Three Dimensional systems. In three dimensions the time-independent Schrödinger Equation becomes

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1 Three Dimensiona systems In three dimensions the time-independent Schrödinger Equation becomes ] [ m r + V r) ψr) = Eψr), r = x + x + x. The easiest case one can imagine in 3D is when the potentia energy is the sum of three independent terms V r) = V x x) + V y y) + V z z). In this case we seek the soution as the product of three independent functions ψr) = Xx)Y y)zz). Substituting this anzats to the Schrödinger Equation and then dividing it by ψ we find [ 1 1 ] Xx) m X x) + V x x) + 1 [ 1 ] Y y) m Y y) + V y y) + 1 [ 1 ] Zz) m Z z) + V z z) Each of the three terms on the.h.s. is a function of a different variabe, either x, or y, or z. This is ony possibe if each of them separatey is a constant equa to E x), E y), and E z), respectivey. In other words, we are deaing with three separate 1D probems, and the tota energy is the sum of three independent terms E = E x) + E y) + E z). The two cases which fa in this category are the paraboic potentias, V = m/) 3 α=1 ω αrα, and the square box, V = α V α sq.box) r α ). [To simpify formuae, it makes sense to use notations x = r 1, y = r, and z = r 3 from time to time.] Soving them is no different from soving onedimensiona probems one-by-one. As an appication, consider partices confined in the infinite 3D square we with inear sizes x 0, L x ), y 0, L y ), z 0, L z ). We immediatey jump to the fina answers for the wavefunctions ψ n1,n,n 3 = ) πnα r α sin, n α = 1,, 3,... L α α L α and spectrum beow n = n 1, n, n 3 ) is a vector with integer components) E n = k m, = E k α = π L α n α. 1) Let us pretend that this is a mode of some compex system where we repace interactions between the partices with the mean-fied potentia which hods them together inside the voume V = L x L y L z. Let N of the partices are fermions with ight mass m and spin 1/. According to the Paui principe each energy eve can be occupied ony with two partices, one with spin up, and the other with spin down. We then ask the same question as before: What is the energy of the ast occupied energy eve, or Fermi energy ɛ F, and tota ground state energy of a fermions in the box?. We proceed aong the same ines as in the discussion of 1D rings, and appy the same approximations when repacing sums over integer index with momentum integras for smooth functions. First, the condition of accommodating a partices is now a factor of is for spin index) N = E n ɛ F n 1,n,n 3 =1 En ɛf 1 ) θn α ) dn, α

2 where the θx) function is fitering positive vaues of x ony; it is unity for x > 0 and zero otherwise. We may aow for both positive and negative vaues of n α in this integra, but then it has to be divided by 8 to have the same vaue N = 1 8 En ɛ F dn = V k /m ɛ F dk π ) 3. In the ast reation we went from integration over n to integration over k the two are ineary reated to each other, see Eq. 1). Taking an integra we get n f = N V = 4π) π ) 3 k 3 F 3, ɛ F = k F m, which defines the reation between the Fermi-momentum and fermion density k F = 3π n f ) 1/3. This reation happens to be protected by theorems to be interaction independent. Aso, in the fina form, this expression is vaid for any shape of the we as ong as the number of fermions is very arge. States with energies E k = ɛ F form a Fermi-surface in momentum space. [It is a sphere here, but in condensed matter it can have other shapes, incuding pumber s nightmare ones with numerous vaeys, pockets, necks ). The tota energy of a fermions siting beow the Fermi surface is k kf k E tot = V m The average energy per partice is then dk π ) 3 = V π 3 E tot /N = 3k F 10m = 3 5 ɛ F kf 5 10m = 3 10mV /3 3π N) /3 N. These are the core resuts for the Fermi gas mode, no matter in what fied of physics it happens to be a reasonabe approximation. In astrophysics, stabiity of white dwarf and neutron stars is based on counter-acting the gravity pu with the Fermi pressure P F = de tot /dv = 3π ) /3 N/V ) 5/3 5m In white dwarf stars the Fermi pressure is mosty due to eectrons, in neutron stars, it is due to neutrons. In both cases this Fermi pressure has to baance the gravity pressure P G = de R gr dv, E G4πR dr ρ)4πr 3 ρ/3) gr = = 3 GM 0 R 5 R = 3 5 ) 4π 1/3 GM 3 V 1/3, where R is the star radius, ρ is its mass density, and M = ρ4πr 3 /3 = ρv is its tota mass. With this P G = 1 ) 4π 1/3 GM 5 3 V 4/3.

