The Syrian Arab army and its allied forces have taken over the strategic town of Al-Sukna, the last major town under the control of Islamic State in the Homs province. With the takeover, the Islamic State is fast-losing ground in the country as they struggle against the offensive launched by Syrian forces backed by Russia. The Syrian forces took over the key Islamic State stronghold in the Homs province according to a UK-based monitor after they launched an offensive earlier last week. Al-Sukna is only 110 km away from another Islamic State stronghold Deir ez-Zor, a province almost entirely controlled by the militant outfit. With the advance reports suggest the Syrian forces will be making a swift advance towards Deir ez-Zor. They will be trying to thwart any Islamic State's attempts of strengthening their defense while they face annihilation in Sukna. Sukna town is located on a key highway that can lead the Syrian forces towards Deir ez-Zor. The IS is also facing offensive in the north of Deir ez-Zor from the US-backed Syrian democratic forces who have also registered a victory against the rebel fighters in Raqqa town. The SDF is only 80 km from the IS stronghold. They are also trying to take over the province which is rich in oil and natural gas reserves.


Hey, good morning everybody This is Alan with Max and I got all my bags packed and you know, really wish I could take everything All my family with me and a good old Maxter here But I really don't think you get along too well with the yak So we're gonna go ahead and let him go. All right Max off he goes So let me tell you a little bit about this climb Obviously, this is climb number three in the seven sevens climb for Alzheimer's memories everything Summited Vincent back in December and Iconcagua in January had a couple of months here about really about six weeks since since Iconcagua Continue the training preparation and it's time to climb So let me tell you a little bit about the Everest climb I'll be climbing with international mountain guides as most people know by now Be climbing on their classic team, which means that I'll be climbing with a Sherpa And so what we're gonna do is first off I'm leaving this afternoon gonna go to Bangkok and then on to Kathmandu It's been a day or two there meeting up with the rest of the team And then we're going to Leucla the little airstrip about 9400 feet And then we're gonna take about ten days to track up to base camp at 17,500 feet and the reason we're gonna take so long is that it's not that far It's only about 30 miles But we need to get our bodies climatized to the altitude and so by taking our time We're gonna gently let our bodies create more red blood cells which carry oxygen to our muscles and That'll help us as we start to climb higher also the kumbu trek is just a fantastic part of the climb is everybody That's read anything. I've ever written about the kumbu nose. It's one of my favorite parts of any Himalayan so once we get to base camp, then we're gonna actually backtrack just a little bit back down near Gork Shep and climb nearby Lobache now, this is a 20,000 foot so-called trekking peak and I love that term trekking peak for a 20,000 foot mountain, but in reality, it's not very technical But the altitude is certainly challenging and that'll be an opportunity again to further Climatize and also to kind of get our mountain feet because we're gonna be climbing. There'll be some fixed ropes There'll be crampons. It obviously can be extremely cold at 20,000 feet. This is full of snow And so I've never chased so it's gonna be an interesting climb for me back to base camp and then Then we're beginning a climatic climatization process going through the the kumbu ice fall Up through the Western Coon up the Loatsey face and then back down to base camp now We're gonna go up and down as everybody is familiar by now with the whole Climatization process and that's gonna take around around three weeks or so to go up and down Several times and that again is to get our bodies to be a climatized back down to base camp and at this point It's really just kind of waiting for the weather window right now The schedule with everything being perfect, which it never is in climbing would be that we would summit around May 17th so we'll see how that goes but coming back to climb is going to be to go back up all the way to Camp 2 up to the South Coal and then that infamous 12 to 18 hour climb to try to get to the top of the world So I'll be dispatching writing post the entire way Sometimes it'll just be a voice dispatch other times will be a written dispatch I hope to be able to upload some videos to bring everybody along in the process and again as everybody knows This is all about helping to raise awareness for Alzheimer's and also to raise money to find a cure through research You know, I've got some very startling facts here on this page that I hope you read but Bears