So, I think there are a couple of very important things that we did yesterday and I want to revisit them before we go to any new topic. One is to calculate these later rules of the type 2, where one of the determinants differs from the other by one occupancy. So, if one of the determinants is psi 0, which has chi 1, chi 2, chi A minus 1, chi A, chi A plus 1, chi N. So, there are N, N spin orbitals. I have specifically marked A, a particular in spin orbital which I am going to replace and hence I have put A minus 1 and A plus 1, otherwise there is no meaning that A can be 2, A can be 3, if A is 3, A is of course 3 then this is 2, if A is 1 then of course this does not exist. I mean I hope you understand just the meaning of this, I have just specifically marked this to make sure that you understand the A. Then, so this is one determinant which I call just a reference determinant with respect to this, the other determinant differs by one occupancy which I called A R and that was chi 1, chi 2, everything is same including chi A minus 1 here, but this chi A is replaced by chi R and the rest is chi R. I hope everybody understand this nomenclature, so it means exactly in that column of chi A I am replacing it by chi R. So, there is a question if I interchange what will happen? So, I could, I can put chi R in some other column and after that I interchange this, so then of course there will be negative sign, so that is a mute point, but at this point what we are saying that wherever the column of chi A is there, I am replacing it by chi R or the row of chi A is there, it depends on how you are doing it. We probably defined the spin orbitals as the row and the column was coordinates, so it does not matter, so either that row or the column depending on how you write the determinant would be replaced by chi R, the coordinates are still 1, 2, 3, 2, n, because that is a dummy coordinate, again I repeat, coordinates are always dummy, so it does not matter, important thing is there are n electrons, there are n spin orbitals, so here also we have n spin orbitals except that the chi A has been replaced by chi R, so when you have these two determinants and these are now arbitrary spin orbitals, they are not necessarily hard to recall when I am using slated rules, so they can be arbitrary spin orbitals, the slated rules now says that psi naught H psi A R is nothing but chi A H chi R plus sum over B equal to n, chi A chi B anti-symmetrize chi R chi B, is it clear? Obviously I have written B equal to 1 to n, but B cannot be equal to actually A, so I can say B naught equal to A and obviously B is not equal to R also anyway, they would become 0, so this is the type 2A, this is the 2B, this is for the one electron part, this is for the two electron part and clearly from the expectation value you saw that it was chi A H chi A sum over all A that is gone here chi A AB chi A AB sum over all B with a half that has changed, so now it is a single term for this and just one single summation for this, so this is the first important thing and this is a slater rule for any arbitrary determinants of this type, we are not suggesting that the spin orbitals are Hartree-Fock or anything, so this slater rule is now applied to show Brillouin's theorem, so Brillouin's theorem now specifically says that these are Hartree-Fock orbitals, so now we are saying that apply slater rule and the next condition is if psi 0 is Hartree-Fock determinant and of course chi R is orthogonal to all chi A, any chi R, I mean in principle we have all unoccupied orbitals which are orthogonal to occupied orbitals, so if chi R is equal to chi A then we are saying that the psi 0 H psi A R which still follows this rule of course, this is a general rule but this will now become 0, so this is the content of the Brillouin's theorem, so this is just a slater rule, however if psi 0 is Hartree-Fock and this spin orbital chi R is orthogonal to chi A and all occupied orbitals actually then it so turns out that this expression becomes equal to 0, so that is the, so you should first separate this, so if I give you any arbitrary determinant and ask you to calculate such, you should always use this, do not write 0, the Brillouin's theorem only applies if one of them is Hartree-Fock. So if I am only using Hartree-Fock orbitals then only Brillouin's theorem applies but this is a general rule for any slater rule just like you had given slater rule for expectation value, so the proof of this why so because in case this is Hartree-Fock then we showed that this expression which is of course a general expression is nothing but chi A F chi R, so this was actually equal to chi A F chi R in case psi 0 is Hartree-Fock written, that is easy to see because we have defined our Fock operator as H plus sum over partial integration over d tau 2 chi B star 2 1 by R 1 2 1 minus 2 chi B 2, so that is exactly this is, please note that this is the expansion, the expansion is just this for this double bar, so that is always there, so that is what we mean by anti-symmetrize, so 1 minus P 1 to maxed anti-symmetrize, so we do not