3 From P F + P G = 0 one determines stabiity conditions, i.e. the radius of the star given its fermion number N and tota mass M. In non-reativistic case P F increases faster than P G as voume shrinks and thus there is aways a soution for the radius. When k F /m becomes comparabe with the veocity of ight one has to start using the reativistic dispersion reation which is ess steep and thus eads to smaer Fermi-pressure for the same partice number and voume. This changes the aw from P F V 5/3 to P F V 4/3 in the reativistic imit and for massive enough stars Chandrasekhar imit) eads to their coapse and back hoe physics. Probem 44. Spin and energy eves in nucear physics a). Consider a heavy nuceus consisting of Z = N p protons and N = N n neutrons so that the tota atomic number is A = Z + N. If the nuceons each have an eementary impenetrabe voume V 0 and are packed so that they are just touching, show that the voume of such a nuceus wi go as V V 0 A, and thus the radius of the nuceus wi be R = r 0 A 1/3. with r 0 = fm this aw works we for most nucei b). Instead of deaing with interactions between nuceons, mode this system as A nuceons confined within the 3D infinite we of voume V = 4πR 3 /3. Using expressions from the notes find the tota ground state energy of a system of A = Z + N nuceons, i.e. EZ, N), recaing that both protons and neutrons are spin-1/ fermions. For simpicity assume that protons and neutrons have the same mass, mc = 940 Mev. c). For fixed A, minimize this energy, and show that equa numbers of neutrons and protons are favored. d). Assuming that N Z A, show that the ground state energy can be approximated by an expression of the form E = E min + E sym N Z) /A +..., and find an expression for E sym. Using the vaues above, show that the numerica vaue of E sym is roughy 1 Mev. Note: This contribution to the tota nucear energy or rest mass) is often caed the symmetry contribution and is part of the we-known semi-empirica formua of nucear physics. We now turn to another important case of sphericay symmetric potentias V r) = V r). To dea with it we introduce spherica coordinates z = r cos θ, x = r sin θ cos ϕ, y = r sin θ sin ϕ. The Lapacian operator in spherica coordinates takes the form r = 1 r [ r r r + 1 sin θ θ sin θ θ + 1 sin θ ] ϕ We wi seek the soution in the form of a product ψr) = Rr)χθ)Φϕ). Substituting this form into the Schrödinger Equation and then dividing Schrödinger Equation by ψ we get [ ] 1 Mr Rr) r [r R 1 r)] + sin θχθ) θ [sin θ χ 1 θ)] + sin θ Φϕ) Φ ϕ) + V r) = E. ) Next we mutipy it by r sin θ M sin θ Rr) r [r R r)] + r sin θ[v r) E] sin θ M χθ) θ [sin θ χ θ)] M Φ ϕ) Φϕ) = 0. 3) 3

