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

Download Solutions for Problem Set #3 in Quantum Mechanics and more Assignments Physics in PDF only on Docsity! PHY–396 K. Solutions for problem set #3. Problem 1(a): First, let us verify eq. (4) for a wave function Ψ(x1, . . . ,xN ) of the form (3), that is, for the quantum state |N, Ψ〉 = |α1, . . . , αN 〉 = ∣∣{nβ}〉. 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) = √ N√ Cα1,...,αN ∫ d3xN ϕ ∗ γ(xN )× ∑ (α̃1,...,α̃N ) ϕα̃1(x1)× · · · × ϕα̃N (xN ) = √ N√ Cα1,...,αN ∑ (α̃1,...,α̃N−1,α̃N ) ϕα̃1(x1)× · · · × ϕα̃N−1(xN−1)× δα̃N ,γ , (S.1) which leads to two distinct situations according to the nγ : (A) For nγ = 0 none of the α1, . . . , αN equals γ, hence for any permutation (α̃1, . . . , α̃N−1, α̃N ) we have α̃n 6= γ. Consequently, every term in the sum (S.1) vanishes and therefore Ψ′ ≡ 0. At the same time, âγ ∣∣{nβ}〉 = 0 for nγ = 0, and therefore ∣∣(N − 1), Ψ′〉 = 0 = âγ ∣∣{nβ}〉 . (S.2) (B) On the other hand, for nγ > 0 we have ∣∣{nβ}〉 = ∣∣α1, . . . , αN−1, γ〉, and permutations with α̃N = αN = γ do contribute to the sum (S.1). Note that we sum over distinct permutations of (α1, . . . , αN ) only, so even if some of the α1, . . . , αN−1 are also equal to γ, restricting the sum to permutations (α̃1, . . . , α̃N−1, α̃N ) with α̃N = γ is equivalent to fixing α̃N = αN and summing over the permutations (α̃1, . . . , α̃N−1) of the (α1, . . . , αN−1). Thus, Ψ′(x1, . . . ,xN−1) = √ N√ Cα1,...,αN ∑ distinct permutations (α̃1,...,α̃N−1) of (α1,...,αN−1) ϕα̃1(x1) · · ·ϕα̃N−1(xN−1). (S.3) Comparing this expression to eq. (3) for N ′ = N − 1, we immediately see that except for the overall normalization, Ψ′(x1, . . . ,xN−1) is the wave function of the (N − 1) particle state 1 ∣∣α1, . . . , αN−1〉. Specifically, ∣∣(N − 1), Ψ′〉 = √N√ Cα1,...,αN−1,αN × √ Cα1,...,αN−1 × ∣∣α1, . . . , αN−1〉 = √ N nγ × Cα1,...,αN−1 Cα1,...,αN−1,αN × âγ ∣∣{nβ}〉 . (S.4) To fix the normalization factor, remember that Cα1,...,αN is the number of distinct permutations of the (α1, . . . , αN ), which is equal to the number of all permutations divided by the number of trivial permutations of identical α’s. In terms of the occupation numbers, Cα1,...,αN = N !∏ β nβ! . (S.5) Likewise, Cα1,...,αN−1 = (N − 1)!