1.1 Problems with Waves Classically, EM radiation was well understood as a wave phenomenon, via Maxwell s theory. Problems arose around

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1 0. Intoduction Recommended textbooks: Phillips, A. C. Intoduction to Quantum Mechanics Whiley Fench, A. P., Taylo, E. F, An Intoduction to Quantum Physics Thomas Nelson Othe books: Rae, A.I.M. Quantum Mechanics IoP Gasiooqioz, S. Quantum Mechanics Whiley 1. Photons, Paticles and Waves 1.1 Poblems with Waves Classically, EM adiation was well undestood as a wave phenomenon, via Maxwell s theoy. Poblems aose aound i The absence of the aethe Michelson-Moley. How can thee be a wave if thee s nothing to oscillate? ii Black body adiation Classical theoy pedicted intensity ising to infinity as the wavelength went to zeo. Ultaviolet catastophe. Max Planck hypothesised that EM oscillatos e.g. electons with fequency have enegy nh whee n = intege and h = constant. Aveage powe adiated pe unit aea fo wavelengths fom to + d : hc I d = d 1 hc 5 kt e 1 Planck Radiation Law The moden value fo h is h = 6.63x10 34 Js and = h = 1.06x1034 Js Moe useful units of enegy ae ev, KeV, MeV etc. Also the combination hc often cops up. hc = 1.4x10 6 evm = 1.4x10 4 evangstom 1 Angstom = m = 0.1nm c = 1.97x10 7 evm = 1980eVA We will tend to use angula fequency athe than. Then E = h = is the enegy associated with fequency. iii Photoelectic Effect the maximum KE of electons emitted fom a metal suface illuminated with adiation of fequency is: K max = h w -- 3 whee w is the popety of the metal the wok function This was evidence fo photons quanta of EM adiation. iv Atomic stuctue and Specta Boh s desciption used a mixtue of classical and quantum ideas the old quantum theoy. This was vey successful fo a hydogen atom. v Compton effect photons scatte elastically fom loosely bound electons. A eal demonstation of photons as paticles. 1

2 Electon at est initially Photon E, P E, P e P e Consevation of momentum and enegy P = P + Pe E = E + Ee Fo any paticle, including photons: E = p c + m c 4-4 Fo a photon, m=0 so E = pc. i.e. h = pc So the momentum of a photon: p = h c = h -- 5 The scatteed photon has p < p and > It tuns out that = = h 1 cos m e c The quantity e = h m e c =.43x101 m is the Compton wavelength of the electon. 1. Paticles, Waves and Uncetainty In 193, de Boglie postulated that matte paticles e.g. electons could have wavelength popeties. The de Boglie wavelength is: = h -- 8 p Fo non-elativistic paticles; p = mv E = 1 mv = p m Theefoe p = me i.e. = h me non-elativistic 9 Fo an electon, m = 9.1x10 31 kg. i.e. = 1.5 nm with E in electon volts 10 E Theefoe electons with KE in the ev to kev ange in the 0.01 to 1nm ange ae diffacted by cystal lattices, since typical atomic sizes and sepaations ae about 0.1nm. Conside a two slit intefeence patten e.g. Young s slits. Each slit behaves as an independent souce of waves in fixed phase elative to each othe egions of constuctive and destuctive intefeence at a sceen. Finges with a spacing given by s = D d whee d = slit sepaation and D = distance fom slits to sceen. In a paticula electon diffaction expeiment D = 35cm, d = x10 4 cm, K = 50keV, = 0.05A. Then s = 10 6 m = 1µm. Closing one slit gives the single slit diffaction patten. With both slits open, any attempt to detect which slit the electon passes though causes the intefeence finges to disappea, leaving the diffaction patten.

3 At vey low intensity, whee only one photon o electon is in the appaatus at any one time, the intefeence patten pesists. Theefoe it no longe makes sense to ask which slit an individual electon passes though. Conside using a single slit to ty to locate the y position of a paticle. Sceen y x Slit of width a p p y x Diffaction Patten Momentum at stat is all p x - p y = 0. Afte passing though the slit, an paticle must have; a < y < + a Thee is equal pobability to be anywhee in this ange, i.e. P y dy = 1 dy so that a a P y dy = 1 Nomalisation condition a P y 1 a a The uncetainty on the y position is the standad deviation of the position pobability distibution. This can be witten as; y = y y NB: Now y = 0 = denotes the expectation o aveage value. + a a yp y dy y + = a y P y dy = a 1 a y = a 1 = 0.9a a Diffaction p y of any individual paticle will be uncetain afte the slit. The standad deviation of the diffaction patten sin whee = asin * y Is = a ~ position of fist minima 3