repeating over and over again that every 70 seconds another family begins that tragic journey And I'd say tragic because you know, there is no cure for this disease and the treatments treat the symptoms But you know, we need we need a lot more and that's what the research is all about So I'm really hoping that the people will sign up to pledge a penny a foot for every foot that I climbed So that's about a hundred ninety six dollars for ever and I calculate that by starting it at Leucla at ninety four hundred feet All the way to the summit twenty nine thousand and thirty five feet So that's where the roughly nineteen thousand feet come from that. I'll be climbing a penny a foot It's hundred ninety six dollars and it's over two months So, you know, if you think about how much you spend on coffee, it may be it's a better better investment over the next the next couple of months Please please follow along. I hope you enjoy the coverage I I'm gonna do my best to kind of cover what's happening on the mountain But you know, it's kind of difficult to cover both sides and climb at the same time So I think you'll understand and I hope you'll enjoy the overall coverage that I do I Think that's about it. So Next climb after this one is to Denali. It's very very quick Get back around the end of May 1st of June Then leave at the end of June for Denali and so lots more about the highest mountain in North America Once once we get Everest under our belt It is a difficult climb. I am well prepared It's gonna be an exciting climb I'm gonna absolutely do my best and I hope you come along but more importantly I hope you do a little bit of research and through the links on the sites Climb4AD.com go to our Facebook page Which is just Climb4AD in Facebook and of course on the blog here on AlanRnet.com But most importantly, I hope that you make a donation to the Cure Alzheimer's Fund or to the National Family Caregivers Alliance in support of family caregivers and they are the silent victims of Alzheimer's Okay, everybody. I gotta go gotta get to the airport gotta catch my flight. All right Climb on. This is Alan and remember, memories are everything


Okay, folks, welcome to a quick round of technical analysis of Sonovous Energy, Inc. ticker symbol CVE. Sonovous is in the energy sector and recently announced as of yesterday that there would be an acquisition of Husky Energy, which is traded on the Toronto Stock Exchange ticker symbol HSE, whereas Sonovous is traded on the New York Stock Exchange, again ticker symbol CVE. Now what we're going to talk about on this video is the news that was announced. We'll spend a few minutes talking about the balance sheet of each company. I'll give you my thoughts on a potential trade here. And then we're going to do a deep dive into the technical analysis of the energy sector. And then we'll leave off with the charts of CVE. We'll do some technical analysis there. And we'll leave it off with what I think the trade is on this announcement, where I believe the share price of CVE is going in the near to short term. And what will the catalyst be to propel the shares up higher? So let's get to it. In the news, off of Reuters, Canada's Sonovous Energy to buy Husky Energy for $18 billion. Now this is a trend that's going on here in the energy sector. And it's during this phase of the consolidation phase of a sector where you really want to begin to take a look at the companies that are acquiring for the long term. Now, earlier this month, Contra Resources Inc. agreed to be taken over by Conoco Phillips for $9.7 billion. And that followed Chevron Corp's $4.2 billion purchase of noble energy. So this is a trend that we're beginning to see. And it's a very positive trend for the energy sector, which has been under a tremendous amount of pressure these past several years. This acquisition would give Sonovous some scale by making it the number three Canadian energy company. And you may be reading this here saying, wait a minute, you just said $18 billion acquisition. Folks, keep in mind that Husky has a tremendous amount of debt. We'll go to that in a moment. But net net, Husky shareholders will receive 0.7845 shares of Sonovous stock, as well as some warrants. And it looks as though this transaction already has the green light by the boards of both companies. So this is for real. It's happening. Expect a lot of volatility tomorrow morning being Monday in the shares of both Sonovous and Husky Inc. Now let's get to some of the numbers of each company. Now on Friday, Husky Energy Inc. closed out the day at $3.17 in Canadian dollars. Now moving on to the balance sheet, Husky has $633 million in cash, which sounds like a lot. But in reality, it's not relative to its debt. They have a tremendous amount of debt over $7 billion in debt, which is going to to be acquired by Sonovous. Now, I suspect on Monday morning, you're going to see the shares of Husky Energy