Physics 862: Atoms, Nuclei, and Elementary Particles

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1 Physics 86: Atoms, Nuclei, and Elementay Paticles Bian Bockelman Septembe 11, 008 Contents 1 Cental Field Poblems 1.1 Classical Teatment Quantum Teatment Angula momentum opeatos in spheical coodinates Geneal Cental Potential Poblem Inteacting Two-Paticle System Boh Model Schödinge Equation Solution fo Hydogen Deive the fist few tems

2 Monday, 5 August 008 Mid-Tem Exam: Octobe 15, 008 Final Exam: 10-1, Fiday, Decembe 19th, 008. URL: Office Hous: M, Tu, W, Th, F, 3:00PM-3:30PM o by appointment 1 Cental Field Poblems We conside poblems with a foce diected towad the oigin. Let be the vecto of a paticle. Let p m v be its linea momentum. In a cental field poblem, we have the foce acting on the paticle, F, paallel to. 1.1 Classical Teatment We fist outline the classical appoach. The potential depends on, not : F v V ( ) V () (1) ( ) dv () d (3) Let L denote the angula momentum, which is defined to be p, whee is the coss poduct. Reminde: i j k a b det a 1 a a 3. b 1 b b 3 So, we can calculate the angula momentum (note that v denotes the component of v pependicula to ; v denotes the component of v paallel to ): L p, whee L mv (4) dl { ( ) } dt τ F dv 0 (5) d (6) The last equation is equal to zeo because we ae looking at the coss poduct of two paallel vectos; we get that angula momentum is conseved in this system. To calculate the enegy, we have: E p m + V () 1 { } v + v + V () (7)

3 But, v v d dt v L m v L m Plugging (8) and (9) into (7), E 1 { } L mv + m + V () (8) L m, whee L L (9) We efe to the backeted tem as V eff (), the effective potential of. (10) (11) 1. Quantum Teatment We want to fom the Hamiltonian of the system, whee Ĥ ˆT + ˆV ˆT ˆp m m (as ˆp i ). We want to solve fo enegy eigenstates of Ĥ (the obsevable values of Ĥ). Bound states ae the eigenstates of Ĥ. The enegy eigenstates ae the wave functions which have finite enegy. A wave function is a supeposition of eigenstates; the coefficients of the wave function ae the pobability of finding it in that eigenstate. A wave function is a solution to Schödinge; we solve fo the eigenstates because they ae time-independent (?). The coefficients depend on the initial conditions o bounday conditions you specify (note: this explanation seems ill-founded and hand-wavy). Wite out the time-independent Schödinge equation: Hψ( ) Eψ( ) (1) [ ] m + V () ψ( ) Eψ( ) (13) In spheical coodinates, 1 { } + 1 { θ + 1 tan θ θ + 1 sin θ } ψ (14) 3

4 We label the second backeted tem ˆL. Substituting (14) into (1), { m {ψ(, θ, φ)} + { }} ˆL (θ, φ) m + V () ψ Eψ (15) If V (, θ, φ) V (), then only dependency on θ and φ is in ˆL. ˆL (θ, φ) has eigenstates itself (the spheical hamonics), which ae seen in many adialsymmety poblems in physics. We denote the spheical hamonics as Y l,m (θ, φ) Sometimes, we efe to them as Y l,m (ˆ), whee ˆ. The eigenvalue is l(l + 1); i.e., ˆL Y l,m (θ, φ) l(l + 1)Y l,m (θ, φ). Hee, is the unit of angula momentum. Wednesday, 7 August Angula momentum opeatos in spheical coodinates Note: vaiables with a hat ae opeatos (such as ˆx). So, ( ˆL x : i sin φ θ + cos φ ˆL y : i ( cos φ θ + sin φ tan θ ) ˆx (16) tan θ φ ) ŷ (17) π ˆL z : i φẑ L zẑ (18) We can expand the tem :, H T + V () (19) 1 p + V () (0) m m + V () (1) () + ˆL ˆL (3) 4