4 small P y P P y = h x a = h a We aleady have y = 0.9a P y y = 0.9h = 1.8 = h 1.3 Heisenbeg s Uncetainty Pinciple It is not possible to know simultaneously the position and momentum of a paticle with any moe pecision than is given by; P x x - 1 No expeiment can eve do bette. It applies to each pai of P y y, P z z Thee is also an enegy-time vesion. Et - 11 Whee E is the uncetainty in enegy, and t the time ove which the paticle stays in an enegy state.. The Schödinge Equation and the Wave Function.1 Schödinge Equation Recall that fo photons and paticles; = h p -7 We postulate that the elation E = hf = applies to paticles as well as photons. If we think of a beam of paticles as a fee pogessive wave of momentum p and enegy E pe paticle, we should descibe the wave using k = = p h = p 14 And = E h = E 15 Fo a non-elativistic paticle; E = p +V V = potential enegy m And = k +V 16 m Non-elativistic only A classical, pogessive wave of wave numbe k k = and angula fequency can be epesented as; y x,t = Asin kx t - 17 You can epesent the wave popeties of a feely popagating paticle similaly by the function; x,t = Asin kx t This does not lead to the wave equation satisfying 16 position diffeentiated twice, time once, howeve if we ty; x,t = Ae i kxt 4

5 We can find a suitable equation. It tuns out that complex solutions ae equied by quantum mechanics they ae not just a useful tool as befoe. This is a new and impotant featue. The wave functions x,t Fo x,t = Ae i kzt 18 x = ik x = k = i t Then; ae necessaily complex. m x +V = i -19 t Would satisfy 16 and 18. This is the 1 dimensional time-dependant Schödinge equation. It is the basic equation of non-elativistic quantum mechanics. If potential V is not a function of time, classically enegy is conseved, one has paticle states of definite enegy e.g. electons in stable atoms. These ae called stationay states. Then we can sepaate out the space and time dependence by witing x,t = xt t. Sepaation of vaiables x = T d dx kdt = dt t dt Putting these into 19 gives; m T d dx +V x T = i dt dt Dividing though by T ; = i 1 T 1 d m dx +V x dt dt On the left, we have a function of x only, and on the left a function of t only. This can only be tue if both sides ae equal to a constant, E. d m dx +V x = E 0 i dt = ET -1 dt Equation 0 is the 1E time-independent Schödinge equation. If we know the fom ofv x Equation 1 is easily solved to give; T t = e iet = e it, we can in geneal solve 0 to get x. if we take E to be the total enegy of the system, since E =.. The Wave Function The quantity x,t is called the wave function. It is essentially complex, so cannot be identified with a single physical popety of a paticle. The wave function contains all the infomation that the uncetainty pinciple allows us to know about the state of the paticle. Accoding to the Bon intepetation 196 the pobability to find the paticle in the egion of space fom x to x + dx at a time t is given by; 5

6 P x,t dx = x,t dx = * x,t x,t dx 3 Remindes: let z = a + ib = Re i = R cos + i sin z* = a ib = Re i = R cos i sin Then z = z * z = R = a + b is the modulus of z. = ag z a = tan 1 b If z 1 and z ae complex, z 1 + z = z1 + z + Re z1 * z. A paticle must be found somewhee so thee is a nomalisation condition; dx x,t = 14 Any physically sensible function must satisfy 4. Equation 0 is linea it does not contain squaes of if 1 is a solution, so also is A whee A is a constant. e.g. a nomalisation constant. Also if 1 and ae solutions, then A 1 + B is also a solution. The phase of A o B is not affected by the nomalisation pocess, so Ae i 1 must also be a solution, whee is an abitay eal constant. This does not affect any physical measuable quantities. Note that the time dependant pat e iet = e it of the wave function fo a stationay state has no physically significant impact on obsevables it disappeas when you take the modulus squaed. A suitable wave function must be a solution of the time-independent Schödinge equation and be nomalised and must satisfy bounday conditions; a Must be a continuous single-valued function of x and t othewise would not be well defined fo all positions b dx must be finite to epesent a pobability. c d dx must be continuous except whee thee is an infinite discontinuity in potentialv x..3 Obsevables, Opeatos and Expectation Values Fist ewite Equation 0 Time Independent Schödinge Equation; d m dx +V x = E We can intepet the left hand side as a mathematical opeato opeating on the wave function. The ight hand side is the wave function multiplied by some allowed value of the total enegy. This has the fom; Ê = E -5 Whee Ê = m d dx +V x -6 Ê is the enegy opeato. Solving 5 involves finding paticula functions n which ae physically acceptable they satisfy all the bounday conditions and which, when opeated on by Ê give allowed values of enegy E n multiplied by n. This is known as an Eigenvalue question. i.e. Ê n = E n n 6

7 Ê is the opeato, n is the wave function Eigenfunction, and E n the Eigenvalue. Othe opeatos exist; Fo the x component of momentum; ˆP x = i x -7 We can also wite; ˆp x = ˆp x ˆp x = x 8 Fo x position opeato it is just the algebaic vaiable x, i.e. ˆx = x. Similaly ˆx = ˆx ˆx = x 30 The use of these opeatos allows us to calculate the expectation values of coesponding dynamical quantities using the sandwich method. Fo an opeato Ô the expectation value of the coesponding physical vaiable, O, is given by; O = * Ôdx 31 Recall that when we deived the Time Independent Schödinge Equation we sepaated out the time dependant pat T t = e i Et. Thus, the full wave function coesponding to an Eigenfunction of the Time Independent Schödinge Equation is n x,t = n xe ie n t Example; A paticle has a wave function given by; x,t = a cos x iet a e fo a < x < a The complex conjugate value of P x is; P x = a cos x iet a e x + * 0 outside this egion a cos x iet a e dx = a 3 a a cos x a dx P x = a 3 Recall that, fo any dynamical quantity e.g. x, p, we defined the uncetainty as; x = x x p x = p x p x Etc..4 Commutatos, Compatible Obsevables Uncetainty Conside two opeatos Ô1 and Ô. The commutato is defined as; Ô,Ô 1 = Ô1 Ô Ô1 Ô 1 33 The opeatos ae said to commute if Ô,Ô 1 = 0. i.e. Ô 1 Ô matte in which ode the opeatos ae applied in. Fo example, conside ˆx and ˆP x and some wave function. Then; = Ô Ô 1, i.e. it does not 7