4 Again, we observe that the first two terms are independent of ϕ and thus the ast term must be a constant Φ ϕ) = const 4) M Φϕ) The soutions of this simpe equation are simpe exponentia functions e ikϕ subject to periodic boundary condition for the ϕ variabe since point ϕ + π is the same as ϕ. Simiary to our soution for the 1D ring, we must then have k = m=integer, i.e. 1 Φ m ϕ) = π eimϕ, m = 0, ±1, ±,... Correspondingy, the vaue of the constant in Eq. 4) is m /M the notation for the partice mass was changed to avoid confusion with the quantum number m for the azimutha motion). With one variabe ess, we return to Eq. 3) and divide it by sin θ to get [ M Rr) ] [ r [r R r)] + r [V r) E] M ] 1 sin θχθ) θ [sin θ χ θ)] m sin = 0. 5) θ Same story as before: the first square bracket depends ony on r whie the second depends ony on θ. Thus both terms have to be equa to the same constant. Let us denote it by λ/m and try to sove first 1 sin θχθ) θ [sin θ χ θ)] m sin = λ. 6) θ It can be simpified by introducing z = cos θ. The derivatives transform as d/dθ)f z) = sin θd/dz)f z) where F z) = χθ). Now we have to sove ) λ m 1 z F + d 1 z )F ) = 0. 7) dz Since θ 0, π) we have z 1, 1) with 1/1 z ) diverging at the interva boundary. Hence, it is expected that F z) tends to zero at the boundary. To understand how, we ook at the equation in the vicinity of the z = ±1 points. Let x = 1 z and fx) = F z) with d/dz) = dx/dz)d/dx) = zd/dx) d/dx). Then, taking care of diverging terms, m x fx) + 4 d xf ) 0 dx whose soution is fx) = x m / checked by eementary substitution and differentiation. This suggests that we use substitution F z) = 1 z ) m / Gz) in Eq. 7) with expectations that Gz) wi be a reguar function at z ±1. The next step is a straightforward substitution with a ot of carefu differentiation: 1 z ) d dz 1 z ) m / Gz) = z m 1 z ) m / Gz) + 1 z ) m /+1 G z), d dz 1 z ) m / [ z m Gz) + 1 z )G z) ] = z m 1 z ) m / 1 [ z m Gz) + 1 z )G z) ] + 1 z ) m / [ m Gz) z m G z) zg z) + 1 z )G z) ]. 4

5 The differentia Equation to sove is then 1 z )G z m + 1)G + [λ m m + 1)]G = 0 8) Let us introduce an auxiiary notation α = λ m m +1) to simpify things. Next, we empoy the same trick we used to sove the harmonic osciator, i.e we seek the soution as power series G = n=0 a nz n which we substitute into the equation nn 1)1 z )z n a n zn m + 1)z n 1 a n + αz n a n = 0, n=0 or, equating coefficients for each power n=0 n=0 n+)n+1)a n+ [nn 1)+n m +1) α]a n = 0 a n+ = The recursive reation can be aso written as nn 1) + n m + 1) α n + )n + 1) a n. a n+ = n + m )n + m + 1) λ n + )n + 1) a n. 9) This immediatey eads to the fina soution because uness the recursion reation it terminated at some finite n and a coefficients are zero afterwards, we woud have Gz) 1 z ) 1 because it is the property of geometrica series in z to have a n+ a n. Since divergent soutions are rued out we must have λ = m + n) m + n + 1), n = 0, 1,,... which is equivaent to λ = + 1) with = m, m + 1,.... Yet another, more canonica, way of specifying quantum numbers λ and m is to have λ = + 1), with = 0, 1,,... and m. For m = 0, we have F z) = Gz) = P z) where P z) are the Legendre poynomias which are conventionay normaized to have P 1) = 1 the reason for this strange normaization is the same as with hermitian poynomias there is a generating function for which P z) are reated to Tayor-expansion coefficients). We easiy get severa functions from our recursion reations P 0 = 1, P 1 = z, P = 1 3z 1), P 3 = 1 5z3 3z), P 4 = z4 30z + 3), etc. The generating function is gz, s) = 1 zs + s ) 1/ = n=0 P nz)s n. For m 0 we have F m z) = 1 z ) m / G m z) = 1 z m / d m ) dz m P z) P m z), 10) which is caed the Legendre poynomia of the second kind. Probem 45. Legendre poynomias Check that 10) is indeed the soution of Eq. 7), or Eq. 8) for non-zero m take for granted that P z) soves this equation for m = 0). I suggest that you actuay verify that the recursion reation 5

