Quantenmechanik II Musterlösung 7. FS 07 Prof. Thomas Gehrmann Übung. [Auswahlregeln] Beweise die Drehimpulsauswahlregeln m = 0, ± und l = ± für Dipolübergänge in wasserstoffähnlichen Atomen unter Verwendung algebraischer Methoden, d.h. mittels der Kommutatoren [ [ [L z, x i ] und L, L, x]]. Lösung. The dipole matrix element is given by f x i and to calculate it we need to know the explicit form of the wavefunctions f and i. However, in order to find the selection rules we do not really need to calculate the full ME, but can determine the MEs for the commutators given on the sheet and use that f and i are eigenfunctions of L i and L, i.e. L i f, i = m f,i f, i L f, i = l f,i (l f,i + ) f, i. (L.) (L.) Using [L i, x j ] = ε ikl [x k p l, x j ] = ε ikl (x k [p l, x j ] [x j, x k ] p l ) = i ε ijk x k = i δ lj =0 (L.3) we find for the first of the given commutators: [L z, x ] = i x (m f m i ) f x i = f [L z, x ] i = i f x i (L.4) [L z, x ] = i x (m f m i ) f x i = f [L z, x ] i = i f x i (L.5) [L z, x 3 ] = 0 (m f m i ) f x 3 i = f [L z, x 3 ] i = 0 (L.6) Combining equations (L.4) and (L.5) we get f x i = (m f m i ) f x i f x i = (m f m i ) f x i, (L.7) (L.8) and from this obviously m m f m i = ± for transitions including non-vanishing f x i and f x i. If on the other hand f x 3 i 0, we see from equation (L.6) that m! = 0. In order to calculate as well as [ [ L, L, x]] we will make use of the identities (L.3), [L i, x j ] = i ε ijk x k = [L j, x i ], L i x i = x i L i [x i, L i ] = ε ijk x i x j p k = 0 (x i x j symmetric under i j) (L.9) =0,(L.3) and [ L, x] ( ) = i L x x L. (L.0) both. Note that f x i = 0 f x i = 0. Furthermore either f x, i or f x 3 i can be non-zero, but not
Identity (L.0) can be shown in the following way: [ L, x] i = L jl j x i x i L j L j = L j L j x i L j x i L j i ε ijk x k L j = L j L j x i L j L j x i i ε ijk L j x k i ε ijk x k L j ( ) = i L x x L where we simply commuted x i to the right in the second term using (L.3). We then have [ [ ([ L, L, x]] = i L, ] [ L x L, x ]) L i (L.0) i i, = i ([ L, L ] x + [ ] [ ] L L, x L, x [ L x L, ]) L =0 =0 = (L.0) ( L ( L x) L ( x L) ( L x) L + ( x L) L )i = ε ijk ε klm (L j L l x m L j x l L m + L l x m L j x l L m L j ) = (δ il δ jm δ im δ jl )(L j L l x m L j x l L m + L l x m L j x l L m L j ) i = (L j L i x j L j L j x i L j x i L j + L j x j L i + L i x j L j L j x i L j x i L j L j + x j L i L j ) = (L.3) (L j x j L i + i ε ijk L j x k L jx i L jx i i ε ijk L j x k x i L j i ε ijk x j L k x i L j + L i x j L j +i ε ijk x j L k ) = ( L x + x L ) i. (L.) With this we can finally calculate [ ]] f [ L, L, x i = f ( L L x L x L + x L L ) i = 4 (l f (l f + ) l i (l i + )) f x i, f ( L x + x L ) i = 4 (l f (l f + ) + l i (l i + )) f x i. Because of (L.) it is (L.)=(L.3) and therefore it must be (L.) (L.3) (l f (l f + ) l i (l i + )) = (l f (l f + ) + l i (l i + )). (L.4) Introducing the quantities L l f + l i and l l f l i one can rewrite this as L(L + ) l = L(L + ). (L.5)
Since l f, l i 0 we have L 0. In the case that L > 0 we immediately find l = ±. On the other hand if L = 0 that means that l f = l i = 0; but then both i and f are spherically symmetric states and f x i will vanish, i.e. the transition is forbidden. Übung. [Rayleigh-Streuung] Ausgangspunkt ist die in der Vorlesung hergeleitete Formel für Rayleigh-Streuung ( ) 4 dσ ω dω = r 0 = r ω 0ω 4 I N A αα N () mit I N = m (E N E A ) 3 A p ɛ α ( k ) N N p ɛ α ( k) A + A p ɛ α ( k) N N p ɛ α ( k ) A. () Betrachte als Anfangszustand A das H-Atom im Grundzustand (A=s). Unter der vereinfachenden Annahme, dass die Summe über die Zwischenzustände N durch N=p dominiert ist, berechne I N I p, N summiert über die Polarisationen, integriert über alle Einfallsrichtungen Ω( k) und gemittelt über die Ausfallsrichtungen Ω( k ), und vergleiche die hieraus resultierende Frequenz