∏ β n ′ β! (S.6) where n′β = nβ − δβ,γ . Consequently, Cα1,...,αN−1 Cα1,...,αN−1,αN = (N − 1)! N ! × nγ ! (nγ − 1)! = nγ N , (S.7) and therefore in eq. (S.4) the numerical factors cancel out and ∣∣(N − 1), Ψ′〉 = âγ ∣∣{nβ}〉 . (S.8) Altogether, combining eqs. (S.2) and (S.8), we find that ∀ |Ψ〉 = |α1, . . . , αN 〉 : ∣∣Ψ′〉 = âγ |Ψ〉 . (S.9) By linearity of eq. (4) 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, 2 eq. (S.16), 〈N, Ψ1| Ô(1)tot |N, Ψ2〉 = 〈N, Ψ1| Ô (2) tot |N, Ψ2〉 for any 〈N, Ψ1| and |N, Ψ2〉 . (S.17) Now we need to extend this result to generic one-body operators. Fortunately, any operator Â1 in the one-particle Hilbert space can be decomposed as Â1 = ∑ α,β |α〉Aα,β 〈β| where Aα,β are the matrix elements 〈α| Â1 |β〉. The definition (6) of first-quantized Â(1)tot of N particles is obviously linear with respect to the Â1, thus  (1) tot = ∑ α,β Aα,β N∑ i=1 ( |α〉 〈β| ) i th particle (S.18) and therefore, thanks to eq. (S.17), 〈N, Ψ1| Â(1)tot |N, Ψ2〉 = ∑ α,β Aα,β 〈N, Ψ1| â†αâβ |N, Ψ2〉 = 〈N, Ψ1|  (2) tot |N, Ψ2〉 . Q.E .D. Problem 1(d): Again, we start with a particularly simple 2-body operator Ô2 = (|α〉⊗ |β〉)(〈γ|⊗〈δ|) which acts on two-particle wave functions according to 〈2, Ψ1| Ô2 |2, Ψ2〉 = ∫ d3x1 ∫ d3x2 Ψ ∗ 1(x1,x2)φα(x1)φβ(x2) (S.19) × ∫ d3y1 ∫ d3y2 φ ∗ γ(x1)φ ∗ δ(x2)Ψ2(x1,x2). Consequently, the first-quantized Ô (1) tot operator constructed according to eq. (9) acts in the N - 5 boson Hilbert space as 〈N, Ψ1| Ô(1)tot |N, Ψ2〉 = = 12 ∑ i6=j ∫ d3x1 · · · ∫ d3xN Ψ ∗ 1(x1, . . . ,xi, . . . ,xj . . . . ,xN )φα(xi)φβ(xj)× × ∫ d3x′i ∫ d3x′j φ ∗ γ(x ′ i)φ ∗ δ(x ′ j)Ψ2(x1, . . . ,x ′ i, . . . ,x ′ j , . . . ,xN ) = N(N − 1) 2 ∫ d3x1 · · · ∫ d3xN Ψ ∗ 1(x1, . . . ,xN−2,xN−1,xN )φα(xN−1)φβ(xN )× × ∫ d3x′N−1 ∫ d3x′N φ ∗ γ(xN−1)φ ∗ δ(xN )Ψ2(x1, . . . ,xN−2,x ′ N−1,x ′ N ) = 12 ∫ d3x1 · · · ∫ d3xN−2 Ψ ′′∗ 1 (x1, . . . ,xN−2) Ψ ′′ 2(x1, . . . ,xN−2) (S.20) where again the second equality follows from the total symmetry of the bosonic wave functions, and where on the the last line Ψ |prime′∗ 1 and Ψ ′′ 2 are (N − 2) particle wave functions constructed according to Ψ′′∗1 (x1, . . . ,xN−2) = √ N(N − 1) ∫ d3xN−1 ∫ d3xN Ψ ∗ 1(x1, . . . ,xN )ϕα(xN−1)ϕβ(xN ), (S.21) Ψ′′2(x1, . . . ,xN−2) = √ N(N − 1) ∫ d3xN−1 ∫ d3xN ϕ ∗ γ(xN−1)ϕ ∗ δ(xN )Ψ2(x1, . . . ,xN ). (S.22) Notice that the double integral in eq. (S.22) is precisely the integral of eq. (4) applied twice, thus in Fock-space notations ∣∣(N − 2), Ψ′′2〉 = âγ âδ |N, Ψ2〉 . (S.23) As to eq. (S.21), it looks like the complex conjugate of eq. (S.22), hence in Fock-space notations 〈 (N − 2), Ψ′′1 ∣∣ = 〈N, Ψ1| â†αâ†β . (S.24) Putting this all together, we arrive at 〈N, Ψ1| Ô(1)tot |N, Ψ2〉 = 12 〈N, Ψ1| â † αâ † β âγ âδ |N, Ψ2〉 for Ô2 = (|α〉 ⊗ |β〉)(〈γ| ⊗ 〈δ|). (S.25) To extend this result to a general two-body operator B̂2, we use matrix-element decomposition 6 in the two-distinct-particle Hilbert space: B̂2 = 1 2 ∑ α,β,γ,δ Bα,β,γ,δ ( |α〉⊗ |β〉 )( 