Quantum Mechanics: Normalization & Commutators - Problem Set #1 Solutions, Assignments of Health sciences

Download Quantum Mechanics: Normalization & Commutators - Problem Set #1 Solutions and more Assignments Health sciences in PDF only on Docsity! PHY–396 K. Solutions for problem set #1. Problem 1(a): As explained in class, the properly normalized wave function of the N particle state |(α1, α2, . . . , αN )〉 = ∣∣{nβ}β〉 is Ψ(x1,x2, . . . ,xN ) = 1√ # distinct permutations distinct permutations∑ α′ 1 ,...,α′ N of α1,...,αN ϕα′1(x1) · · ·ϕα′N (xN ) = 1√ # distinct permutations × 1(# redundant permutations ) all permutations∑ α′ 1 ,...,α′ N of α1,...,αN ϕα′1(x1) · · ·ϕα′N (xN ). (S.1) Combinatorically, the number of redundant i.e., trivial permutation of the set of α1, . . . , αN is∏ β nβ! while the number of distinct permutations is therefore N !/ ∏ β nβ!. Furthermore, any permutation of the functions ϕα1 , . . . , ϕαN in a product ϕα1(x1) · · ·ϕαN (xN ) is equivalent to an opposite permutation of the coordinates x1, . . . ,xN . Substituting these facts into the second line of eq. (S.1) immediately yields eq. (2) of the problem set. Q.E .D. Problem 1(b): First, let us assume |N,Ψ〉 = |(α1, α2, . . . , αN )〉 and hence Ψ(x1, . . . ,xN ) of the form (2). Con- sequently, ∣∣N + 1,Ψ′〉 def= â†α |N,Ψ〉 = √nα + 1 |(α1, α2, . . . , αN , α)〉 , (S.2) so Ψ′ is also of the form (2). Specifically, Ψ′(x1, . . . ,xN+1) ∝ distinct∑ ν1,...,νN+1=1,...,(N+1) ϕα1(xν1) · · ·ϕαN (xνN )ϕα(xνN+1) 〈〈renaming νN+1 → i and regrouping terms〉〉 = N+1∑ i=1 ϕα(xi) distinct∑ ν1,...,νN=1,...,6 i,...,(N+1) ϕα1(xν1) · · ·ϕαN (xνN ) ∝ N+1∑ i=1 ϕα(xi) Ψ(x1, . . . , 6xi, . . . ,xN+1), (S.3) 1 exactly as in eq. (3) we are trying to prove. As to the overall numerical factor in eq. (3), we need the ratio between the coefficients in eqs. (2) for the Ψ and the Ψ′: coeff.(Ψ′) coeff.(Ψ) = ( N ′!× ∏ β n ′ β! )−1/2 ( N !× ∏ β nβ! )−1/2 = √ N ! N ′! × ∏ β √ nβ! n′β! = 1√ N + 1 × 1√ nα + 1 . (S.4) Combining this ratio with the √ nα + 1 factor in eq. (S.2) we arrive at the 1/ √ N + 1 factor in eq. (3). Note that the last coefficient depends only on the total particle number N but not on any coincidencies between α1, . . . , αN . Consequently, eq . (3) applies not only to wave functions of the form (2) but also to all linear conbinations of such wave functions (note linearity of the creation operator â†α). And since the states |(α1, α2, . . . , αN )〉 comprise a complete basis of the N–boson Hilbert spaceHBN , eq. (3) applies to all (properly symmetric) N–particle wave functions Ψ. Q.E .D. Problem 1(c): The action of an annihilation operator âα follows from that of its hermitian conjugate — the creation operator â†α. Indeed, for any N–particle state |N,Ψ〉 and any N − 1 particle state |N − 1, Ψ̃〉, one has 〈N − 1, Ψ̃| âα |N,Ψ〉 = 〈N,Ψ| â†α |N − 1, Ψ̃〉 ∗ , (S.5) which in terms of the wave