write it but whenever I am writing double bar it has a 1 by R 1 2 into 1 minus P 1 2 and of course if we write for the spatial orbitals then we have to specifically write 2 times 2 minus this, so since you have a chi B star 2 chi B 2 with this stuck in the middle, quite clearly if I integrate this with chi A star on the left, chi R on the right, so it becomes chi A star 1 chi R 1 then the entire quantity is nothing but an integral chi A star F chi R, this is a number, this is also a number, it is a complete integration, whereas a fork operator is an incomplete integration remember fork operator is an operator, so it depends on coordinates 1, this is a number, so the further integration is done between 1 coordinate 1 which was lying unintegrated in the fork operator, so that gives you a number, so please try to understand this how to manipulate this, so this become this is a 2 electron integral, this is a 1 electron integral but when I write fork operator it is just H of 1 no integration and then only the partial second integration is done in the fork operator, so after I present this chi A star F chi R it is a complete integration and this is nothing but this, then we say that since these are Eigen functions of F or even if they are not Eigen functions F acts on chi A and transforms within the occupied set and the virtual orbital chi R is orthogonal to all the set, this quantity is equal to 0, so we say that even if it is non canonical F of chi A is nothing but sum over B lambda B A chi B, chi B which is actually 1 to n, so obviously your chi A and F is Hermitian operator is nothing but chi B lambda B A star sum over B and now we can quite clearly see if you close the integration with chi R then all there is a sum over B but these are only occupied so all chi R, chi R is orthogonal to chi B, this is a number which goes out so it becomes 0 either way and if it is canonical then of course it is trivial to show, so either in the case of a canonical or a non-canonical Hartree fork this quantity is 0, so these are the two essential things that we did yesterday, so I want to again write that very clearly the first part was a general Slater rule for any arbitrary determinant, second part was application of this Slater rule to the case where sinot is Hartree fork and the chi R's are of course orthogonal to this spin orbitals, in such a case you get the simple relation which can be derived from this general rule and that is what I did and this is called the Brillouin's theorem, so whenever now Brillouin's theorem says whenever I start from Hartree fork determinant, if I calculate the matrix element of the Hamiltonian between Hartree fork and another determinant which is generated by one change of spin orbitals and this is what we will call now singly excited determinant, the matrix element vanishes starting from Hartree fork, so that is the final statement, so then we will move forward. The Brillouin's theorem is a very important content which will be actually used when you go to the post Hartree fork theory, but remember the Brillouin's theorem is a consequence of Hartree fork, we have not done anything, we have not done any more new theory, it is a consequence of Hartree fork because this is something that is given I cannot help it and from here the definition of fork operator gives me this and automatically I get 0, so the Brillouin's theorem is not a new theorem, it is a consequence of Hartree fork and many times Hartree fork is actually stated as a Brillouin's theorem, in fact that is also interesting interpretation of Hartree fork, it is you say we said that Hartree fork is a determinant which minimizes the energy for a single determinant, you can say given a single determinant can you find a set of spin orbitals such that this is equal to 0, it is also a Hartree fork condition, so it is a way of putting it, so in a way this is also a Hartree fork condition, the Brillouin's theorem is just a consequence of Hartree fork theory, so that philosophy should understand that we are not doing anything new, it is a direct outcome of the Hartree fork theory, but you recognize this when you move forward, so that is what we will see later. At this point I have also introduced the term called the singly excited determinants, please note that they are different from singly excited state, many people get confused, we are still talking everything about ground state, exact ground state, of course exact ground state we are far from reaching there, Hartree fork is an approximation, so Hartree fork determinant is an approximation to ground state and these are singly excited determinant with respect to a reference, whatever is the reference, so our reference that we are usually bothered is the Hartree fork reference, so reference can be Hartree fork, but it can be any other reference, an arbitrary reference like any size 0, so with respect to that because whenever I say singly excited it must be with respect to something, so there is a given determinant and then only you are exciting, so this is a class of singly excited determinants and of course