S[Ψ(χ , t)]=∫dt (T - V)=∫dt dx L°(ψ , ∂Ψ/∂χ , ∂Ψ/∂t)) | 𝑴𝒂𝒙 𝑷𝒍𝒂𝒏𝒄𝒌 𝒏𝒆𝒘𝒔𝒑𝒂𝒑𝒆𝒓

S[Ψ(χ , t)]=∫dt (T - V)=∫dt dx L°(ψ , ∂Ψ/∂χ , ∂Ψ/∂t)) (1-31)

L°(ψ , ∂Ψ/∂χ , ∂Ψ/∂t))=(ρ/2) (∂Ψ/∂t)²+ (Tₑ/2)(∂Ψ/∂x)² (1-32)

0=δS/δψ=∂L°/∂Ψ + (d/dx)[∂L°/∂(∂Ψ/∂x)] + (d/dt)[∂L°/∂(∂Ψ/∂t)]

0=δS/δψ=0 - Tₑ (∂²Ψ/∂x²) + ρ(∂²Ψ/∂t²). (1-33)

S=∫ d⁴x L°(Φ , ∂ᵩΦ) (1-34)

∂²Ψ/∂x²=(1/v²)(∂²Ψ/∂t²)

δS/δΦ=∂L°/∂Φ - ∂ᵩΦ [∂L°/(∂(∂ᵩΦ))]=0 (1-35)

L°=(1/2)(∂ᵩΦ)² - (1/2)m²Φ² (1-36)

∂L/∂Φ= - m²Φ , ∂L°/∂(∂ᵩΦ)=∂ᵠΦ (1-37)

(∂² + m²)Φ= 0 (1-38)

exp[iS/ħ] (1-39)

δS/dx(t)=0 (1-40)

الفصل الثانى : المتذبذبات التوافقية البسيطة(البندول البسيط)
The simple harmonic
oscillattor
قوانين الفصل الثاني :


[- ℏ²∂²/∂x² + (1/2)Kx²]Ψ = E Ψ (2-3)

Ψₙ(ξ)=[1/√(2ⁿn!)](mω/ℏπ)¼ Hₙ(ξ) exp(- iξ²/2) (2-4)
mω/(ℏχ)

Eₙ = (n + 1/2)ℏω (2-5)


Ĥ=p²/(2m)+(1/2)mω²x² (2-6)

(1/2)mω²(x - ip/(mω))(x + ip/(mω)) (2-7)

(mω²/2)(x - ip/(mω))(x + ip/(mω))=p²/(2m)+(1/2)mω²x²+(iω/2)[x , p] (2-8)


â=√(mω/(2ℏ))(x + ip/(mω)) (2-9)

â†=√(mω/(2ℏ))(χ - ip/(mω)) (2-10)

[â ، â†] =1 (2-11)

x=√(ℏ/(2mω))(â + â†) (2-12)

p=- i(ℏmω/2) (â - â†) (2-13)

Ĥ = ℏω(â†â + 1/2) (2-14)

n=< n|â†â|n > =< ân|ân >=||â|n >|² ≥ 0 (2-15)

n^ = â†â (2-16)

n^ | n > = n | n > (2-17)

Ĥ = ℏω(n^ + 1/2) (2-18)

Ĥ| n >=(n + 1/2)ℏω| n > (2-19)

n^â†|n> = â†ââ†|n> (2-20)

n^â†|n> = (n + 1)â†|n> (2-21)

n^â|n> = â†ââ|n> (2-22)

n^â|n> = (n - 1)â|n> (2-23)

==n =n (2-25)

|â†ln>|²==c² < n+1|n+1>=|c|² (2-26)

== n +1 (2-27)

â| n>=√n|n-1> (2-28)

â†| n>= √(n+1)|n+1> (2-29)

Ĥ|0>=ħω (n^ + 1/2) |0>=(1/2) ħω |0> (2-30)


â†| 0 > = | 1 > (2-31)

â†| 1 > =√2 | 2 >⇒ |2>=[(â†)²/√2]|0> (2-32)