rip through the ceiling. Anybody saying, you know, that's no brainer, Bob. They're being bought out. Well, what's going to make it even worse is the fact that you have a lot of shares being held short. If you do the math here, you have about, I'm just going to round this off to say four days of short covering that is going to happen on Monday, Tuesday. So the poor son of a gun that is short of Husky Energy, I feel for you. And I also feel for the hangover you're going to have tomorrow morning after you get done with your second bottle of scotch today. So very bad news for shorts in Husky Energy Inc. Now, taking a look at Sonovous, you know, this is their balance sheet. This is a company that has been through really, really rough times since 2015. They've seen a huge cash burn rate. They were carrying cash of $4.11 billion in 2015. As of the end of 2019, that dwindled down to $186 million. Now, the five-year trend on sales and revenue has increased dramatically since 2015 from 13 spot 06 billion up to a 2018 high of $20.84 billion in revenues. You saw a tick down in 2019 by about 3.18%. But for a couple of years there, you saw some really dramatic growth in sales. In terms of liabilities, I mean, really for an energy company, not bad. They've done a really nice job of dropping down their debt from a 2018 high of $682 million down to $196 million in 2019. And when you take a look at their return on equity, that's pretty good. They're in the double digits on return on equity, nearly 12%. So apparently we have a fairly good management team at Sonovous. And given how they're managing this company, they appear to know what they're doing. So in terms of confidence in their ability, it's pretty high. Now, taking a look at the share price of Sonovous, you can see that we are in a long-term bear market. And what do I think the outcome is going to be tomorrow of the impact of the news on the shares of CVE, considering the fact that they are accumulating a lot more debt than what they were carrying as of the close on Friday. My prediction is that you're probably going to see a gap down. Now, if we don't see a gap down, well, then these shares are priced for a worst case scenario. Now, we're going to do some technical analysis here based upon my thought process that in all probability, you're going to see weakness in the shares of Sonovous on Monday, Tuesday, the remainder of the week. But let's talk about what the opportunities are here. We're seeing consolidation in the energy sector as a whole. Let's go to that chart of energy and we'll come back to Sonovous in a moment. Here is the chart of the XLE monthly timeframe. And you can see that energy peaked out in June of 2014. And ever since that point in time, we have been in a nasty, nasty bear market. But I'm of the opinion because we have been buying energy stocks. I'm of the opinion that the bear market in all probability saw a bottom back in February of 2020 when you saw the baby get thrown out with the bath water on energy stocks. We since rallied seeing a peak on the XLE in June of 2020. Now, we're pulling back to what I believe to be a higher low on the XLE, which is representative of the energy sector. Here's our lower band of support. You can see for the month of October, we are up at current with a few trading days left to go. And here is our upper band of resistance. So the deal was announced at a point where we may be seeing an inflection point in the performance of the energy sector moving from a clear and defined downtrend to a respectable sideways consolidation and perhaps with a bias now to the upside. And with the U.S. dollar weakening, that should only help the price of oil longer term. And if the price of oil continues to move up higher, well, then energy stocks should follow suit. Now taking a look at a chart of CVE, same timeframe as the XLE monthly time frame, you could see that the two look very much alike. We saw a bottom on CVE back in March. We saw a counter trend rally. And now we're pulling back to put in potentially a higher low. Let's draw some trend lines. So here's our triangle formation on CVE. The question is now, after the news, do the shares of CVE being the choir and the choir usually going down? Initially, when an acquisition is announced, do they hold this lower band of support or do they break? That question we have no idea about, right? The market's going to tell us whether or not the share price is priced for the worst case scenario. And we'll know that right off the bat. If we gap up tomorrow or we move lower on the day only to close up higher, well, then that's a stock that needs to be bought. That's the market telling you, yeah, this is a nothing burger. Don't worry about it. We want this. This is good. But what if there's a delayed reaction to what we're seeing as longer term bullish for the shares because of what we're seeing in the overall sector, meaning technical bottoming