Solutions for Problem Set #3 in Quantum Mechanics, Assignments of Health sciences

Download Solutions for Problem Set #3 in Quantum Mechanics and more Assignments Health sciences in PDF only on Docsity! PHY–396 K. Solutions for problem set #3. Problem 1(a): First, let us verify eq. (3) for a wave function Ψ(x1, . . . ,xN ) of the form (1), that is, for the quantum state |N,Ψ〉 = |α1, . . . , αN 〉 = ∣ ∣{nβ} 〉 . Note that the number of distinct permutations of N one-particle modes α1, . . . , αN corresponding to occupation numbers nβ is precisely ( N ! / ∏ β nβ ! ) , so the wave function (1) is properly normalized. Let us define Ψ′(x1, . . . ,xN−1) according to eq. (3). Using orthonormality of the 1-particle wave functions φβ(x), we have Ψ′(x1, . . . ,xN−1) = √ ∏ β nβ ! N − 1! ∫ d3xN φ ∗ γ(xN ) ∑ (α̃1,...,α̃N ) φα̃1(x1) · · ·φα̃N (xN ) = √ ∏ β nβ ! (N − 1)! ∑ (α̃1,...,α̃N ) φα̃1(x1) · · ·φα̃N−1(xN−1)× δα̃N ,γ , (S.1) which leads to two distinct situations: (A) If nγ = 0 i.e., if none of the α1, . . . , αN equals γ, then for any permutation (α̃1, . . . , α̃N ) α̃n 6= γ and every term in the sum (S.1) vanishes, thus Ψ′(x1, . . . ,xN−1) ≡ 0. Clearly, this agrees with âγ ∣ ∣{nβ} 〉 = 0 for nγ = 0. (B) On the other hand, if nγ > 0 then without loss of generality we let αN = γ (remember that the list (α1, . . . , αN ) id un-ordered). Consequently, the δα̃N ,γ factor in eq. (S.1) restricts permutations (α̃1, . . . , α̃N ) to α̃N = γ = αN , and since we count distinct permutations only, we have Ψ′(x1, . . . ,xN−1) = √ ∏ β nβ ! N − 1! ∑ distinct permutations (α̃1,...,α̃N−1) of (α1,...,αN−1) φα̃1(x1) · · ·φα̃N−1(xN−1). (S.2) Comparing this expression to eq. (1), we see that Ψ′ is the wave function of the state |α1, . . . , αN−1〉 = ∣ ∣ ∣ {n′β = nβ − δβ,γ} 〉 , except for the normalization factor √ ∏ β nβ! N − 1! = √ nγ × √ ∏ β n ′ β ! N − 1! instead of √ ∏ β n ′ β ! N − 1! . (S.3) 1 In other words, ∣ ∣(N − 1),Ψ′ 〉 = √ nγ |α1, . . . , αN−1〉 = âγ |α1, . . . , αN 〉 , for αN = γ, (S.4) and we already saw that for α1, . . . , αN 6= γ, |Ψ′〉 = 0 = â |α1, . . . , αN 〉 as well. Thus, ∀ |Ψ〉 = |α1, . . . , αN 〉 : ∣ ∣Ψ′ 〉 = âγ |Ψ〉 , (S.5) and by linearity of eq. (3) it follows that |Ψ′〉 = âγ |Ψ〉 for any linear combination of the |α1, . . . , αN 〉 states. As we saw in class, such states form a complete basis of the N -boson Hilbert space, therefore ∣ ∣Ψ′ 〉 = âγ |Ψ〉 ∀ |Ψ〉 ∈ HN bosons . Q.E .D. Problem 1(b): First, consider a one-body operator of the form Ô = |α〉 〈β|. For this operator, the second- quantized Ô (2) tot is simply â † αâβ while the first-quantized Ô (1) tot has matrix elements 〈N,Ψ1| Ô(1)tot |N,Ψ2〉 = = N ∑ i=1 ∫ d3x1 · · · ∫ d3xN ∫ d3x′i Ψ ∗ 1(x1, . . . ,xi, . . . ,xN )φα(xi)φ ∗ β(x ′ i)Ψ2(x1, . . . ,x ′ i, . . . ,xN ) = N ∫ d3x1 · · · ∫ d3xN ∫ d3x′N Ψ ∗ 1(x1, . . . ,xN−1,xN )φα(xN )φ ∗ β(x ′ N )Ψ2(x1, . . . ,xN−1,x ′ N ) = ∫ d3x1 · · · ∫ d3xN−1Ψ ′∗ 1 (x1, . . . ,xN−1)Ψ ′ 2(x1, . . . ,xN−1) where the second equality follows from the total symmetry of the wave functions Ψ1 and Ψ2, and the (N − 1)-particle wave functions Ψ′1,2(x1, . . . ,xN−1) on the last line are defined according to