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Lecture 16. Energy Level Diagrams, Spin-Orbit Coupling and the Pauli Principle


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Title Lecture 16. Energy Level Diagrams, Spin-Orbit Coupling and the Pauli Principle
Alternative Title Lecture 16. Quantum Principles: Energy Level Diagrams, Spin-Orbit Coupling and the Pauli Principle
Title of Series Chemistry 131A: Quantum Principles
Part Number 16
Number of Parts 28
Author Shaka, Athan J.
License CC Attribution - ShareAlike 4.0 International:
You are free to use, adapt and copy, distribute and transmit the work or content in adapted or unchanged form for any legal purpose as long as the work is attributed to the author in the manner specified by the author or licensor and the work or content is shared also in adapted form only under the conditions of this license.
DOI 10.5446/18894
Publisher University of California Irvine (UCI)
Release Date 2014
Language English

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Subject Area Chemistry
Abstract UCI Chem 131A Quantum Principles (Winter 2014) Instructor: A.J. Shaka, Ph.D Description: This course provides an introduction to quantum mechanics and principles of quantum chemistry with applications to nuclear motions and the electronic structure of the hydrogen atom. It also examines the Schrödinger equation and study how it describes the behavior of very light particles, the quantum description of rotating and vibrating molecules is compared to the classical description, and the quantum description of the electronic structure of atoms is studied. Index of Topics: 0:00:51 Cesium 0:19:37 Another Look at Spin-Orbit Coupling 0:23:42 The Energy Shift 0:25:00 Two-Electron Atoms 0:37:06 The Pauli Principle 0:41:31 Spin and Space
hello today were gonna continue on that infamous cesium problem we had all those transitions that we were supposed to assign the quantum numbers for and we were asked if we could estimate the ionization energy or ionization potential saying things and of course we can otherwise we would ask that today then we're going to finish that problems in the energy level diagram problem we're going to talk about spin orbit coupling a little bit and then we're going to talk a little bit about multiple electron Adams and the Polly principle as kind of an introduction to much more complicated systems that we're going to be dealing with in the future recalled that we had assigned the 2 resonance lines for cesium and those of the 2 strongest lines in the emission spectrum which were from "quotation mark P 3 Halverson double P 1 half to the ground state "quotation mark test 1 hat and the splitting between the peace states was a very large value of 554 wave numbers which is a spin-off coupling for the 6 p electrons the let's pick up then where we left off we had this fragment of the energy level diagram that drawn here on slide and we had these 2 transitions assigned 11 178 an 11 7 32 now we look at the other sets of life and what we look for is whether there's a difference of and we start with a set of lines that there's just too a woman and there is a son Michael so this problem the key Kia's especially if you get this on an exam is that the same Collins means something that means that they're giving you at the end and ordering them if they just give you a list of lines without any kind of ordering the problem is far far harder to do and they didn't want to be that hard but it's plenty hard isn't it if we look at the other 2 lines up 7 3 5 7 and we subtract 6 8 0 3 from those 2 went together that difference is 554 and what that means is that there must be some level that is going to both those key states if it's just too that means it's investing and so what we think is that down here we've got success and that he is coming down here so that's the 554 from both and that's quite a big energy difference and then closer To the P is the 7th now up here and it can make a transition of both of those that's allowed just like run either way so for about a week a minute and we met those 2 lines go over there and the nose to both go to the park so these things can happen sequentially and of course will we look at the spectrum we don't have any kind of time resolution usually were just looking at
all of the stuff that comes out but we don't keep track is just developing other things on a CCD camera if we assume that the 7 as than we have now this picture on slide 416 with the success 6 p splits into 2 levels the 7 as a single level again WAS ONE have we have those 2 lines 7 357 to the lower state and 68 all 3 to the upper stage now what we have by adding these numbers are as we have the energy difference between 6 hours and 7 minutes and that's how we're going estimate the ionization energy and a little better the other sets are grouped into 3 and what I've done then is I've just taken the 3 and I've taken all possible differences between 1 minus 2 1 minus 3 to minus-3 arranged the most positive numbers 33 21 minus 28 65 as 456 numbers I haven't seen it 33 21 minus 27 67 is 5 54 that I have seen so that's the hand that whatever is involved in these 3 is going to the same to there the central guys in this scheme that my linchpin for figuring out how this stuff works and then the other 1 28 65 minus 27 67 is 98 so that's another does as far as I'm concerned right now I don't know exactly what the