action and acquisitions as recently as this month? Well, what I do here is I would set up a strategy. Where do I think that these shares, if the market sells them off, where are they going to land? Let's take our crayon out. I think that there's good chance that we retest this support level here, $8.17. That marked the low of May. We had rallied off the lows in April, pulled back in May. That was the market sending us a signal that there were buyers here. We continued to rally off of that level. Now we're probing for a higher low. Maybe that low will be here at $8.56. I tend to doubt it. And the ideal entry point from my perspective is one where we see $8.17 get hit and hold. And if we happen to rally off of that level and close back above $8.56, well, then you'd better back up the truck and begin buying these shares, especially if you have the dollar tanking and oil moving up higher. Let's do a deeper dive weekly chart. You can see that moving into the weekend. The share price here was looking very good. Here's our defined higher low. Here are our four mentioned support levels. And it appears as though we were consolidating very, very nicely at the $8.56 level. And we're ready to move up higher as energy has shown some signs of strength of late. Even if you take a look at a daily chart, while down on Friday, we're still putting in a higher low, holding support, not horrible price action. Now, if we do gap down, you may be saying, you know, why would anybody want this company with all the debt? The sector is a mess. What's going to be the catalyst to send the share prices up higher? Simple. When you take a look at the shares held short, there are 19 million shares held short, which as a percentage of the outstanding shares, not much. Let's face it, one spot, five, five percent, no big deal. But when you take a look at it from the perspective of the average three month daily volume, which is only 4.78 million shares per day, relative to 19 million shares held short as of September the 30th. Well, then add into the potential short number, those who lean into the short side on Monday, assuming that we move downward, and you get a setup when we hit support levels back to a monthly chart, you get the potential for a short covering rally off of one of these two support levels. If in fact they do hold, that is the $800,000 question, right? We don't know if they're going to hold yet. So this is a trade I'd be interested in putting on for the short term, not necessarily for the long term, but for the short term as a swing trade, but the share price must come to us. The bet is here is that we're going to see a pullback on CVE. The question is where do we land? And of course we'll send out a trade alert to members. Folks, if you got any value out of this video, if I could ask you to please smash that like button, I'd really appreciate it. It helps me out with the YouTube algorithms. Everyone have a very profitable trading week and be well.


I think she's accidentally revealing how her psychology works here she describes this kind of character of her drawn by media as cruelest meanest horrible list person and in her mind I think she disagrees with this and I think this is honest and she was probably kind and polite while she was recruiting these very very young girls and I think the character description is what she's so disgusted by I think she reassures herself about the crimes by justifying that she was nice and polite and kind to them while doing so


This is a tutorial on Boolean Searching from Bellevue College Library. In this video, you will learn to become a power searcher. Library databases have scholarly articles and the information you need for your assignment, but they don't work the same as a search engine like Google. So you can use the techniques in this video to become a power searcher. Let's upgrade our search skills with Boolean operators. So, what are they? Boolean operators are short, connecting words typed in all caps that help you optimize your database search, and narrows your search, or broadens your search, and not excludes specific search terms. Let's try a search. Search with and to tell the database that all search terms must be present. Here I am using the ProQuest database to search for crows and communication, and I get a lot of results. Search with or to tell the database that any of your search terms can be present. I add the or operator to search for crows, or ravens, and communication, and I get even more results. Search with not to tell the database to ignore concepts that may be implied by your search terms. It looks like a lot of the results in my previous search were about communication and the football team, the Baltimore ravens. So I'm using the not operator to exclude Baltimore or football from my search results, and you can see I get quite a few fewer than before. And that's it. Have questions? Ask a librarian for help.