functions becomes ∫ dx1 · · · ∫ dxN−1 Ψ̃ ∗(x1, . . . ,xN−1) Ψ ′′(x1, . . . ,xN−1) = [∫ dx1 · · · ∫ dxN Ψ ∗(x1, . . . ,xN ) Ψ̃ ′(x1, . . . ,xN ) ]∗ = ∫ dx1 · · · ∫ dxN Ψ(x1, . . . ,xN )× 1√ N N∑ i=1 ϕ∗α(xi) Ψ̃ ∗(x1, . . . , 6xi, . . . ,xN ) 〈〈by symmetry of the Ψ〉〉 2 Consequently, [â†αâ † β âγ âδΨ](x1, . . . ,xN ) (S.14) = N∑ i=1 ϕα(xi) ∑ j 6=i ϕβ(xj)× ∫ dx′i ∫ dx′j ϕ ∗ γ(x ′ i)ϕ ∗ δ(x ′ j) Ψ(x1, . . . ,x ′ i, . . . ,x ′ j , . . . ,xN ); comparing this formula with eq. (S.12) immediatly proves eq. (9). To complete the proof, consider the special cases of the vacuum and of the one particle states. In both cases, there are no two-particle operators and hence Âtot ≡ 0. In the second quantized formalism,  |N < 2, whatever〉 = 0 follows from all N < 2 states being annihilated by all products of two annihilation operators: Indeed, âδ |0〉 = 0 while âδ |N = 1,Ψ〉 ∝ |0〉 and hence âγ âδ |N = 1,Ψ〉 = 0. Q.E .D. Problem 3(a): This is a simple exercise of the Leibnitz rule for commutators: [â†αâβ, â † γ ] = [â † α, â † γ ]âβ + â † α[âβ, â † γ ] = 0 + â † αδβ,γ = δβ,γ â † α , [â†αâβ, âδ] = [â † α, âδ]âβ + â † α[âβ, âδ] = −δα,δâβ + 0 = −δα,δâβ , [â†αâβ, â † γ âδ] = [â † αâβ, â † γ ]âδ + â † γ [â † αâβ, âγ âδ] = δβ,γ â † αâδ − δα,δâ † γ âβ . (S.15) Problem 3(b): Given  = ∑ α,β 〈α| Â1 |β〉 â†αâβ and B̂ = ∑ γ,δ 〈γ| B̂1 |δ〉 â†γ âδ , 5 we immediately have [Â, B̂] = ∑ α,β,γ,δ 〈α| Â1 |β〉 〈γ| B̂1 |δ〉 [â†αâβ, â † γ âδ] 〈〈using (S.15)〉〉 = ∑ α,β,γ,δ 〈α| Â1 |β〉 〈γ| B̂1 |δ〉 ( δβ,γ â † αâδ − δα,δâ † γ âβ ) = ∑ α,δ â†αâδ × ∑ β=γ 〈α| Â1 |γ〉 〈γ| B̂1 |δ〉 − ∑ β,γ â†γ âβ × ∑ α=δ 〈γ| B̂1 |α〉 〈α| Â1 |β〉 = ∑ α,δ â†αâδ 〈α| Â1B̂1 |δ〉 − ∑ β,γ â†γ âβ 〈γ| B̂1Â1 |β〉 〈〈renaming summation indices〉〉 = ∑ α,β â†αâβ × ( 〈α| Â1B̂1 |β〉 − 〈α| B̂1Â1 |β〉 ) = ∑ α,β â†αâβ × 〈α| ( [Â1, B̂1] = Ĉ1 ) |β〉 ≡ Ĉ. (S.16) Problem 3(c): Again we use the Leibnitz rule: [↵âν , â † αâ † β âγ âδ] = [â † µâν , â † α]â † β âγ âδ + â † α[â † µâν , â † β]âγ âδ + â † αâ † β[â † µâν , âγ ]âδ + â † αâ † β âγ [â † µâν , âδ] = δναâ † µâ † β âγ âδ + δνβ â † αâ † µâγ âδ − δµγ â † αâ † β âν âδ − δµδâ † αâ † β âγ âν . (S.17) Problem 3(d): Proceeding as in 3(b), we calculate 6 [Â, B̂] = ∑ µ,ν,α,β,γ,δ 〈µ| Â1 |ν〉 〈α⊗ β| B̂2 |γ ⊗ δ〉 [↵âν , â†αâ † β âγ âδ] 〈〈using (S.17)〉〉 = ∑ µ,β,γ,δ ↵⠆ β âγ âδ × ∑ ν 〈µ| Â1 |ν〉 〈ν ⊗ β| B̂2 |γ ⊗ δ〉 + ∑ α,µ,γ,δ â†αâ † µâγ âδ × ∑ ν 〈µ| Â1 |ν〉 〈α⊗ ν| B̂2 |γ ⊗ δ〉 − ∑ α,β,ν,δ â†αâ † β âν âδ × ∑ µ 〈α⊗ β| B̂2 |µ⊗ δ〉 〈µ| Â1 |ν〉 − ∑ α,β,γ,ν â†αâ † β âγ âν × ∑ µ 〈α⊗ β| B̂2 |γ ⊗ µ〉 〈µ| Â1 |ν〉 〈〈renaming summation indices〉〉 = ∑ α,β,γ,δ â†αâ † β âγ âδ × Cα,β,γ,δ , (S.18) where Cα,β,γ,δ = ∑ λ 〈α| Â1 |λ〉 〈λ⊗ β| B̂2 |γ ⊗ δ〉 + ∑ λ 〈β| Â1 |λ〉 〈α⊗ λ| B̂2 |γ ⊗ δ〉 − ∑ λ 〈α⊗ β| B̂2 |λ⊗ δ〉 〈λ| Â1 |γ〉 − ∑ λ 〈α⊗ β| B̂2 |γ ⊗ λ〉 〈λ| Â1 |δ〉 = 〈α⊗ β| ( Â1(1 st)B̂2 + Â1(2 nd)B̂2 − B̂2Â1(1 st) − B̂2Â1(2nd) ) |γ ⊗ δ〉 = 〈α⊗ β| [( Â1(2 nd) + Â1(2 nd) ) , B̂2 ] |γ ⊗ δ〉 ≡ 〈α⊗ β| Ĉ2 |γ ⊗ δ〉 . (S.19) Consequently, [Â, B̂] = Ĉ. Q.E .D. 7