Midterm 2 Review. Drew Rollins

Transcription

1 Midterm 2 Review Drew Roins 1 Centra Potentias and Spherica Coordinates 1.1 separation of variabes Soving centra force probems in physics (physica systems described by two objects with a force between the objects that is a function ony of the distance between them) is handed by taking advantage of reduced mass coordinates. In this kind of system we take a probem of 2 partices m 1 and m 2 at positions r 1 and r 2 and use the foowing change of variabes: 1 µ = 1 m m 2 (1) r = r 1 r 2 (2) R = m 1 r 1 + m 2 r 2 m 1 + m 2 (3) By doing this we are abe to reduce the probem of soving the motion of two bodies about each other, to the probem of soving the motion of the reduced mass about some stationary point. 1 Note, this is an exact method. However if as is the case in soving the hydrogen atom, if one of the masses is a ot bigger than the other (m p 1800m e ), then the reduced mass is approximatey equa to the smaer of the two masses. So in soving the hydrogen atom the approximation is typicay made µ m e. This is a very good approximation, but it s worth noting that when you do this you are now thinking about the proton being fixed at the origin, and the eectron moving around it, and this physica picture is an approximation. If we appy this reduced mass coordinate system to the Schrödinger equation, where the force between the two bodies is described by V (r), we have the equation ) ( 2 2µ 2 + V (r) ψ = Eψ Since the potentia is ony a function of r, it s natura to choose a coordinate system where one of the coordinates is the distance from the origin. If we do this and choose spherica poar coordinates, then we need to write the Lapacian in this coordinate system: 2 = 1 r 2 r 2 r + 1 r 2 ( 2 θ 2 + cot θ θ ) sin 2 θ φ 2 1 Stricty speaking the reduced mass does not exist and if you want to retrieve the information about the position of m 1 and m 2 you have to use eqn s 1-3 again. The advantage is that you end up soving a differentia equation of 1 variabe instead of 2. 1

2 Note, you may see a different form of this in some books, but they are equivaent (trigonometric identities and such). So now that we have the Schrödinger equation in spherica coordinates, we ook for a soution to the wavefunction in this coordinate system ψ(r, θ, φ). Because we chose spherica coordinates, and V is ony a function of r, it s possibe to separate the probem into three different 1-D probems. This is done by the typica separation of variabes technique. We assume that ψ is a product of three functions, each ony of one variabe: ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ). Now we are faced with the probem: [ 2 2µr 2 r 2 r 2 2µr 2 ( 2 θ 2 + cot θ θ + 1 sin 2 θ 2 φ 2 ) + V (r) R(r)Θ(θ)Φ(φ) = ER(r)Θ(θ)Φ(φ) By going through the separation of variabes, the strategy is to first move a of the r s and derivatives with respect to r on one side of the equation, and a of the anguar parts on the other side. Then we say that because the equaity has to be true for arbitrary vaues of r, θ and φ, both sides must equa a constant (this is the same exact way that we separated the spatia and tempora parts of the time dependent S.E. out in the first week of cass, reducing it to a spatia probem ony). First though, et s see what happens when we fix the vaue of r. That is to say, we imagine a system where r can not change. The physica picture of this is two masses fixed in space reative to each other that can ony rotate. 1.2 spherica harmonics Fixing r, the radia function is forced to be a constant which wi not matter as the wavefunction wi be mutipied by whatever constant is required to normaize it. The r derivatives wi be zero, and the V (r) is just a constant, changing the ground state energy (which is arbitrary because it is aways reative to something ese), so et s set it to zero. This eaves us with the Schrödinger equation for the rigid rotor µr 2 ( 2 θ 2 + cot θ θ + 1 sin 2 θ 2 ) φ 2 Θ(θ)Φ(φ) = EΘ(θ)Φ(φ) (4) The term µr 2 is nice because it has the cassica definition of moment of inertia, so we ca it I. From cassica mechanics we know that the rotationa kinetic energy and anguar momentum are reated by E = L2 (5) 2I So if the stuff on the eft side of (4) is operating on the anguar wavefunction and returning E, this eads us to want to make the identification ( ˆL 2 = 2 2 θ 2 + cot θ θ ) sin 2 θ φ 2 (6) This is starting to ook nice. Soving for the wavefunctions in (4) takes some effort, but to do it we use the separation of variabes strategy. The first thing that we find out is that when we put φ on one side and θ on the other we have a reativey simpe D.E. to sove for Φ(φ). If the separation constant is chosen to be m 2 then we have 1 d 2 Φ Φ(φ) dφ 2 = m2 2

