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Lecture 03. Transformation Matrices.

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it is sorry everybody let's see what's going and get started trying to get this feeling my microphone is not working so well is that correct are a case have now that it's still quite effectively making movies that located the parade so last time we managed we talked about transformation matrices and I want you to be really really careful about making sure that you know what basis Syrian and be consistent and then I made a mistake involving not doing that so what's on what's put up the slide again when that happens the corrected version so basically what I did was I had half of it in a different basis than the other half of which was which is not and so thanks to the person who pointed out that use a corrected version of how we set up these transformation matrices matrices were right it's just that the answer was not consistent with the vector that I find you so sorry about that mistakes do happen but I was 1 Beckham's side you want to point out like if things like that happening glass I really do want to tell me and so I appreciate that when when people on sometimes will have time to stop and then cleared up right away at other times will just have to to move on and deal with it later fled nearly always wanna make sure you get that cleared up stuff is clear on this anybody have any questions about this or other stuff there that has gone on so far something that's good then yes or no to the rejection of the notion of what it was it was all that was needed well so you just hear just ending up with Europe you just want to production on the x-axis so that's the only so when 1 is the only solid you want to have anything in it this is the 1st element on the diagonal technical so we're going to go and talk about that theory and today it is going to be cold because we're going to learn how to take the 2 skills that we've learned so far and put them together to do something useful so so far we've learned how to to put molecules and actually any kind of different shapes into pointers and we also learned set transformation matrices and now we're going to see how this actually has an application in chemistry the 1st application were going to do is to chemical bonding and I like to do these examples because we learn the general formalism for going through with it and using the group theory but the advantage is that the answer you get something that you already know from general chemistry so it's really easy to check the work if you get the wrong answer you're going to know right away that it's wrong this is going to be helpful because later on we're going to be doing these applications to spectroscopy worry won't be so obvious what's happening so it's it's good to make sure that that you get what you expect when we talk about this in relation bonding another source of general strategy that thing that I want to point out before I forget his some with respect to putting things in the right places and making sure you're consistent when he gonna do these things on the exam we give lots of partial credit for different things and also there are just different ways to get the right answer please make sure that whatever you're doing you write down what basis using drop pictures drawing coordinate system just make sure that that the tears and I understand what you're doing with whatever it may be OK so let's say let's go through talking about .period groups soaring started out talking about water which is in the sea to the point group so of course we could be talking about any object that fits in that group and if we look at our character table we see that we have these operations and last time we went through all the various things there in the character table and we talked about the irreducible representations which are these things that are called ABC E. etc. and I said that they describe how things transform under these operations were in fact what that means so we need to introduce the the idea of being able to look at something and say OK when I perform this operation it's a valid symmetry operation but we have to be able to distinguish between whether it leaves the object completely unchanged lot and so in order to do this I will be hydrogen is red and blue in reality of course with real water molecule we can't tell the difference between those hybrids are the same but in our minds weakening union designate them as red and blue and and keep track of where they go OK so we performing identity operation it doesn't change that's what the at any means but if we do a C arbitration itself final cemetery operator believes that the molecule unchanged as far as hydrants being indistinguishable In but it does swap them now we look at our reflections in the X z plane which notice will coordinate system here that's 1 of the plane of the screen that leaves unchanged where's the other 1 which goes through the middle of a molecule swaps them so but a simple enough right we want to be able to to look at how we can use this so now we need to set up
