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Hermitian and unitary operator An operator is Hermitian if it is self-adjoint: A+ = A Or equivalently: < ψ|A| φ> = (<φ|A|ψ>)* and so 〈A〉= < ψ|A|ψ> is real. An operator is skew-Hermitian if B+ = -B and 〈B〉= < ψ|B|ψ> is imaginary. In quantum mechanics, the expectation of any physical quantity has to be real and hence an operator corresponds to a physical observable must be Hermitian. For example, momentum operator and Hamiltonian are Hermitian. An operator is Unitary if its inverse equal to its adjoints: U-1 = U+ or UU+ = U+U = I In quantum mechanics, unitary operator is used for change of basis. Commutators Operators do not commute. Commutator: [A, B] = AB-BA Anti-commutator: {A, B} = AB+BA Algebra of commutators: 1. Antisymmetric: [A, B] = -[B, A] 2. Bilinear: [αA+βB+γC+…. , X] = α [A,X]+ β[B,X]+ γ[C,X]+…. [X, αA+βB+γC+…. ] = α[X, A]+ β[X, B]+ γ[X, C]+…. 3. Adjoint: [A, B]+ = [B+, A+] 4. Distributive: [A, BC] = [A,B]C+B[A,C] [BC, A] = [B,A]C+B[C,A] 5. n −1 [A, B ] = ∑ B j[A, B]A n − j−1 n j= 0 n −1 [A , B] = ∑ A n − j−1[A, B]B j n j= 0 6. Jacobi identity: [A, [B, C]] + [B, [C, A]] + [ C, [A, B]] = 0 7. If A and B are Hermitian, [A, B] is skew-Hermitian and {A, B} is Hermitian. 8. Note that [A, B]=0 and [B, C]=0 does NOT imply [B, C]=0. Function of operators If f(A) is a “function” of A. We can Taylor expand f(A) in a power series of A: ∞ f(A) = ∑ a n A n n =0 ∞ ∴[f(A)] = ∑ a n (A + ) n = f * (A + ) + n =0 For example, ∞ e aA * =∑ n =0 a nAn 1 1 = I + aA + a 2 A 2 + a 3 A 3 + L n! 2 6 (e A ) + = e A + and (eiA ) + = e -iA + Commutators involving function of operators : 1. [A, f(A)] = 0 but e A e B = e A + Be[A,B]/2 1 1 3. e -A Be A = B + [A, B] + [A, [A, B]] + [A,[A, [A, B]]] + L 2! 3! 2. e e ≠ e A B A+B Commutators and uncertainty relation Define operator δA=A- 〈A〉 where 〈A〉=<ψ|A|ψ> and A2 =<ψ|A2| ψ> Let δA|ψ>=|χ> and δB |ψ>=|φ>. Note that if A and B are Hermitian, so are δA and δB. δA=δA+ ⇒ 〈δA2〉 = 〈δAδA〉 =<ψ|δAδA|ψ> =<ψ|δA+δA|ψ> =<χ|χ> ⇒ 〈δA2〉 = <χ|χ> Similarly, 〈δB2〉= <φ |φ> and 〈δAδB〉 = <χ| φ> Schwarz inequality: <χ|χ> <φ |φ> ≥ | <χ| φ>|2 ⇒ 〈δA2〉〈δB2〉 ≥ 〈δAδB〉2 Now, δAδB = 1 2 (δAδB - δBδA) + 12 (δAδB + δBδA) [δA, δB] + 12 {δA, δB} = 1 2 = [A, B] + 12 {δA, δB} 1 424 3 14243 1 2 Re al Im aginary ∴ δAδB 2 = 14 [A, B] δA 2 δB2 ≥ δAδB 2 2 + 1 2 {δA, δB} ⇒ δA 2 δB 2 ≥ δA 2 = (A - ∴ ∆A 2 ∆B2 ≥ 14 [A, B] A 2 1 4 [A, B] δA 2 δB2 ≥ ⇒ Define ∆A = 2 ) 2 ⇒ = A2 - A 1 4 2 2 + [A, B] 1 2 {δA, δB} 2 and ∆B = ∆A∆B ≥ 12 [A, B] δ B2 2