3 This probem is easiy soved: Φ(φ) = A m e imφ (7) Where the A m is used to normaize the function, and the aowed vaues for m are found by considering the boundary conditions (Φ(θ) = Φ(θ + 2π)), and we come up with m = ± integers. Soving the θ part is a bit messier, but in the end we make the identification Y m (θ, φ) = Θ(θ)Φ(φ) = Θ(θ)A m e imφ (8) where is aowed to be 0, 1, 2,... and m = 0, ±1, ±2,...±. The spherica harmonics are eigenfunctions of 2 operators: ˆL 2 Y m (θ, φ) = ( + 1) 2 Y m (θ, φ) (9) ˆL z Y m (θ, φ) = m Y m (θ, φ) (10) With this knowedge, we can go back and rewrite the 3-D Schrödinger equation (aowing for dispacements in r now). If we do this and make the substitution R(r) = u(r) r, we have... [ µ r r 2 r + ˆL 2 2µr 2 + V (r) u(r) r Y m (θ, φ) = E u(r) r Y m (θ, φ) Because the spherica harmonics are eigenfunctions of the ˆL 2 operator and they pass through the derivatives in r, we can aow the ˆL 2 operator to do it s work, and then divide out the spherica harmonic function, eaving the radia equation: [ µ r r 2 r + 2 ( + 1) u(r) 2µr 2 + V (r) r = E u(r) r and canceing out the r s we have: [ 2 2 2µ r ( + 1) 2µr 2 + V (r) u(r) = Eu(r) (11) This is a manageabe probem. We are now eft with a differentia equation that is a function of r ony. If we make the identification V eff (r) = 2 ( + 1) 2µr 2 + V (r) (12) then we see that this is just the 1-D S.E. So why does it make sense physicay to make the identification (12)? We, maybe we shoud be thinking about what a potentia energy is in the first pace. In soving the S.E. we are considering energy eigenstates. This means that as the partice moves around in its wavefunction it aways has a tota energy E that is distributed between its motion (kinetic energy) and the cost of occupying certain regions of space (potentia energy). If the anguar momentum of the system of interest is conserved (which is a reasonabe picture to have because we aready described its anguar motion by a spherica harmonic, an eigenfunction of ˆL 2 ), then the kinetic energy associated with rotation, as described by (5), wi vary as r changes. More rotationa kinetic energy means there is ess avaiabe for motions in the r direction and vice-versa. So it makes sense that we think of the rotationa kinetic energy term as a potentia energy as far as the radia 3