matrices to perform a transformations In this case the basis that we're looking at is the hydrogen so the basis is that that's the thing that that we're talking about and what we're going to see is that the transformation matrix for particular operation it depends on what the system using so I can't say making a I can't make a transformation matrix for a C to recreation completely in general I have to specify what I'm talking about so last time we talked about little unit vectors 21 the axis Europe player in 3 space now a basis is the hydrogen is on this water molecules and so for of the identity matrix we need to look at the thing that keeps our hydrogen is unchanged so now notice and calling and each in a crime and so are identity matrix is always going to be were always going to get something that has ones on the diagonal and zeros everywhere else but how big it is depends on where this year we have 2 things we get at 2 by 2 matrix for that and notice that its character too so that's the hero that's the character Keister remember that's what you get when you add the elements on the diagonal and it's working with character tables so we know that that has something to do with those those elements that go into the irreducible representations emergency how in a little bit so in men .period groups there's a shortcut that we can use all we can say ah I don't need to set up this whole matrix I can just say you know what we have to 2 things and neither 1 of them changed and so we can add them up and get the character too so this is a shortcut that often works I want to point out that it doesn't always so if we have things that have trick all kinds of symmetry so like it for a minute d 3 H C 3 .period this becomes a lot harder and we'll will see some examples that later but in many cases we can use the shortcut we can discount the things that change and find the character right away and all point out we were able to today to do that OK so now we can look at our C 2 operations and we know we're here with that does swaps hydrogen is so that's our basis if 1 changes places with the other than that character is going to be 0 and we can use that shortcut and we can also set up the actual matrix so that the matrix that makes these things for the position so I'm not going to do the other 2 for this thing right this minute I will leave it as an exercise questioned why you should use the phone this only thing are turned how out about that's just as my picture is wrong there has to be at least 1 every lecture thanks so the colors are bright and the vectors rates but the time should be on the red 1 that correct OK so now let's look at what happens if we're talking about objects that can change the sign so for hydrogen atoms are only option here is they can stay in the same place they could move to different places and so we get ones and zeros for the character question and it's not exits Ky and that's that's the trace of the matrix of the character of the matrix that's what you get when you add up all the elements on the diagonal so that is that is this shortcut that we can sometimes using sometimes not and I'm going to keep giving examples of this so in the 1st example B. We have 2 hydrogen atoms and neither 1 of them changed places or or change at all and so that's so we can say OK there are 2 things and they didn't change so that means the character of this thing is to In this case the swap places and so the character that is 0 so we're saying you know in the cases where the shortcut works on show you some examples where it doesn't you can say if something stays the same it gets 1 if it changes places with another element it gets 0 and if it changes sign it gets minus 1 yet so of course and it's me again this is that this is a heuristic it's a shortcut that you can use sometimes it doesn't always work and it doesn't work when you have things where you can't you to where you can't match 1 element onto another very easily with the cemetery operations we have and also you cases where that's true yes it's true that there's just the transformation the purpose is to test the character that that goes under that transformation so what we're doing here is we're making a reducible representation she how a particular thing which is our basis transforms under all the symmetry elements and we're going to build up to had actually use that to find out something about the molecules but for right now we're just practicing we're just seeing how different things transformed under the cemetery elements guests filled with the that's so that's a good question so the characters in those tables are the characters of the irreducible representations so those are there very basic representations of how objects transforms under the cemetery operations and worried that and I don't think that's clear right now but it will be as we go through more examples of what we're making now is called reducible representation and so we can we can add at different combinations of those irreducible representations in the table and make a reducible representation which is what were were making up now and we're going to see how that works in Britain on a formula for reducing it and going to enable us to find how you know for instance how many and which kind of orbitals can be involved in making a particular kind of bond and again that's why I like these kind of examples because you already know the answer to that involve learned in general chemistry so it's it's very the procedure that we learn is little that involved but it's very easy to check their work because the answer is that you get is something that you were you know for right now later we're going to Harvard thanks but OK so here's another example where a look at the same kind of thing so now we're still in the to the point where parade but now