how many of them you can generate, we did that in the CI, how many of you can generate? N times M minus N, so let us say there are M number of spin orbitals, this is your total basis or whatever number of spin orbital that you generate, N is the electrons, then of course the number is N times M minus N, so N is the occupied, M minus N is the virtual spin orbitals, so that many number of, we are only talking of one of them, but just to say that any one of them will have the same property, any one of those N into M minus N determinants if I, with the Hamiltonian expectation value with respect to Hartree fork and the matrix element between Hartree fork and any of these singly excited determinants will be equal to 0, is it clear? So let us do a little bit of practice, so I gave some simple examples, has any of you verified this? Please do that, I told you just take a 2 electron system and please verify this, the rule itself, any arbitrary 2 electron determinant, so now we are going to do an application, so I said let us say chi 1, chi 2, H, chi 1, chi 3, so I have 2 sets of special orbitals from where I generate 4 spin orbitals, so let us say my special orbitals are like beryllium like which is 1s alpha, 1s beta, do not confuse the 1s with hydrogen 1s, it is just 1s, it can be any 1s, it does not matter phi 1, I could have said phi 1 and then the chi 2 may be 2s alpha, a chi 3 may be 2s alpha, so this is chi 1 and chi 2, 1s alpha, 1s beta and now your chi 3 is 2s alpha, so basically what you are doing, you have an electron which you are exciting from one of the orbitals, one of the 1s electron, 2 or 2s in a simple system of 2 electron system of let us say helium, so then I have a determinant which is 1s alpha, 1s beta, H and let us say one of them I, you decide, let us say I have put up 1s alpha, 2s alpha, 1s beta, so I ask you to calculate this matrix element, this is the helium ground state, I mean in a very simple term, this is an excited determinant with respect to this, I will not call it excited state but many chemists will call it excited state, I mean no problem, it is an approximation to an excited state, to be honest it is an excited determinant just like this is also a ground state determinant, so now tell me what will be the result, applies later rule, let us do a practice, I have replaced 1s alpha by 2s alpha, so the first one itself I have replaced, see please remember this order really does not matter whether you write chi A, chi B or chi A but make sure that the order is same because if you interchange both sides it is identical, so do not be particular about writing, the point here is that the ones which are replaced must come in the same place, so if this is coordinate 1 this should be coordinate 1 and the rest should be all other of it, so it is very simple do not even look at the formula because after that it does not matter whether you write that as coordinate 1 or coordinate 2, so that is identical, do you agree with this, that is it, do you agree because this summation over B is only 2 but B cannot be equal to A, so there is only one summation, A is 1s alpha, so it has to be only 1s beta and this should be 1s alpha and 2s alpha which are replaced, so this B remains as 1s beta 1s beta, I mean I am not understanding what is the problem, so that is the reason I am little bit perturbed, I mean it is just a trivial example, what is the problem, no problem but just confidence, lack of confidence, that is not bad, if there is no problem and lack of confidence it is okay, eventually you will gain confidence but if there is a more serious problem I am worried of conceptual problem, everybody agrees with this, of course if it was 3 electron there would have been 1 motor because then summation over B you would have taken you further, I am just giving a simple thing, now somebody can do spin integration, now it is trivial just do the spin integration and write it in terms of spatial orbitals, 1 electron, there is a 2 electron, how there is only 1 electron, no, no, no, but oh my God you are doing separately, that is 2, if you are writing exchange it should be minus, if that becomes 0 then you do not have to worry about the other one, if one of them is 0 you do not have to worry, 0 multiplied by anything is 0, okay good, thank you, see it is very easy, there is a think of it, there is a coulomb and exchange term, exchange term will be just think mentally, when there is a coulomb term this is 1, this is 1, this is 2, this is 2, so just write the same thing erase alpha, alpha, beta, beta, that is doing nothing, so simply write 1s, 1s, 2s, 1s as it exists and when there is exchange you do not have to do it, moment this becomes beta 1 it is 0, so I am still not understanding why you are not able to do fast, so if I were you I would simply write this as 1s, 1s, 2s, 1s, I remove alpha, beta, alpha, beta and you have to be smart, street smart sometimes to write this, you do not have to do that exercise and this is exactly why we say that if there is a opposite spin there is no exchange and we have said this, so exchange cannot