4 wavefunction is concerned. Now, as gets bigger, this term gets bigger. What does this mean? We, this wi make V eff get big as r 0 faster. So the radia wavefunction wi be pushed out a bit, i.e. R R wi be bigger for bigger vaues of r. If you reate this to twiring in space a ba tied to the end of a rubber band, you notice as you twir faster (more anguar momentum) the ba moves out because of what you ca centrifuga force. In an anaogous way then, we ca the part of V eff associated with anguar motion the centrifuga barrier because it forces R(r) away from the origin. A few points about the spherica harmonic functions themseves: The spherica harmonic functions that you see in a book are normaized. This means that if you are to do Y m Y m you shoud get 1. Furthermore, they are orthogona, so Y m Y m = δ m mδ. The radia part of the wavefunction has it s own, independent normaization constant, but when you see the radia and anguar parts of the wavefunction mutipied together and isted in a tabe, these constants are typicay aready mutipied together. The spherica harmonics are defined on the domain 0 < θ < π and 0 < φ < 2π. The number of anguar nodes corresponds to the quantum number. So = 0 is the ground state and there are no anguar nodes. As you traverse the spherica harmonic on the domain of either one of its coordinates it wi not go to zero. For = 1 because this is the first excited state there wi be one anguar node Anguar Momentum As aready stated, the Y m are eigenfunctions of the ˆL 2 and ˆL z operators. You have aso seen the raising and owering operators and their action on the Y m. ˆL 2 Y m = 2 ( + 1) Y m (13) ˆL z Y m = m Y m (14) ˆL + = ˆL x + iˆl y (15) ˆL = ˆL x iˆl y (16) ˆL z ˆL+ Y m = (m + 1) Y m+1 (17) ˆL z ˆL Y m = (m 1) Y m 1 (18) Don t forget where we came up with these operators in the first pace. The anguar momentum operators are derived from their cassica anaogs... L = r p So to generate the quantum mechanica operators we do the cross product: ˆL = ˆr ˆp = i j k ˆx ŷ ẑ = iˆl x + j ˆL y + k ˆL z ˆp x ˆp y ˆp z 4

5 giving us the three anguar momentum operators in cartesian coordinates: ˆL x = ŷˆp z ẑ ˆp y ˆL y = ẑ ˆp x ˆxˆp z ˆL z = ˆxˆp y ŷˆp z with the momentum operators given by ˆp x = d i dx ˆp y = d i dy ˆp z = d i dz The components add in quadrature ˆL 2 = ˆL 2 x + ˆL 2 y + ˆL 2 z A change of coordinates can be used to express the anguar momentum operators in spherica poar coordinates. The simpest one of these is ˆL z = i This makes sense when we consider its action on the known functiona form of the Y m, (8). 1.4 isotropic osciator ˆL z Y m = i d dφ d dφ Θ(θ)A me imφ = i Θ(θ)A m d dφ eimφ = i (im)θ(θ)a me imφ = m Y m You saw the isotropic osciator a few times. This means that we set V (r) = kr 2. This probem is actuay more natura to sove in Cartesian coordinates, because r 2 = x 2 + y 2 + z 2 and the probem easiy separates as a product of three harmonic osciator wavefunctions. This is just ike soving the P.I.B. in 3D. We can however do it using (20). You approximated the soution this way as a homework. The energy eves are E = ω(3/2 + n x + n y + n z ) where each quantum number can be 0, 1, 2, hydrogen atom The hydrogen atom is a pretty important probem. This probem uses the couomb interaction between the eectron and proton: e2 V (r) = 4πɛ 0 r 5 (19)