we're looking at a different molecule so we have C O 2 and our basis is 2 particular appeal rules on the use chlorine atoms and the wager the picture you can see that they have opposite phase from each other and 1 thing that people get confused about is why did picked this as my basis what about all the rest of the pure rules on the chlorine atoms What about the pure rules on the oxygen atom and the answer is I don't care I'm not looking at those right now so you get to set up the basis you're making up the year the problem that that we're looking at in this case by making up so this is this is what would happen to look at the choice of the bases here is arbitrary because you just practicing seeing how things transformed in a real scenario where you're actually trying to figure something out you're going to be looking at which direction bonds are pointing because you might want to learn which organized going making up those bonds when we start talking about molecular motions are basis is going to be a little x y and z axis every atom in a molecule so that we can learn about its notions those are examples of realistic things that we might come up with to solve a problem this is just for practice because once the whole works OK so again we look at the identity we can use the short cut here and say alright we have these 2 few orbitals and I know that they're not going to change would do the identity so Our character equals 2 you can also make up the transformation matrix that we always know when it's going to be for the identity of it's just once on diagonal on zeros everywhere else and now look what happens when we do see 2 rotation so a and B. swap places but their signs don't change and this is important look to to realize the signs of the orbitals are I'm I'm saying there are going with the positions so the sign of the finger on the left is the same as it was before however the identity of these orbitals swap places so something change places the character of that matrix is 0 and it looks just like it did in the hydrogen atoms placed it tightened the Haydn case and we can see how the transformation matrix works OK so now let's look at what happens with a reflection so again my XT planes industry in the plane of the screen so if I do that reflection these orbitals I'm going to a chain of reflect through the screen and they stay in the same place the identities of a and B don't change but they both change sign so that tells me that the character now is minus 2 using a little touristic that we have 2 things enables change and then we can also set up the appropriate transformation matrix so what I want to get out of the city is it is kind of a toy example there there's not necessarily anything useful about knowing about these particular P orbitals what I want to see is you know just how we can bring our 1st of all talk telling difference between the elements changing places things flopping around and changing sign and the fact that there is a separate from each other and I also want to point out that exactly this problem was on the exam last year so this is just a good example of things that that you might need to be able to do yes I did today reflected that said the X z plane as the plane of the screen and I reflected about it true that yet guess what would you do if you really want to 2 of it is up to that you would that I'm so glad you asked that because of soaring OK so I'm OK so as we really are as we anticipate this really shows you that how these matrix and how these matrices and coming out depends on both what's the operation and what are you doing it to the basis is really important and so this is what I am saying when you gotta do this on the exam draw lots of pictures tell us what you're using for the basis because there might be equivalent ways of doing the same thing and also you want to be able to get partial credit for your work so make sure you make the water during a base over this case we are looking at the office we have the identity matrix that's always going to be the same so we don't even talk about it again but now as we do I see creation not only do these things change signs were not only to be changed places but they also change science because now we turn it around the things that the the doctor lobes that were pointing forward are now pointing back so in this case we get negative ones In the octagonal moments so that transformation depends not only on the operation that were doing but also on the objects that were doing it too and so again just to bring it back to what's in the character table although the irreducible representations are objects that transformed in different ways under all of the cemetery operations in the .period yet again this the things that we're making here are reducible representations that can be generated by adding up different irreducible representation and you're going to see how OK so again what's reflected in the air in the exceed playing the same thing nothing changes places but they do change sign so that 1 is the same as it was in the other examples so OK so now we've illustrated how we can look at these things and make up a transformation matrices for operations in a particular basis we have learned that the basis is really important to figure out what these matrices are and you were starting to get a sense of how this goes together with the things that are in the character table a case announced