6 9) is satisfied. We have determined the fu ange dependence of our soution at this point χθ)φϕ) Y m θ, ϕ), where Y m θ, ϕ) is caed the spherica harmonic. With normaization factors in pace it reads factors of 1) m are a matter of convention, not principe) For m < 0 we use Y m Y m θ, ϕ) = 1) m + 1) m)! 4π + m)! e imϕ P m cos θ), for m 0. 11) θ, ϕ) = 1) m Y m θ, ϕ) ). The normaization condition is such that π 0 π dϕ sin θdθ Y m θ, ϕ) ) Y m θ, ϕ) = δ, δ m,m. 0 We proceed now with the formuation of the remaining radia equation for Rr) which is easy since a we have to do is to take Eq. 5) and equate the radia part to λ /M = +1) /M. So we do just that and divide this equation by 1/r to have M [ ] 1 r r [r R r)] + V r) + + 1) Mr Rr) = ERr). 1) As before, we seek the soution in the form Rr) = ur)/r to cast it in the form of a 1D Schrödinger Equation [ ] u + 1) + M E V r) Mr ur) = 0, 13) It is the same as for a 1D partice moving on the 0, ) interva in the potentia V r)++1)/mr which incudes the centrifuga-type piece. The normaization condition for the wavefunction ψ = Rr)Y m θ, ϕ) is π π r dr Rr) dϕ sin θdθ Y m θ, ϕ) = dr ur) = Anguar momentum To proceed we have to specify the potentia. But before we do that, et us discuss a itte the physica meaning of quantum numbers m and. Consider an operator which is a vector product of the coordinate and momentum, which in cassica mechanics stands for the anguar momentum In components α = 1,, 3 we have ˆL = r p = i r. ˆL α = i βγ ɛ αβγ r β, r γ 6

7 where ɛ αβγ in an anti-symmetric tensor such that ɛ 13 = 1 and a other vaues are obtained from the property ɛ βαγ = ɛ αβγ = ɛ βγα, i.e. it changes sign when we change any pair of indexes paces. This, in particuar, means that a indexes must be different to have a non-zero eement. Let s take α = 3. Then ˆL 3 = ˆL z = i x y y ). x In spherica coordinates ϕ = α r α ϕ = y r α x + x y = i L z, meaning that I wi stop using ˆ on the operators) L z = i ϕ. With this we immediatey check that L z Φ m ϕ) = i 1 ϕ π eimϕ = m Φ m ϕ), i.e. Φ m ϕ) is an eigenfunction of the anguar momentum projection on the z-axis with eigenvaue m. This expains puzzing SG-machine experiments: no matter what axis z is seected by the magnet, one aways finds that the measured anguar momentum projections form equidistant ines at vaues m. Next we ook at the operator L = L L = 3 i=1 L il i : L = ijkβγ ɛ ijk ɛ iβγ r j r β. r k r γ But i ɛ ijkɛ iβγ = δ jβ δ kγ δ jγ δ kβ because i) j and k must be different from i and from each other, and ii) β and γ must be different from i and from each other. Hence L = r j r j r j r k = r k r k r k r j jk rj r r j r k + 1 3) r j = r r rr ) r ). jk k r k r j r j j In spherica coordinates r = r / r) and we find L = r r r r r ) = r r r r r r ) = r r ) r r r = r = r r 1 ) 1 r r r = r sin θ θ sin θ θ + 1 sin θ ) ϕ. 7