8 ˆP x, ˆx = i x,x i.e. = i x x * +, - x. / x i d = i i + ix x x = i ˆp x, ˆx = i 34 If the opeatos do not commute, it is not possible fo a quantum state to be an eigenstate simultaneously with both opeatos uncetainty elation between the two dynamic vaiables. Fo X position and y-momentum; ˆp y, ˆx = i y x x i * * y +,, + = ix y + ix y = 0 Thus it is a possible fo a paticle to have exact values of p y and x simultaneously. 3. One Dimensional Potential Wells 3.1 Infinite Potential Well V x x = 0 L x Conside a paticle confined within an infinite One Dimensional Potential Well V x = 0 fo 0 < x < L, V x = elsewhee The paticle in a box Classically a paticle of enegy E would bounce backwad and fowad within the 1D egion 0 < x < L. Outside of the potential well, the wave function must be zeo. = 0. It is impossible fo the paticle to be found thee the paticle would need infinite enegy. i.e. 0 = L = 0. We use the Time Independent Schödinge Equation with V = 0 fo 0 < x < L to find x d m dx = E i.e. d dx + me = 0 The geneal solution is; = c 1 e ikx + c e ikx k = me Whee c 1 and c ae constants and possibly complex. Bounday conditions; 0 = L = 0. 8

9 c 1 + c = 0 i.e. x = c 1 e ikx e ikx = ic 1 sinkx solution of the fom = Asinkx whee A is a nomalisation constant. But L = 0 sinkl = 0 i.e. kl = n n = 1,,3,... The values of k ae esticted to a discete set; k n = n L thee ae an allowable set of enegies given by; k n = me n = n L E n = n ml = n h 8mL -35 Note that E n is popotional to n and 1 m and to 1 L. The complete set of allowed wave functions enegy eigenfunctions fo the stationay states of a paticle in a 1D infinite potential well is theefoe; n x,t = A n sin n x L e Fo 0 < x < L. n = 1,,3,.... If n = 0, x,t ie n t - 36 = 0 no pobability of finding anything in the box no paticle. n = 3 n = n = L Nomalisation constants A n can be found using; + dx = 1 The quantisation of enegy levels has aisen fom the bounday conditions. Position pobability distibutions ae obtained, as usual, fom and geneally have points of zeo pobability, i.e. places whee the paticle will neve be found. Although a paticle in a stationay state has a definite value of enegies, it does not have a definite value of momentum. The uncetainty pinciple p x x the naowe the well the lage the uncetainty in momentum. In tems of epesenting a paticle by a wave packet, the naowe the packet the moe wavelengths ae equied to epesent it. Fouie analysis is needed as naow wave packet needs an infinite amount of components to make up the wave. But a wave packet can have a unique, fixed fequency and also a fixed enegy. 3. Non-Stationay Time-Dependant States We have Equation 36; L 9

10 n x,t = A n sin n x ie n t L e Fo a set of allowable enegy eigenfunctions fo stationay constant enegy state fo an infinite potential well. These functions ae othonomal othogonal i.e. n * n dx = 1 m = n 37 0 othewise The Time Independent Schödinge Equation is linea. Any linea supeposition of these n s will also be a solution. e.g. = C C This will be a solution. Nomalised if C 1 + C = 1. It must be possible fo a paticle to exist with this new wave function = C C. What does this physically epesent? It clealy no longe coesponds to a fixed, definite E. x,t = C 1 sin x L e ie 1 t + C sin x L e ie t = f 1 e ie 1 t + f e ie t Whee f ae the spatial pats of the wave functions. Position pobability density is; ie 1 t = f 1 + f + Re f 1 f e e ie t i E 1 E t = f 1 + f + Re f 1 f e = f 1 + f + cos E E 1t Using cos = cos Position pobability oscillates with fequency depending on E E 1, the enegy diffeence. The paticle isn t moving in the classical sense, but the pobability is changing. If the paticle is an electon, also epesents the distibution of chage i.e. in this case the chage oscillates with fequency = E E 1 i.e. E E 1 =. This is the fequency of a photon emitted in a tansition fom the two enegy eigenstates. Fom enegy eigenstate E to state E 1 1. E C C E 1 1 How can a paticle with = C C appea to have moe than one enegy? We conside the equation in tems of measuement. If we measue the enegy we always get eithe E 1 o E, i.e. one of the allowed eigenvalues E n. We cannot know befoehand which one is going to happen. Et t E The pobability of E 1 is C 1. The pobability of E is C. Immediately aftewads the paticle will be in the enegy eigenstate coesponding to the measued value. In othe wods, the measuement has collapsed the wave function. 3.3 The Hamonic Oscillato Potential Classically, this is vey impotant. Foce is popotional to small displacement fom an equilibium position, diected towad the equilibium point. 10