So, this is what we discussed last time, we had new definitions, we redefined momentum which we said P is equal to m times u, where u is the velocity of the particle, m is actually m naught multiplied by gamma u which is has been given here, this particular expression which has been given there, this is the gamma u. Now, this we said that this in the classical limit which means when the velocity of particle is much smaller in comparison to the speed of light will reduce to the classical expression of momentum. We went around very very hurriedly last time, but it is rather easy to see that all that will happen that gamma u will because when your u is much smaller than c, this factor m will be essentially equal to m naught and this expression reduces to the classical expression of momentum which is m naught times u. Then we defined the energy is equal to m c square and then that is the time we said that this energy on the other hand if you make speed very small in comparison to speed of light will not reduce to the classical energy because this is a totally new concept of energy. We said that actually the kinetic energy should be regarded as the difference of m c square and m naught c square because according to this particular definition of energy even if the particle is at rest, this particular particle possesses the energy which is equal to m naught c square and this m naught which is rest mass is supposed to carry all other forms of energy and therefore, this e is also supposed to carry all other forms of energy. For example, if the particle becomes hot in principle it gains energy and if it gains energy so its m naught should go up. So, this is what we discussed last time. Now, there is one particular point which I would like to connect to this particular thing which probably will become obvious why I am doing it and these are the relationship which we discussed you know relationship between momentum and the energy and relationship between kinetic energy and the momentum. In any problem these relationships are very very important. One of the most important concept which comes from special theory of relativity which has no classical analog is that this gives you a possibility of a particle which has a zero rest mass. Classically if a particle has a zero rest mass it does not exist it is no longer a particle but here there is a possibility that the particle could have a zero rest mass. These are the expressions which I have just now written e is equal to m c square remember m was m naught about under root 1 minus u square by c square. Similarly, your momentum was m naught u divided by under root 1 minus u square by c square this is the same thing which I have written in the expanded form. Now, if m naught happens to be zero in principle energy and momentum would be zero because m naught appears in the numerator. So, in principle that should be zero however there is a possibility that if denominator also becomes zero then this energy and momentum will both become zero by zero and zero by zero need not be equal to zero it could also have a finite value. So, it means that there is a possibility that if there is a particle which has a zero rest mass then this particular particle could possess energy as well as momentum ok. But in that case relative will allow that only if this denominator also becomes zero and that will become zero only when u becomes equal to c. It means that particular particle must travel with speed of light. So, it gives you a possibility the relativity gives you a possibility of a particle with zero rest mass which could possess energy which could possess momentum but may but must travel with speed of light. It cannot travel with any other speed because otherwise it will be inconsistent with the special theory of relativity. And as all of you know that this particular particle is generally called a photon actually one of such particles as photons as we all know that photons in principle I mean photon is we always believe it to be light and it has to travel with speed of light because this is light. Therefore, this scan this particular photon will have energy which we know from our old classical photoelectric effect that it has a energy as nu. The only thing which I would like to say that if you go to this particular expression which is e square is equal to p square c square plus m naught square c to the power 4. If m naught happens to be zero then the relationship reduces to e square is equal to p square c square it means the momentum of the particle must be given by e by c. So, the particle momentum must be equal to e by c. So, therefore, if everything is consistent with special theory of relativity then this photon can be treated as a particle with zero rest mass of course it has a energy as nu which we know from other experiments then it should also have a momentum which is equal to as nu by c and we believe that light can support on. In fact, it is a slightly a dilemma because when we started our relativity lessons we always start by looking at the light and realizing that light is actually an electromagnetic wave and therefore it must travel with a speed which is given by under root 1 upon under root epsilon naught mu naught and then suddenly we come back and talk of a totally different language and try to say that well light actually consists of particles. This looks contradictory but this is unfortunately what is true in the nature that light does seem to have a dual nature. Light sometimes behave like a particle it behaves sometimes like a wave as all of us know this is what we call as a dual nature of light ok. There experiments like interference, diffraction, polarization which can be much better understood on the basis of the wave theory but there experiments like photoelectric effect and the Compton effect that we are going to describe today which are much better understood only on the particle nature of the light it shows a dual nature and that is what is basically the culprit for us to land in quantum mechanics. Now I am sure all of you would have