8 In other words L is identica to the anguar dependent part of the Lapace operator times r which has Y m θ, ϕ) as eigenfunctions with eigenvaues + 1) [see Eqs. ) and 5)]. Thus L Y m θ, ϕ) = + 1) Y m θ, ϕ). With this we understand the origin of the L /Mr centrifuga potentia in the radia equation. We aso understand that the Hamitonian symmetry is such that the partice anguar momentum is conserved cassicay. Strangey enough, in quantum mechanics this transates into a counterintuitive aw that the anguar momentum projection on one, and ony one!) axis is conserved aong with the anguar momentum squared. This is because anguar momentum operators for different axis form an agebra of non-commuting operators and thus can not assume we defined vaues at the same time. We wi dea with it ater, after we finish with the radia equation for the hydrogen atom. Hydrogen atom Consider the most important potentia in Nature for human beings at east) V r) = Ze /r. As usua we first introduce dimensioness variabes we ook for bound states with negative energies E = E ): m E ρ = r, γ = Ze m E, to bring this equation to the universa form [ ] γ u + 1) ρ) + ρ ρ 1 uρ) = 0. We demand that proper soutions go to zero as ρ and ρ 0 otherwise they cannot be either normaized or made with finite energy. Again, we check for the asymptotic behavior to factor it out. For ρ 0 we ook at u + 1) ρ) ρ uρ) = 0, whose soution is uρ) ρ +1 we disregard the ρ soution). For arge ρ we have u ρ) uρ) = 0, whose soution is uρ) e ρ. Thus we propose to seek the soution in the form uρ) = ρ +1 e ρ vρ). Next we have to differentiate this construction twice, pug it in, and attempt to find the soution for v as Tayor series expansion in ρ same protoco as before for the harmonic osciator and spherica functions. Foowing the dri u = + 1)ρ e ρ v ρ +1 e ρ v + ρ +1 e ρ v u = + 1)ρ 1 e ρ v + 1)ρ e ρ v + ρ +1 e ρ v + + 1)ρ e ρ v ρ +1 e ρ v + ρ +1 e ρ v = { } ρ +1 e ρ v + 1) + v + 1) + 1) v ρ ρ v + v v. ρ 8

9 With this sma torture the equation for v reads ) + 1) v ρ) + 1 v ρ) + ρ γ + 1) ρ vρ) = 0. Next we empoy v = n a nρ n to get [nn 1) + + 1)n]a n ρ n + n n [ n + γ + 1)]a n ρ n 1 = 0 [n + 1)n + + )]a n+1 + [γ + n + 1)]a n = 0 a n+1 = + n + 1) γ n + 1)n + + ) a n. Next we notice that if the series are not terminated then asymptoticay a n+1 /n)a n which is characteristic of the exponentiay divergent soution e ρ. Thus the ony option we have is to demand that the series are terminated by having γ = ν, with integer ν = + 1, +, + 3,... 14) The origin of the periodic tabe of eements is expained! We found that the spectrum of the hydrogen atom is determined by three quantum numbers: Principa quantum number: ν = 1,, 3,... Orbita Anguar momentum: = 0, 1,,..., ν 1 Anguar momentum projection: m =, + 1,..., 1, For this set of quantum numbers, the energy depends ony on ν: Ze m = ν E ν = Z e 4 m E ν ν, reproducing the famous Bamer spectra ine series. The degeneracies of the spectrum are such that for a given ν there are ν 1 + 1) = νν 1) + ν = ν =0 eves with the same energy which can accommodate N ν = ν eectrons. We find N 1 =, N = 8, N 3 = 18, etc In spectroscopy, ν = 1 orbitas are known as s-states one can pace eectrons on the s-orbita); ν =, = 1 orbitas are known as p-states one can pace eectrons in the = 0 state and 8 = 6 eectrons on the p-orbitas); ν = 3, = orbitas are known as d-states one can pace, and 6 and eectrons in = 0, 1 states, and on the d-orbita); other orbitas foow the aphabet f, g, h etc. It becomes evident that the periodic tabe is refecting these magic numbers corresponding to fied energy orbitas. Of course, in reaity eectrons coupe to each other by the 9