11 F x = Cx Use C to avoid confusion with k =. Potential enegy is; = 1 Cx - 39 V x Thus the potential is time independent. The poblem is 1D use the Time Independent Schödinge Equation in 1D. d m dx + 1 Cx = E - 40 We expect 40 to apply to, fo example, diatomic molecules. Classically o = C m C = m o whee o = natual fequency of oscillations. So ewite 40 as; m d dx Whee o = C m. + 1 m x = E - 41 o We need solutions to 41 such that 0 as x ± so as to be nomalisable. It is found tat fo all non-infinitev x, thee is a finite pobability to find the paticle outside the classically allowed egion. This leads to quantum tunnelling and adioactive decay. It is tedious to solve 41, so we quote the solutions. We define a constant length a as; a 4 = mc = m o i.e. a = m - 4 The enegy eigenfunctions ae; n x = 1 1 n n a Nomalisation facto x H n a Hemite polynomial e x a - 43 Gaussian exponential function Gaussian tem detemines the behaviou as x a ± The Eigenvalues ae E n = n + 1 fo n = 0,1,, o The fist fou solutions ae; 0 x = 1 x = x = 1 a 1 e x a 1 1 e x a a E 0 = 1 * o E 0 = 3 * o x a a 0, - / x 0 e a E 0 = 5 * o We veify that one case, 1 x, is an enegy eigenfunction with eigenvalue 3. o 1 Now, 1 = xe x a a 3 = Axe x a whee A = 1 = const. a 3 11

12 d 1 dx = 1 x x a 1 d 1 dx = 1 x d 1 dx 1 x 1 a x d 1 a dx = 1 x 1 a 1 x 1 a 1 a + x a 4 1 = x 1 a a 4 d 1 = 3 1 m dx ma x 1 ma 4 And V x 1 = 1 m x o 1 = ma 4 x 1 since a 4 = Theefoe d 1 +V x m dx 1 = 3 1 ma = 3 o 1 i.e. E 1 = 3 o QED. m o = x even paity o = x odd paity. Fo any symmetic potential, V x, this has to be tue. We note that the allowed eigenfunctions ae eithe even x odd x Fo any continuous odd function, = 0. Fo any continuous even function, d dx = 0 at x = 0. Fo any sensible w 0 fo x ± This allows one to sketch possible wave functions, especially if the ode of polynomial is known. The enegy levels fo stationay states of a SHO ae equally spaced by o. The gound state has zeo point enegy 1 o. Non-stationay states can be constucted as fo infinite squae wall e.g. = 1 o e ie o t e ie 1 t. 3.4 Diatomic Molecules To a good appoximation, the foce between atoms of a diatomic molecule e.g. O, N, CO, HCl etc can be descibed by SHO fo small displacements about the equilibium position. The total enegy is E = p 1 + p + 1 m 1 m Cx. In the cente of the mass fame of the two paticles; p 1 = p i.e. equal and opposite. Then; E = p m 1 + p m + 1 Cx whee p = p 1 + p i.e. E = p m 1 m + Cx With educed mass µ = m 1 m m 1 + m

13 Then E = p µ + 1 Cx - 46 Fomally this is equivalent to a single paticle of mass µ in a 1D potential V x = 1 Cx solutions 43 will apply fo vibational states of diatomic molecules. Example; obseved fequency of EM adiation fom tansitions between adjacent vibational levels in CO Cabon Monoxide is 6.43x10 13 Hz. Estimate the foce constant C. Level spacing is E = o = h = 4.6x10 0 J = 0.66eV Infaed Now o = C µ C = µ o Fo Cabon Monoxide µ = m 1 m m 1 + m = 1x = 1.14x106 kg Theefoe sping constant C = µ o = 1680Nm 1 In pactice, such molecules also have otational enegy. It is difficult to deduce the vibational enegy levels fom obseving the specta of the adiation. 4. Quantum Mechanics in Two o Thee Dimensions Think of a paabola otated aound to fom a bowl o dish. 4.1 D Simple Hamonic Oscillato Suppose a bound paticle can move in the xy plane in a D paabolic potential. = 1 C x + y - 47 V x,y i.e. V x,y = 1 m x + y o An example may be an atom in a plane in a cystalline stuctue. The kinetic enegy is now; E k = p x m + p y m The 1D Time Independent Schödinge Equation can natually be extended; m x + y + 1 m* x + y o = E - 48 Now = x,y dxdy aound x, y and dxdy = pobability to find a paticle within an elemental aea. We look fo solutions that ae poducts of solutions to the 1D equations fo the X and Y components of the motion. i.e. x,y = f xg y whee f and g ae solutions to 41. Then d f x = g y dx x = f x d g dy Since g is constant with espect to x, and f to y. Then 48 becomes; g y d f m dx + 1 m x f x o + f x d g m dy + 1 m y g y o = ef xg y i.e. g y E nx f x + f xe ny g y = Ef xg y since f x and g y the 1D poblem. i.e. E nx + E ny f xg y = Ef xg y. ae any eigenfunctions of 13

