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Lecture 04. Group Theory Applications.
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so last time we're talking about using theory to understand chemical bonding and we're going to continue that today we will see if we can get through the rest of it OK so we talked about how to set up matrices for some of the operations and this spring the suckers as we did last time so we talked about looking at these things or we can swap places between elements on the basis and we can set up matrices this way and we also talked about a heuristic were weekend without writing the character with or without writing down matrix we can get the character of the matrix by just looking at how many of these elements either stayed the same exchanged places or change sign and some people were asking me after class how do you know when that works and that doesn't and the answer is it works in the case where every operation that you do in that In that basis resulting in 1 of those things happening so in other words the elements you changed places status saying for swaps or change site so for example in this basis were talking about these 3 vectors and so if we do see 3 workstation then 1 of the maps onto the other 1 in every case when we do that but now imagine that are basis was unit vectors and the x y and z directions out so now my basis is the explained the unit vectors and if I rotate 120 degrees they don't not exactly onto each other rights in that case we have to write down the full matrix and we can't use that shortcut to get the character so that's how you tell when you're able to do that when you're not OK so this in this case we can do that so let's talk about a C to rotation and you know we have to be careful about how we set it up so in this case it's we set up such that 82 and 83 change places of course there are a couple of other of C 2 operations we can do as well but some years the majors operation for that and it's character equals 1 which again we can get using the shortcut so we see that anyone stays the same source said it contributes 1 to the character and then a 293 swap places them contribute 0 How about sigma age This is a reflection through the horizontal plane which in this case is the plane screen and here we're not talking about anything that can change sign when you do that we just have these pointing out and so the matrix we for that is the same as the identity and it's character equals 3 at the time the anybody have any questions about how we got here before we go on to remember what we're doing is we're setting up irreducible representation is is going to tell us something about bonding in the small Kewell and embroidery and hopefully come up with an answer that is intuitive OK so let's move on been setting up the characters for all these operations and so so far we have this sort of this resource representation retaining gamma for bonding and I couldn't characters for of these operations that we found so far so we still have a couple of them left so improper rotation remember that's when 4 S 3 innings barricade 120 degrees and then reflected through a plane perpendicular to the for to the principal access and so we can visualize that by looking at power will actors not onto to each other and then set up the corresponding matrix that gives us that and so the character of that is 0 and then the last 1 3 signaled the start of the of course the vertical planes that that cut through each 1 of these vectors and when we do that we get a threeday tour swapping and so the character that is 1 question here is whether the results of the well so that the horizontal like this so I don't say the horizontal plane is the same as the identity has said that the matrix you get ends up being the same because were reflecting its so the molecules flat rates were reflecting in the plane that's horizontal so perpendicular to the principal axes and remember that it depends on the basis are basis as these vectors pointing outward and so few reflect that in a horizontal plane it doesn't change anything right there's nothing that can change signed and so you just get you get the same thing as the identity because nothing changed yes I think that the reluctantly decided to reduce the risk of going to be that this is the only veteran of the report With right was brought to the the dollar was the what is the you are and for the improper admissions it's hard to remember you rotated by 360 degrees over and reflected in the horizontal planes so in this case you just irritation and reflection doesn't do anything whereas we already discovered that so it's the same as rotation in this case and that means you might just needs some more practice visualizing these things and but it's it's a good question you are definitely going to have to go through this entire process yourself once you get through some more practice I'm going to give you examples where I tell you OK we're looking at this particular set of bonds now you set up the basis and and make all these representations but you know all walking through reducing examples in class so that you see having yes you on the Austrian clockwise and did tonight I In we should do it counterclockwise and I will change this before it was the slides and the moral of the story is you try to always be consistent with with with what you're doing and write down what you're doing just about any of this will you yeah so it will be at the heart of the well it's reflected through a plane that's perpendicular to power rotated at which may or may not be the principal access in this case it happens to be but it doesn't have to be so it's so the S the S 3 access means that you rotated about that access and reflect so whatever the access is it's perfectly you reflect perpendicular to hide repeated a mail may not be the same as the principal axis yes is 1 of the major the pages of the cost of that will the molecules is flat and the way I drew it could in the plane of the spring rate the principal axis pointed out that you so know that horizontal plane is in the plane of the screen the way strong here OK so here's our reducible representation and I'm giving you some some practice problems that you can practice setting up these things and in the meantime we're going to go on and on initially we reduce use but questions 1st but if the region is on hold during the it's well also so I rotated in the wrong way I said We're going clockwise so a should actually know that's OK isn't so a 3 2 were 81 previously was and then we fighting and that's alright OK so that 1 is fine