6 so (20) becomes [ 2 2 2µ r ( + 1) 2µr 2 + e2 u(r) = Eu(r) (20) 4πɛ 0 r Soving this is aso messy. After soving u(r) we write out the soutions to the hydrogen atom ψ nm (r, θ, φ) = u(r) r Y m (θ, φ) = R n (r)y m (θ, φ) The aowed quantum numbers come out of finding a series soution for R(r) and reaizing that this series must terminate to yied a physicay reasonabe soution. From soving the radia equation we get that n = 1, 2..., = 0, 1,...n 1. The condition that m = 0, ±1, ±2,... ± comes out of the spherica harmonic soution. The quantum number n tes us how many nodes the function has to have. n = 1 means no nodes, n = 2 means one node... tes us how these nodes wi be partitioned between the anguar, and radia degrees of freedom. = n 1 means that a of the nodes wi be anguar. = 0 means a of the nodes wi be radia. When evauating an inner product or expectation vaue of the hydrogen atom wavefunction, you do π 2π ψ f ψ = r 3 ψ fψ = r 2 dr sin θdθ dφψ f ψ And if f is ony a function of r, the anguar parts of the integra separate out, and you get ψ f ψ = Y m Y m r 2 drr(r) f(r) R(r) Of course, to do this correcty you have to keep the normaization constant associated with the radia part of the wavefunction with R(r), and not get this constant mixed up with the Y m. 2 Approximation Methods 2.1 Perturbation Theory Perturbation theory is a usefu way to approximate energy engenvaues and eigenfunctions to probems that are very simiar to probems we aready know the answer to. It is deveoped by assuming we have a Hamitonian which is the sum of a Hamitonian with known soutions Ĥ(0), and some sma perturbing hamitonian Ĥ(1). The way that I deveop it here is sighty different then the way it was done in cass, but the foowing is a very common and usefu way of getting to the resuts. If you aren t interested that s OK. You can just remember the basic idea (eqn s 21, 22, 23) and the first order correction to the energy (eqn 28). The size of the perturbation is kept track of using the size parameter λ. So, athough we woud ike to sove the probem: Ĥ ψ n = E n ψ n (21) we reaize that this probem is too hard to sove. So we take advantage of the form of the hamitonian and if we can break it up into a part we know how to sove and some sma part, the probem is recast as: Ĥ = Ĥ(0) + λĥ(1) (22) 0 6

7 where Ĥ0 is a hamitonian that we do know how to sove exacty, with eigenvaues and eigenkets given by Ĥ (0) ψ n (0) = E n (0) ψ n (0) (23) Because the perturbing hamitonian scaes with λ, as λ 0 the perturbed eigenvaues and and eigenfunctions must reduce to the unperturbed ones. For this to happen we set up the soutions as power series in ambda. E n = E (0) n + λe (1) n + λ 2 E (2) n +... (24) ψ n = ψ (0) n + λ ψ (1) n + λ 2 ψ (2) n +... (25) If we operate on the expanded ket with the fu hamitonian, keeping ony terms that are first order in λ... (Ĥ(0) ( ) Ĥ ψ n = + λĥ(1)) ψ n (0) + λ ψ n (1) + λ 2 ψ n (2) +... ( E n (0) + λe n (1) E n ψ n = Ĥ(0) ψ n (0) + Ĥ(0) λ ψ n (1) + λĥ(1) ψ n (0) + O(λ 2 ) ) ( ) ψ n (0) + λ ψ n (1) = E (0) n + λĥ(0) ψ (1) n + λĥ(1) ψ (0) n + O(λ 2 ) E (0) n + λe (1) n + λe (0) n ψ (1) n + O(λ 2 ) = E (0) n + λĥ(0) ψ (1) n + λĥ(1) ψ (0) n + O(λ 2 ) Where O(λ 2 ) mean terms that are proportiona to higher powers of λ (remember because λ is a sma number, higher powers wi be increasingy sma so we can ignore them to a good approximation). Now if we coect terms of equa powers in λ we have two equations E (0) n = E (0) n (26) E (1) n + E (0) n ψ (1) n = Ĥ(0) ψ (1) n + Ĥ(1) ψ (0) n (27) Note here that by keeping higher powers of λ we coud have generated more equations, corresponding to higher order corrections. The first thing that we are interested in doing with these perturbation equations is soving for the first order correction to the energy, E n (1). Taking the inner product of (27) with ψ n (0)... E (1) n ψ (0) n + E (0) ψ (0) n ψ (1) n = ψ (0) n Ĥ(0) ψ (1) n + ψ (0) n Ĥ(1) ψ (0) n E (1) n ψ (0) n + E (0) ψ (0) n ψ (1) n = E (0) ψ (0) n ψ (1) n + ψ (0) n Ĥ(1) ψ (0) n Now, we know that the eigenkets in the power series must be orthogona, so terms ike ψ (0) n ψ (1) n are zero. If we assume that the eigenkets of Ĥ are propery normaized, so that ψ n ψ n = 1, then we have the reationship that we were ooking for E (1) n = ψ (0) n Ĥ(1) ψ (0) n (28) So in perturbation theory, the first order correction to the energy eigenvaue is just the expectation vaue of the perturbing hamitonian taken with respect to the eigenkets of the unperturbed hamitonian. In cass Prof. Yang deveoped the theory further. One main distinction to note is that if the eigenstates of Ĥ(0) are degenerate, this causes a probem when we try to find corrections to the energy eigenstates. So in degenerate perturbation theory we end up doing some matrix diagonaization to find a new set of basis eigenstates to use with our perturbed hamitonian. This was covered in cass. 7

