Quantum Mechanics Problem

 Quantum Mechanics Problem

 

[1] Eigen Value Of Hermitin Operator Are Real :

 

- Let  Is Hermitin Operator

   Â†=Â.......(1)       (Dagger Symbol[†] - Complex Conjugate And

                                                                                        Transpose)

- Â|ψ⟩ = a|ψ⟩.......(2)

- Taking Scalar Product |Ψ⟩

   ⟨Ψ|Â|Ψ⟩ = a⟨Ψ|Ψ⟩

   ⟨Ψ|Â|Ψ⟩ = a......(3)  (⟨Ψ|Ψ⟩ = 1)

- (Â|Ψ⟩)† = (a|Ψ⟩)†

   ⟨Ψ|† = ⟨Ψ|a*

- From Equation (1)

   ⟨Ψ|Â = ⟨Ψ|a*

- Taking Scalar Product |Ψ⟩

   ⟨Ψ|Â|Ψ⟩ = a*⟨Ψ|Ψ⟩

   ⟨Ψ|Â|Ψ⟩ = a*......(4)  (⟨Ψ|Ψ⟩ = 1)

- From Equation (3) & (4)

   a*=a

- So,Eigan Value Of Hermitin Operator Are Real

 

[2] Eigan Value Of Â^-1 Operator Is Inverse Of Eigen ValueO Of  Operator

 

- Let  Opertor Operate On State |Ψ⟩ Give 'A' Eigen Value And

  Same State

   Â|Ψ⟩ = a|Ψ⟩

- Multiply  Both Side By Â^-1

   Â^-1 (Â|Ψ⟩) = (a|Ψ⟩)Â^-1

   (Â^-1 Â)|Ψ⟩) = a (Â^-1|Ψ⟩)

   I|Ψ⟩) = a (Â^-1|Ψ⟩)........(I = Identy Opertor )

   |Ψ⟩) = a (Â^-1|Ψ⟩).........(I=1)

-  (Â^-1|Ψ⟩) = 1/a |Ψ⟩

 

[3] Eigan State Corresponding To Different- 2 Eigen ValueO OfHermitin Operator Are Orthogonal

 

- let,

  †=Â.......(1) 

- Â|Ψ_m⟩ = am|Ψ_m⟩.......(2)

   Â|Ψ_n⟩ = a_n|Ψ_n⟩.......(3)

   a_m ≠ a_n

- Taking Scalar Product With |Ψn⟩ Of  Equation 

   (2)

   ⟨Ψ_n|Â|Ψ_m⟩ = a_m ⟨Ψn|Ψ_m⟩.....(4)

- Taking Both Side † Of  Equation (3)

  (Â|Ψ_n⟩)† = (an|Ψ_n⟩)†

  ⟨Ψ_n|† = ⟨Ψ_n|a_n*

  ⟨Ψ_n|† = ⟨Ψ_n|a_n    (a_n*=a_n)

- From Equation (1)

  ⟨Ψ_n|Â = ⟨Ψn|a_n

- Taking Scalar Product With |Ψ_m⟩

   ⟨Ψ_n|Â|Ψ_m⟩ = an ⟨Ψ_n|Ψ_m⟩.....(5)

- Equation (4)-(5)

  a_m ⟨Ψ_n|Ψ_m⟩ - a_n ⟨Ψ_n|Ψ_m⟩ = 0

  (a_m-a_n) ⟨Ψ_n|Ψ_m⟩ = 0

- (a_m-a_n) = 0 or ⟨Ψn|Ψm⟩ = 0

- But (a_m-a_n) ≠ 0 Because  a_m ≠ a_n

  So ⟨Ψ_n|Ψ_m⟩ = 0 Is Orthogonal

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