Schoedinger equations

Schoedinger equations Time Dependant (Derivation )

A particle is moving along X direction represented by wave function

𝒙
−𝒊𝝎(𝒕−𝒗)
ψ = 𝑨𝒆
2𝜕 ψ
2
2
ω= 2πf v=λf 𝑝 ψ= −ħ 𝜕𝑥 2

𝑥 𝜕2 ψ
−𝑖2π(𝑓𝑡− ) = −𝑝2 /ħ2 ψ
ψ= 𝐴𝑒 λ 𝜕𝑥 2

ℎ 2𝜋ħ 𝜕ψ 𝑖𝐸
𝐸 = ℎ𝑓 = 2𝜋ħ𝑓 and λ = = = -− ψ
𝑝 𝑝 𝜕𝑡 ħ

ħ 𝜕ψ
E ψ= 𝑖
ψ = 𝑨𝒆−𝒊/ħ(𝑬𝒕−𝒑𝒙) 𝜕𝑡

𝑝2
Total energy of particle is E= + 𝑈 𝑥, 𝑡
2𝑚

𝑝2 ψ 𝝏ψ ħ𝟐 𝝏𝟐 ψ
Eψ = 2𝑚 + 𝑈 𝑥, 𝑡 ψ 𝒊ħ = +𝑼 𝒙, 𝒕 ψ
𝝏𝒕 𝟐𝒎 𝝏𝒙𝟐
Operators
In quantum mechanics, operators are tools used to extract information about physical properties, such as
momentum, energy, and position, from a system's wavefunction.
Instead of just using numbers to describe these properties (as in classical physics), quantum mechanics
uses operators to act on wavefunctions (which describe the state of a system) to get measurable outcomes.

For example:
- The **position operator** gives the location of a particle.
- The **momentum operator** tells us the particle’s momentum.
- The **Hamiltonian operator** represents the total energy of the system.
 Expectation value

expectation value represents the average value of a physical quantity that you would expect to measure
over many experiments on a quantum system. It is calculated using the wavefunction of the system and
the operator corresponding to the physical quantity (like position, momentum, or energy).
 Schoedinger equations Time Independent (Derivation )-study state

ψ = 𝑨 𝒆−𝒊/ħ(𝑬𝒕−𝒑𝒙)
ψ = 𝑨𝒆−𝒊𝑬𝒕/ħ) 𝒆−𝒊𝒑𝒙/ħ
Wave function consist position dependent term(x)
and time dependent term (t)

ψ = ψ𝒆−𝒊𝑬𝒕/ħ) ψ = 𝑨𝒆−𝒊𝒑𝒙/ħ

Substituting the operator corresponding

to energy and momentum

𝝏²ѱ 𝟐𝒎
+ 𝟐 𝑬−𝑼 ѱ=𝟎
𝝏𝒙² 𝓱
 𝝏𝟐 ψ 𝟐𝒎
+ 𝟐 ( E- 𝑼)ψ = 0
𝝏𝒙𝟐 ħ
Problem solving
E>U E<U U=0
2𝑚 2𝑚 2𝑚𝐸
2 ( E- 𝑈) is positive = 𝑘 2
2 ( E- 𝑈) is negative = -𝑘 2
2 = 𝑘 2
ħ ħ ħ
𝜕2 ψ 𝜕2 ψ 𝜕2 ψ
2 + 𝑘 2 ψ=0 − 𝑘 2
ψ=0 + 𝑘 2
ψ=0
𝜕𝑥 𝜕𝑥 2
𝜕𝑥 2

