Physics 505 Homework No. 11 Solutions S11-1

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1 Physis 55 Homework No 11 s S This problem is from the My, 24 Prelims Hydrogen moleule Consider the neutrl hydrogen moleule, H 2 Write down the Hmiltonin keeping only the kineti energy terms nd the Coulomb intertions of ll the onstituents nd omitting terms whih use fine nd hyperfine struture Clerly, the Hmiltonin ontins four kineti energy terms nd six Coulomb terms We let indies 1 nd 2 be the protons nd 3 nd 4 be the eletrons H = i p 2 i 2m i i,j,i j q i q j r i r j However, this is probbly not the best wy to write the Hmiltonin The eletrons, 2 times less mssive thn the protons, n djust their onfigurtion to the onfigurtion of the protons very rpidly So better ide is to write the Hmiltonin for the eletrons seprtely H e = p2 3 2m e p2 4 2m e e 2 r 3 r 1 e 2 r 3 r 2 e 2 r 4 r 1 e 2 r 4 r 2 e 2 r 4 r 3 Assuming the protons re t the fixed positions r 1, nd r 2, the ground stte of the eletrons n be found with totl energy U r 1 r 2 Then, the Hmiltonin for the protons my be pproximted by H p = p2 1 2m p p2 2 2m p e 2 r 1 r 2 U r 1 r 2 Some things to notie re tht when the protons re very fr prt, the ground stte of the eletrons is tht of two Hydrogen toms with binding energy of ev When the protons re right on top of eh other the ground stte of the eletrons is tht of Helium with binding energy of 79 ev We expet tht the eletron ground stte energy U r 1 r 2 monotonilly dereses s the proton seprtion dereses On the other hnd, the Coulomb energy of the protons inreses s their seprtion dereses At some seprtion r, the eletron energy plus proton Coulomb energy is minimum This will be the equilibrium seprtion of the protons We n mke Tylor expnsion of the energy round the Equilibrium position nd the result is hrmoni osilltor potentil So the Hmiltonin of the protons n be written H p = p2 1 2m p p2 2 2m p 1 2 k r 1 r 2 r 2 As rough guesses, we might tke r, the Bohr rdius, nd k e 2 / 3 We n seprte off the enter of mss motion whih yields the following Hmiltonin for the motion of the protons, where r is the distne between the two protons H p = p2 2µ 1 2 kr r 2, Copyright 212, Edwrd J Groth

2 Physis 55 Homework No 11 s S11-2 nd µ = m p /2 is the redued mss of the protons Finlly, L 2 ommutes with this Hmiltonin, so we write it in terms of the rdil momentum, H p = p2 r 2µ h2 ll 1 2µr kr r 2 Other onsidertions: The Hmiltonin for the entire system is even so ll nondegenerte eigensttes re even or odd nd degenerte eigensttes my be hosen to be even or odd In prtiulr the eletron ground stte ssumed the eletrons were in n even onfigurtion, so they must be in singlet spin onfigurtion Wht is the degenery of the ground stte? Give ll quntum numbers nd symmetries of the ground sttes, inluding the eletron nd proton spin degrees of freedom Eletrons re in singlet stte, S e = The protons re in singlet stte, S p = The degenery of the ground stte is 1 The ngulr momentum quntum number for the protons is l =, nd the vibrtionl quntum number is ν = This mkes the sptil prt of the stte symmetri under interhnge of the protons, whih is why they must be in singlet stte The eletons re in hydrogen-like ground stte with orbitl ngulr momentum nd prinipl quntum number n = 1 This mkes the sptil prt of the eletron wve funtion symmetri whih is why they must be in singlet stte b Wht is the degenery nd wht re ll the quntum numbers of the first exited stte of this H 2 moleule? Explin The first exited stte will involve hnge in the ngulr momentum of the protons We n see this s follows Eletroni exittions re the order of e 2 / where = h 2 /me 2 is the Bohr rdius Vibrtionl exittions re the order of hω, where ω = k/µ Plugging in our guess for k, we find tht the exittion energy of vibrtionl mode is e 2 / m e /µ or bout 1/3 times the eletroni exittions The rottionl exittions will be the order of h 2 /2µr 2 = h2 /2µ 2 = e 2 /2m e µ or nother ftor of 1/3 down from the vibrtionl exittions Sine the sptil stte must be even, the first exited stte nnot be l = 1, but must be l = 2 whih hs degenery of 5 The sptil stte must be even beuse we ssume the nuler spins sty in the singlet stte In ft, the nuler spins ouple very wekly to nything else, so trnsitions involving hnge of the nuler spin from singlet to triplet nd vie vers re rre If the nuler spins were in the triplet stte, then the sptil stte would hve to be nti-symmetri nd the lowest energy stte is the l = 1 stte Although simple rditive trnsitions do not ouple this stte to the ground stte, one ould tke this s the first exited stte Then Copyright 212, Edwrd J Groth