look at some objects that transformed in ways that give us reducible representations so again here's a point that people often get confused about when the 1st learning this this concept so here we have a PX or law and you might say if we had just have a PX off by itself in space it doesn't belong to the sea to the point you know yet I'm doing these examples in the to the pointer that's because we decided that that's the basis that were working and because of something about the problem so we were looking at 0 C O 2 that molecules in the sea to the point that's what we're doing so now we can we can consider various wobbles and objects that are associated with that that molecule and look at how transforms under operations net .period group but we're not starting over in assigning the orbital to appointed makes sense to everybody but I just I bring that up because that is something that people tend to get confused about yes you read that the 2 sides of the pure or different colors because pure rules have noted the middle so changes fine so that's the that's the phase of the overall and you know of course that's that's going to be important as things might change signed under some of the operations OK so now we have a PX or old were in the 2 were in the sea to the point where Because I said so I know that if you orbital alone in space does not belong to that point group but were using this to set up a problem that involves a molecule that belongs to .period group and we're going to consider how that adorable transforms into each cemetery operation and it helps if you have your character table out and opened to the right .period routinely do that's OK so if I did the identity nothing changes so it gets the character of 1 if I do this each irritation it moves around and so changes and so that means it gets a character of minus 1 if I do a reflection in the act of the play in which again as the plane of the screen nothing happens doesn't change so it's a character of what if we do the same thing In the wisely now it does change science and so it gets a character of minus 1 and so that's how we set up a representation in this group and so if we go to the character table and look at the scene to me table with you'll notice it is you have this this pattern of characters 1 minus 1 1 minus 1 and that belongs to they're corresponds to the B-1 symmetry species so that's an irreducible representations and then if you noticed in the 1st column on the right there it says In that column so that means that the character table is telling you how something that has the cemetery even think of it as a PX or or as a little unit vector pointing in the extraction were describing how that transforms under all the operations and if you look the lower under me to hear that 1 is it that 1 has a wide and so that tells you that that's how acute why orbital is going to transform these representations are under the cemetery operations so all these things are collected for you here in the character table and this is this contains a lot of useful information OK so what else is in here we have Ozzie or wine racks anybody know what those are yeah there's a rotations so those rotations about those Accies so it takes a little bit of much of the work to to visualize had was signed up rotation to a symmetry group but you have if you think about it for a while you can probably visualize it will talk about it later we get into 2 talking about spectroscopy and then these other things the OMB X Y X z y z etc. these are quite quadratic terms in terms of the Cartesian coordinates you can think of them as d orbitals for now or will see that they are that they belonged to vibrating of the vibrational motions that changeup ability of a molecule so hopefully you get an idea of just how much information is already tabulated for you in the pointer table question was what was present at those are just the names of the the reducible representations and so all you really need to know about is a and B are a single Arlanda degenerate so there's only 1 1 0 symmetries pieces about energy there it's called even it's going to be doubly degenerate and if it's called T but it's strictly desire and will see more examples of that the other thing that think you should know is that every point group has something sometimes it's called a sometimes it's called a 1 sometimes it's called a 1 prime etc it's always called a something where the Kerry-Edwards where the characters are ones for everything so there's always a species where it remains invariant under any operation and that's a sphere greatest fear always has the property and so for talking about chemical bonding you can think about that as an ass orbital and every point brutal house that representation otherwise you don't really need you to worry about what those things are there just the just the names of the irreducible representations in that particular
group case it's look at why orbitals some meaning that the missile that fast because it's are you don't need writing down it's already on the character table which is that the joy of this whole thing is all this information already tabulated for you and all you have to do is figure out how to read it you don't have to generate yourself every time OK so we already went through facts we know how that 1 behaves so if we look at the PC or all we can see that that 1 doesn't change and the identity 2 doesn't do anything to it and neither do these reflections so in this particular case there's the orbital the PC orbital belongs to that symmetry group that you know that doesn't change under any transformation which in this case is a 1 OK so now I know some people might be thinking but what about in that case we were we were looking at the O C O 2 we had these orbitals you pointing in different directions and when we got differences for these that's because in that case my basis was the set of the 2 orbitals together and so that means that you want what we have so we have our own orbital isolation like this it in this case they belonged to the irreducible representations created few start building up sets of love than that might transform differently and you can make a producible representations for the England had a deal with that OK so if we look at why orbital we can do the same thing you again I'm just going to go through it quickly you can do it yourself for extra practice of the need to end then now at the next thing is let's look at this thing that's called a to here so that doesn't have anything that belongs to the x y and z unit vectors but
