Quantum Mechanics Homework Problem Solutions, Assignments of Quantum Mechanics

Download Quantum Mechanics Homework Problem Solutions and more Assignments Quantum Mechanics in PDF only on Docsity! Physics 580: Homework 1 1. a) L b ⇒ θ is small enough. Then, λ/2 = b sin θ ≈ bθ, δθ = 2θ ≈ λ/b b) i) 〈px〉before = 0 ii) ∆xbefore = ∞ iii) ∆xafter ≈ b iv) ∆px ' 2pz sin(δθ/2) ≈ pzδθ = h/b, so ∆px ∆xafter ≈ h & ~ 2. Let the four momentum for each object be P µγ , P µ γ′ , P µ e and P µ e′ P µγ = ( h λ , h λ , 0, 0) P µ γ′ = ( h λ′ , h λ′ cos θ, h λ′ sin θ, 0) P µe = (mc, 0, 0, 0) P µ e′ = ( √ m2c2 + p2, p cosφ, p sinφ, 0) Using the four momentum conservation, P µγ + P µ e = P µ γ′ + P µ e′ ,( P µγ + P µ e − P µ γ′ )2 = ( P µe′ )2 Remember (P µγ ) 2 = (P µγ′ ) 2 = 0 and (P µγ′ ) 2 = (P µe′ ) 2 = m2c2. Then, we have P µγ Peµ − P µe Pγ′µ − P µ γ′ Pγµ = 0 ⇒ hλmc− h λ′mc− h2 λλ′ (1− cos θ) = 0 Hence, λ′ − λ = hmc (1− cos θ) 3. Photon energy hf = hcλ ≈ 62eV . If the proton acquires kinetic energy from the incident photon, 1 2Mc 2(v/c)2 ≤ hf ⇒ v/c ≤ √ 2hf Mc2 = √ 124eV 938MeV  1. This means that we can regard this problem entirely in nonrelativistic regime. Then, energy conservation and momentum conservation tells us hf = hf ′ + P 2 2M + ∆Ee hf c = hf ′ c cos θ + P cosφ 0 = hf ′ c sin θ + P sinφ where, M denotes the mass of hydrogen and ∆Ee = En − E1. After little algebra, we find h∆f = −∆Ee − 1 2Mc2 ( (h∆f)2 + 2(hf)(hf ′)(1− cos θ) ) ∆f ≡ f ′ − f The (r.h.s) of the above equation is negligible since (h∆f) 2 Mc2 < (hf)(hf ′) Mc2 ∼ 10 −6eV , while ∆Ee ∼ 1eV . Therefore, ∆f = −En−E1h , the frequency of the photon does not depend on the scattering angle, though quantized. (Or, at least, the dependence is almost negligible). To a good approximation, we can obtain the following equation. h∆f = −∆Ee − (hf)2 2Mc2 sin2 θ ' −∆Ee If the electron is not excited at all in the scattering process, then we will find the Compton relation, which we have obtained in the previous problem, with electron’s mass replaced with the mass of hydro- gen. Please remember this approach is semiclassical, in that we did not quantize the electromagnetic field. 1 4. LQM1-1 For λ = 700nm, and E = 10J , the number of incident photon N = Ehc/λ = 10J 1240eV ·nm/700nm ≈ 3.5 × 1019. The probability of passing p = | cos(π/6)|2 = 34 . Since the event of passing follows the binomial distribution, the average energy Ē = N̄ hc/λ = 7.5J and the fluctuation in the number of photons passed is ∆N = √ Np(1− p). Then, the fluctuation in the energy ∆E = ∆N hcλ =√ Np(1− p)EN = √ p(1− p) E√ N ' 7.2× 10−11E = 7.2× 10−10 J . Note that ∆E/E ∝ 1√ N ∝ √ ~ 5. LQM1-2 a) |π4 〉 = 1√ 2 (|x〉+ |y〉), | 3π4 〉 = 1√ 2 (−|x〉+ |y〉) b) Let {|a〉, |b〈}, such that |a〉 = 1√ 2 (|x〉+eiθ|y〉) and |b〉 = 1√ 2 (−|x〉+eiθ|y〉), where, 0 < θ < π, θ 6= π/2. Note that basis is complete and orthonormal. 6. LQM1-3 a) Mxy→RL = ( 〈R|x〉 〈R|y〉 〈L|x〉 〈L|y〉 ) = 1√ 2 ( 1 −i 1 i ) b) MRL→ab = ( 〈R|a〉 〈R|b〉 〈L|a〉 〈L|b〉 ) = 1 2 ( 1 + ie−iθ 1− ie−iθ −1 + ie−iθ −1− ie−iθ ) c) Mxy→ab = ( 〈x|a〉 〈x|b〉 〈y|a〉 〈y|b〉 ) = 1√ 2 ( 1 e−iθ −1 e−iθ ) You can easily show Mxy→ab = MRL→ab ·Mxy→RL d) Let {|αi〉}, {|βj〉}, {|γk〉} be orthonormal complete basis sets.( Mα→β ) jk = 〈βj |αk〉( Mβ→γ ) ij = 〈βj |βj〉 For an arbitrary state |ψ〉, |ψ〉 = ∑ i |γj〉〈γj |ψ〉 = ∑ ijk |γi〉〈γi| (|βj〉〈βj |αk〉〈αk|) |ψ〉 = ∑ ik |γi〉 ∑ j 〈γi|βj〉 〈βj |αk〉  〈αk|ψ〉 From the above, we can read off( Mα→γ ) ij = ∑ j ( Mβ→γ ) ij ( Mα→β ) jk = ( Mβ→γ ·Mα→β ) ik ⇒ Mα→γ = Mβ→γ ·Mα→β 7. LQM1-4 a) 〈ψ|M†|φ〈= 〈Mψ|φ〈= 〈φ|Mψ〈∗= 〈φ|M |ψ〉∗. Or, 〈φ|M |ψ〉∗ = ∑ ij〈φ|ai〉∗〈ai|M |aj〉∗〈aj |ψ〉∗ =∑ ij〈ψ|aj〉〈aj |M†|ai〉〈ai|φ〉 = 〈ψ|M†|φ〉. b) For an arbitrary state |φ〉, 〈φ|M |ψ〉 = λ〈φ|ψ〉 ⇒ 〈ψ|M†|φ〉 = 〈φ|M |ψ〉∗ = λ∗〈ψ|φ〉. Therefore, 〈ψ|M† = λ∗〈ψ| 2