ
= 𝑠𝑖𝑛𝛼𝑥 + 𝑖𝑐𝑜𝑠𝛼𝑥
Boundary condition: 𝜓 = 0 at x=0 and x =L, the imaginary part should be zero
2𝑚𝐸
𝐿 = 𝑛𝜋
ℏ
n=1,2,3,....
𝑛2 𝜋 2 ℏ2
𝐸𝑛 =
2𝑚𝐿2
Wave function:
𝜓𝑛 𝑥 = 𝐴𝑠𝑖𝑛
2𝑚𝐸𝑛
ℏ
𝑥
𝜓𝑛 𝑥 = 𝐴𝑠𝑖𝑛
𝑛𝜋𝑥
𝐿
Probability
∞
𝐿
𝜓𝑛 2 𝑑𝑥 =
𝜓𝑛 2 𝑑𝑥 = 𝐴2
−∞
0
= A2
=
=
𝐴2
2
sin2
0
𝐿1
(1
0 2
− 𝑐𝑜𝑠 2
𝑛𝜋𝑥
𝐿
𝐿
𝑑𝑥
0
−
𝐿
𝑐𝑜𝑠
0
2
𝐴2
2
=
Normalization
𝐿
𝑥−
𝐿
2𝑛𝜋
𝑠𝑖𝑛
𝑛𝜋𝑥
𝑑𝑥
𝐿
)𝑑𝑥
𝑛𝜋𝑥
𝐿
𝑛𝜋𝑥
2 𝐿
𝑑𝑥
𝐿
0
𝐴2 𝐿
2
∞
−∞
𝜓𝑛 2 𝑑𝑥 =1
𝐴=
2
𝐿
Solution of wave function of particle in a box with infinite potential energy in the boundary
𝜓𝑛 𝑥 =
2
𝑛𝜋𝑥
𝑠𝑖𝑛
𝐿
𝐿
Box with finite potential energy higher than electron energy
E
Regions I and III (x<0 and x>L)
U
I
-x
III
II
0
𝜕2
2𝑚
𝜓
+
(𝐸 − 𝑈)𝜓 = 0
𝑑𝑥 2
ℏ2
L
Remember: U>E
+x
Or
𝜕2
2𝜓 = 0
𝜓
−
𝑎
𝑑𝑥 2
Solution:
𝜓𝐼 = 𝐶𝑒 𝑎𝑥 + 𝐷𝑒 −𝑎𝑥
Boundary condition:
Solution:
With
𝑎=
𝜓𝐼𝐼𝐼 = 𝐹𝑒 𝑎𝑥 + 𝐺𝑒 −𝑎𝑥
𝜓 should be finite everywhere
𝜓𝐼 = 𝐶𝑒 𝑎𝑥
2𝑚
(𝑈 − 𝐸)
ℏ2
𝜓𝐼𝐼𝐼 = 𝐺𝑒 −𝑎𝑥
Region II (0<x<L)
E
𝜕2
2𝑚
𝜓 + 2 𝐸𝜓 = 0
𝑑𝑥 2
ℏ
U
I
II
0
Solution:
III
L
+x
𝜓𝐼𝐼 𝑥 = 𝐴𝑠𝑖𝑛
2𝑚𝐸
𝑥
ℏ
+ Bcos
2𝑚𝐸
𝑥
ℏ
Tunneling The Barrier Potential
U
𝜕2
2𝑚
𝜓
+
𝐸𝜓𝐼 = 0
𝑑𝑥 2 𝐼 ℏ2
+
𝜓𝐼𝐼𝐼
𝜓𝐼+
𝜓_𝐼𝐼
𝜓𝐼−
𝜕2
2𝑚
𝜓
+
𝐸𝜓𝐼𝐼𝐼 = 0
𝑑𝑥 2 𝐼𝐼𝐼 ℏ2
x=0
x=L
Solutions
𝜓𝐼 𝑥 = 𝐴𝑒 𝑖𝑘1𝑥 + 𝐵𝑒 −𝑖𝑘1𝑥
𝑘1 =
𝜓𝐼𝐼𝐼 𝑥 = 𝐹𝑒 𝑖𝑘1𝑥 + 𝐺𝑒 −𝑖𝑘1𝑥
2𝑚𝐸
ℏ
=
2𝜋
𝜆
𝑝
=ℏ
The incoming wave
𝜓𝐼 𝑥 = 𝐴𝑒 𝑖𝑘1𝑥
The reflected wave
𝜓𝐼 𝑥 = 𝐵𝑒 −𝑖𝑘1 𝑥
The transmitted wave
𝜓𝐼𝐼𝐼 𝑥 = 𝐹𝑒 𝑖𝑘1𝑥
Transmission Probability
+ 2 +
+
𝜓𝐼𝐼𝐼
𝑣𝐼𝐼𝐼 𝐹𝐹 ∗ 𝑣𝐼𝐼𝐼
𝑇=
=
𝜓𝐼+ 2 𝑣𝐼+
AA∗ 𝑣𝐼+
Inside the barrier
𝜕2
2𝑚
𝜓
+
(𝐸 − 𝑈)𝜓 = 0
𝑑𝑥 2
ℏ2
U
+
𝜓𝐼𝐼𝐼
𝜓𝐼+
𝜕2
2𝑚
𝜓
−
(𝑈 − 𝐸)𝜓 = 0
𝑑𝑥 2
ℏ2
𝜓_𝐼𝐼
𝜓𝐼−
x=0
x=L
Solution
𝜓𝐼𝐼 = 𝐶𝑒 −𝑘2𝑥 + 𝐷𝑒 𝑘2𝑥
Wave number inside the barrier
𝑘2 =
2𝑚
(𝑈 − 𝐸)
ℏ2
Since U>E
Boundary Conditions
At x=0
𝜓𝐼 = 𝜓𝐼𝐼
𝜕𝜓𝐼
𝜕𝑥
At x=L
=
𝜕𝜓𝐼𝐼
𝜕𝑥
𝜓𝐼𝐼𝐼 = 𝜓𝐼𝐼
𝜕𝜓𝐼𝐼𝐼
𝜕𝑥
=
𝜕𝜓𝐼𝐼
𝜕𝑥
How tunneling effect can occur in p-n juction?
Wikipedia.com
Quantumaniac.tumblr.com
Scanning Tunneling Microscope
Nobel Prize 1986
Harmonic Oscilator
𝜕2
2𝑚
𝜓
+
(𝐸 − 𝑈)𝜓 = 0
𝑑𝑥 2
ℏ2
𝜕2
2𝑚
1 2
𝜓 + 2 (𝐸 − 𝑘𝑥 )𝜓 = 0
𝑑𝑥 2
ℏ
2
Solusi:
2𝑚𝑓
𝜓𝑛 =
ℏ
1
4
1
−2
𝑛
2 𝑛! 𝐻𝑛
𝑦 𝑒 −𝑦
2 /2
1
𝐸𝑛 = 𝑛 +
ℎ𝑓
2
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