09:13
are so here's our full reducible representation for a binding question because when we do the C 3 rotation all although little vectors swap places right and so they each contribute 0 2 the character you can also write on the full matrix you see that you get 0 for the character various so let's move on OK so I want to point out here's how we would do this if we didn't have to deal with the basis being on the x y and z axis I'm doing this for completeness if so so it's a little bit confusing right now don't worry about it we're going to go through it again later on but the point is we have to use those rotation matrices and actually put in the signs and coastlines of the angles in order to deal with the the case where the elements that you're using do not exactly not onto to each other so I drew this this molecule you differently with respect to the access system or my basis were the unit vectors I would have to set up my rotation matrix In this way will still get the same answer for the character but we have to to do a little bit differently we can't use the heuristic if we don't have an object that were everything maps onto another element of the basis when you do the operation OK so again that's just for completeness of it we're going to go on if it's not 100 per cent clear right now don't worry we'll do more examples like that later so now let's go back to our original problem that was we want to understand the bonding in the small terms of symmetry so we have a
11:09
reducible representation that we made up by going through the characters of Bob basis which is the bonds and the last time we learned of the reduction formula that is going to tell us which during usable representations band together to make this reduced representation and so we're just going to go through music just like we did last time so each is 12 in this .period remember that's what you get when you just add up the total number of operations that in this .period and so we're just going to go through stock character sofa for we have Our on character in the reducible representation the irreducible representation and we just had these things out and so I just want to really quick because we did this before sites so in case it was too fast I got 3 4 E I got 0 4 C 3 1 4 c to etc. just going through all of these operations 1 by 1 and so for the number of anyone we get 1 now we need to go through the same for the other irreducible representations and Ingrid yes I Kiai irreducible representation that's on the character table so so when you look at the character table the 1st irreducible representation is anyone prime rate if you look at the character under the say it's 1 Arab rights and so just follow through the characters for each of the operations there on the table of it's not really the same thing freight to prime and again I'm going to go fast because we learn how to do this last time it's just they're seeking following and checker your work so same thing we go through each operation and multiply the character in the reducible representation which is what we made up just now the character and irreducible representation which is from the point of cable and a number of operations in the past and we get that there were no 8 to primes in this particular race for representation and now we have to do a 1 double Prime which we do using the same procedure and we don't have any of those and then looks like a to double crime again we get 0 for that so remember if we get 0 for some of these things that's completely fine if we get fractions that means we did something bad and we need to look at the prime if we go through the formula we get 1 of those Annan and last won in double crime we had 0 so we now have the are reducible representation broken down in terms of which irreducible representations we can add up to get that and believe it or not that is until something useful so here we go are gonna be for bonding single bonding to be specific it is 81 prime plus the prime so now we think about what this actually means so we were talking about you what what we wanted to learn in the 1st place is which orbitals in this molecule have great symmetry to contribute to the signal and think again to the answer that we all know from General chemistry which is that that central atom is as P 2 hybridized and still look at what we got here we got 1 cemetery species that is not degenerate A1 crimes that tells us already that you maybe that something like the US or and then we got something that is doubly degenerate the practice of software that's consistent with what we now know what's what's going ahead and
15:41
I look at this and a little bit more detailed so if we look at 81 prime noticed that that is the representation that invariant all transformations that means it has a 1 under every operation in the character table and I told you that every point groups will have 1 of these and that the objects that transforms in that way a sphere so that's like as yes ,comma of you that's just it's name that's the name of that particular irreducible representation you could name at Joe if you want but it's it's just that's just what's called and they have different names in different .period groups but if you ask a mathematician that question you will get a very different but it's so that for fervor solving chemistry problems it's fine to consider those those that's that's just the name of the representation OK don't get confused the the identity on the EU on the top of the character table is the identity operator the ease of the along the the lefthand side those aura symmetry species for irreducible representations that also happen to be called something you'd so sorry I didn't name but it's you you just have to know why context so where it's located in the table will tell you that OK so anyone crime in this group is the thing that's invariant under all transformations and so this is this 1 is important because you have to remember that so for the other ones yeah there's a little xyz and so you can you can determine really