there's there still some characters for it if we think about it as a DXY orbital we can see that that's going to behave like this so he and Yancy to organize do anything to it but it will change signed under those reflections and so we get that under the 8 2 operation and so sometimes if you look at at these different .period groups there might be irreducible representations that are included for completeness but you might notice that there's nothing listed under that year in terms of linear quadratic terms of the Cartesian coordinates or occasions the other might just be nothing in those categories that's OK it it just means that there's nothing of chemical interests belonging to that symmetry species there yes he would be on the right of the people as you can get all hold of money all there will be able to transform according to that cemetery species and yet again how this is going to work and where these things go totally depends on the pointer so if you if you look around at different ones don't belong to different representations and so that's why it's so important when you start doing these problems to assign a molecule that to the correct .period OK
so now we have made up that we have looked at this we know what irreducible representations are we have the least scene some practice examples of how to make up reducible representations let's look at how we get from 1 to the other and there's a formula that we can use that so here you have to keep track of what you're doing but it's conceptually not really hard OK so what we want to know is we have some reducible representation that we're going to make up because it has to do with the bonding or molecular motion or some property of the molecule that were interested in and we want to see how can we add up the various irreducible representations and sorry in and get that reduce or representation so what we're looking for is for example the number of a ones that appears in Europe representation so sometimes we can do this by inspection but often it's not really that easy particularly when we weakening the problems we need to use this formula OK so boring and do it is add the following things so 1st of all we do we have a problem 1 over each so what's H. remember that some of what you get when you add up the total number of operations it's also called the order of the group so the character table I gave you unfortunately doesn't give you HC half-dead up yourself but that's a fact and then what's in this song here priority is the character in the reducible representation kind Is the character in the irreducible representation so remember the reducible representation is whatever you generated that's going to help you solve the the molecular problem the character and irreducible representation is the 1 from the table under that particular operation and the end is the number of equivalent operations in the class so that means the coefficient in front of the the operation at the top the table so let's let's look at this late hour the
yesterday give questions about interested asserted today did you absorb everything around our contacts you OK so each is the number as the total number of operations in the class so like for 4 c to be we have arm so there's the identity there CQ and then there are at the signal the insignia the prime so H is for for that group some character tables give you the value of age but this 1 doesn't see it that way OK so let's go through an example of how to use this so here is irreducible representation again if we're doing this in a realistic scenario we're going to have made this reducible representation of shelves because it's telling us something interesting that we want now in this case for the sake of not taking forever I'm just giving you an example of 1 of that works OK so now we're in the C 3 .period group you know why again this is an example OK so for this 1 H equals sex because we have the identity we have to see 3 easily of 3 sigma these so we add up all those on and it's 6 and so now we have to look at the character table and see how many irreducible representations we have there 3 years a 182 in and we have to figure out how how many of each kind are added up to give us this reducible representation OK so we have 1 over 6 times now we're going to start adding the out for each operation in the table we take the character in the reducible representation that's the problem that we're trying to solve I see ICU question on to when I when I finished explaining this really quite OK on so here the character in the reducible representation is for under again and again that's what we were given 1 is the character In the irreducible representation that we looked up from the table and then no 1 is the number of symmetry elements in that class there's only 1 identity and now we have to go through and do this for every operation that's possible in the C 3 b . grew so now for C 3 we have won In our readers full representation that's what we got from the problem 1 from the character table that we looked out and there are 2 C 3 operations and then we go through the same thing for this the plants and questions about OK cool OK so again this has picky things that you have to keep track of but it's not that hard yet so we just have the character that were given in the reducible representation we have the character in the irreducible representation in this case a 1 because that's what we're trying to figure out which we look up from the table and a number of operations and then we just have to go through an ad these things for each element in the table questions here that's a