Solutions to Physics Homework Problems: 325, #6 and #16, Assignments of Quantum Mechanics

Download Solutions to Physics Homework Problems: 325, #6 and #16 and more Assignments Quantum Mechanics in PDF only on Docsity! 1 Physics 325 Solution to Homework Problems #6 Winter 2004 1. Your textbook Prob. 9.5 We start with: ċa = 1 ih̄ H ′abe −iω0tcb ċb = 1 ih̄ H ′bae iω0tca with ca(0) = a, cb(0) = b For the first order result, we substitute ci(0) into the right side of the equations: ċ(1)a ≈ 1 ih̄ H ′abe −iω0tcb(0) = b ih̄ H ′abe −iω0t => c(1)a (t) ≈ b ih̄ ∫ t 0 H ′ab(t ′)e−iω0t ′ dt′ Similarly c(1)b (t) ≈ a ih̄ ∫ t 0 H ′ba(t ′)eiω0t ′ dt′ For second order, we use the first order results for ci(t) in the right hand side of the exact equations: ċ(2)a (t) ≈ 1 ih̄ H ′abe −iω0tc (1) b = 1 ih̄ H ′abe −iω0t a ih̄ ∫ t 0 H ′ba(t ′)eiω0t ′ dt′ = − a h̄2 H ′abe −iω0t ∫ t 0 H ′ba(t ′)eiω0t ′ dt′ => c(2)a (t) ≈ − a h̄2 ∫ t 0 H ′ab(t ′)e−iω0t ′ [ ∫ t′ 0 H ′ba(t ′′)eiω0t ′′ dt′′ ] dt′ Similarly, c(2)b ≈ − b h̄2 ∫ t 0 H ′ba(t ′)eiω0t ′ [ ∫ t′ 0 H ′ab(t ′′)e−iω0t ′′ dt′′ ] dt′ To second order, ca(t) ≈ a+ c(1)a + c(2)a , cb(t) ≈ b+ c(1)b + c (2) b . 2. Your textbook Prob. 9.16 (a) We start with the exact equation, Eq. 9.82: ċm = 1 ih̄ ∑ n cnH ′ mne iωmnt where ωmn = (Em −En)/h̄ And next compute H ′mn where H ′ = V0(t): H ′mn = 〈ψm(x)|V0(t)|ψn(x)〉 = V0(t)〈ψm(x)|ψn(x)〉 = V0(t)δmn because H ′ is constant over x, and the energy eigenstates are orthonormal. Therefore, ċm = 1 ih̄ cmH ′ mm = 1 ih̄ cmV0(t) Because we started in state N , for m 6= N , cm(0) = 0 and ċm = 0, leaving cm(t) = 0 for all t. Therefore no transitions take place and the only non-vanishing equation is then: ċN = 1 ih̄ cNV0(t); d(ln cN) dt = ċN cN = 1 ih̄ V0(t) => ln cN (t) = 1 ih̄ ∫ t 0 V0(t′)dt′ => cN (t) = e 1 ih̄ ∫ t 0 V0(t ′)dt′ Therefore only the phase of cN changes. The total phase change is : φ(T ) = 1 ih̄ ∫ T 0 V0(t′)dt′