3 Physis 55 Homework No 11 s S11-3 the degenery is 9, 3 for l = 1 times 3 for the triplet spin stte The energy is the sme order of mgnitude s in prt below ll 1 = 2 insted of 6 Wht is the energy differene between ground nd first exited sttes? Estimte it first through formul, in terms of properties of the moleule s ground stte, nd then in eletron-volts ev From the solution to prt b, the formul is ll 1e 2 /2m e /m p /2 or bout 8 ev 2 With Hund s rules, determine S, L, nd J for nitrogen whose onfigurtion is 1s 2 2s 2 2p 3 Rule number 1 sys, S should be mximl whih mens S = 3/2 Rule number 2 sys, L should be mximl However, the spin stte is symmetri, so the sptil stte must be ntisymmetri With three l = 1 sttes, the possibilities re L =, 1, 2, 3 The esiest wy to see whih re llowed is to onsider m l for eh eletron Cn we find n ntisymmetri stte tht dds up to m L =, 1, 2, 3? So, we ll write out the sttes where the position from left to write orresponds to eletrons 1, 2, 3 For exmple 1, 1, 1 mens eletron 1 hs m l = 1, eletron 2 hs m l = 1, nd eletron 3 hs m l = 1 We need to ntisymmetrize this stte; we tke ll six permuttions with the pproprite sign: 1, 1, 1 1, 1, 1 1, 1, 1 1, 1, 1 1, 1, 1 1, 1, 1 =, so we n t hve L = 3 How bout L = 2? To get m L = 2, we would hve to dd up two m l = 1 long with n m l = The stte would be 1, 1, 1,, 1, 1, 1, 1, 1 1,, 1 1, 1, =, so we n t hve L = 2 How bout L = 1? There re two wys we ould ome up with m L = 1 We ould dd up two m l = 1 long with n m l = or we ould dd n m l = 1 long with two m l = In either se, two would be the sme nd there s no wy to form non-vnishing ntisymmetrized stte How bout L =? If we n t hve this, then nitrogen n t exist! There re two wys to ome up with m L = One wy is ll three m l = This n t led to n ntisymmetri stte Another wy is m l = 1, m l = nd m l = 1, 1,, 1 1, 1, 1, 1, 1,, 1, 1, 1, 1, 1, so L = works! Then J = S nd the term is 4 S 3/2 Copyright 212, Edwrd J Groth

4 Physis 55 Homework No 11 s S Two prtiles of mss m re onfined to retngulr box of sides < b < They re in the lowest energy stte omptible with the onditions in the ses below For eh of these ses, determine the lowest energy stte nd its energy nd lso use first order perturbtion theory to determine the orretion to the energy if there is n intertion between the prtiles of the form V = V bδ 3 r 1 r 2 Two non-identil prtiles The single prtile wve funtions will be of the form n, n b, n = 8 b sin πn x sin πn by sin πn z b with ll quntum numbers greter thn nd with energy E n,n b,n = h2 π 2 n 2m 2 nb b 2 n 2 The lowest energy single prtile stte is 1, 1, 1 with energy 1 E 1,1,1 = h2 π 2 2 2m 2 1 b 2 1 Sine is the lrgest side of the box, the next lowest energy single prtile stte is 1, 1, 2 with energy 1 E 1,1,2 = h2 π m b We will write two prtile sttes s n, n b, n n, n b, n where the first ket refers to prtile 1 nd the seond to prtile 2 For the se t hnd, non-identil prtiles, both prtiles n be pled in the lowest energy single prtile stte, so the stte ψ = 1, 1, 1 1, 1, 1,, with energy 1 E 1,1,1,1,1,1 = 2E 1,1,1 = 2 h2 π 2 2 2m 2 1 b 2 1 Copyright 212, Edwrd J Groth