10 same Couomb force and the picture is not that simpe. It get scrambed in the many-eectron system, but the basic structure of atomic energy shes remains vaid. Thus in one stroke QM expains the order among eements in Nature and many of their chemica properties by cacuating energies of the ast occupied states in atoms. The dimensioness coordinate ρ is reated to r by ρ = m E r = Z ν r a B, where a B = me is the Bohr radius Let us now get famiiar with the soutions a itte coser. For = 0 states with compete spherica symmetry we have Y0 0θ, ϕ) = 1/4π, whie for states with = 1 the wavefunction becomes more eaborate Y 0 1 θ, ϕ) = 3/4π cos θ, Y 1 1 θ, ϕ) = 3/8π e iϕ sin θ, Y 1 1 θ, ϕ) = 3/8π e iϕ sin θ, The other way to present them is Y1 0 ±1 z/r, Y 1 x ± iy)/ r. For arge and m = the spherica harmonic is sharpy peaked in the θ = π/ pane. The radia functions Rr) = u ν r)/r r e Zr/νa B v ν Zr/νa B ), are determined by the Laguerre poynomias studied by mathematicians we before quantum mechanics same with Hermite and Legendre poynomias). Just for the reference with generating function v ν ρ) = L +1 ν 1 ρ) F z, s) = 1 z) p 1 e sz/1 z) = n=1 L p nz) n + p)! sn. The normaized soutions for the hydrogen atom in terms of standard functions are [ k) 3 ] 1/ ν 1)! ψ νm = ν[ν + )!] 3 kr) e kr L +1 ν 1 kr) Y m θ, ϕ), k = Z, 15) νa B with d 3 rψ νm r)ψ ν m r) = δ mm δ δ νν. For exampe: ) Z 3 1/ Ground state ψ 100 = e Zr/a B The expectation vaue of r, or the atom size is ) 4πZ 3 r = r 3 e Zr/a B dr = a B 4Z πa 3 B 0 πa 3 B Probem 46. Variationa principe and the Hydrogen atom 0 x 3 e x dx = 3! a B 4Z = 3 Z a B. 10

11 a). Use the variationa principe to estimate the ground state energy of the hydrogn atom by using the tria wavefunction ψr, b) = e r/b, where b is the variationa parameter. [Note, the state as written is not normaized yet.] You shoud, of course, get the correct answer since this happens to be the right functiona form of the exact soution. a). Repeat the cacuation, but now use the gaussian form ψr, b) = e r /b, and find the fractiona error made in the energy. Show numericay) that the overap of the true ground state wavefunction and the optimized variationa estimate is roughy ψ100 ψr, b opt ) Probem 47. Momentum space wavefunctions for Hydrogen a). Cacuate the momentum space wavefunctions for the two owest-ying s-wave states of hydrogen, ψ νm, i.e. for ν = 1 and ν = and = m = 0. When doing the 3D Fourier transform pay attention that there is no preferred direction for s states and thus p-direction can be set to be the z-axis of the integration, i.e. p r = pr cos θ. To check that you are on the right path compare your answer to ψ 100 p) = 64πk5 p + k ) 4, where k = Z a B. b) Do not forget to do ψ 00 p). Probem 48. Tritium decay The nuceus of two neutrons and a singe proton hence with Z = 1) is caed triton, t by anaogy with the deuteron, d), and the system consisting of a singe eectron and a triton is caed a tritium atom. The triton is unstabe against radioactive β-decay via the process t 3 He + e + ν e where 3 He consists of one neutron and two protons i.e. Z = ). On an atomic time scae, the decay process and the ejection of the decay products e and ν e happens amost instantaneousy. After such a decay, an eectron initiay in the ground state of the tritium atom suddeny finds itsef in a Couomb potentia with twice the nucear charge. Assuming that after decay the 3 He nuceus remains static a). What is the probabiity that an eectron that was originay in the ground state wi remain in the ground state of the new system? b). In the n = s state? c). In some = 1 state? Hint: this is essentiay the same setup as in the moving was probem you did before. Probem 49. Wavefunction at the origin Consider a partice bound to a fixed center by a sphericay symmetric potentia V r). a). prove that m ) ψ0) dv = π dr 11