easily those with pure also for the US orbital you have to remember that it spherical and that that's the 1 that I did that identical under old transformations are it's analogy prime is our species that's doubly degenerate so that means there are 2 orbitals and if we look at the table and look at what's listed there were prime it says X and Y so that means that it's the PX and PY Robles that are involved in the morning and again let's go back to our general chemistry in tuition we know this because this is your role we said You have easy direction is defined as the principal axis which is sticking out of the plane and we're talking about the bonding that's going on in that plane perpendicular to the axis so we knew it had to be the PX in PY orbital and it's comforting that we got the right answer by going through this whole reducible representation so hopefully that gives a good understanding of the process and makes it more intuitive for later on what we do things were the answer is not to be obvious before we start OK so let's talk about some other things that that could happen we also see that 2 of them on a character table under the prime the operator for sorry the new prime symmetry species we also have a DX squared minus the y squared and DXY and so by symmetry along it's equally probable that those on within the body so how do we know it's not yet the answer is we know some chemistry and so we know that there aren't any d orbitals that available for bonding In war on there not filled that would be a really high energy state and it's unlikely so symmetry alone tells us a lot but it doesn't tell us everything we still have to know some chemistry we still have to think about it so I emphasizing the so much because I want you to go back and check your answer make sure you got something that makes sense when were doing these problems because it will be really important to making sure that you understand it for later we do things that they're not as obvious OK so as the 2 would be perfectly financer of her looking at cemetery alone but he said he had to know something about rule that out OK so what if we had another molecule that had the same cemetery but it could make pipe bombs we can use the same argument to look at which orbitals might be involved in the pipe bombs so before we do that everybody is happy with the signal 1 example on how we get to where we are it's fine if you're not completely ready to make all these representations of yourself or abuse practice examples that you can go through as homework will try to get those posted maybe later today OK let's talk about what I want OK so now
20:47
let's say we have an 0 3 minus so that not only has the signal was playing but it's got some pie bonding perpendicular to it we know it's perpendicular to it is we know some chemistry OK so here's archive bond and I'm going to go through this because it just did it again it helps you build intuition from some of these problems but they still want talk about the symmetry of the pipeline so we want set this up in a way that it's easy to visualize saw pipelines is it has a at the age the nuclei and we know that it's it's got intensity above and below and it's it's it's a positive on 1 side negative on the other and so we can represent its symmetry it's just a little arrow pointing perpendicular to the signal and so now are basis is going to be 6 directors representing the possible orthogonal so why do we have 6 matters because it's in a threedimensional space right we could have the pipeline's perpendicular in 1 of 2 ways you could imagine that sticking up out of the plane and molecule which of course we know is the right answer or by symmetry arguments we can also look at the Taiwan's that perpendicular to the single bonds but in the plane so here's where basis looks like again were doing this not because we expect to be surprised by the answer but as practice for some of these kind of problems OK so we wanna know by symmetry which orbitals
23:01
conform our and so we're going to consider the In Plain and the of outofplane sat separately because it can be both at once right so 1st wearing a look at the ones that are pointing out in this picture out of plane the reducing thing in go through cemetery operations and set up matrices so here is our identity operation unsurprisingly that gives us a character of 3 and looks like it always does and now we do see 3 rotation which in this case everything changes places and we did it in the correct direction it's counterclockwise and so we know that it's going to have a character of 0 because everything changed places c 2 now we have to be careful because were working with objects that can change sign and so we have to pay attention to that so rumor we we represented the direction of the bond with the Alero and so we have to pay attention to the fact that we do see each irritation not only did to haven't switched places but you're turning it over so the change sign and we have to keep track of that In case so if we had that up its character is minus 1 and again we can get that using the shortcut we can say to haven't changed places so they contribute 0 come and the 3rd 1 changes signs that contributes minus 1 yes pursuant to the decision to use the also I have my little arrows all pointing out right they're all the same direction you well so it was so my basis it is those orbitals better going to make a pipe bomb inside represented that symmetry of the pipe bombs as just boleros pointing in the same direction because if the orbitals or out of phase with each other it could make a pipeline rate they have to be all the same direction for that to work and so what we're doing is we're starting with the symmetry of the object that we want to see that being the pipe bombs and were saying which orbitals can contribute to that symmetry and so the other the other thing that's important is remember that I said OK in principle we could have perpendicular ones in the plane but we can't have both at once right because something like this in something like that can make a pipeline together so we're considering 1 said at the time just to make our lives easier I could set up a 6 by 6 matrix for all them but it would be harder that needs to be so we're not going to OK so we need to consider are signs and again if we do a Sigma H. reflection now so again that's perpendicular to the principal access if I do that all 3 of unchanged sign so again we see that the matrix you get totally depends on the basis of depends on what you're applying the operation to so the last time we had these sigma bond vectors but couldn't change when we did that because they don't really have a sign now we're dealing with something that structural and so for minus once instead of ones on the diagonal there so its characters minus3 OK so we're almost there to setting up reducible representation for the army out of place and then