problem that's what I gave you so that that thing here that I'm calling 1 that's given that's that's the question in real life you're going to make it up yourself because it's going to help you answer a question that that you want to know the answer to I'm just giving you some time to practice will do a real example later on if we don't get to at this time will do it next lecture so yeah this this capital gamma it's just that the reducible representation that there were looking at OK and so after all of this on but we go through and find that there are no way ones in particular representation that's fine but I want to point out here that if you do this and you don't get an integer remember something went badly wrong and you need to check your work OK so now let's look at 8 2 words going on the table we have to check out all the possible irreducible representations and so we knew the same thing so again be identity the character in the reducible representation is for that's given the character and the 8 2 is 1 that we look up on the table and there's only 1 event and then we just go through and follow the the characters under a 2 and do the same thing for that symmetry species and we get there too long in this case and then we have to do the same thing for a so this is 1 of the things that that's a little bit more challenging don't get confused about e the cemetery species and the identity operator is there they have the same name which is not so ideal but you can tell from context
OK so we're looking forward the number of the cemetery space is and again we go through a news our reduction formula for each operation in the table and we get 1 and so we went through all of the possible irreducible representations in this .period group and we use this reduction formula and so we're gone and here's what the former the insertion of a place we should say Demel 1 equals 2 8 2 was so then it's not it's not hard it's just you just have a lot of stuff to keep track of and you know it's fine to get 0 for these somethin you'll find that some reducible representations don't contain a particular irreducible representation what you don't want to get his fractions you you really do need to get the whole numbers for these things that we don't protect the work yes so I'm sorry Saigon negative to below the 3 sigma B. that's a problem that's what I gave you the problem is here's this reducible representation demo 1 reduces and the idea that I want to point out we do real ones that's going to come from something that we do as part of setting up a problem another question this that's right and the reason that that we want to do that is because we have all sorts of information about that on the character table and we can use it to learn stuff OK let's try to bring this back to 1 discussion instead of many because it's really hard to hear what's going on that question yes the 3rd term is the number of operations in that class the equivalent so that's how efficient informed about operations it's on the character table so if you look at C 3 there's a 2 in front of it if you look at that said Sigma Beta 2 3 in front of it OK so everybody clear on our terminology and and what we're going to do here is what's so let's move on OK let's see another example something to give you another example were in the same .period group there were going to go through this again and we're going we're going to do it fast because I we've already gone through red power if you get confused I'm happy to answer questions about it OK so we need to go through the number of a 1 again and so now we have a different reducible representation which recalling to and so in this case when we go through this we find that the number of a ones equals 1 and I guess what I want to point out here is that in this reducible representation the same to the 1st 2 characters are the same and only the 3rd 1 is different we can get quite differences from so it's not always easy to do them by inspection sometimes you can if you can just see what stands out that's a fine way to do it but it's not always easy and the formula always work so we definitely want to do that OK so if we don't fearing that the number of a choose now again using this formula we get 1 and the same thing for the asymmetry species and now we get 1 for that 1 and so we can write Gamache to A-1 plus a 2 plus OK so now we know how to use this reduction formula to reduced rates for representations and get some arbitrary representation in terms of on the irreducible representations as you can imagine you could make up ones that don't work right see if you if you make up a reducible representation without a corresponding to some to something real In terms of chemical bonding or something like that you can imagine coming up with sets of numbers that don't give you images in terms of the pointer so you know if you're going to look for practice examples you don't just make up arbitrary ones it may not work OK so let's arms I had another 1 in
here but I don't think we need to to do it let's quickly go through how we're going to set this up In