come and that is obvious, see try to mentally think that this is 1, 2, 1, 2, this is 2, 1, so 1, 2, 1, 2 everything survives because 1, 1 is alpha, 2, 2 is also beta, moment it becomes 2, 2, 1 none of this will survive, so the exchange is gone and when it survives alpha, alpha is 1, beta, beta is 1, so you do not bother about them, so simply remove this, you have the j term, you have the Coulomb term, unfortunately this term cannot be interpreted as a j, j is different, so you note that you have a 1s, 1, 2s, 1, so it is not a density, this is 1s density but this is not a density, it is a transition density, so this cannot be written as j minus k, that is very specific for expectation values, that is the only difference, so please do not try to interpret in j and k, it is a Coulomb term in some way that it is a regular term but it is really not Coulomb in the sense of expectation value, because average value of energy has a Coulomb term, this is not energy, why are you not getting it because this is not energy, this is a matrix element between one determinant to another determinant and that is the reason what you are getting is like a transition, 1s to 2s, which is what you expect, so do not interpret them as Coulomb in exchange but write in terms of this integral, so the result is this, so this result is finally 1s, so this result is finally 1s h 2s plus 1s 1s, 1 by R1 2, 2s 1s or 1s 2s, it does not matter how you write it, you can write it 1s 2s, it does not matter because both sides are identical, so that is the result of the spin integration, exchange vanishes for the same result, I could have done another integral, remember do not interpret these as whether the transition takes place or not, selection rules and all that, because these are not excited states, they are only determinants, we are only following rules to write later rules, so let us say my chi 3 is, so my let us say I am extending this from 1s alpha to 2s beta, I am excited, so then my integral is 1s alpha, 1s beta, h 2s beta, 1s beta, here after the spin integration, spin integration is very easy, just see here, see here, even here you can make out, you do not have to know that, from the 2 determinants you can make out, 1s alpha, 1s beta, 2s beta, 1s beta, good, agreed or all agreed, what is 0 now, 1 beta, why that is 0, oh that is alpha and beta, good, very good, everything is 0, fantastic, no exchange also, good, so I hope you learn something, that what happens to such an integral, if I expand 1s alpha to 2s beta, the matrix element becomes completely 0, just that happens because of spin integration, okay, all right, good, so I hope you can do many practices, of course you can excite, you can excite both 1s alpha, 1s beta to something, but that will be type 3, which you will do later, yeah, because of spin integration, but those are all rules that will come, but actually when spectroscopy come, they are dipole selection rule, that is completely different, in the place of H, so those are different rules, but here it just turns out because of spin integrations, for the closed shell, yeah, because spins will, that is no other reason here, physically it means that when you have a closed shell singlet, you are trying to make a state which is triplet, so basically singlet and triplet do not mix, so in a way, in a way this is something that we expect, okay, so that is a pure triplet, this is a pure singlet, so they do not mix, with Hamiltonian, Hamiltonian is spin commutes to the square operator, all the physical Hamiltonians, so that is the reason it happens, another closed shell triplet state, then they will not mix, that is the meaning of this, matrix element, yeah, I mean I can, I can always have a matrix element of any operator between two basis functions, right, this is a N electron basis, this is another N electron basis, because all of these together will be full CI, when I do actual ground shed web function, it is not Hartree-Fock, Hartree-Fock plus singlet excited plus W excited will be also there, I told you MCN, so that time when I construct the total energy, I will have to construct the entire Hamiltonian, which will also involve such off diagonal terms, these are off diagonal terms of Hamiltonian, so they will be shifting my energy, so physically they have an important role, if these were very large, that means ground state will be shifted by these determinants in a very large way, significant way, so what it shows is that if I start with Hartree-Fock, then the singlet excited have very little role, because they cannot shift it significantly, but you will see that there is a role, it is not absolutely that there is no role, they have a role because they are matrix element with W excited determinant, like this is W excited, then that will not be 0, so they will shift into the higher order, they will not directly shift the ground state, but they will shift in a different way, so that I mean we will discuss when it comes to CI, I do not want to worry about it, but it is important at this point I want to say that this is important to evaluate the off