5 We can wite out the entie ˆL opeato, and use the fact that ˆx, ŷ, and ẑ ae othogonal to compute L : The key point hee is that ˆL L xˆx + L y ŷ + L z ẑ (4) L ˆL ˆL L x + L y + L z (5) [H, ˆL] 0, fo V V () (6) So, H and ˆL commute. If two opeatos commute, then you can measue the physical quantities associated with those simultaneously. If [H, ˆL] 0, then when you measue enegy (putting it in a well-defined state), then angula momentum is put into an unknown state of enegy - even if you had peviously put angula momentum into a well-defined state by measuing it. In any physical system, it is useful to find the maximal set of opeatos which commute with each othe. In ode to compute the commutato, the fist pat of H only has adial tems, so it will commute. The computation beaks down to: Does [L, ˆL] 0? Recall the following elation between the components of angula momentum: whee To illustate cyclic/anti-cyclic ode: [L i, L j ] ɛ ijk i L k, (7) i, j, k {x, y, z} { +1, fo cyclic ode ɛ ijk 1, fo anti-cyclic ode. ɛ xyz 1 ɛ zyx 1 ɛ yzx 1 ɛ xzy 1 We ae eady to pefom the commutato computation: [L, ˆL] [L, L x ]ˆx + [L, L y ]ŷ + [L, L z ]ẑ (8) Conside: [L, L x ] [L x + L y + L z, L x ] [L y + L z, L x ] (9) (as [L x, L x ] 0 because L xl x L x L x). Then, [L y, L x ] L yl x L x L y (30) L yl x L y L x L y + L y L x L y L x L y (31) L y (L y L x L x L y ) + (L y L x L x L y )L y (3) L y [L y, L x ] + [L y, L x ]L y (33) 5

6 Fom (7): [L y, L x ] i (L y L z + L z L y ) (34) We now conside [L z, L x ]. We can follow the same deivation pocess to get: [L z, L x ] L z [L z, L x ] + [L z, L x ]L y (35) Plugging (34) and (36) into (9), we get i (L z L y + L y L z ) (36) [L, L x ] [L y + L z, L x ] i (L y L z + L z L y ) + i (L z L y + L y L z ) 0 (37) Similaly, [L, L y ] [L, L z ] 0 (38) Also, [V (), ˆL] 0. If V was a function of ˆ and not, then these would not commute. and We know that the eigenstates of the angula momentum opeatos ae L Y l,m (θ, φ) l(l + 1) Y l,m (θ, φ) L z Y l,m (θ, φ) m Y l,m (θ, φ) Comment: The maximal set of commuting opeatos is H, L, and L z. We could have chosen L x o L y instead of L z ; we choose L z because it has the simplest fom in spheical coodinates. Because of this, if we make measuements of E, l, and m, then we collapse the wave function entiely. In descibing atoms with one electon, the inteaction with the nucleus only depends on the Coulumb potential, which is spheical symmetical. Without loss of geneality, the wave function fo an electon in a potential may be witten ψ E,l,m (, θ, φ) R E,l ()Y l,m (θ, φ) (39) The set of all these states is complete in the sense that any wave function can be witten in the above basis. Hence, Hψ(, θ, φ) Eψ(, θ, φ) implies: } { l(l + 1) + m m + V () R() ER() R() does not depend on m. Fiday, 9 August 008 Possible times fo makeup classes: Tuesday 1:30 - :30 PM 6

7 Thusdays 5-6 PM o 1 - PM Fidays 3-4 PM Discussing HW poblem: The time-dependent Schödinge equation is: Hψ i ψ t If ψ(x, t) φ(x)e Et, then ψ(x, t) is a stationay state - Why? We say it is a stationay state because: ψ(x, t) φ(x). Hence, the pobability of finding the paticle at a place is independent of time. We know solutions of Hφ n (x) E n φ n (x) ae complete. Meaning that any function, f(x) in the domain 0 x L can be expanded in the φ n (x). So, f(x) c n φ n (x) n1 Fom the sum, the pobability that we find enegy state E n whee n is c. Hemitian opeatos have eigenfunctions φ n such that φ n, φ n δ n,n Hee, the inne poduct is: Thus, φ n, f φ nφ n (x)dx c n φ n, φ n c n n1 What is the wave function fo this poblem fo t > 0? Thus, The linea momentum is The expected value is f(x, t) c n φ n (x)e i/ Ent n1 ˆp x i x f(x, t) ˆp x f(x, t) χ(t) L 0 f (x, t)ˆp x f(x, t)dx χ(t) 7