98 where the 456 me but I can get us if the 3 of but it involves a turned 5 because I've got a double P 1 half and 3 halves when I go to Dublin D I have by the clutch caught 3 halves and 5 while the 3 halves can go to both 3 have says Delta Jake was 0 and the 1 that's still J minus 1 the final can only go To the Dublin P-3 head because it's not allowed to go inside pass 1 and so I expect to see 3 lines I expect to see 2 of them from the the state the 2 those 2 should have a difference of 5 5 4 while there is a pair with a difference of 5 5 for the Minnesota match what that is and then the others have to come out to be the other differences and so on let's have a look at them and what that means well let's do the same thing with the other with the other set of 3 1 what why did they have a much larger value while they must be from another city state that's higher energy coming down again and if I take the 11 thousand 411 minus 10 thousand 900 get another value of seen 11 but the fighting the 11 thousand 411 minus 10 867 bingo fireflies there is a level again coming down those 2 two-piece stick and if I take the 3rd difference I get a splitting of 43 the about 5 5 4 suspects orbit splitting in the peace state I expected the state to be lower then that and so I'm gonna sort of assume that the small value is the splitting the the state and that's I'm going to try to assignment now we don't know what the number of the D level is we've got 6 S 6 p 7 hours and that's pretty much set and then we've got 5 D and made 60 but if we had to figure out if I the His below 6 p and energy so it could be gone this way down or if it's above energy and the way we do that we so we had to do a bit of thinking here but this probably 5 but we don't know for sure if it's below or not but if we do put them the 5 deep along and then we assigned the transitions to get the 554 than what we find is that the spin orbit splitting the duties that there would have to be 456 wave numbers that would be it and that's far too big for anti-state have to know a little bit of trivial knowledge in order to get it for sure and that is that it shouldn't be that big and therefore it's the other way around it's it's the 1948 the splitting in the state and in fact both at the levels are higher than the 6 p and if you make that assumption that we get the well complicated diagram on slide here we have the ground state success we've got the 6 key states those 2 the 7 answer than over here we've got they 2 states in the 5 and then we've got the 2 states in the 60 and we've got all these transitions assigned and we've got all the quantum numbers assigned and so assigned all the quantum numbers you can means assigning every single 1 of them for every single transition in this case now the question is How are we going to do figure out the ionization potentials well if we didn't know about that formula miners are over and Delta quantity square we would never get anywhere because we if we assume that it goes like hydrogen with 6 and 7 we get totally the wrong answer in this case we have 2 ways of estimating the ionization energy and I'm going to estimate the ionization energy both ways I'm 1st and he is the guest the success of 7 and then as a check I'm going to use the 5 B in the 16 and we'll learn a little bit of such a book about how to do that because these displays and so the question is which 1 do you use and the answer is neither of them you have to figure out how to figure out what the center of gravity units for those transitions if we add the 11 178 and the 7th 357 what we get is 18 thousand 535 wave numbers for the difference between success and 7 s so that's Delta III In wave numbers and now we have that formula and their differences that must be minus are times won over 7 -minus Delta L quantities square -minus 1 over 6 don't mind Deltaville quantities square and this is a quadratic equation basically sulfur Delta L we know the value of R 1 0 9 7 3 7 we should actually use the corrected value but for cesium that the reduced mass is essentially the mass of the electron and so the Ripper constant as so-called are Infinity which is very close to 1 0 9 7 3 7 soldiers use that and then we have a constraint on what Delta Elkin B. because I think we have to be careful because we can get different routes of core of a quadratic equation it has to be real it has to be positive and it has to be less than 6 and with those constraints on the possibility of the solution what we find is that Delta for the cesium and is 4 . 1 5 this is an enormous defects and compared 6 it's subtracting off this huge number well but that that actually makes sense when you consider all electrons that are around there In the penetration to the To the
nuclear and then we can figure out the ionization potential from the six-state because then we can just put on any equals infinity for the state take the difference masters are over 6 minus the quantum defect quality square to take 6 minus 4 . 1 5 square and divide 1 0 9 7 3 7 by that and we get 32 thousand 63 where numbers that's our estimate for the ionization potential of cesium based on just 7 2 levels there and that's a little bit dicey because usually you like to have many more just to see what the trend now let's also try to get an estimate from the Mideast states for 1 thing that's going to confirm that the quantum numbers that we've assigned the corrected for another thing it's probably going to give us a better value because there's less distortion in the state's they more ideal but in order to do so we can't we got these 2 sets here at this month's but by 90 something the missus 40 we we needed to levels which are a sort of the unperturbed state energies before we have to stay in orbit couplings Internet to deal with that with the yesterday and the way you do that is you have to wait the 2 states by their multiplicity and you have to take the weighted mean the 3 halves has more sub-state them US excuse me the 5 have more substance than the 3 have and we get a ratio of 6 to 4 for the state's or 302 and therefore I shown on these 2 equations the difference between the 5 the state and the success today because I only have the difference I don't have the absolute is 2 times the difference to the lower level which is 1 1 1 7 8 plus 33 21 plus 3 times the difference to the upper level which is the same thing plus the 98 divided by 5 Parker taken 604 divided by 10 actually stay there for the energy difference that be the correct 0 of energy for the 550 versus the success is 14 thousand 557 . 