14 So E = E nx + E ny is an enegy eigenvalue of the D oscillato, coesponding to eigenfunction f xg y. Total enegy E = E nx + E ny = n x n + 1 o y = n o + n + 1 x y o fo n x = 0,1,,... n y = 0,1,,... The D wave functions ae; x +y x nx x,y n y = H nx a H y n y a e a - 49 Now we can have moe than one state with the same enegy. E.g. 1,0 x,y as the same as 0,1 x,y. This is known as degeneacy. enegy o In the D case the enegy of the stationay state is no longe enough to specify the eigenstate; we need two quantum numbes. These could be n x and n y, but could also be enegy i.e. n x + n y, and some othe equation numbe. In pactice, n x and n y have little physical significance. 4. The Z-Component of Angula Momentum We take a cue fom classical physics, expeiment e.g. fine stuctue in atomic specta, and symmety. A paticle moving in the xy-plane has angula momentum about the z-axis i.e. L is in the ±z diection. Chaged paticles with obital o spin angula momentum give ise to magnetic moments expeiments show that these ae quantised. Classically, angula momentum of a paticle with momentum p and position is; L = xp = ˆx ŷ ẑ x y z 50, 51 p x p y p z Fo motion in a plane = x,y,0 p = p x,p y,0 magnitude L z = xp y yp x 5 and L points along the z axis with We postulate that the Quantum Mechanics opeato fo L z is just; ˆL z = ˆx ˆp y ŷ ˆp x = x i x y i y = i x y y y 53 We will use plane pola coodinates, x = cos whee y = sin The chain ule of implicit diffeentiation says; = x x + y = sin y x + cos y = x y + y x i.e. ˆL z = i 54. We wite wave function as, The D Simple Hamonic Oscillato potential depends only o whee = x + y and not. Theefoe we assume solutions of the fom;, = R 55 14

15 Conside applying the ˆL z opeato. ˆL z = i Eigenfunction of ˆL z would satisfy; = ˆL z R i R i.e. ir = ˆL z R i = ˆL z - 56 Simila to 1 i.e. i ˆL z = 0 i ˆL z The solutions = e - 57 Compae this with T t functions. = e iet fo the Time-Dependant pat of stationay state wave The bounday conditions ae that * and have to be single-valued at evey point in. Fo example + ˆL e i z + ˆLz = e i ˆL e i z = 1 = i.e. L z = m fo m = 0,±1,±,±3,... L z = m fo m = 0,±1,±,±3,... Then, = R e im 58 could be a solution. This stiking esult is bone out by expeiment evey measuement of one component of obital angula momentum conventionally the z component always yields an intege multiple of. i.e. is the fundamental quantum of obital angula momentum. 4.3 Angula Momentum Eigenfunctions fo D Simple Hamonic Oscillato Can we find linea combinations of the degeneate enegy eigenfunctions of the type 49 such that the esults ae of the type 58? We take wave functions labelled by n x and n y with the same enegy eigenvalues and combine to poduce new wave functions labelled by n = n x + n y i.e. total enegy and m giving L z. Combined states will have the same time-dependant facto e iet so combined states will still be solutions of the time-dependant Schödinge equation. Recall 49; nx n y x,y And 58 = H nx x a H y n y a e x +y a 15

16 = R, Clealy e e im x +y a = e a and so this is pat of R. Fo simplicity, we take the case n x + n y = 1. Two possibilities; n x,n y i.e. fom 43 et seq. 1,0 = Cxe a = 1,0 o 0,1. 0,1 = Cye a Whee C is an oveall constant. 1,0 = C cos e a 0,1 = C sin e a Note by inspection that; 1,0 + i 0,1 = C cos + i sin e And 1,0 i 0,1 = Ce a e i. a = Ce a e i = Ce z Thus, we have two functions of the fom R e im whee R a and m = ±1. These ae clealy eigenstates of total enegy E = n + 1 o and component of angula momentum, with L z = ±. 16

17 4.4 Angula Momentum in 3D Recall classically; ˆx ŷ ẑ L = xp = x y z Equations 50, 51 p x p y p z L = yp z zp y,zp x xp z,xp y yp x In tems of opeatos; ˆL x = ŷ ˆP z ẑ ˆP y = i y z z y 59 ˆL y = ẑ ˆP x ˆx ˆP z = i z x x z 60 ˆL z = ˆx ˆP y ŷ ˆP x = i x y y x 61 Can a paticle be simultaneously in an eigenvalue state of any two of these opeatos? Conside e.g. ˆL z, ˆL x. Look at the commutato ˆLz, ˆL x acting on some wave function. Then; ˆL z, ˆL x = L L L L z x x z = x y y + * x, - y z z + * y, - + y z z + * y, - x y y + * x, - = x z z + * x, - = z x x + * z, - = i ˆL y i.e. ˆLz, ˆL x = i ˆL y. The same applies to the othe diections, i.e.; ˆL z, ˆL x = i ˆL y ˆL x, ˆL y = i ˆL z 6 ˆL y, ˆL z = i ˆL x Commutatos ae non-zeo; no states fo which any pai of angula momentum components have simultaneous eigenvalues. Thus it is only possible to know one component pecisely. Conventionally taken as L z i.e. so we can t know the diection of vecto L. But can we know L? We look at opeato fo squae of total angula momentum. L = L x + Ly + Lz = Lx Lx + Ly Ly + Lz Lz 63 i.e. L = y d dz z d dy y d dz z d dy + z d dx x d dz z d dx x d dz + x d dy y d dx We will teat mainly spheically symmetic poblems, we use symmetical polas:- 17