5 Physis 55 Homework No 11 s S11-5 like To tke ount of the ontt intertion we need the produt of three integrls x = 4 dx 2 1 = 4 dx sin 2 πn x dx 2 sin 2 πn x 1 sin 2 πn x sin 2 πn x 2 δx 1 x 2 = δ n,n For this se, n = n = 1, nd the integrl is 3/2 Putting it ll together, we find b Two identil prtiles of spin E = V x y z = 27 8 V In this se, the wve funtion must be symmetri under the exhnge of the two prtiles Sine spin prtiles must neessrily be in symmetri spin stte, the sptil stte must be symmetri The stte desribed in prt is symmetri so it works for this prt The stte, the energy, nd the ontt intertion energy re the sme s for prt Two identil prtiles of spin 1/2 in the singlet stte Now the wve funtion must be ntisymmetri Sine the singlet stte is ntisymmetri, the sptil stte must be symmetri So the wve funtion, energy, nd ontt intertion energy re gin, the sme s in prt d Two identil prtiles of spin 1/2 in the triplet stte The wve funtion must be ntisymmetri under exhnge of the two prtiles, the spin stte is symmetri, so the sptil stte must be ntisymmetri The lowest energy stte is ψ = 1 2 1, 1, 1 1, 1, 2 1, 1, 2 1, 1, 1, with energy E 1,1,1,1,1,2 = E 1,1,1 E 1,1,2 = h2 π 2 2 2m b 2 1 Copyright 212, Edwrd J Groth

6 Physis 55 Homework No 11 s S11-6 Now we need the mtrix element of the perturbtion for this stte The integrls for x nd y will be the sme s before The integrl for z is, z = = 2 π = π dz 1 dz 2 sin πz 1 sin 2πz 2 dzsin z sin 2z sin 2z sin z sin 2πz 1 sin πz 2 2 δz 1 z 2 Not surprisingly, the prtiles in the triplet stte re prevented from being right on top of eh other by the Puli exlusion priniple, so the ontt intertion is zero 4 Bet dey rerrngement Bsed on problem in Shwbl In β-dey, the nuler hrge number Z of Z 1 times ionized tom so it s hydrogen-like! hnges suddenly to Z 1 Clulte the probbilities for the trnsition of the eletron to the 2s or 3s sttes given tht the eletron ws in the ground stte before the β-dey Do you hve ny omments on energy onservtion? We use the sudden pproximtion, nd we need to lulte the overlp integrl between the 1s stte of n eletron orbiting nuleus with hrge Ze nd the 2s nd 3s sttes of n eletron orbiting nuleus of hrge Z 1e We ignore everything but the Coulomb intertion The ngulr prt of the wve funtion is Y in ll ses, so we don t need to worry bout it Also, we n only wind up in n s-stte sine Y is orthogonl to ll higher ngulr momentum sttes! The rdil prt of the wve funtion before the dey nd the two hoies post dey re 3/2 Z R 1 = 2 e Zr/ 3/2 Z 1 R 2 = Z 1 R 3 = 2 3 Z 1r e Z 1r/2 2 3/2 2Z 1r 1 3 2Z 1r e Z 1r/3 Copyright 212, Edwrd J Groth