8 numbers and if we do the same thing with the 60 versus the success we get a difference of 22 thousand 614 . 8 wave numbers and now what we can do is we can take the difference between those 2 numbers and then we have difference between 60 and 5 because whatever the differences that as drops out of the equation and using that difference of 80 57 we can then solved for the quantum defect in the deceased state by solving miners are times won over 6 -minus Delta sub D quantities square minus 1 over 5 and again we have a constraint that this solution be real that the positive and that is the lesson 5 and allowed value than of the quantum defect in that the state is 2 . 4 3 which is less than asthma recall said the quantum defect and gets smaller as you go out becomes more ideal using that we can get the ionization potential we 1st take are over 5 minus the quantum defects square but then what we have to do is we have to have the energy to to get out there from the success
so we now get an estimate from here from this formula and then we've got we gotta get up there so we take that lost 1 1 1 7 8 plus 33 21 and we get 31 thousand 198 wave numbers which is a little bit different than the best but closed very close considering the way we're doing this with this approximate formula and 2 levels and so really comforting if you get a result like this 1 accounts and you're going to get a great on in fact the literature values is 3 . 8 9 4 electron volts which Converse 231 thousand 324 wave numbers therefore the estimate from the G states it with their smaller bond defect is closer to the accepted value what I would say is that if you happen to have been trained to look at problems like this and then you just give a problem like this cold and you just know something about the hydrogen as a problem like this is pretty much nearly impossible and once you know how to do them and how to look for these common differences then pretty much you can get everything done although if it's not an alkali metal it's going to be much much harder and it just depends whether it would be able to estimate the ionization potential not you may need more data if there's probably had more data while we tried to do His we tried to organize the data rather than just solving 2 equations to and 1 equation 1 and rather than the and inserting ionization energy we try to make where the ionization energy would be interests and we will try to organize our variables so that the plot were linear and then what we do is we checked whether the plot is linear era 1st so that we know we're going to protecting and we will come soon and and then get the ionized section OK let's take a little bit more detailed look it's been orbit coupling this magnetic interaction we never included in the Hambletonian we had the kinetic energy but the electrons we factored out the nuclear motion the potential energy electrostatic energy and always said later there's this magnetic but how could we included if we want to include but we argued that the electron sees a magnetic field from the nucleus going around the other way and that's why there's only a spin orbit splitting and honest but we can do a little bit better than that because we know but the energy of a bar magnet for any died electric dipole on an electric field magnetic dipole in magnetic field is minus new be I or -minus million-dollar the and if insist that the magnetic moment is proportional to ask the spin and if the magnetic field is proportional to the the orbital angular momentum that we get energy term which has some number which I'm going to call data timescale .period because that should be the dark products of the 2 and energies are often .period products and forces are often cross products and that's the way things play out then we can cast the adopt product in terms of things that we actually know we want to get it in terms of quality numbers we can do a trick that we can take J . J that's a scalar product is the the total angular momentum of the atom with itself adjacent and substituting for JAL process in each place we get elsewhere which is held out but the word which is estimate close to know the answer and then we just saw that for all the which was going to be our energy terms and we get 1 J. squared minus elsewhere might successful and now we know the quantum numbers we just put in numbers for those things and more we don't have to worry about what the operators are we just put a number let's apply that to the double P 3 halves and double P 1 half states we know all the 1 numbers we know as we know well we know J. so we're going to take the expectation value of the spin orbit energy in the States well what we find that we put in the quantum numbers we get paid over 2 there are some things that has to do with how strong interaction is we'll get to that in a 2nd and we get further chased where J. little J times little J. plus 1 or at some other works -minus little Elton's little outpost 1 mines last was 1 we do that double P 3 have we simply get 15 over for minus to minus-3 over 4 data over to an end so we get paid over 2 times a are square and if we do it for a 381 have but 1st Excuse me Dublin P 1 half what we get is -minus beta age spots where you can see 1 of them is moved up by a certain amount and the other is moved down and that's why when we did the problem we took the weighted mean because they're moved by different amounts depending on the multiplicity so that the center of gravity remains the same the shares whatever it is this energy term data should depend on CD atomic number because