â†| 3 > =√3 | 3 > ⇒ |3>=[(â†)³/√(3x2)]|0> (2-33)

|n>=[(â†)ⁿ/√n! ]|0> ( 2.34)

Ĥₖ=Σₖ₌₁ᴺ Ĥₖ (2-35)

Ĥₖ=pₖ²/(2mₖ) + (1/2) mωₖ²χₖ² (2-36)

âₖ†|n₁ ، n₂ ، ...، nᵣ ، ...> α |n₁ ، n₂ ، ...، nₖ₊₁ ، ...>،

âₖ |n₁ ، n₂ ، ..، nₖ ،...> α |n₁ ، n₂ ، ...، nₖ₋₁ ، ...> (2-37)


[âₖ ، âᵣ]=0 (2-38)
[âₖ† ، âᵣ†]=0 ، (2-39)
[âₖ† ، âᵣ]=δₖᵣ (2-40)

Ĥₖ=Σₖ₌₁ᴺ Ĥₖ ℏωₖ(âₖ†âₖ + 1/2) (2-41)

âₖ | 0 >= 0 (2-42)

|n₁ ، n₂ ، ... ، nᴺ> =[1/√(n₁! ، n₂! ، ... ، nN!)]â₁†ⁿ¹ (â₂†)ⁿ² ... (âN†)ⁿᴺ| 0 ، 0 ، ... 0 > (2-43)

|{nₖ}> =Πₖ[1/√nₖ](âₖ†)ⁿᵏ | 0 > (2-44)

xⱼ= Σₖ x'ₖ eⁱᵏʲᵃ (2-46)

pⱼ= Σₖ p'ₖ eⁱᵏʲᵃ (2-47)

و التحويل المضاد

x'ₖ=Σⱼ xⱼ e⁻ⁱᵏʲᵃ (2-48)

p'ₖ=Σⱼ pⱼ e⁻ⁱᵏʲᵃ (2-49)


Σⱼ eⁱᵏʲᵃ = Nδₖ,₀ (2-50)

(1/N)Σⱼ Σₖₛ pₖ pₛ eⁱ⁽ᵏ⁺ˢ⁾ʲᵃ (2-51)

Σₖₛ pₖ pₛ δₖ,₋ₛ (2-52)

Σₖ pₖ p₋ₖ (2-53)

[xⱼ , pⱼ،]= iħδⱼⱼ، (2-54)

[xₖ , pₖ،]=(1/N)ΣⱼΣⱼ، e⁻ⁱᵏʲᵃ e⁻ⁱᵏ'ʲ'ᵃ [xⱼ , pⱼ،]

[xₖ , pₖ،]=(iħ/N)Σⱼ e⁻ⁱ⁽ᵏ⁺ᵏ'⁾ʲ'ᵃ

[xₖ , pₖ،]= iħδₖ,₋ₖ، (2-55)

Σⱼ pⱼ²=Σₖ p'ₖ p'₋ₖ (2-57)

â=√(mωₖ/(2ħ))(xₖ+(i/(mωₖ))pₖ) (2-60)

â†=√(mωₖ/(2ħ))(xₖ-(i/(mωₖ))pₖ) (2-61)

χₖ=√(ħ/(2mωₖ))(âₖ + â₋ₖ†). (2-62)

pₖ=i√(ħmωₖ/2)(âₖ† - â₋ₖ†) (2-63)

Ĥ=Σₖ₌₁ (ħωₖ/2)(âₖâₖ† +â₋ₖ†â₋ₖ) (2-64)

Ĥ=Σₖ₌₁ (ħωₖ/2)(âₖâₖ† +âₖ†âₖ) (2-65)

Ĥ=Σₖ₌₁ᴺ ħωₖ(âₖ†âₖ +1/2) (2-66)

الفصل الثالث : تمثيل رقم الشغل
Occupation number representation
قوانين :


p = - i ∂/∂x (3-1)

[p^ψ(x) = - i(dΨ/dx)=pψ(χ)]


eⁱᵖᴸ = 1
cos(pL)+isin(pL)=1

ψ(x + L)=ψ(x)


pₘ=2πm/L (3-2)