diagonal matrix elements, and one consequence is for the Hartree-Fock that we get is a Bravais theorem, which just shows that if I start with the Hartree-Fock, these off diagonal matrix elements are completely 0, and this has one significance that if I do CI only with Hartree-Fock and singlet excited determinant, the ground state energy will not be changed, because all the singlet excited determinant block will not couple with the Hartree-Fock, so that entire block will become 0, so this is called a block diagonal structure, so if I diagonalize one of them will be energy, so basically if I do a CI I told you that CI is a Hamiltonian matrix element in that space, so one of them is Hartree-Fock, and the rest of them are all singlet excited states, so they are all 0, these are all 0, here of course there will be values, these are Hamiltonian between one singlet excited and another singlet excited, all values will be there, but then you can see that this quantity is already decoupled from the rest of the matrix, because this is 0, this is 0, so this is what is called block diagonal structure of matrix, and one of the eigenvalues will remain Hartree-Fock, so if I do what is called CI s, the ground shed energy remains Hartree-Fock, you do not change anything, what you will get is the other states excited states, when you diagonalize the rest of them, so we will come to that when you CI, so its significance will be more understood when you come to CI, at this point you need not worry about it, but since you ask the question I thought I would just answer, but the significance will be discussed in greater detail when you come to CI, so please practice more such thing with 3 electrons and so on, the basic result will be matrix element of the Hamiltonian and eigenvalues of that, so there it will have a block diagonal and matrix element, matrix will become block diagonal, so one of the eigenvalues will be unshifted, the Hartree-Fock part will never couple with anything, if you do CI s, so that is the reason for ground shed energy we never do CI s, if you remember, everywhere you do CI sd or DCI, CI d, double CI, you have never heard of single CI, single CI is done only for excited states, so those who have run Gaussian and so on you realize that for ground shed energy you never do CI s, because it does not change the result and this is the reason is the Hartree-Fock, of course all CI s are done from the Hartree-Fock as a reference, that is one thing I want to tell, so Brillou's theorem holds good, if of course CI 0 is not Hartree-Fock then it will be different result, but this particular thing that happened was because of spin, that is nothing to do with Brillou's theorem as per say, so because we are trying to make singlet and triplet and it should become 0, so that is fine.


Samantha Ruth Prabu ...  I'm Samantha Ruth Prabu. Incorrect. And you can follow us on Instagram and Twitter. Ah! Say content. Oh! That's his orgasm noise. Yep. Every time. At least you get a good laugh in. And subscribe if you haven't hit the like button.  Ah! Well. That was a good one. Thanks. I have practice. What are we doing? Today we got a new trailer that just dropped out. Does it have a Samantha Ruth? No. The reason we did that is because the one we just did had her. You guys saw it already. You don't know. Anyways, we came to this morning to watch the KRK trailer that dropped without subs. That dropped without subs so we had to wait, but then apparently like 30 minutes after that the new CBI 5, which as you guessed correctly, I don't know if it's the fifth. It's the fifth. Ah! No one told me. I wasn't looking. Yeah. But yeah, it's the fifth of this iconic kind of series that apparently is all done by one director. He's in all of them. It's kind of an iconic series. Cool. Great. Like an MI series. So you guys can tell us if this is a series we should. I'm assuming it's not going to. And again, yeah, like if it's a standalone thing where you don't have to see the others. Yeah. I'm assuming it's not going to spoil anything.  I know Tom Cruise is not going to die in Mission Possible. It's like James Bond. James Donne's not going to, unless you know it's like their last film, they might die. Yeah. If it's the character portraying them in their last one, it might. You know, but like if it's an iconic action series, I just figure the lead star is not going to die. That's just not how people make films. That's typically, you know, like it's called John Wick. If he dies, you can't call it John Wick anymore. So. John Wick adjacent. That's all the ones after him are going to. Although they did that with Bourne, they kind of did that with Bourne. I've never seen Bourne. You haven't seen any of the Bournes? No. They're good. Yeah, I know. It's one of the series that I've never seen. Yeah. No, they're good. Anyways. So yeah. Brian Cox is in the first one. You guys can tell us if that's one of the. I'm a moody is one of the actors that we need to explore a lot of because we've only seen, I think, two of his films, I think, right? Two? Yes. It's. You're going to ask. Yeah. And then. There was another one, right? Yeah. I can't remember.  