after all we saw was cesium that was much much bigger than sodium and we would expect a bigger nucleus to generate a bigger magnetic field but it it's going to take us far too far afield to actually calculate from 1st principles with Thomas Fermi procession and all this other stuff that comes in and so I'm just going to quote the answer that age R square data the energy terms in front of the all-time best go sightseeing at the 4th and then a magnetic conversion factor and then the chief factor of the electron bore magnet on and then this term comes from doing some angular momentum Algebra 1 over and times a knock due to unstable times almost 1 and selfless 1 one-half that's quite a long and involved formula but what it shows is that as Annandale increased the splitting decreases we saw that again in the cesium we saw 1 of was nice the was 47 out so cents OK let's talk about 2 electron the we did hydrogen with alkali metals which is kind of a Dodger because they're just basically 1 electron and clouded charges
inside and not the 1 actually due to electron and we have to start doing some serious work because of this we have to take into account what's going on with the 2 electrons and that isn't so easy to see and to defend some experience with With the hydrogen atom we noted that we had a 6 plus one-dimensional problems before when we started at 6 dimensions in space the 3 accordance of the protons the 3 quarters of the electron and then 1 dimension time and what we decided is well we are solving overtime dependents if we just want the static ground state ,comma time independent answer we could get rid of the time the time-dependent showing quotations and just calculate the energy I could states because the energy Oregon State's article states that don't change in time that's why the special and then how did we get rid of these the other 3 quarters it's well recall we factored out the center of mass motion which is just the whole Adam drifting around and they're not really but we were left with is just the relative distance between the Proton and the electrons and then at that point we can make a mental Dodge and say Well look the 6 the distances to the electron and at that point we just start thinking about the cordon in terms of where the electron is in the hydrogen but for hydride let's H minus that has a proton 2 electrons fluffy thing very strong reducing agent or helium not quite so big and classy and not very reactive the wave function is 6 dimensional even after we take out the motion of the nucleus wiped couldn't get to electron and so as I remarked earlier we can't comprehend how to apply 6 dimensional things very well and so we probably get rid of all that complexity and we say Look we have to take some kind of a simpler approach that's more tractable we are going right away functions as a function of 6 coordinates and then try to figure out what's going on even if we can solve that sort of thing which even 3 would think separated nicely wasn't so easy and in this case there is a separate because even for a three-body-problem not separate all we can disentangle the electrons from each other so simple so what we do here is what we always do what we don't like suddenly thrown away that's the 1st step I don't like that turned the equation well OK let's assume it's small and that it is this fall the stratosphere out how to fix it later but 1st let's get an answer but we can get and so in this case the term that's a pain Is the electron electron repulsion the fact that his 2 electrons are buzzing around somehow and whenever they are close to each other but they really repel like crazy because they can in principle get very close after all they're almost geometric points that would give the extremely high energy and that can happen anywhere in states where they are so they try to avoid each other and then they tried also cluster around the nucleus and if we just ignore the repulsion we can just treat them separately I don't see you you don't see be we both see this guy but stride again estimate that and then correct in that case if we ignore the repulsion then the energy is just the sum of the energy of this electron interacting with the nuclear and the summer this electron interacting with the nuclear and if the energies have that means the wave function as a product because that's basically what we did with the particle on box we said Look the EX and Emeline energies just as so with functioned factorize into a product because those guys had nothing to do with each other and if we turn off the repulsion between electrons they had nothing to do with each other personal forces between and so for these 2 electrons instead of always function that lets say a function of our 1 come on Oct 2 which is 6 Courtney we break it up right away into a products sigh of our ones time the site 1 of our times site 2 of our too we just right away as soon and that's our starting once we assume a product of wave functions that depends only on 1 set of course what we're doing essentially is worth putting each the electrons into its all this is an important concept mentioned the term for approval before for a hydrogen atoms there is only 1 electron but for multi electron Adams there are lots of electrons and the way function is a big mess but we don't deal with the big mess so what we do is is we put each electron into its own orbital this is an approximation it's a pretty good 1 and a lot of cases and it's called the orbital approximation it let us consider each wave function for each electron to be only dependent on the coordinates of that electron you can see right away that that can't be right because how do you know what but this one's doing is you don't know where this 1 is when they're repulsive but nevertheless it can be a pretty good approximations and it's the 1 we use now we don't