8 2.2 Variationa Principe Consider an arbitrary Hamitonian that has a set of basis functions: Ĥ ϕ n = E n ϕ n If we can t sove for these eigenkets exacty, we can approximate the ground state energy by taking advantage of the variationa principe. Consider an arbitrary function expanded in terms of the (unknown) eigenkets of the hamitonian: ψ = n c n ϕ n If we cacuate the expectation vaue of the energy for this function with the unsovabe hamitonian, we have ψ Ĥ ψ = m c m ϕ m Ĥ n c n ϕ n = m,n c mc n ϕ m E n ϕ n = m,n c mc n E n δ mn = n c n 2 E n Because the energies are increasing in vaue (i.e. E n E 0 ) we have the inequaity: ψ Ĥ ψ c n 2 E 0 = E 0 c n 2 n n And dividing by the inner product: ψ ψ = n c n 2 ψ Ĥ ψ ψ ψ We can use this to find an approximate wavefunction, and upper imit on the ground state energy by guessing a wavefunction, cacuating ψ Ĥ ψ ψ ψ and then minimizing this expectation vaue with respect to some variationa parameters (to do this you have to eave some arbitrary parameters in your guess wavefunction that you wi use to minimize the energy). There are a coupe of typica ways this principe is appied. You might try to sove a simpe 1-D probem (ike V (x) = V 0 x ) by guessing a soution that seems reasonabe (ike a gaussian ψ(x) = e αx2 ) and then minimizing the expectation vaue of the energy with respect to the fitting parameter (in this case, α). This isn t a bad practice probem. The gaussian is a decent guess because it satisfies the boundary conditions (it goes to zero at ± ) and as Mier might say, to a topoogist, the absoute vaue function is the same as the harmonic osciator. E 0 8