18 z x, y, z x y x = sin cos y = sin sin z = cos = x + y + z cos = z 64 = z x + y + z tan = y x We aleady have; ˆL z = i Equation 54 It can be shown See Fench Taylo, box 11-1; L = + cot + 1 sin * - 65 If L,L z = 0 we can have states with simultaneous eigenvalues, and so definite values of L and L z but not L x and L y Note that only appeas as and, and =. L,L z = 0. By symmety, L,L x = L,L y = Eigenfunctions and Eigenvalues of L We shall be looking fo wave functions,, which ae simultaneously eigenfunctions of L and L z. Note that the L opeato does not depend on ty sepaating out the dependence. i.e.,, = R F, The eigenvalue equation will be; L = L i.e.; L R F, R L F, = L R = R F, L F, 18

19 Theefoe L F, = L F, since R is constant with espect to L Using equation 65 gives; + cot + 1 sin * F = L F - 67 We know that the dependence of F must include the tem e im since F is assumed to be an eigenfunction of L z. We ty to find F, = P e im. We can then substitute fo F, and divide by eim, to get; d d + cot d d + m sin P = L P Now L is the squae of an angula momentum so we wite it as whee is some constant This could in pinciple be a faction, not a eal numbe. So we have; d P d + cos dp sin d + m sin P = 0 68 Solving to 68 will be functions P 0. that ae consistent with the eigenvalue equation 67, such that L = and L z = m. Let s ty vaious possible solutions fo 68. a P = const. This woks if; m sin = 0 This has to be tue fo all = m = 0. all values of b P = sin. dp This gives; d = cos and d P d = sin sin + cos m + sin sin sin = 0 sin + 1 sin + sin m sin sin = 0 sin + sin + 1 sin m sin = 0 This woks if = and m = 1 m = ±1 c P = cos. This gives; cos + m sin cos = 0 This woks fo =, m = 0. d P = sin This woks fo = 6, m = 4 i.e. m = ± Etc Systematic exploation shows that solutions to 68 exist fo a seies = 0,,6,1,0,... These ae the solutions fo of the fom = + 1 Fo each value of L thee ae + 1 L is the obital angula momentum quantum numbe. whee = 0,1,,3,... solutions coesponding to m =,...,0,...,+. The simultaneous quantisation of L and L z has lead to; 19

20 = 0,1,,... L = L z = m m Figue 11.3 of Fench and Taylo gives a semi-classical pictoial epesentation. L z + z + L = z 0 6 L = 6 The eigenfunctions satisfying 68 ae called associated legende function P,m The combination F, = P,m e im is called a spheical hamonic, usually witteny,m, See tables in textbooks The complete eigenfunctions would be;,, = R Y,m, Rotational States of Molecules We can teat molecules as igid bodies e.g. neglecting vibations which have vey small amplitude elative to the equilibium sepaations. Fo a igid body, E k of the otation is; E ot = 1 I Whee I is the moment of inetia about the axis though the cente of mass.. = angula fequency of otation. But L = I Theefoe E ot = 1 I L I = L 71 I But L is quantised with L = + 1 fo = 0,1,,... and we expect otational enegy levels E = I Fo a diatomic molecule with atoms of mass m 1, m sepaated by o, educed mass is; µ = m 1 m m 1 + m And I = µ o Then; E = µ o 0

21 Example; hydogen chloide HCl o o 1A, µ 1.67x10 7kg Note that since m C/ >> m H, µ m H = Then E E 1 = + µ o µ o The vibational enegy level spacings incease with in contast to the even spacings in the vibational specta. E.g. fo HCl ; = 3 3 µ o 0.144eV = = 1 = 0 µ o µ o 0.077eV 0.06eV 0eV 5. Cental Foce Poblems and the Hydogen Atom 5.1 The Schödinge Equation in 3D We simply extend fom 1D and D to wite; m x + y + z +V x, y,z = i t E k tem Whee = x,y,z,t function. E p tem. The last pat of the equation is the time evolution of the wave The time independent vesion of this can be witten in tems of Laplacian and = x,y,z and fo V = V time-independent potential as; m +V = E 74 Whee,t = e iet In pola coodinates; = cot + 1 sin * - 75 Note altenative fom fo fist tem of ; + = 1-76 Fo D poblems, no dependence. sin = 1. = = Motion unde a Cental foce 1

22 v t V P v O Classically, we can esolve the velocity vecto into adial, v, and tansvese, v t, components. Total enegy is; E = 1 mv + 1 mv +V t - 77 since V is a function of adial distance fom the oigin. The angula momentum L = mv t. Classically L is conseved fo cental foce motion. This gives; v t = L m. We wite adial component of linea momentum as P = mv Then 77 becomes; - 78 E = p m + L m +V Moving fom classical to quantum; P p = + Fom 75 L L = + cot + 1 sin + Fom 65 * So the Time Independent Schödinge Equation would be; m + m * + cot* * + 1 sin * + +V = E 79 It tuns out to be useful to wite 79 in tems of athe that just. Fist we note that; = + = + = + Multiplying 79 by gives; L +V = E 80 m m This is the 3D Time Independent Schödinge Equation fo a cental foce. We will be looking fo solutions to 80 of the fom;,, = R Y,m, since we aleady know that Y,m, and L z. Thus we wite the function R = u R 81 so that = u Y,m, = R ae eigenfunctions of L as; Y,m, Note that since must emain finite as 0 then we must have u We will also use the esult; 0 as 0.