have to stick with hydrogen orbitals for the solution we can because the hydrogen orbitals former complete set remember what we learned about the Eigen functions of any mediation operator that their orthogonal to each other so we can consider them as a different directions and we can make up any function we lie as combinations of these hydrogen functions and since we know what they are and we can write them all down that can be really attractive but we don't have to stick just with them so but we can take any kind of functions that we like doused him functional or any other thing and then what we can do it is just the electrons in sequence going round and round and round until we get a better solution will talk more about that later on in this series of lectures that's the so-called self-consistent field model but you can understand what it's going to amount to pretty easily I've got all these electrons they're all over the place their repellent there and so forth and they're attracted to the nucleus and I don't know what's going on and it's a 20
7 dimensional problem what I do is I put all the electrons into shells and orbitals like they would be in the hydrogen atoms
but with an appropriate value received and then I take 1 electron and I take all the others and ice near the mouth it's charge I just forget the fact that the real wave function there and I just stared amounted to some charge distributions and then I take the 1 electron I've got and get this weird charge distribution from all the other ones and the nucleus and I take this electron and I try to optimize its way function until it lowers the energy so that it's more ideas and then I freeze it and make it into the charge and I pick another electrons are always and minus 1 and I have 1 leftovers go round and round and round and when I can prove it any more than I say I've got a self-consistent solution doesn't mean it's correct in fact it's not correct because the problem and smearing it into a charge is pretty good but it's not the same as taking into account how things are actually moving so-called electron correlation and infect other theories and much better too but once we're done it was stated that go back to 2 electron animal forget the 27 national 1 if we go back and we've got to electrons Indiana we can treat the electron electron repulsion as per patient and then we can try to adjust the energy to see if we get something that's more acceptable than measures better with the spectroscopic lines which of course can be be measured too many many digits which is a great test between theory and experiment if you look back at the slides on perturbation theory remember that we started we we introduce this parameter that I called lambda and when land that was 0 0 we had our but undeterred solutions that we had the exact solution for and then when land is equal to 1 we have what we're trying to get and we've tried to connect pieced together 0 1 1 by taking a power series
and land on both sides of the equation and matching the powers because it is going to match all the way through the various powers should manage and after we manage the powers we get some equations meant to get rid of land that we just that landing equals to 1 and then it's out of there and then we have some equations to solve and if you look back at that then there is a correction to the energy right away when you have perturbation as long as it doesn't matter as long as it has non-zero matrix elements and the correction of the wave functions is higher up so you need a higher order calculation if you're actually going correct the wave function itself for the from this prior to something better than that but there is another player in the game now and we had to take a look at that before the Stan OK 1 electron spin on couple but now the state is going to play a major role forget to electrons because that's going to dictate which I'm states we can even have we had to be very careful and this thing kind of an annoyance because it doesn't have these variables are the fighters justice quantum number and so that sponsor minus a half but because they can't have the same quantum numbers dictates everything the the 2 principles here at the Pauli exclusion principle is usually stated in the following terms that 2 electrons in the same atomic orbital or with all of the same spatial quantum numbers that same spatial orbital after have opposite spin they cannot have the same spirit if 1 of the the other it's not quite literally true but we'll see what I mean and there's this turns out to be a result of a much cheaper some symmetry principle which is this but just the Polly principle which is a principle that applies to all fermions firm eons electrons the formula and the idea is that if you take the total weight function of Iraq and the swap any 2 identical particles To firming then the wave function changes sign that's OK that the wave function changes sign because when you square the probability density as the saying of course if you're exchanging mentally 2 identical particles physically you could argue you've done nothing at all so of course the electron density the probability distribution of charges shouldn't change of course it doesn't change but the way function can change sign and it does if we swapped so we write a function and we swap on what we call electron 1 with electron too everywhere in the function then it should change sign Of the wave function if it doesn't it's not allowed OK as I recall as mentioned the probability density stays the same and then there is another 1 of those on which is of which a photon is an example and that 1 if you if you swallow it doesn't change 2 states the same side and it appears that these 2 particles have completely different properties mathematically in terms of our equations 1 of orange always changes