〈γ| ⊗ 〈δ| ) where Bα,β,γ,δ = (〈α| ⊗ 〈β|)B̂2(|γ〉⊗ |δ〉). (S.26) Consequently, similarly to eq. (S.18), the first-quantized form of Ŝtot can be written as B̂ (1) tot = 1 2 ∑ α,β,γ,δ Bα,β,γ,δ ∑ i6=j ( |α〉 〈γ| ) i th particle × ( |β〉 〈δ| ) j th particle , (S.27) and therefore 〈N, Ψ1| B̂(1)tot |N, Ψ2〉 = 12 ∑ α,β,γ,δ Bα,β,γ,δ × 〈N, Ψ1| â†αâ † β âγ âδ |N, Ψ2〉 ≡ 〈N, Ψ1| B̂ (2) tot |N, Ψ2〉 Q.E .D. Problem 2(a): This is a simple exercise of the Leibniz 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.28) Problem 2(b): Given  (2) tot = ∑ α,β 〈α| Â1 |β〉 â†αâβ (S.29) and B̂ (2) tot = ∑ γ,δ 〈γ| B̂1 |δ〉 â†γ âδ , (S.30) 7 and therefore |ξ〉 def= eξâ †−ξ∗â |0〉 = e−|ξ| 2/2eξâ † e−ξ ∗â |0〉 = e−|ξ| 2/2eξâ † |0〉 . (S.37) (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 3(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.38) hence ∆n = √ 〈n̂2〉 − n̄2 = √ n̄. In a similar manner, q̂ = √ h̄ 2mω (â + â†), q̂2 = h̄ 2mω ( (â)2 + (â†)2 + 2â†â + 1 ) =⇒ 〈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.39) Q.E .D. 10 Problem 3(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| 2/2eξâ † |0〉 ) = dξ dt ↠|ξ〉 = 1 ξ dξ dt â†â |ξ〉 = −iωâ†â |ξ〉 (S.40) 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.41) Q.E .D. Problem 3(d): In question 3(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.42) Consequently, the quantum overlap of the coherent states |ξ〉 and 〈η| is 〈η|ξ〉 = e−|η| 2/2e−|ξ| 2/2 〈0| eη ∗âeξâ † |0〉 = e−|η| 2/2e−|ξ| 2/2e−η ∗ξ 〈0| eξâ † eη ∗â |0〉 = exp ( −12 |η| 2 − 12 |ξ| 2 − η∗ξ ) × 1 (S.43) because eη ∗â |0〉 = |0〉, 〈0| eξ↠= 〈0| and 〈0|0〉 = 1. In terms of the probability overlap, |〈η|ξ〉|2 = e−|η−ξ| 2 . (S.44) 11 Problem 3(e): Generalization of coherent states to multi-oscillatory systems and further to the creation / an- nihilation fields is completely straightforward: |coherent〉 def= exp ( F̂ † − F̂ ) |0〉 = e−N̄/2 eF̂ † |0〉 (S.45) where F̂ † = ξ↠→ ∑ α ξαâ † α → ∫ d3xΦ(x)Ψ̂†(x). (S.46) Similar to the single-oscillator theory, ( Ψ̂(x)− Φ(x) ) eF̂ † = eF̂ † Ψ̂†(x), hence Ψ̂(x) |Φ〉 = Φ(x) |Φ〉 . (S.47) Problem 3(f): Using eq. (S.47) and its hermitian conjugate, we have 〈Φ| Ψ̂†(x1) · · · Ψ̂†(xk)Ψ̂(y1) · · · Ψ̂(y`) |Φ〉 = Φ∗(x1) · · ·Φ∗(xk)Φ(y1) · · ·Φ(y`) (S.48) for any normal-ordered product of the quantum fields. Specifically, for the particle-number operator N̂ we have eq. (12), while for its square — whose normal-ordered form N̂2 = ∫∫ d3x d3y Ψ̂†(x)Ψ̂†(y)Ψ̂(x)Ψ̂(y) + ∫ d3x Ψ̂†(x)Ψ̂(x) (S.49) generalizes eq. (S.38) — we have 〈Φ| N̂2 |Φ〉 = ∫∫ d3x d3y Φ∗(x)Φ∗(y)Φ(x)Φ(y) + ∫ d3xΦ∗(x)Φ(x) = 〈Φ| N̂ |Φ〉2 + 〈Φ| N̂ |Φ〉 , (S.50) and hence ∆N = √ N̄ , Q.E .D. 12