7 Physis 55 Homework No 11 s S11-7 The mplitude to go to the 2s-stte is 2 1 = 2 1/2 ZZ 1 3/2 1 3 = 2 1/2 ZZ 1 3/2 1 3 = 2 7/2ZZ 13/2 3Z 1 3 = 2 7/2ZZ 13/2 3Z 1 3 = 2 11/2ZZ 13/2 3Z 1 4 x 2 dx r 2 dr r 2 dr Z 1 3Z 1 Similrly, the mplitude to go to the 3s-stte is, 3 1 = 4 ZZ 13/2 3 3/2 3 = 4 33/2 ZZ 1 3/2 4Z 1 3 = 8 33/2 ZZ 1 3/2 4Z 1 3 r 2 dr Z 1r 2 Z 1r 2 Z 1x 3Z 1 2Z 1r 1 3 = 8 35/2 ZZ 1 3/2 2Z 1 4Z 1 5 So, the trnsition probbilities re e Z 1r/2 Zr/ e 3Z 1r/2 e x 2Z 1r2 e 4Z 1r/ x 2 2Z 1x dx 1 2Z 12 x 2 4Z 1 34Z 1 2 6Z 1 8Z Z 1 4Z 1 2 P 1 2 = 211 ZZ 1 3 3Z 1 8 P 1 3 = ZZ 1 3 2Z 1 2 4Z 1 1 As n exmple, we might onsider the dey of tritium, 3 H 3 He β ν Tritium is just extr-hevy hydrogen, so Z = 1 nd, P 1 2 = 1 4 = 25 P 1 3 = = e x With respet to the onservtion of energy question: Agin, onsider tritium for onrete exmple Before the dey, the binding energy of the eletron ws 136 ev After the dey, the eletron in the 2s or 3s sttes hs energy 136 ev n ident of our exmple or 6 ev, respetively So, in the dey with the eletron going into the 3s stte, 76 ev hs pprently been gined by the exited Helium tom left behind Presumbly, this mens the dey produts: the β, the ν, nd the reoiling nuleus hve 76 ev less kineti energy thn they otherwise would Copyright 212, Edwrd J Groth

8 Physis 55 Homework No 11 s S Coulomb exittion Somewht bsed on problem from Shwbl Consider hydrogen in its ground stte t t = It s ted on by n eletri field in the z-diretion of the form Et = E e z 1 t 2 /τ 2 This field n be represented by the potentil φ = Etz This is n pproximtion to wht hppens when hrge prtile psses nerby If it s not reltivisti, we n ignore its mgneti field Wht is the probbility tht the eletron winds up in the 2p stte t t =? One you ve produed formul, mke some resonble ssumptions nd lulte numeril estimte of the probbility to see if Coulomb ollision is resonble wy to exite n tom With smll extension letting the perturbtion strt t t = rther thn t = to wht ws disussed in leture, the mplitude to go from stte m to stte n is n, m, = 1 dt e iω nmt n V t m, i h where ω nm = E n E m / h nd where the perturbing potentil is V t = eetz The potentil n be thought of s ting on the eletron Now, z = r os θ The 2p sttes involve Y 1,1, Y 1,, nd Y 1, 1 For the m = ±1 sttes, the integrl over φ will give So, we need to evlute sin θ dθ dφ Y 1, θ, φ os θ Y,θ, φ = 1 3 The rdil prt requires the evlution of r 2 dr R 21 r r R 1, r = 215/2 3 9/2 Next we need to evlute the time integrl whih is relly the Fourier trnsform of the potentil t the frequeny orresponding to the energy differene dt eiω nmt ee 1 t 2 /τ 2 = πee τe ω nm τ We multiply the three results together to get the trnsition mplitude, nd the trnsition probbility is n, m, = 215/2 π 3 5 i ee τe ω nm τ h, P mn = 215 π ee τ 2 e 2 ω nm τ h 2 Copyright 212, Edwrd J Groth

9 Physis 55 Homework No 11 s S11-9 It will be interesting to mke numeril estimte of this probbility However, there re lot of prmeters! To proeed, we note tht τ exp ω nm τ is mximum when τ = 1/ω nm So, let s ssume this Also, suppose the field is used by one unit of hrge e pssing t distne of just grzing the tom We note tht hω nm = 3/4136 ev, so τ = s The veloity of the inident prtile is the order of /τ whih works out to bout 3, so it s sfely non-reltivisti Sine the prtile pssed t, the pek field is E = e/ 2 Plugging it ll in, P nm = 215 π 2 e e 2 / 2 hω nm 2 = 215 π 2 exp e 2 / 2 3/4e 2 /2 2 = 221 π 2 exp = 53, just bit bigger thn 1! Not to worry The min reson to do this numeril lultion ws to see if this is vible wy to exite n tom The nswer is yes! For the prmeters we ve hosen it ppers tht the ollision is too violent for perturbtion theory However, it the olliding prtile pssed t 4 rther thn the probbility would be 33, perfetly resonble Copyright 212, Edwrd J Groth