eq. (3) (but using âα instead of âγ for the Ψ ′ 1 and âβ for the Ψ2). In light of problem (a), this means 〈N,Ψ1| Ô(1)tot |N,Ψ2〉 = 〈 (N − 1),Ψ′1|(N − 1),Ψ′2 〉 = 〈N,Ψ1| â†αâβ |N,Ψ2〉 ≡ 〈N,Ψ1| Ô(2)tot |N,Ψ2〉 . (S.6) Now we need to extend this result to generic one-body operators. Any operator R̂1 in the one- particle Hilbert space can be decomposed as R̂1 = ∑ α,β |α〉Rα,β 〈β| where Rα,β are the matrix 2 Problem 1(e): First, let us calculate the Fock-space commutator [↵âν , â † αâ † β âγ âδ]. Using the commutators [↵âν , â † α] = δν.αâ † µ and [â † µâν , âβ ] = −δµ.β âν (cf. previous homework) and applying the Leibniz rule, we have [↵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) Next, given a second-quantized one-body operator  = ∑ µ,ν 〈µ| Â1 |ν〉 â † µâν and a second- quantized two-body operator B̂ = 12 ∑ α,β,γ,δ 〈α| ⊗ 〈β| B̂2 |γ〉 ⊗ |δ〉 â † αâ † β âγ âδ, we calculate [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. 5 Problem 2(a): Use product-of-exponentials formula ∀Â, B̂ : eÂeB̂ = exp ( Â+ B̂ + 12 [Â, B̂] + 1 12 [(Â− B̂), [Â, B̂]] + · · · ) . (S.20) In particular, for  = ξâ†, B̂ = ξ∗â and [Â, B̂] = ξξ∗ being a c-number, eξâ † e−ξ ∗â = exp ( ξ↠− ξ∗â + 12ξξ ∗ ) , exactly, (S.21) and therefore |ξ〉 def= eξâ†−ξ∗â |0〉 = e−|ξ|2/2eξâ†e−ξ∗â |0〉 = e−|ξ|2/2eξ↠|0〉 . (S.22) (Note e−ξ ∗â |0〉 = |0〉 since â |0〉 = 0.) Next, [â, â†] = 1 implies that for any function f(â†), [â, f(â†)] = f ′(â†). In particular, [â, eξâ † ] = ξeξâ † or in other words, (â − ξ)eξ↠= eξ↠â and hence (â − ξ) |ξ〉 ∝ eξ↠â |0〉 = 0. Q.E .D. Problem 2(b): For any normal-ordered product of creation and annihilation operators — i.e., a product in which all creation operators are to the right of all annihilation operators — one has 〈ξ| (â†)k(â)` |ξ〉 = (ξ∗)kξ`, simply because â |ξ〉 = ξ |ξ〉 and 〈ξ| ↠= ξ∗ 〈ξ|. In particular, 〈ξ| (n̂ = â†â) |ξ〉 = ξ∗ξ. On the other hand, n̂2 = â†ââ†â = â†â†ââ + â†â =⇒ 〈ξ| n̂2 |ξ〉 = (ξ∗)2ξ2 + ξ∗ξ = n̄2 + n̄ (S.23) hence ∆n = √ 〈n̂2〉 − n̄2 = √ n̄. In a similar manner, q̂ = √ h̄ 2mω (â + â†), q̂2 = h̄ 2mω ( (â)2 + (â†)2 + 2â†â + 1 ) =⇒ 6 〈ξ| q̂2 |ξ〉 = h̄ 2mω ( (ξ + ξ∗)2 + 1 ) = 〈ξ| q̂ |ξ〉2 + h̄ 2mω and likewise 〈ξ| p̂2 |ξ〉 = mωh̄ 2 ( (−iξ + iξ∗)2 + 1 ) = 〈ξ| p̂ |ξ〉2 + mωh̄ 2 , thus ∆q = √ h̄ 2mω , ∆p = √ mωh̄ 2 , ∆q∆p = h̄ 2 . (S.24) Q.E .D. Problem 2(c): In the Schrödinger picture, ↠is time independent, hence (d/dt)eξâ † = (dξ/dt)â†eξâ † . Using time independence of the magnitude |ξ|, we then have d dt ( |ξ〉 = e−|ξ|2/2eξ↠|0〉 ) = dξ dt ↠|ξ〉 = 1 ξ dξ dt â†â |ξ〉 = −iωâ†â |ξ〉 (S.25) where the last equality comes from ξ(t) = ξ0e −iωt. In other words, ih̄ d dt |ξ(t)〉 = h̄ωâ†â |ξ(t)〉 ≡ Ĥ |ξ(t)〉 . (S.26) Q.E .D. Problem 2(d): In question 2(a) we saw that [â, â†] = 1 implies eξâ † â = (â− ξ)eξ↠for any c-number ξ. Iterating this identity gives us eξâ † f(â) = f(â − ξ)eξ↠for any function f(â) of the annihilation operator, and in particular eξâ † eη ∗â = eη ∗(â−ξ) eξâ † = e−η ∗ξ eη ∗â eξâ † . (S.27) Consequently, the quantum overlap of the coherent states |ξ〉 and 〈η| is 〈η|xi〉 = e−|η|2/2e−|ξ|2/2 〈0| eη∗âeξ↠|0〉 = e−|η| 2/2e−|ξ| 2/2e−η ∗ξ 〈0| eξâ†eη∗â |0〉 = exp ( −12 |η| 2 − 12 |ξ| 2 − η∗ξ ) (S.28) 7