Cơ học lượng tử 2.3

I. Time-dependent Perturbation Theory II. Fermi's Golden Rule Read Chapter 23 Sections 1, 2, 3 1" I. Time-dependent perturbation theory Consider again the effect of H=H 0 + H 1 If H 1 ≠ f (t) we can study H with the time-independent Schrodinger equation, then just multiply by e − iEt  later. But if H 1 = H 1 (t) , this causes H to be H(t), so we must solve the time-dependent Schrodinger equation from the beginning. Assume: (i) the eigenfunctions ϕ n and eigenvalues E n of H 0 are known: H 0 ϕ n = E n (0) ϕ n and ϕ n = e − iE n ( 0 ) t  ˆ ϕ n ˆ ϕ m ˆ ϕ n = δ mn ( ) (ii) the eigenfunctions ϕ n of H are not yet known, but they solve H Ψ n = i ∂ ∂t Ψ n (iii) The ϕ n can form a basis in which Ψ n can be expanded: Ψ n (t) = c n (t) ϕ n n ∑ c n (t) = ϕ n Ψ n Goal: find c n (t) To solve c n (t), plug Ψ n (t) directly into the time-dependent Schrodinger equation has known eigenfunctions ϕ n and eigenvalues E n arbitrary short-cut to solving the time-dependent Sch. Eq. this E must include perturbative corrections E (1) , E (2) , etc. this is what we just did for H 1 small this is what we will do now 2" H c n (t) ϕ n n ∑ = i ∂ ∂t c n (t) ϕ n n ∑ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ H 0 + H 1 ( ) c n (t) ϕ n n ∑ = i c n (t) ∂ ∂t ϕ n + ∂ ∂t c n (t) ( ) ϕ n { } n ∑ =i c n (t) −iE n ( 0 )  ( ) ϕ n + ∂ ∂t c n (t) ( ) ϕ n { } n ∑ Rewrite: c n (t) H 0 − E n (0) ( ) ϕ n n ∑ + c n (t)H 1 − i ∂ ∂t c n (t) ( ) ϕ n n ∑ = 0 = 0 because H 0 ϕ n = E n (0) ϕ n multiply by ϕ k c n (t) ϕ k H 1 ϕ n − i ∂ ∂t c n (t) ϕ k ϕ n ⎡ ⎣ ⎤ ⎦ n ∑ = 0 Plug in: ϕ n = e − iE n ( 0 ) t  ˆ ϕ n ϕ k = e − iE k ( 0 ) t  ˆ ϕ k ϕ k = e + iE k ( 0 ) t  ˆ ϕ k c n (t) ˆ ϕ k H 1 ˆ ϕ n e − i E k ( 0 ) −E n ( 0 ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ t  n ∑ − i ∂ ∂t c n (t) ˆ ϕ k ˆ ϕ n n ∑ e − i E k ( 0 ) −E n ( 0 ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ t  = 0 call this "H 1 kn " call E k ( 0 ) − E n ( 0 ) ( )  " ω kn " 3" 4" I. Time-dependent Perturbation Theory (continued) II. Fermi's Golden Rule III. The Variational Method Read Chapter 23 Section 1, except Traut from com to lab 5" c n H 1 kn e i ω kn t − n ∑ i ∂ ∂t c k = 0 This is an exact equation that relates each coefficients (the c k 'th) to all the other coefficients (the c n ) n ∑ It is impossible to solve analytically for arbitrary H 1 If H 1 is "small", assume: (1) The c n (t) are almost constants, not really functions of t (2) @ t=0, the state of the system is known, so one coefficient (call is c j (t = 0)) = 1 all the rest =0 (3) Then since the c n are constants, c j remains ≈ 1 even at later t, and the other c n≠ j remain ≈ 0 So in this approximation the equation becomes H 1 kj e i ω kj t − i ∂ ∂t c k ≈ 0 Solve it: c k = δ kj −i  d ′ t e i E k ( 0) −E j ( 0) t ⎛ ⎝ ⎜ ⎞ ⎠ ⎟  ˆ ϕ k H 1 ˆ ϕ j 0 t ∫ So if the system began with c j = 1, c n≠ j = 0 @ t=0, the probability that the system is in state k @ t= ′ t is c k ( ′ t ) 2 6" Example solution of c k if H 1 (t) = 0 t<0 consant if t ≥ 0 ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ then c k = δ kj − i  H 1 kj d ′ t e i ω kj ′ t 0 t ∫ = δ kj − i  H 1 kj 1 i ω kj ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ e i ω kj ′ t 0 t = δ kj − i  H 1 kj i ω kj e i ω kj t −1 ( ) c k = δ kj + H 1 kj  ω kj 1− e i ω kj t ( ) Note k indexes the