9 The other way that this principe can be appied is by guessing a soution to a probem as a inear combination of known soutions, and then using the coefficients as the fitting parameters. An exampe of this woud be approximating a diatomic H 2 bond as a inear combination of 2 1S orbitas): ψ H2 = c 1 ψ1 1s + c 2 ψ2 1s. In this case we woud write out the correct Hamitonian for H 2, cacuate the expectation vaue of E with this Hamitonian and our guessed wavefunction. Finay we woud find the best vaues of the c n by setting E c n = 0 for the various c n. 2.3 Hartree-Fock This method is a computationay intensive method for deveoping accurate wavefunctions for compicated probems. As is customary, I describe the basic idea using Heium as an exampe, and indicate how the method is generaized. Suppose you want to do the heium atom. The Schrodinger equation woud have kinetic energy terms for each eectron, potentia energy terms for each eectron, and a term representing the eectron - eectron repusion. Ĥ He = 2 2m e m e 2 2 Ze2 4πɛ 0 r 1 Ze2 4πɛ 0 r 2 + e 2 4πɛ 0 r 1 r 2 (29) where the subscripts 1 and 2 refer to the operators and coordinates with respect to eectrons 1 and 2. Z is the number of protons, which woud be set to 2 for this probem. Just as a note, if we were to ignore the ast term, we woud just have a hamitonian for 2 independent eectrons. We woud sove this by spitting the wavefunction into a product of the two 1-eectron wavefunctions, Ψ = ψ 1 ψ 2, and we aready know that the soution woud be 2 hydrogen-ike orbitas (hydrogen ike because Z 1). So how were you tod to think about this probem in chem-1? You think of the first eectron providing some shieding around the nuceus, so that the second eectron experiences an effective nucear charge that is something ess than 2. Because some of the time one of the eectrons is getting between the nuceus and the other eectron, the eectron you picture as the outer eectron experiences a nucear charge on average that is somewhat decreased. So how woud you account for this effective nucear charge rigorousy? We, a decent way to do it is to think of eectron 2 as fixed in space at r 2 and think of eectron 1 as smeared out over space, with a charge density given by e ψ1(r 1 )ψ 1 (r 1 ). Foowing this ine of reasoning through, we woud say that eectron 2 experiences a tota potentia from this charge density given by dr 3 1ψ1(r e 2 1 ) 4πɛ 0 r 1 r 2 ψ 1(r 1 ) (30) We, that s nice but Ĥ 2 = Ze2 + 2m e 4πɛ 0 r 2 e 2 dr 3 1ψ1(r 1 ) 4πɛ 0 r 1 r 2 ψ 1(r 1 ) (31) isn t exacty an easy probem to sove either. To do the probem then we turn to the variationa principe. If we guess that ψ 1 and ψ 2 are both 1S hydrogen-ike orbitas, we can cacuate an effective (mean-fied) potentia for eectron 2 (by doing the integra (30)), and then then use the variationa principe to optimize the wavefunction for this eectron. To do this of course we need some parameter with which to optimize the energy, so we choose the ony parameter in the equation, Z. Now that we have a better guess for the wavefunction for eectron 2, we might as we use it in the same way 9

10 to improve our guessed wavefunction for eectron 1. Once we ve done this we wi focus on eectron 2 again, and optimize its wavefunction... OK, so this is a ot of work, and we re ony taking about Heium. This is what computers are for, but the basic idea of this strategy is usefu to understand. As ong as the wavefunctions eventuay converge on something, we ca this method a sef-consistent, mean-fied method. It can be appied to physica systems other than soving the S.E. as we. So this is the sef consistent - mean fied theory approach to probems. What if we want three eectrons? We, we have to start with a good guess for the wavefunction. In genera we use sater determinants to generate an antisymmetric muti-eectron wavefunction, and do the mean fied thing on the wavefunction one eectron at a time. 2.4 Commutators You have seen some commutation reationships between these various operators [ˆx, ˆp x = i (32) [ˆL2, ˆL z = 0 (33) [ˆLx, ˆL y = i ˆL z (34) Because the choice of axes is arbitrary, ˆL 2 commutes with ˆL x and ˆL y as we, and we can permute the coordinates in (34) to come up with [ˆLz, ˆL x = i ˆL y (35) [ˆLy, ˆL z = i ˆL x (36) 2.5 Spin + Paui Principe So you know that the Paui Principe says you can ony have one eectron in each quantum state at a time. Mathematicay the way that this principe is handed is to make the tota wavefunction (incuding spatia and spin components, everything we know about the eectron) antisymmetric under exchange of any of the eectrons. So if you write out a wavefunction, abeing the parts with numbers indicating which eectron they refer to, when you switch around any two of the numbers in the function it wi be negative of what you started out with. Sater Determinants can be used to generate these antisymmetric wavefunctions. We start by isting a of the possibe position/spin states that we want to consider (1sα, 1sβ, 2sα,...) and make each coumn in a matrix correspond to one of these (there wi be one for each eectron). Then we make each row correspond to an eectron that is identified by a number (eectron 1, 2,...). This gives us: 1sα 1sβ 2sα ϕ n 1 1sα(1) 1sβ(1) 2sα(1) 2 1sα(2) 1sβ(2) 2sα(2)..... n 1sα(n) 1sβ(n) 2sα(n) ϕ n (n) The determinant of this matrix is the Sater Determinant, an antisymmetric mutieectron wavefunction. 10