You're far. No. Was he? No. I don't remember. Anyways, it was very little and he has like probably over a hundred credits or 200 or some 200,000. Like Lara. I think that was Leland. Hey! Yep. He's in everything, man. You know, sir, he's highly conducted. I like that. It's like disco. I thought there was going to be more. Shout out to the music people. I love the music in that. I agree. Yeah, the overall trailer, maybe it's just because we haven't seen any of the other ones. It kind of fell almost flat for me a little bit. It did. I was just going to say, I remember this is reactions. So we react honestly the way we see it. It's nothing personal against anyone. It's just the reaction of the moment. And my reaction to that was, huh, I bet if you knew the franchise, you would understand every little nuance of this that's going to be giving you what you expect for a standalone thing I'm not exposed to. I agree. Yeah, it was kind of just a crime drama. Yeah, with that at first I was engaged because I thought it was going to be a little dark and then it got kind of campy. And I thought, okay, is this going to go pink panther? Is it going to go clue? Maybe it's none of the above. Yeah. But either way, there was just nothing in it that had me like fully engaged. Yeah, obviously, the music did. Obviously, I mean, if it's a continuation and you guys can tell us if like this series is like one that's one we should watch. Obviously, we would start with the first one. We wouldn't have to hop into the fifth one. Unless you say it's okay to do that because I would tell people with MI, you can. You can jump. Yeah, but would you say that, though? You wouldn't. I don't think you would. I would be like for the new release, like if someone said to me the next Mission Impossible, I would say if you can at least watch these two before you go in, but if you don't, it's no big deal. Still go see it. Yeah, you don't have. It's like the Bond films. You don't have to see another Bond film before you go see another Bond film. I feel like you should. Nah. I feel like you should. Why?  He's shaken, not stirred, and his name is Bond. James Bond. He's 007. He has a different Bond girl every time. He's going to have awesome trinkets and gadgets, and he's going to get the bad guy. And the girl. And great clothes. And there's always boobs in the intro. There's always naked women in the intro. You know that? Shadowy naked women's. You know that? Oh, yeah. It's like a, it's like a staple. There's a shadowy naked woman in the intro of, like when they do the big, the song, the Bond song. There's always a woman's breasts. Breast goat. Because Bond loves women. Like that? Yep. There's always a, a breasticle. If you called my phone right now, you know what would ring? One, you're showing your age.  No, I was. People of my generation. No, it's on, it's on ring tone. It's on vibrate. Yeah. People of my generation never turn on the sound. Ever. That's a cardinal sin. Tell, tell me I'm wrong. I am not wrong.  You don't care about people calling you. I hope you miss a really important call. No, no, you're, you're misunderstanding. I'm saying the sound part. I have vibration. It's just we don't, what does that mean for you? If your phone is on the couch and you're in the kitchen and someone calls you, you can't hear it because your sound isn't on. Yeah. Yeah, that shows you what you feel about people trying to reach you. Oh, I don't want them to. I know. I hope you miss an important call. You know what I'll do? I'll call them back. Yeah. We'll see. The generational gaps are fun. Missed opportunities are fun. No. Oh yeah. I've never had an opportunity that's just like you missed my call. You don't get it now.  What's the opportunity that I'm going to miss? There are a lot of, there are a lot of people with a lot of regrets because they missed a call. I'm here for five minutes, man. You need to get over just telling you there's people with lots of regrets because they missed a call. Well, I'm sure your wife feels really encouraged to know that if she's ever out and about and she can't reach you, it's probably because you didn't have the ringer on. That's great to know. I, I promise you if you ask my wife right now, she does not have her ringer on. And it's not because of the reason you're thinking. It's because we don't want anybody else to know that somebody's calling us. Why? I don't want you in my business.    One, people don't call me right. Don't ever call me. I'll block you. Don't call me. What do you say whenever I call you? Why are you calling me? Right, but I answer because I got a ringtone. Vibration works just as well. You know what else vibration works for kids? Ask Rick's mom. Anyways, yeah. So, you guys could let us know what, what you think about the trailer and uh, is this a series of Mammootis we should explore and also you like it into Mammootis himself. Yeah, and if you're, if you're a fan of the series, do you like the trailer? Are you excited? Yeah. Please let us know. Yeah. Down below.