heard about photoelectric effect experiment which I am not going to describe there is only one aspect of photoelectric effect which I want to tell. So, photoelectric effect was one of the key experiment one of the direct evidence that light behaves like particles, behaves like it consists of photons which have energy h nu but on the other hand this particular experiment cannot really prove that it has a momentum equal to h nu by c ok. Let me just try to explain this particular aspect little bit more. See first thing I made a statement rather really last time which I will probably like to emphasize again that for example if you take a free electron let us suppose I mean if at all we can create an electron at rest ok we can show that it can never absorb a photon. It means see what is photoelectric effect is basically a photon gets absorbed and its energy is transferred to electron alright. Now is it possible that there is a free electron and it absorbs a photon and whatever is its energy this particular electron gains ok it can it is always possible because this is a conservation of energy but we can show that if that happens and if this particular photon has a momentum h nu by c then momentum conservation will not be valid. And as we have said that for any process to be valid both energy and momentum has to be conserved. So, you can show very very easily ok you try it yourself if you have difficulty I can probably help you but it is very very simple way of looking at that. First write energy you write h nu is equal to kinetic energy of the electron then write momentum h nu upon c and try to solve this equation you will never be able to solve these equations ok. It means it is not possible for conservation of energy and conservation of momentum both to hold good for a free electron. So, that it could absorb a photon it means photoelectric effect is never possible with respect to a free electron. However, there are two things which are possible one thing which is possible that this particular electron is actually bound to an atom ok which is actually the way actual photoelectric effect experiment takes place because certain amount of energy of the photon is used for releasing this particular electron and the balance of the energy is given to kinetic energy that protocol process becomes possible. There is another possibility which we will describe today is that the photon gives only part of its energy to the electron and not complete energy to the electron. That is also a process which is possible which is called content effect ok. So, we have two possibilities that either we let this particular electron get bound ok. And therefore, photon gives uses part of its energy to make this particular electron free and only remaining balance of the energy it gives as a kinetic energy to the electron or number 2 the photon does not get completely absorbed it gives only part of its energy to the electron and as a result you can still get a photon, but with a reduced energy and electron with a higher energy which is what is called content effect. So, these two processes are possible. But the point which I was trying to discuss it because this particular electron if we have to observe in photoelectric effect has to be bound to an atom and atom is much heavier in comparison to electron. Therefore, you will never be able to see conservation of momentum.  To you it would appear as if the momentum is not conserved because the ball momentum initially was in this particular direction it has changed its direction ok. So, it appears that momentum is not conserved strictly speaking momentum is conserved, but the fact is that the difference of this particular momentum has been taken to by this particular ball and this ball is rigidly connected to earth and it is almost impossible to measure the change in the speed of the earth which is so heavy which is so massive ok and really test whether conservation momentum took place or not. So, a precisely similar situation lies when I am talking of photoelectric effect because this particular electron is bound to a much more heavier atom ok. So, if photon gets absorbed and if I want to really test whether momentum conservation has taken place or not ok the entire atom will require and it will probably change its energy very very little and therefore, it will be very difficult to measure how much require has taken place and therefore, you will not be able to test the conservation of momentum. So, unfortunately photoelectric effect though it is a very good experiment to tell that photons carry a energy h nu, but to test the second aspect about the momentum which should come from relativity because if it has a energy h nu, if it can be treated as a particle 0 rest mass it should also have a momentum which is equal to h nu by c ok. By the way let me just tell you that you know probably you are aware that the fact that light carries momentum was known much earlier much before relativity was discovered people knew that light does carry momentum. There have been experiments which were done earlier and first realized that you know light does carry momentum, but here we are treating into a particular formalism using special theory of relativity when we are saying that if it has a energy h nu it must have a momentum of h nu upon c if this particular particle has a 0 rest mass. Fortunately for us it is a Compton effect which can give this particular information it can test that photon is really having a momentum of h nu by c because then basically you are requiring a lighter particle and if we having a lighter particle scattering then as I said unfortunately you cannot absorb the photon completely you have to only absorb the photons energy partially. So, this is basically what is called as a Compton effect experiment. Here you have an electron which you are assuming to be free and or rather at rest and there is a photon which is coming incident from the left hand side with energy h nu and once this particular photon interacts with this particular electron as a outcome you will