Ψ (x) = A Cos Kx +B Sin Kx Ψ (x) = A Cos Kx +B Sin Kx

Ψ (x)= A 𝑒 𝑖𝑘𝑥 + 𝐵𝑒 −𝑖𝑘𝑥 Ψ (x)= A 𝑒 𝑘𝑥 + 𝐵𝑒 −𝑘𝑥 Ψ (x)= A 𝑒 𝑖𝑘𝑥 + 𝐵𝑒 −𝑖𝑘𝑥

ID Box

𝜕2 ψ 2𝑚
+ ( E- 𝑈)ψ = 0
𝜕𝑥 2 ħ2
2𝑚𝐸
Here in Region where the particle exist U=0 = 𝑘2
ħ2
𝜕2 ψ
+ 𝑘 2 ψ=0
𝜕𝑥 2
Ψ (x) = A Cos Kx +B Sin Kx
Take the 1 st Boundary condition ψ = 0 At x=0
0= A cos 0 +B sin 0
This equation is Possible only when A=0
Ψ (x) = B Sin Kx
 Take the 2 nd Boundary condition ψ = 0 At x=L
0= B Sin KL Ψ (x) = B Sin Kx

KL = nπ K = nπ / L 2𝑚𝐸
= 𝑘2
ħ2
2𝑚𝐸𝐿2
= 𝑛2 π2
ħ2

𝑛 2 π2 ħ 2
𝐸𝑛 = Ψ (x) = B Sin
𝑛𝜋
x
2𝑚𝐿2 𝐿

∞
𝑊𝑎𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑟𝑚𝑎𝑙𝑖𝑠𝑒𝑑 𝑖𝑒 න ψ ∗ ψ 𝑑𝑥 = 1
−∞
1
𝐿 2 𝑛𝜋𝑥 𝐿
1 𝑛𝜋𝑥
‫׬‬0 𝐵2 𝑠𝑖𝑛2 𝐿 =1 𝐵2 2 =1 B= න 𝑠𝑖𝑛2
0 𝐿
=
2
𝐿

2 𝑛𝜋
Ψ (x) = Sin x
𝐿 𝐿
 Degenerate energy states

Degenerate energy states happen when different quantum states have the same energy. In simple terms, it
means multiple ways to describe a system, but all these ways share the same energy level.
 Harmonic oscillator

F=-kx 𝑑2 𝑥 −𝑘 𝑑2 𝑥 𝑚
= 𝑥 = ω2 𝑥 X=a Sin(ωt-φ) T= 2π/ω T=2π
𝑑2 𝑥 𝑑𝑡 2 𝑚 𝑑𝑡 2 𝑘
F=m 𝑑𝑡 2

𝑑𝑉 𝑥 𝑥 1 1
F=-𝑑𝑥 V=-‫׬‬0 𝐹
2 2 2
𝑑𝑥 V=‫׬‬0 𝑘𝑥 𝑑𝑥 𝑉 = 2 𝑘𝑥 = 2 𝑚𝜔 𝑥
1 1 𝑑𝑥2 1
𝐾 = 2 𝑚𝑣 2 = 𝑚 𝑑𝑡 K= 2 𝑚𝜔2 ( 𝑎2 − 𝑥 2 )
2

1 1
E =K+V= 2 𝑚𝜔2 𝑥 2 + 2 𝑚𝜔2 ( 𝑎2 − 𝑥 2 )
𝟏
E =𝟐 𝒎𝝎𝟐 𝒂𝟐
 𝝏𝟐 ψ 𝟐𝒎
+ 𝟐 ( E- 𝑼)ψ = 0
𝝏𝒙𝟐 ħ

1
𝑉 = 2 𝑚𝜔2 𝑥 2

𝝏𝟐 ψ 𝟐𝒎 1
𝟐 + 𝟐 ( E- 𝑚𝜔2 𝑥 2 )ψ = 0
𝝏𝒙 ħ 2
𝑚𝜔
y = 𝛼𝑥 𝛼2 =
ħ
𝜕2 ψ 2𝐸
2 +(λ-y 2 )ψ = 0 λ=ħω
𝜕y

2𝐸
λ=(2n+1) (2n+1)= ħω

ħω 1
𝐸𝑛 =(2n+1) =(n+ )ħω
2 2

1
𝐸𝑛 =(n+2 )ℎ𝑣
Bohr Correspondence Principle states that the predictions
of quantum mechanics must agree with classical physics in
the limit of large quantum numbers (i.e., when the system is
large or when energy levels are closely spaced)
Harmonic oscillator wave function is given by

Where H is called Hermite polynomial