Alan was three years older than I was, quintessential big brother. I tried to do everything he did. He was the one who taught me how to ride a bike, drive a car. He called my parents and said, I'm going to join the army. I want to be the best. And he was selected to be Special Forces. We got a call that they came to our house and told my parents, actually very little other than he was missing. And so for the first couple of years we were just on our own. The anguish you suffered, families of brave men, of listed prisoners of war or missing in action. The misery and anguish is unspeakable. Something the rest of us can really never know, even in a partial measure. If they could be here today, millions of Americans from every corner of this land who have only a glimmer of your pain would say to you, we want with all our hearts to share your burden, to realize our common quest, the return of all POWs, the fullest possible accounting for the still missing, and the repatriation of the remains of those who died. Until they all come home. A creed of the American military to never leave a man behind. For the Defense POW-MIA Accounting Agency, it's their singular objective. I think it speaks volumes to the character and the values of the United States, that here we are decades after the complex ended, that we are still inherently fulfilling our role and our promise made, that when we send a servicemember off to war, that should they become missing, that it's our responsibility as a nation to bring them back home. To meet their goal, they travel anywhere U.S. troops have served, conducting historical investigations and archeological digs. Time is our biggest enemy, so we're talking decades that remains have been in the ground. You have high speed crashes, where the impact created a massive field investigation site. Witnesses are dying. In many cases, you have a remote, rural, dense, jungle-infested territory. It's difficult for our teams to excavate the site in an expeditious and more importantly, diligent manner. And even when remains are recovered, identifying them in the lab presents challenges of their own. In this laboratory, we're working with what in the forensic realm people might call residual remains. These are mostly the hard tissues that are still here with us after so many years of being missing. And this would be bones and teeth, for example. And so we have a heavy reliance on anthropology. That's the profession that actually studies how you can say things about a person from their bones. We also have dental records from most of our missing servicemembers, thankfully. And so our forensic dentists here in the lab are able to compare what we see in the teeth to dental records that we have retained in the Department of Defense for all these years. But we also use DNA extensively. The majority of our cases involve some type of DNA testing. But to make good use of DNA results, you have to have something to compare it to. Though it's now common practice to retain DNA samples from service members, before the early 1990s, DNA testing essentially didn't exist as a forensic tool. This means that to be able to use DNA as a method of identification, they need samples from close family members to use as a reference. But it's not easy. Some of them we run into dead ends with, and we can't find anybody. As you can imagine, when you're dealing with literally tens of thousands of missing service members, particularly from World War II, the services have an enormous burden to try finding all of these family members, each and every one of them, with their individual unique stories. What happened to the family, where the family moved. We always like to take opportunities to say if there's anyone out there who has a family member who's missing to reach out to that service casualty office and help us get the DNA samples that we need. On the front end, conducting archeological digs in dense jungle, harsh terrain and occasionally underwater, requires support from outside of DPAA. Dozens of service members from every branch of the military are asked to augment recovery teams. We can not only support the recoveries, we actually can see it accomplished. And it's one of the few times in our service that we can start to finish and really feel good about what we've done. What's amazing to me is when we have soldiers go that volunteer to go for whatever reason, may not have been a reason other than they wanted to stop the monotony of the day here or take a mission, they come back a different person and they have a different respect for our army and for what we do. Bringing our missing home can be a decades-long journey that takes a small city worth of people to accomplish. Actually it was March 7th of 2016. I picked up the phone. It was the night before what would have been Alan's 70th birthday. I picked up the phone and there's a deep voice asked for Judy and I said, this is she. And he said, I'm calling from the army. My legs literally went out from under me and I said, you found Alan. And he said, we did. I mean, I could barely breathe. The funeral itself was just such a joyful homecoming. And again, I'm one of the very fortunate ones. I hope that the other family members will finally receive their accounting. That is such a sacred commitment that the United States has to do. Americans need to be made aware that POWs and MIAs are still missing from all wars and that DPA is making a difference in fulfilling that mission for the fullest possible accounting and bringing all of them home.