12 for a s-states, ground and excited. Hint: Consider Schrödinger Equation for ur), mutipy by u and integrate over r 0, ). b). Check this reation for the ground state of a three dimensiona isotropic osciator and the hydrogen atoms. Historicay, this reation has been usefu in guessing the form of the potentia between a quark and an antiquark. Probem 50. Poarizabiity of the Hydrogen atom When an eectric fied E 0 is appied to a system of charges, a deformation takes pace which eads to an eectric dipoe moment P = α E E 0, where α E is caed the inear eectric poarizabiity. This eads to an energy shift E = 1 α EE 0. Actuay, this reation may be used to define inear poarizabiity as α E = d E/dE 0 ) 0. This is exacty what we wi be doing in this probem. According to the second-order perturbation theory the energy shift of a hydrogen atom in a constant eectric fied E 0 oriented in the z direction is E = n 1s 1s ee 0 z n n ee 0 z 1s E n E 1s, thus α E = e n 1s 1s z n n z 1s E n E 1s a). Cacuate the eading p ) contribution to this answer and show that a ower bound is given by α E > e 1s z p = 19 E p E 1s 3 11 a3 B. b). Show that the upper bound is given by use of competeness reation) α E < e E p E 1s 1s z 1s = 16 3 a3 B. c). Demonstrate that bonus part: extra 10 points to the homework if done correcty) z 1s = [Ĥ0, ˆF ] 1s where ˆF = ma B r/ + a B )z and Ĥ0 is the unperturbed hydrogen atom Hamitonian. Note, this is not an operator identity; this is equaity vaid when acting on a particuar state. d). Use the proof estabished in c) to derive the exact resut bonus part: extra 10 points to the homework if done correcty) α E = e 1s z ˆF 1s = 9 a3 B. Compare with the two bounds obtained in previous parts. 1

13 Three dimensiona harmonic osciator Consider now an isotropic harmonic osciator V r) = mω We have two ways to sove it by considering three separate 1D probems for x, y, and z variabes separatey, as for the square box potentia described at the beginning of this set of notes, or by using spherica coordinates, as for the hydrogen atom. The former way is straightforward, and can be done for the anisotropic osciator easiy too different frequencies for different directions). The resut for the spectrum is E n1,n,n 3 = ω n ) +ω n + 1 r. ) +ω n ) = ω n 1 + n + n ), n α = 0, 1,,... and for the wavefunctions we have the product of three harmonic osciator soutions ψ n1 n n 3 = ψ n1 x)ψ n y)ψ n3 z). We ceary see that excited states are degenerate since the spectrum depends ony on the sum ν = n 1 + n + n 3 of three quantum numbers. In spherica coordinates we wi find that the ange-dependent part of soutions is given by spherica harmonics Y m θ, ϕ), and the radia function Rr) = ur)/r satisfies an equation as before ξ = mωr and ɛ = E/ω) u + [ ɛ ξ ] + 1) ξ u = 0. I wi not go through yet another exercise of estabishing an asymptotic behavior to motivate the form uξ) = ξ +1 e ξ / vξ), and showing next that by ooking for the power-series soution for v we arrive at the condition that power series must be truncated to have a meaningfu soution. Utimatey this eads to the condition ɛ = n = ν + 3 E = ω n ), n = 0, 1,... or E = ω ν + 3 ), with ν = 0, 1,,.... and additiona rue that ν and are both either even or odd. Probem 51. Degeneracies in the 3D harmonic osciator Cacuate degeneracy factors for the energy spectrum of the three dimensiona harmonic osciator from two approaches, using E ν = ω ν + 3 ), with ν = n 1 + n + n 3 and integer non-negative n α, and using E νm = ω ν + 3 ), 13