26:56
we have to deal with the S3 3 so let's check that I actually do it counterclockwise and I did so were in good shape but again when we do that workstation a couple of them swap places where they may also places sorry and then only do the reflection perpendicular To that they'll change sign then again that is the heart of the improper rotation is the hardest when visualize so it it's going to take practice OK so now we're left with just the signal being so the same thing as before to them change places but because we're doing a reflection not turning it over now nothing changes sign and so that gives us a character at once so we here's are complete reducible representation pipelines In out of playing set saw next step is to reduce it and again reviews the reduction formula that we weren't and I'm going to go through quickly yeah actually looks like I opted to not go through so that is left is as an exercise in student but again news the the formula I had the sliding here I think I decided it was 94 OK so what we get for Dennis Michael that O'Keefe rout of plane we get to double prime and the double prime and I encourage you to you know go through and do this and confirm that you get that make sure that you understand how to use the reduction formula
28:59
and now it's considerably in playing set so these are the blue ones so these are the ones that are perpendicular to the sigma bond but they're in the plane of the molecule and we're going to go through quickly so here a C 3 rotation doesn't change the sign of anything it just makes stuff switch places so it's character is 0 again c 2 and then we have to foot the worst thing is signed Sigma each would he think anything change sign no right is now everything is on the same plane as reflected and so that gives us a character of 3 and yes on the other hand if you think of way we need to let's see well so you have to flip it over and so the errors that were pointing this way enough pointing out way the principal axis is the principal axis movie the axis the year that sticking out at you if you do well so you have to you have to rotate about the sea to access which is perpendicular to that yet got so even these these are now in in In that plane when you do this each irritation yourself over so the ones that were this way another when the C 2 is not the principal axis that's right the C 3 is the principal axes the highest water access is always the principal axis OK so let's
31:14
style let's try to finish up this example sleeping at least see what the answer is to for practice purposes OK so again U.S. 3 so we rotated 120 degrees counterclockwise and reflected In that horizontal plane which doesn't change the sign of everything so that just gives us a position swap and are vertical planes following a change society because stuff is the plane that were there were reflecting about and so we get minus 1 for the vertical planes and so again we need you to use the reduction formula and reduce said and I'm not going to do it because I think I correctly calculated that we would run out of time if we if we did all that class and so what we find is that we get free the inflamed 8 2 prime plus the prime so homework number 1 next class is actually do those so go through reducing yourself verify that this is what you get and then proved yourself that the orbitals that are involved are the ones that you expect to see involved in that once said OK so I think that's where we are now is worth finished up with what we're going to do with group theory for the moment we're going to see it later in the quarter because it's going to come up we talk about vibrational spectroscopy actually will come up we talk about selection rules preparing regional officer consists of a few minutes left it's distracting when everyone is is packing up and and rustling around all OK Serena secret .period again we talk about selection rules were going to talk about how you know that enables them to 0 in a particular place so 1 way to prepare for that if you don't remember it is know look up the Wikipedia page on even on rule that that's 1 case of this on that's something that is very useful in terms of being able look at intervals and to say that they go to Zurich revealing that's it's something that is very helpful to be able to do so next time when you start talking about rotational spectroscopy so we moved on to the next chapter on right I will see you either at the office hours of exports
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Hydroxybuttersäure <gamma>
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09:13
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15:38
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23:01
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Metadaten
Formale Metadaten
Titel  Lecture 04. Group Theory Applications. 
Serientitel  Chem 131B: Molecular Structure & Statistical Mechanics 
Teil  4 
Anzahl der Teile  26 
Autor 
Martin, Rachel

Lizenz 
CCNamensnennung  Weitergabe unter gleichen Bedingungen 3.0 Unported: Sie dürfen das Werk bzw. den Inhalt zu jedem legalen und nichtkommerziellen 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  10.5446/18912 
Herausgeber  University of California Irvine (UCI) 
Erscheinungsjahr  2013 
Sprache  Englisch 
Inhaltliche Metadaten
Fachgebiet  Chemie 
Abstract  UCI Chem 131B Molecular Structure & Statistical Mechanics (Winter 2013) Lec 04. Molecular Structure & Statistical Mechanics  Group Theory Applications. 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: 0:00:07 Bonding 0:15:38 Which Orbitals Form the Sigma Bond? 0:20:44 Which Orbitals Can Form Pi Bonds? 0:26:53 Out of Plane 0:28:57 Which Orbitals Can Form Pi Bonds  Consider the InPlane Set 0:31:13 InPlane 