coefficient being examined (c k = ϕ k Ψ ) while j indexed the one coefficient (c j ) which has non-zero @ t=0 II. Fermi's Golden Rule Recall c k when k ≠ j [where initially c j was the only non-zero amplitude] for H 1 (t) = H (t>0) 0 (t<0) c k ≠ j = H 1 kj  ω kj 1− e −i ω kj t ( ) 7" Probability (observing state k @ time t) = c k 2 = H 1 kj  ω kj 2 1− e −i ω kj t ( ) 1− e + i ω kj t ( ) = H 1 kj  ω kj 2 1− e + i ω kj t − e −i ω kj t + 1 ( ) = H 1 kj  ω kj 2 2 − 2 e + i ω kj t + e −i ω kj t 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ = H 1 kj  ω kj 2 2 1− cos ω t ( ) { } = H 1 kj  ω kj 2 4 1 2 − 1 2 cos ω t ( ) { } use cos2x=cos 2 x − sin 2 x = (1− sin 2 x) − sin 2 x = 1− 2sin 2 x so 1 2 cos2x = 1 2 − sin 2 x so 1 2 − 1 2 cos2x = sin 2 x Probability of transition from ϕ j → ϕ k = 2H 1 kj E k (0) − E j (0) 2 sin 2 ω kj t 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Now we want to know Probability (system that begins in state ϕ j makes a transistion to any other states, given infinite time) 8" Probability (system that begins in state ϕ j makes a transistion to any other states, given infinite time) = Probability (j→ k) k ∑ = 2H 1 kj E k (0) − E j (0) 2 sin 2 ω kj t 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ k ∑ If states ϕ k are continuously distributed (i.e. scattering states) rather than discretely distributed (i.e. bound states), then ∑ → dn ∫ = dn k 2H 1 kj E k (0) − E j (0) 2 sin 2 ω kj t 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ∫ If there is a degeneracy @ E=E k 0 , so there is a density of states there, ρ E k 0 ( ) = dn dE k 0 , then dn= ρ ⋅dE = d E k 0 ρ E k 0 ( ) 2H 1 kj E k (0) − E j (0) 2 sin 2 E k (0) − E j (0) ( ) t 2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ∫ Let x ≡ E k (0) − E j (0) ( ) t 2 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ E k (0) − E j (0) 2 = 4 2 x 2 t 2 and dx=dE k (0) ⋅ t 2 9" Prob tot = 2 t dx ρ x ( ) 4 H 1 2 t 2 4 2 x 2 sin 2 x ∫ consider the case where ρ x ( ) = ρ , average density over all final states, a constant and H 1 2 = H 1 2 , average value of ϕ k H ϕ j over all ϕ k , also a constant Then Prob tot = 2t  H 1 2 ρ dx sin 2 x x 2 −∞ +∞ ∫ π Probability(system transitions out of ϕ j due to H 1 ) = 2 π t  H 1 2 ρ So transition rate = dProb dt = 2 π  H 1 2 ρ this is Fermi's Golden Rule we worked this out for H 1 = const if t > 0 0 if t < 0 but it is true for any H 1 10" I. The Variational Method II. Intro to Scattering Theory III. Probability current of scattered particles . Ψ 2b π ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1 2 dxe −bx 2 − 2 2m d 2 dx 2 + 1 2 m ω 2 x 2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ −∞ +∞ ∫ e −bx 2 H =  2 b 2m + m ω 2 8b Minimize: 0= ∂ ∂b H =  2 2m − 8m ω 2 64b 2 =  2 2m − m ω 2 8b 2 So 8b 2 = 2m 2 ω 2  2 b = m ω 2 Plug. e −i ω kj t 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ = H 1 kj  ω kj 2 2 1− cos ω t ( ) { } = H 1 kj  ω kj 2 4 1 2 − 1 2 cos ω t ( ) { } use cos2x=cos 2 x − sin 2 x = (1− sin 2 x) − sin 2 x = 1− 2sin 2 x so 1 2 cos2x = 1 2 − sin 2 x so 1 2 − 1 2 cos2x. ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ =  2mi f 2 r 3 +ikr −1+ ikr + 1 ⎡ ⎣ ⎤ ⎦ =  2mi f 2 r 3 2ikr ⎡ ⎣ ⎤ ⎦ = k m f 2 r 2 So the total # particles at any particular radius r is  J scat ⋅ dArea   = ˆ r ⋅  Jr 2 dΩ = k f 2 m r 2 r 2 dΩ