14 with orbita momentum quantum numbers restricted by ν and m pay attention to even/odd rue for ). Make sure that the two cacuations agree with each other. Free partice in spherica coordinates You may wonder why woud one do a free partice with pane wave soutions using spherica coordinates? This is done i) in preparation for the future discussions of scattering probems, when partices moving in free space are incident on the scattering center, and ii) to dea with probems which have free moving partices subject to certain sphericay symmetric boundary conditions. Thus instead of ψr) e ik r we seek soutions in the form ψr) Y m θ, ϕ)ur)/r with ) ) u r) + k + 1) r ur) = 0 u + 1) ρ) + 1 ρ uρ), where ρ = kr. This is the differentia equation of the spherica Besse ρj ρ) and Neumann ρn ρ) functions. In the imit of ρ 0 we have to dea with u + 1) ρ uρ) = 0. Substituting u = ρ a we get aa 1) = + 1) which soves by a = + 1 and a = ; the former corresponding to the asymptotic behavior of ρj ρ) and the atter to the asymptotic behavior of ρn ρ). More specificay j ρ) =! + 1)! ρ + Oρ + ) ρ 1 and more expicity n ρ) = )!! ρ 1 + Oρ +1 ) ρ 1 j ρ) 1 ρ cos ρ + 1) π ) ρ 1 n ρ) 1 ρ sin ρ + 1) π ) ρ 1 j 0 = sin ρ ρ, j 1 = sin ρ ρ cos ρ ρ, j = 3 ρ ) sin ρ ρ 3 3 cos ρ ρ, n 0 = cos ρ ρ, n 1 = cos ρ ρ sin ρ ρ, n = 3 ρ ) cos ρ ρ 3 3 sin ρ ρ. You immediatey see that in the imit of arge ρ the two functions can be combined into one exponentia h 1) = j + in 1 ρ ei[ρ +1)π/], ρ 1, h ) = j in 1 ρ e i[ρ +1)π/], ρ 1, 14

15 where h 1) and h ) functions are known as spherica Hanke functions of the first and second kind. Three-dimensiona square we: The Deuteron As a practica appication of free-partice-in-spherica-coordinates soutions we consider bound states in the sphericay symmetric potentia we { V0 r a V r) = 0 r > a. 16) Let k = mv 0 E ), and κ = m E, as usua. Then the soution in the we is that of a free partice which is reguar at the origin ψ I r) = Nj kr)y m θ, ϕ). For the underbarrier region we need an exponentiay decreasing soution with compex momentum, which is h 1) iκr). Thus ψ II r) = NAh 1) iκr)y m θ, ϕ). where N is the normaization factor. Now we match the soutions j ka) = Ah 1) dj kr) iκa), dr = A dh1) iκr) r=a dr, r=a to fix the aowed vaues of k and A. There are numerous physica exampes of systems which are reasonaby we described by this setup. One of them is a deuteron which is a bound state of a proton and a neutron. They form a shaow bound state, E.36 Mev, which is sphericay symmetric, i.e. = 0 and m = 0. If we approximate the potentia between partices by Eq. 16) we can estabish a reation between the binding energy and strong force V 0 the reative motion of two partices is described by the reduced mass, which is, in this case, approximatey 1/ of the proton mass). For = 0 we have and the matching conditions read j 0 z) = sin z z, h 1) z) = i eiz z, sin ka = A ie κa sin ka = Ak/κ)e κa ka iκa dsin kr/kr) dr = A de κr /κr) r=a dr cos ka = Ae κa. r=a By dividing two equations we find the quantization condition tan ka = k κ, which is exacty the same as for antisymmetric soutions in the 1D square we!) reca that here we are ooking for sphericay symmetric soution in 3D). Thus, we can immediatey state from notes and homework) that in the imit of a deep we V 0 1/ma the ground state energy is E V 0 + π ma ma V 0 ).

16 If the we is shaow, i.e. V 0 < V c = π /8ma then there are no bound states at a!) in contrast with square wes in 1D and in D. For a = 1.4 fm one finds V c 50 Mev, meaning that the bound state energy for a deuteron is much smaer than V c and thus the vaue of V 0 has to be sighty arger, but cose to V c. This Probem is not part of Homework! Do it if you want to. Semicassica quantization of the Hydrogen atom The semicassica picture of the hydrogen atom considers ony panar orbits with variabes r and ϕ, and their corresponding momenta. The semicassica quantization condition on the anguar momentum is that of free partice motion on a circe of ength π: π 0 Ldϕ = π or L = The quantization condition for the radia coordinate is rmax r min p r dr = rmax r min m E + Ze r Evauate the integra and show that the resuting answer is E = mz e 4 n r + ). ) mr dr = πn r. Hint: If you do it by residue anaysis, do not forget the residue at infinity. 16