8 Fo the last poblem of the homewok, ˆL x φ ˆL x φ Note about the book: Equation 7-8 is not quite coect. The 4π only comes in because the book has not nomalized the angula pat of the wave function. Book uses Θ lm (θ) and Φ m (φ) instead of Y lm (θ, φ). Schödinge equation fo an enegy eigenstate of a paticle in a cental potential V (), ψ(, θ, φ)e iet satisfies: Hψe iet i t ( ψe iet ) Eψe iet Thus Hψ Eψ may be witten { } 1 d ˆL ( ) + m d m + V () ψ(, θ, φ) Eψ(, θ, φ) (40) We know ˆL Y lm (θ, φ) l(l + 1) Y lm (θ, φ) whee Y lm (θ, φ) is a spheical hamonic and ˆL z Y lm (θ, φ) m Y l,m (θ, φ) So wite ψ(, θ, φ) R nl ()Y lm (θ, φ) and substitute into (40), then { 1 d } l(l + 1) + m d m + V () R nl () E n R nl () (41) whee n labels the enegy eigenstate and eigenvalue. Nomalization of an enegy eigenstate ψ nlm R nl ()Y lm (θ, φ): 1 π π ψnlm sin θ ddφdθ (4) ( ) ( ) R nl () d Y lm sin θdθdφ (43) 0 Theefoe, since we have 0 Y lm Y lm 1, The adial pobability is then P nl () R nl (). Monday, 1 Septembe 008 Labo Day, no class. Wednesday, 3 Septembe 008 R nl () 1 (44) 8

9 1.4 Geneal Cental Potential Poblem If ψ Elm (, θ, φ) R El ()Y lm (θ, φ) then R El satisfies: { } l(l + 1) + m m + V () R El () ER El () (45) To simplify the equation and educe it to an equation simila to the Schödinge equation fo one-dimensional poblems, let R El () U El() (46) and substitute (46) into (45). Then, U El () satisfies d { } U() l(l + 1) m d + m + V () EU(), (47) whee the tem in the backets is V eff (). Note that if V () is negative (attactive) and l 0, then V eff () < 0 always. Howeve, if l 0, then the effective potential is positive close to zeo - epelling electons away fom the oigin! This is called the centifugal baie. As l inceases, the centifugal baie becomes geate. 1.5 Inteacting Two-Paticle System Conside an inteacting two-paticle system; fo example, the hydogen atom, which is one poton and one electon. The classical teatment: Conside paticle 1 to be located at 1 and paticle at. Then, the distance between them is (going fom paticle to paticle 1). Then, we get 1 (48) p 1 m 1 1 (hee, p 1 is the linea momentum of paticle 1; 1 is the fist deivative of 1 ) and p m The cente of mass (o cente of momentum is: Take M m 1 + m. Hence, What is cm? (m 1 + m ) cm m m. p cm p 1 + p 9

10 Adding (48) and (49), we get Hence, m cm m m m m 1 m 1 + (49) M m cm + M m 1 (50) Hence, 1 cm + m M (51) cm m 1 M (5) We want p cm m cm m m p 1 + p (53) { p µ µ { 1 } p1 µ p } (54) m 1 m p µ + p cm M p 1 m 1 p m Expanding the left hand side and subtacting the ight: ( 1 µ p1 p ) + ( p 1 + p ) ( ) p m 1 m M 1 + p 0 m 1 m Expand and collect the tems: ( µ 0 p 1 m M 1 ) m 1 ( µ + p m + 1 M 1 ) m ( + p 1 p µ + 1 ) m 1 m M (55) (56) (57) We want to pick a µ such that the equation above holds tue; futhe, as the dot poduct and the magnitudes ae independent of the pick of µ, we want the coefficients in the paenthesis to be equal to 0. The pick of µ m 1m (58) M will wok. Hence, we have educed the two-paticle poblem into a one-paticle poblem with educed mass. 10

11 Quantum Teatment: The poblem with going fom the classical to the quantum teatment is that the coodinates and conjugate momenta do not necessaily commute. Then, Fo example, does the commutation elation [x cm, p x cm] i? Recall: p x i x [x, p x ]f(x) (xp x p x x)f(x) (59) ( x( i ) f(x) ) x ( x( i ) f(x) x + i (xf(x)) (60) x ) + i f(x) + i x f x (61) i f(x) (6) So, poof of the oiginal statement: [ ] m1 x 1 + m x, p x 1 + p x m 1 M M [x 1, p x 1] + m M [x, p x ] (63) ( m1 M + m ) i (64) M i (65) We can do the same thing fo the y and z components. Poof. We now claim [x, p x ] i : [ x 1 x, m p x 1 m 1 p x ] M The same can be done fo the y and z components. Fo the Hamiltonian, So, we get H m M [x 1, p x 1] + m 1 M [x, p x ] (66) m 1 + m i (67) M i (68) p 1 m 1 + p m + V ( 1 ) (69) p cm M + p + V ( ) (70) µ H H cm + H el (71) If the two commute ([H cm, H el ] 0), then we can independently measue the cente of mass of the system and the elative coodinates. We now attempt to show these do indeed commute. 11