23 L = + 1 so thee is no longe any, to get; dependence in the equation, and we can cancel though the Y,m, d m d + 1 +V u = Eu - 8 m We have educed the poblem to a 1D equation in the adial coodinate. 5.3 The Coulomb Potential and the Hydogen Atom Coulomb Potential V = e 4 o be looking fo solutions to 80 of the fom;,, = R Y,m, since we aleady know Y,m, We use the Boh Radius; a o = 4 o e o = 0.53 A m e and ewite = a o i.e. = a o fo an electon in a z = 1 atom e.g. hydogen. We will ae eigenfunctions of L and use the Rydbeg enegy E R = 13.6eV 4 o a o and ewite E = o E R i.e. = E E R Equation 8; d m d + 1 e m 4 o u = Eu V has been eplaced with the electostatic potential d u d u u = u - 83 e and L z. i.e. d u d +V u = u - 84 eff whee the effective potential defined in tems of dimensionless quantity is; = + 1 v eff - 85 O in tems of ; v eff = + 1 e m 4 o - 86 Centifugal potential Coulomb Potential 5.4 Radial Wave Functions fo Hydogen We fist look at the behaviou of 83 fo some limiting cases. Fo vey lage distances, Then we have d u d u 3

24 Bound states coespond to negative total enegy i.e. need to put in enegy to fee electons i.e. is positive. We set b = b is eal and positive Then, u ~ e +b and u ~ e b would be possible solutions, but e +b is not acceptable because the wave function would not be nomalisable. Fo small distances, 0 The leading ode tems ae d u d u = u Because the thid 1 tem becomes much smalle than 1 The equation has two foms of solutions; u But R = u ~ +1 and u ~ p. u 0 as 0 to keep R finite Howeve does not go to zeo as 0 see equation 81 Thus, we assume u ~ +1 fo the ange 0. tem as 0 Theefoe we wite solutions to 83 as; u = +1 e b f 87 is a function that does not involve the equied small and lage behaviou. whee f We can substitute 87 into 83 and obtain afte much algeba and calculus [execise]; b d f d df d + 1 b + 1 f = 0-88 Solutions exist only fo cetain values of enegy paamete b ecall b = = E E R Simplest solution is f = cons tant - 89 = 0 which woks fo 1 b + 1 i.e. b = This solution coesponds to enegy E R E = b E R = + 1 since + 1 is an intege, E is quantised. We know that, in the Boh modes, enegy levels ae given by E n = E R n n = 1,,3,... So w can identify this solution as belonging to a pinciple quantum numbe n such that n = + 1 o = n 1. We go on to look fo solutions to 88 whee f = polynomial. The simplest such solution is; 4

25 = c o c 1 +1 e b - 90 u This woks execise if; b = 1 + E R i.e. E = + This coesponds to n = +, o = n. We can classify states in tems of a paticula value of so states coesponding to 89 and 90 coespond to diffeent enegies. Altenatively, we could take states as belonging to the same value of n and so they have the same enegy, but diffeent. = n 1and l = n. Systematic analysis of possible of possible solutions shows that polynomials f exist fo all values of with 1 b = n = + 1, +,... Altenatively, fo a given value of n = 1 b, solutions can be found fo intege values of such that n 1. It is usually moe convenient to classify states fist with n, the pinciple quantum numbe. The geneal fom of the full solutions is; u n, = +1 e n1 i c i i n 1-91 i=o Notes: Solutions ae indexed by n and. We have set b = 1 n. The coefficients, c i, ae positive, and 1 The numbe of nodes in f n is called the adial quantum numbe. i factos ensue altenating signs. is given by n = n 1 not counting the nodes at 0 and. Fom 81, the complete adial Coulomb wave functions ae; R n, = e n i c i i n 1 9 i=0 whee n = n 1 and = a o. Some functions R n, ; n = 0 = 1 = 1 e a o 1 a e a o a o 7a o e 3a o e a o 1 a o 6a e o a o 3a o a o e 3a o 5.5 Complete Wave Functions fo Hydogen Fom equations 70 and 9, the complete stationay state wave function fo an electon in a coulomb potential: 5