some of the other 1 doesn't change son and he could ask well why is it always wanna minus 1 maybe maybe you could change by the the the I think the and it would still have the same probability density it's the other 1 change by you to the miners and everything would come out the watch and the answer is maybe there are some things like that called any onsen you're if you're intrigued about those you can look them up but as far as we're concerned with Adams in changes assigned dozens and since were only concerned with electron what we're really concerned about this this change of science and the way function when we swapped the thing now it's the total weight function that changes sign and the wave function when we've got more than 1 electron what we have to worry about is the spatial part which depends on the coordinates and the spent part which is just grafted on but the spend part change signed 2 because of the fact that 2 electrons if they're both then if I stop nothing changed if one's out once down and I swap of my get something different so that's no good but if there were both down and ice wall nothing changed and so what's going to have to happen is if we have the 2 electrons with the same spent then the spatial part is going to have to be the part of the changes the spatial parts as the entire symmetric if the spin imparted symmetric and the spin part has to be antisymmetric if the spatial part of some entry if the 2 electrons are the same orbital station parties matter because what never the same orbital that has to be completely symmetric and so that's why they have to have opposite that's the extent of exclusion principle let's have a look at them and how this plays out we never included the spend explicitly with hydrogen we will did say a In this single electron out was it was the back and figure out if it's been up and spend down and that because unless you look extremely closely but it does matter whether been upper spend only a few worried about protons spin well that's that's the matter but when we got to electrons is the major thing to keep track of and you have to keep track of it and you have to learn how to do it let's have a look at how to do this let's abbreviated away function here on slides 437 as just sign of 1 ,comma 2 and 1 is is electron 1 and 2 is electron to and it's going to depend on spatial part and and the spin part then the entire cemetery constraint of the tollway function for the Adam means that sigh of 1 ,comma too is equal to minus so I of 2 ,comma 1 that's it if I have a wave function and it doesn't satisfy them it's no good I can use it and that's the throttle on possible solutions so that the total wave functions now supposed to electrons have the same at part let's call that I want and side too at the same function for each electron could be you the miners are over in order whatever it's the same that part symmetric does arise swap amidst the same and so the spent part has to be of entire symmetric in the case because typesetting books especially with arrows historically is incredibly tedious this is especially for computer types that usually instead using these arrows while we uses alfalfa once-banned state and data for the other we just right in order and the order is the order of the electron so what we can do is we can write the 4 combinations like this Alford 1 2 both also 1 beta too data 1 Alpha 2 and there 1 dated to those of the 4 combinations N but the alpha alpha and beta beta states a symmetric isn't the other ones are neither symmetric nor an isometric and that means the other ones are no good and as we saw when we took 2 electrons and coupled what we have to do is we have to make a proper value of big guests for those and remember 1 value what with the and the other 1 with the singlet and that's exactly what we have to do here that taking the same thing that we did when we took to use to electron spin 1 has to get bigger we get us a magic combination which is rude to over 2 alpha-beta plus the Alpha and the other combination which is antisymmetric which is route to over alpha-beta minor spit out and only the US those 1st combination symmetric seconds antisymmetric and you can see that if you actually substitute swap of you have to swap of around and then look carefully at the the 2nd 1 changes signs shown that Iran slide 440 since the combination is symmetric than 3 out of 4 of the spin wave functions are no good the Office of data data and cultivate a plus prospect also are all logo so those with as equals 1 the triplet states and it's only be antisymmetric single at stake vesicles 0 so it's not really that the 2 electrons 1 is up and 1 is down remember this is quantum mechanics so it's always weird than we think it's a 50 50 next year of I'm up and you're down -minus Europe the I'm down that's what it is not just 1 combination because that would mean that the rights ETA and that part than say what I call signal minus here can pair with the spatial part will get the overall wave function for this to electron system if they're in the same orbital which they would be let's say for healing this someone cited time signal minus 1 2 that'll be where we pick up next time only actually trying to figure out OK we've got these electrons let's try to actually calculate some energies for the Adams let's have a look at that how it plays out and how we can take into account the repulsive terms between electrons and that that's quite an interesting little exercise there's a lot of mathematics but most of it we can serve bludgeoned way through with the help of some friends that know how to do a lot of integral so we're going to have to be able to do To get them down to a number at the end of year what that number is and how much energy shift so pick it up there next time through
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