11 2.6 Term Symbos Term symbos te us the components of orbita, spin, and tota anguar momentum for a given eectronic configuration. Using Hund s rues we can order the various terms energeticay, and find out how many transitions wi be aowed from one state to another. To generate a term symbo we have to figure out how the contributions to orbita, spin, and anguar momentum from the individua eectrons sum up vectoray to give us the sums of these components. In genera, the goa is to find term symbos that woud be possibe for a given eectron configuration. Like in cass, we did the probem of finding the possibe term symbos given the ground state eectronic configuration of carbon. Finding the tota spin is easy. Spin can ony be ±1/2 for each eectron. Paired spins (2 eectrons occupying the same orbita) cance, and the rest of them can either be pointing a in the same direction, or in some combination of up and down. The maximum spin for n unpaired eectrons wi be n/2. For 2 eectrons as an exampe (ike the carbon atom), the aowed vaues of S are 1 (both spins in the same direction) or zero (one up and one down). S is an absoute vaue so 2 spin down eectrons gives us S = 1 just ike 2 spin up eectrons. The mutipicity is 2S + 1, the number of different microstates with this vaue of S. The tota orbita anguar momentum is found by considering how the individua s ine up vectoray. For two eectrons the possibe vaues of L are 1 + 2, , Once we find a vaue of L for a term, we equate this to a etter just ike we do with the. So L = 0 corresponds to S, L = 1 corresponds to P, L = 2 D... Using carbon as an exampe we coud have and S term where L = 0 (the eectrons with their orbita anguar momenta opposed, ike m 1 = 1 and m 2 = 1), P terms where L = 1 (ike m 1 = 0, m 2 = 1 or m 1 = 0 and m 2 = 1 for exampe), D terms (both eectrons in the same p orbita, or course this wi mean that S = 0 for this term because the eectrons are necessariy spin paired). The vaue of the tota anguar momentum J is given by how S and L ine up vectoray. So if we had for some reason a D term with one eectron, we coud have either J = L+S = 2+1/2 = 5/2 or J = 2 1/2 = 3/2. Because shes that are competey fied necessariy have a eectrons spin paired, and a orbita anguar momenta quanta equay opposed, these fied shes wi not contribute to the term. That is, the contribution from a fied she wi be S = 0 and L = 0. So in generating a term symbo we ony have to consider the unfied shes. The term symbo itsef is given by 2S+1 L J Where L is repaced by the corresponding etter. Hund s rues te us how the different terms wi stack up energeticay. They are based on thinking about how arger or smaer quanta of spin and orbita anguar momenta transates to the physica separation of the eectrons, and spin - orbit magnetic interactions. Hund s first rue says that for a given eectronic configuration (ike 2p 2 ) the term with the highest mutipicity wi be owest in energy. This is because higher mutipicity means the most possibe number of eectrons have their spins aigned, and by the paui principe this means they wi be in different spatia orbitas. Eectrons occupying different spatia orbitas have ess couomb repusion than eectrons in the same orbita. 11

12 Hund s second rue says that for states with the same mutipicity, the state with the highest orbita anguar momentum wi be owest in energy. This rue aso comes from considering eectron eectron repusion. Hund s third rue says that for configurations with the same mutipicity and orbita anguar momentum, if the vaence eectrons are in a subshe that is ess than haf fied, the smaest vaue of J is ower in energy whie for subshes that are more than haf fied, higher J is ower in energy. This comes from considering the spin-orbit interaction of the eectron s intrinsic magnetic moment, with the magnetic moment that arises from orbit about the proton. Reca Ĥ so = Ŝ ˆL The hamitonian which describes the energy that arises from spin-orbit interactions in an atom. 12