26 n,,m,, = R n, Y,m, - 93 im = R n, P e whee P,m, ae associated Legende functions. R n, = a e n1 na o 1 c i o a i=0 o The eigenfunctions ae chaacteised by thee quantum numbes; n - the pinciple quantum numbe. n = 1,,3... E = E R n - the obital angula momentum quantum numbe. = 0,...,n 1 m = azimuthal o magnetic quantum numbe. m =,..., i L = + 1 L = m. z How to deduce n,,m by inspecting the wave function, and to check if it is a possible wave function. ~ a e o 1 n a o is the powe in the leading e.g.; sin cos = 3cos 1 = polynomial P,m e im* a o tem = the sum of the sin and cos tems in P.,m n is the diviso of a o tem in the exponential. n = n 1= numbe of nodes in the polynomial = ode of polynomial. m = intege facto in the e im tem. The usual nomalisation condition is; =0 d n,,m,, sind d =0 =0 whee dv = d sindd = infinitesimal volume element. Note that taking fom 0 while taking fom 0 sweeps out all of angula space. No dependence in because e im = 1 pobability is unifom in azimuthal angle. The adial pobability density fo position and so also chage is got by integating ove and : P d = 4 R d which is the pobability to find an electon within a spheical shell between and + d. Inceasing n gives a tendency fo the electon to be futhe fom the nucleus. Thee is a maximum pobability at = n a o fo = n 1. Fo = 0, the electon can be found in the vicinity of the nucleus. This leads to electon captue by a poton, thus adioactive decay. States with n = n 1= 0 ae the most Boh-like, i.e. no nodes in adial pobability distibution no positions of zeo pobability. We look also at the dependant modulation of the pobability distibutions. We use pola diagams; 6

27 Distance fom oigin P Fo = 0, no dependence. Fo = 0, m = 0. Y,m ~ cos. m = 0 L z = 0 This agees with the semi-classical pictue of electon close to z-axis. Fo = 1, m = ±1, Y,m ~ sin z x, y m = ±1 L z = ± Paticle is moe likely to be in the xy plane. 5.6 Othe single-electon atoms We extend the analysis of hydogen z = 1 atom, to simila one-electon atoms, e.g., Li ++ z = 3. Nuclea chage becomes +z e. He + z = Potential isv = Ze 4 o, and the Boh adius becomes a = a o. The enegies o z become E n = z E R n. Thus fo He+, enegy levels ae times 4 compaed with Hydogen and the electon obits a facto of close to the nucleus. Fo Li ++, factos ae x9 and x3. We have assumed that the mass of the electon will be negligible compaed with the nuclea mass. Stictly, we should used educed mass, which will give small shifts in enegy levels, and diffeences between H, He + and Li ++. 7

28 6. Multi-Electon Atoms 6.1 Electon Spin The Sten-Gelach expeiment 191 showed that electons have intinsic spin angula momentum which gives ise to magnetic moment. The total spin angula momentum is given by S = s s in analogy with L, whee s = spin angula momentum quantum numbe. In fact, s can take only one value s = 1. Theefoe S = 3 4 fo an electon and poton and neuton. Thee is s + 1= possible values fo the z component of spin, labelled by the quantum numbe m s. Sometimes m fo obital angula momentum components is labelled m. m s = + 1 spin up m s = 1 spin down Theefoe to completely specify quantum states of an electon in an atom 4 quantum numbes: n,, m, m s. Then fo each value of n, n 1 values of, + 1 values of m, values of m s. Thus degeneacy is n, since to fist ode enegy depends only on n. 6.3 Combined states of two identical paticles Conside two non-inteacting electons in the same potential e.g. He atoms. Fo simplicity, conside a 1D case. It can be shown that the wave function can be witten as a poduct: = A x 1 x 1,x,t whee; A x 1 B x e iet - 97 = wave function fo electon A at x 1 = wave function fo electon B at x B x E = E A + E B E A, E B ae the enegy eigenvalues The position pobability density is pobability to find electon A nea x 1 at the same time as electon B nea x is; P x 1,x dx 1 dx = A x 1 dx 1 B x dx The electons ae indistinguishable we neve know whethe we ae obseving A o B. Pobability must be the same if intechangeable labels i.e. P x 1,x dx 1 dx = P x,x 1 dx dx 1 i.e. A x 1 B x = A x B x 1 A and B ae, in geneal, diffeent functions and this condition will not be satisfied. We can constuct functions that do satisfy the intechange of labels equiement: S x 1,x = 1 x A 1 B x + A x B x 1-98 is the symmetic combination, since S x 1,x = S x,x 1. = 1 x A 1 a x 1,x B x = a x,x 1 A x B x 1-99 Now a x 1,x This is the antisymmetic combination. A full analysis would include the spin pat of the wave functions. 8

29 6.4 Pauli Exclusion Pinciple It tuns out that natue always chooses antisymmetic wave functions fo combinations of two o moe spin- 1 paticles known as femions. Paticles with intege spin occu in the symmetic combinations. These ae called Bosons. Since, fo electons, the oveall function must be of type 99, A B o a would be zeo. Theefoe each electon must have a diffeent wave function i.e. diffeent set of quantum numbes in an atom Pauli s Exclusion Pinciple. No two electons in the same atom can have all quantum numbes the same Multi-Electon atoms The enegy of a level in a multi-electon atom depends on oveall chage distibution of the nucleus and othe electons. Inne electons patially shield oute ones fom nuclea chage, educing z to z* o z eff effective. Atoms of alkali metals Li, K, Na,Cs ae paticulaly hydogen-like. Most loosely bound electon is in = 0 state well shielded fom the nucleus. Conventional spectoscopic notation is to label the angula momentum states; = s p d f g h and pecede with values of n. Some gound state configuations ae; H : 1s He: 1s electons, m s = ± 1 in 1s level. Li: 1s s 1 Na: 1s s p 6 3s K: 1s s p 6 3s 3p 6 4s K L M N Shells In k, the 4s electon penetates close to nucleus than 3d and so sees a lage z eff and so has lowe moe negative enegy i.e. moe tightly bound. 9