12 Poof. Key point is that [p cm, V ]: [p cm, V ] [ p 1 + p, V ( 1 )] (7) [ p 1, V ] + [ p, V ] (73) i V + i V (74) 0 (75) Fiday, 5 Septembe 008 Review: Discussed geneal popeties of the solutions of the Schödinge equation fo a paticle in a cental potential. Noted that we can educe this to a Schödinge equation in one-dimension. Discussed the poblem of inteacting paticles, and how to educe this to a cente-of-mass Hamiltonian and a elative Hamiltonian. H atom (p + e ). Recall we defined H cm p cm M, whee M m 1 + m, (76) H el p µ + V ( ), whee µ m 1m M, (77) whee µ is the educed mass and V ( ) is the inteaction potential enegy between two paticles and is thei elative coodinate. We know [H, H cm ] [H, H el ] [H cm, H el ] 0 which means the system s enegy, E, the CM enegy, E CM, and the elative enegy, E el, can all be known and measued simultaneously. Hence, we can sepaate the two coodinates: whee Ψ( R cm, el ) Φ( R cm )ψ( ) H CM Φ( R cm ) i Φ( R cm, t) E CM Φ( R t cm )e iecmt/ We can wite Φ( R cm, t) e iecmt/ Φ( R cm ). Then, H CM Φ( R cm ) p CM M Φ( R cm ) cm M Φ( R cm ) E cm Φ( R cm ) 1

13 ( ) cmφ cm ( R MEcm cm ) Φ cm ( R cm ) Take Kcm ME cm. Then, the solution fo Φ cm is Φ cm ( R cm ) e i K cm Rcm (78) So, the H cm hamiltonian poduces constant motion, which is not inteesting. Hence, we want to just conside the cente of mass fame. Fo the H-atom, Hence, µ m e. 1.6 Boh Model µ m pm e m e m e m p + m e 1 + me m p Assumption: Cicula motion of adius 000 Hence, E T + V 1 µv e ( F µ a µ( V ) µ ˆ e ˆ ( ) 1 e ˆ F e ˆ µ a ( e )) We know fo cicula motion with constant velocity that a v ˆ, so Hence, Thus, e v ˆ v µ e E 1 µv e so E 1 µ 1 e The quantization condition that Boh intoduced was that the action is a constant, i.e., 13

14 pdl nh whee h is Planck s constant, and the line integal is taken aound a cicle. Hence, as p is a constant pdl (µv)π nh Define h π, so O, Ou angula momentum is quantized. Monday, 8 Septembe 008 µv n. L n. Note: 1st make-up class is on 4PM Thusday. In summay, we have: E 1 e Using (79)-(105) to solve fo n, we get µv e (79) µv n (80) (81) n n µe n a 0, (8) whee a, which is the adius of the n 1 obit. µe Fom (99), V n n n µ n µ(n a 0 ) e nµa 0 n Fom (105) e e (83) E n 1 1 n n E I a 0 n (84) whee E I e a 0, which is defined to be 1 Rydbeg, which woks out to be ev. This is the binding enegy of the gound state (n 1) of the H atom. We know that a 0 µe 0.5Å m 14

15 1.7 Schödinge Equation Solution fo Hydogen Let R kl () u kl() whee { The full solution is d l(l + 1) + µ d µ h e ψ klm ( ) u kl() Y lm (θ, φ). } u kl () E kl u kl (). (85) whee k is some index identifying the eigenstate (and the enegy). Let us switch to scaled units so that all vaiables ae dimensionless. Using these scalings on equation (85): { a 0E d l(l + 1) I + d ρ (86) a 0 λ kl E kl (87) E I a 0E I a 0E I We set a 0 1 ρ and divide both sides by E I to get } U kl () E kl U kl () (88) { d l(l + 1) dρ ρ + } U kl (ρ) λ ρ klu kl (ρ) (89) Now, we facto the asymptotic dependence. Fo ρ, (89) becomes The solutions ae of the fom d dρ U kl(ρ) λ klu kl (ρ) u kl (ρ)ẽ ±λ klρ fo ρ. In ode to get physically meaningful solutions, we must equie 0 R kl (ρ)ρ dρ 1 Hence, we dop the positive solution. Theefoe, we have u kl (ρ) y kl (ρ)e λ klρ (90) 15