terms of a real problem we won't have time to finish this problem but I respond set up so that you can see it for next time and has had a chance to think about it OK so let's think about bonding in that triggered Oakland a molecule so to give you a concrete example what say it's BCL 3 and what I want to know is which riddles make up the arms the signal bonds the seal free and we all know the is answer what's the hybridization of the central as the 2 so at the end we should get something that's consistent with that answer but we're going to go through the process because works will see example later that are not so simple OK so in order to figure out what's which orbitals are involved in the bonding in the small at all I'm going to take a basis that has to do with the problem that I want solved so I wanna know about those Sigma bonds and so my basis is 3 vectors I'm just calling them a 1 a 283 pointing in the direction from the central atom to each other Adam and that's why basis and I'm going to figure out by symmetry which orbitals have the appropriate symmetry to belong to those bonds and I'm going to do that by making a reducible representation that corresponds to the spaces and then I'm going to reduce it and look at the character table and see which irreducible representations fit with that cemetery so this is what we do so here are basis 53 vectors and now what I need to do is go through and figure out the character of my reducible representation under each of the cemetery operations and so I can do that yeah a number of different ways I can users are heuristic if I'm careful Hi can also set up the actual matrix for transforming this and then take its characters don't always work let's just look at it how we do this if we set up a vector that contains a the are 3 representations of bonds and then just look at how that gets transformed under these operations so where know with the identity is going to be that doesn't change anything so we just get one-to-one on zeros everywhere else and so the character of the identity matrix is 3 In case announced you see 3 creation counterclockwise and you know those who 1 look for mistakes as a prime waited this is a prime place to make them so make sure I didn't do it clockwise and in the matrix OK so the elephant-headed messing up right so we have our own vectors I didn't see 3 rotation and we can look at how they changed places and now I need the matrix that generates that and so this is what we get and it's character is 0 then again we can use the heuristic that all of these things swap places so we know our character has to be 0 OK so now however seek to question the role of the yeah I understood I'm going through you know have what's that what's the process for thinking about this and say you know sometimes when you get a really good at it can just kind of see in your head that's fine but you know for you know it it's kind of hard when you 1st start learning how to do it so I wanna make sure to go through step by by step for all these things question although the beaches the the right of how would you clockwise would be you know a 2 would take the place of a 1 and Southern life my matrix would be reflected about the diagonal if that's 1 you know you just have to do it set up the matrix and multiply by here original thing and see what gives you the the result that you want I don't think that the the picture going have yet that is the yesterday OK this is a great so everybody was and is an important question the question is whether the axis of of the axis is always the principal axis that you determine when you put in a pointer and then we will do did so in this in this case it's coming out at you OK maritime and this is a decent place to stop we're going to finish this next time also know although quizzes are unannounced but this is kind of a good point to maybe think about having won so that's something that you should keep in mind the next week
Chemische Forschung
Metallmatrix-Verbundwerkstoff
Hydrierung
Fülle <Speise>
Screening
Metallmatrix-Verbundwerkstoff
Wursthülle
Quellgebiet
Wasser
Reflexionsspektrum
Brillenglas
Erdrutsch
Formaldehyd
Wassertropfen
Expressionsvektor
Chemische Bindung
Spektralanalyse
Baustahl
Operon
Molekül
Hybridisierung <Chemie>
Chemie
Chemisches Element
Expressionsvektor
Metallmatrix-Verbundwerkstoff
Phasengleichgewicht
Screening
Wursthülle
Oktanzahl
Wasser
Orbital
Computeranimation
Atom
Chlor
Sense
Zündholz
Chemische Bindung
Optische Aktivität
Linker
Operon
Molekül
Systemische Therapie <Pharmakologie>
d-Orbital
Hydrierung
Metallmatrix-Verbundwerkstoff
Mähdrescher
Base
Kalisalze
Ordnungszahl
Reflexionsspektrum
Gelöster organischer Stoff
Bewegung
Biskalcitratum
Chemische Formel
Farbenindustrie
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Metadaten

Formale Metadaten

Titel Lecture 03. Transformation Matrices.
Serientitel Chem 131B: Molecular Structure & Statistical Mechanics
Teil 03
Anzahl der Teile 26
Autor Martin, Rachel
Lizenz CC-Namensnennung - Weitergabe unter gleichen Bedingungen 3.0 Unported:
Sie dürfen das Werk bzw. den Inhalt zu jedem legalen und nicht-kommerziellen Zweck nutzen, verändern und in unveränderter oder veränderter Form vervielfältigen, verbreiten und öffentlich zugänglich machen, sofern Sie den Namen des Autors/Rechteinhabers in der von ihm festgelegten Weise nennen und das Werk bzw. diesen Inhalt auch in veränderter Form nur unter den Bedingungen dieser Lizenz weitergeben.
DOI 10.5446/18911
Herausgeber University of California Irvine (UCI)
Erscheinungsjahr 2013
Sprache Englisch

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Dauer 49:45

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Fachgebiet Chemie
Abstract UCI Chem 131B Molecular Structure & Statistical Mechanics (Winter 2013) Lec 03. Molecular Structure & Statistical Mechanics -- Transformation Matrices. Instructor: Rachel Martin, Ph.D. Description: Principles of quantum mechanics with application to the elements of atomic structure and energy levels, diatomic molecular spectroscopy and structure determination, and chemical bonding in simple molecules. Index of Topics: 0:00:10 Transformation Matrices 0:02:44 Group Theory 0:06:27 Everything is About the Basis 0:19:59 Characters: C2v 0:29:56 Reduction Formula 0:43:36 Bonding

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