mit der Frequenz von Licht im sichtbaren Bereich. Lösung. we use The first thing we can do is to rewrite I N in terms of x instead of p. As in the lecture im [H 0, x i ] = i [p jp j, x i ] = i (p j[p j, x i ] + [p j, x i ]p j ) = i ( i δ ji)p j = p i (L.6) N p A = im N [H 0, x] A = im (E N E A ) N x A = imω NA N x A. (L.7) We then find for I N : I N = m = m A x ε α ω ( k ) N N x ε α ( k) A + A x ε α ( k) N N x ε α ( k ) A NA (L.8) ε α ω ( k ) i ε α ( k) j A x i N N x j A + A x j N N x i A. (L.9) NA We need to square this expression, sum over the polarisations, integrate over the ingoing momenta (diffuse scattering, i.e. photons from all directions) and average over the outgoing momenta 3
(there is no fixed reference direction in the problem, so we take the average). dω dω 4π (π) 3 (π) I N = α,α averaging m dω 4π ωna (π) 3 ε α ( k ) i ε α ( k dω ) l (π) 3 ε α( k) m ε α ( k) j α α A x i N N x j A + A x j N N x i A A x m N N x l A + A x l N N x m A. (L.0) Using the identities proven in exercise 4, series 6, dω ( ε α( k) i ε α ( k) j = dω δ ij k ik j k α ) = 8π 3 δ ij (L.) this reduces to 4π dω (π) 3 m 4π ω NA dω (π) 3 I N = α,α [ ] 8π (π) 3 A x i N N x j A + A x j N N x i A 3 A x j N N x i A + A x i N N x j A = m 4π ω NA [ 3π (L.) (L.3) (L.4) ] N x Σi A N x Σj A (L.5) + A x i N A x i N N x j A N x j A ],, (L.6) where we can identify N x Σi A = XNA as in the lecture. The subscript Σi indicates that we still have to sum over i. We know XNA for the p s transition: XNA = 3768 3 a 0 9683 δ m l,m δ m l s,m, (L.7) s where a 0 = /mcα. For the crossterm we have to think a bit: the ME A x i N in general is a complex number and we have to consider this when we conjugate some of the MEs in order to relate the crossterm to the known X NA. In our case A represents the s state, which is real. So any complex phase can only enter through the exp(im l ϕ) term in the angular part of N (look at the precise from of the spherical harmonics). But this means complex conjugation is the same as letting m l m l. We then can write A x i N A x i N = N x i A ml m l A x i N (L.8) = XNA ml m l = 3768 3 a 0 9683 δ m l,m δ m l s,m. s (L.9) 4
Using this we find I = 4π α,α dω( k) (π) 3 dω( k ) (π) 3 I p = 9π 5 m a4 0 ω NA ( ) 3768 (δ 9683 ml,m + δ l m l,m )δ m l s,m s (L.30) Summing over m l, m l, m s, m s yields a factor of. With ( ) ω NA = E p E s = mc α = mc 3 α 4 (L.3) we finally find... = ( 3768 9π 5 9683 = ( 3768 6π 5 9683 =: ω 4, ) a 4 0 m ( 8 3 ) 64 9 a4 0 c 4 α 4 ) m c α (L.3) (L.33) (L.34) with ω. 0 7 s λ.5 nm. (Here we used a 0 0.53 0 0 m, c = 3 0 8 m s, α = 37.) Visible light has a wavelength of λ 400 600 nm, which corresponds to a frequency of ω vis 3 4.7 0 5 s. Übung 3. [Übergangswahrscheinlichkeiten bei großen Zeiten] Die Zeitentwickulung eines Systems werde durch folgenden Hamiltonoperator bestimmt: H(t) = H 0 (t) + V (t), (3) wobei ( V (t) = V 0 exp t ) τ (τ > 0) (4) eine kleine Störung beschreibt und H 0 (t) der Hamiltonoperator des ungestörten Systems ist. Das System befinde sich im Grundzustand 0 von H 0 (t). Zeige, dass die Wahrscheinlichkeit dafür, das System für t im Zustand zu finden, gegeben ist durch V 0 0 P 0 (t ) = /τ + ( ε 0 ). (5) Dabei ist ε 0 = ω 0 die Energiedifferenz zwischen und 0. Erkläre dabei explizit, welche Annahmen in den einzelnen Rechenschritten getroffen werden. 5
Lösung. Since we assume V (t) to be a small perturbation, we can restrict ourselves to first order PT. The transition probability is then given by P a b (t) t dt e iω bat b V (t ) a For the given V (t) the Integral can easily be calculated: t 0 0. (L.35) (( dt exp iω ba ) )t b V 0 a = exp(iω bat) exp( t/τ) b V 0 a τ iω ba /τ For t this reduces to (the oscillatory part of the exponential function is suppressed by the exp( t/τ) part) b V 0 a /τ iω ba. (L.36) Therefore P 0 (t ) = V 0 0 /τ iω 0 = V 0 0 /τ + ( ε 0 ). (L.37) 6