16 Substitute (90) into (89) to find the equation satisfied by y kl (ρ). du dρ d u dρ Substituting (96) into (89), { d y dρ λdy dρ + λ y The final esult is that whee y kl (ρ) satisfies d y kl dρ Wednesday, 10 Septembe 008 d dρ { ykl e λ klρ } (91) dy kl dρ e λ klρ λ kl y kl (ρ)e λ klρ (9) ( ) dykl dρ λ kly kl e λ klρ (93) d { } dy dρ dρ λy e λρ (94) ( d y dρ y dy ) ( ) dy e λρ λ dρ dρ λy e λρ (95) { d } y dρ ρdy dρ + λ y e λρ (96) l(l + 1) ρ y + } ρ y e λρ λ ye λρ (97) u kl y kl (ρ)e λ klρ λ dy kl l(l + 1) kl dρ ρ y kl (ρ) + ρ y kl(ρ) 0 (98) Note: Make-up class tomoow, 4:00-4:50. We use a powe seies expansion of y: y(ρ) ρ s n0 c n p n, (99) whee s is the lowest powe o ρ and hence c 0 0. Fom (??): dy dρ d y dρ (n + s)c n ρ n+s 1 (100) n0 (n + s)(n + s 1)c n ρ n+s (101) n0 (10) 16

17 n0 Substituting back into (), we get ((n + s)(n + s 1) l(l + 1)) c n ρ n+s + ( 1 λ n+s) c n ρ n+s 1 0 (103) We now attempt to combine the summations; poceed nomally by eodeing. s(s 1) l(l+1)c 0 ρ s + {(n + s)(n + s 1) l(l + 1)} c n + {1 λ(n + s 1)} c n 1 ρ n+s n1 We want to solve each tem fo 0. Fist tem: n0 {s(s 1) l(l + 1)} c 0 0 (104) whee c 0 0. The solutions ae s l + 1 o s l. We don t want to use s 0 because then R u/ρ goes to as ρ 0; we end up with a singula wave function, which is physically impossible. Hence, we have s l + 1. Hence, R kl (ρ) ρ l, whee ρ 0 Fo n 1, we can combine the two sums to get a -tem ecusion elation: So, {(n + l + 1)(n + l) l(l + 1)} c n + [1 λ(n + l)]c n 1 0 c n [λ(n + l) 1] c n 1 n(n + l + 1) Thusday, 11 Septembe 008: If you look at n in (105), then the limit is: (105) c n c n 1 λ n Thus, as n, the coefficient c n has leading ode behavio of: c n (λ)n n! This is bad! This leaves to the powe seies fo e λρ, which is not a physically feasible solution. So, we need a way to get n N max 0. This is tue povided λ(n max + l)

18 o λ So, the enegies ae now detemined: 1 N max + l E kl E I E I E kl λ E I (N max + l) We want N max 1 and l 0. Define n N max + l to be the pincipal quantum numbe. Note that { N max 1, sincec 0 0 (106) l 0 So, the complete wave function (up to nomalization) is N max 1 R nl (ρ) ρ l k0 E n E I n (Note that E n is degeneate in l). c k ρ k e ρ/n (107) 1.8 Deive the fist few tems Gound state has n 1 implies E n E I 13.46eV. n 1 implies N max 1. Why? n N max + l whee N max + l; as N max 1 and l 0. So, n 1. Witing out R 10 : R 10 (ρ) ρ 0 0 c 0 ρ 0 e ρ/n c 0 e ρ k0 We can detemine c 0 by nomalizing R 10 (ρ). So, ψ 100 R 10 Y 00 If we integate, ψ 100 sin θddθdφ Because Y 00 was aleady nomalized, this comes down to picking c 0 such that: 0 R 10 (/a 0 ) d 1 18

19 Changing vaiables a 0 ρ and d a 0 dρ, we get a ρ dρ c 0 e ρ dρ 1 Keep handy the following integal table fomula: Applying this: So, 0 x n e µx dx n!, Reµ > 0 µ n+1 c 0 a 3! 0 3 c 0a R 10 () a 3/ 0 e /a0 (108) ψ 100 (, θ, φ) R 10 ()Y 00 (θ, φ) (109) 19