find that another photon comes out which has a reduced energy which gets scattered at an angle theta from the original direction and electron also gets scattered at an angle of phi this electron which was originally at rest is no longer at rest but starts moving with a particular kinetic energy. Now we will apply conservation of energy and conservation of momentum to this particular problem and work out what should be the energy of the photon okay and if we can experimentally verify whatever we get the energy from this particular thing then we have sort of understood that really it is true that photon carries a momentum okay it carries energy and whatever expressions we have obtained relativistically those expressions seem to be correct alright. So, I will not go into the detail about mathematics as I said but let us just discuss the physics out of it. So, what we will do that first we will conserve momentum and because momentum is a vector quantity and each component has to be separately conserved. So, let us call this as x direction this is the y direction. So, we will write the momentum along the x direction the original momentum along the x direction was only because of this particular particle because this electron was assumed to be at rest. So, original momentum was only h nu upon c because only particle which was scanning momentum was the photon. So, original momentum must be h nu by c which is in x direction original momentum along the y direction was 0. Now, as a result of interaction as a result of scattering you have another photon which is coming out with energy h nu prime which is here. So, let us take x component of that particular momentum which will be h nu prime by c if whatever I am saying is correct whatever relatively says is correct then its momentum should be h nu divided by c cos theta along the x direction. Similarly, whatever is the momentum of the electron let us assume that particular momentum is p e. So, that component along the x direction should be p e cos theta. So, we will get one equation which is h nu upon c is equal to the momentum component of the photon and momentum component of the electron. The second equation that I will get will be along the y direction because the y direction there is no momentum. So, we take h nu prime divided by c into sin theta which is the component along the y direction that must balance with the y component of the momentum of the electron which is in this particular direction. So, it must be balancing p e sin theta. The third equation is conservation of energy of course, we have to use relativistic equations if you use relativistic equation you know this initial energy even though electron was at rest it had a energy equal to m naught c square. So, original energy was h nu plus m naught c square. Now, once it starts moving m naught c square changes it becomes m c square m c square plus the energy which is h nu prime because energy is not a vector quantity. So, we will just take the total sum of the energies. So, this is what I have written in the second equation. This is the first equation is the conservation of momentum along the x direction thing which you just now described. The second equation is the conservation of momentum along the y direction and third is the conservation of energy which is m naught c square plus h nu is equal to h nu prime plus c. Of course, energy and momentum are not independent of the electron. If I know the energy of the electron momentum has to be related and that has to be related by this equation which we have described earlier e square should be equal to p square c square plus m naught square c to the power 4. Once we describe experiment, we will describe that it is very very difficult to know about the recoil of the electron to experimentally measure the recoil of the electron. Therefore, any parameter which is related to electron we sort of eliminate from this equation. So, first thing that we do is to eliminate phi. Phi is the angle at which the electron gets scattered which as we just now described is little difficult to measure. So, let us get rid of this particular constant. So, what we do from this particular equation you have p e cos phi on the first equation here. Second equation you have p e sin phi. So, you square and add. So, sin square phi plus cos square phi will become 1. So, you will get rid of sin square phi. Then you manipulate with other equation and then you will finally get the result. So, this is what I am doing first thing eliminating phi. So, I get this particular equation. Then next thing that I do is use this particular equation p square c square is equal to e square minus m naught square c to the power 4. So, I substitute p square c square which I have got from the earlier equation here. See here I can get multiply by c square. So, I get this equation and this is the equation which I have got and equate from the energy conservation which I get from here. These equations are equated here. As I said I will not go into the mathematics.  Remember theta is the angle by English direction the photon has got scattered. So, basically it tells you that if a particular photon is observed after scattering at an angle theta from the incident direction of gamma ray. By gamma rays that we will discuss later. From the photon incident direction of the photon velocity then it is changed wavelength. Remember lambda prime has to go up if the energy has to go down. So, h nu prime will be lower than h nu. It means its wavelength will be larger. Therefore, the increase in the wavelength of the photon will be given by this particular expression h upon m naught c 1 minus cos theta. It is the famous Compton effect expression. Let us just look at this particular thing. The first thing that we realize that the wavelength increase is independent of wavelength and depends only on the scattering angle. You look at this particular expression. This increase does not depend on the lambda. It just depends on just theta. H m naught c are fundamental constants. This lambda c is equal to h upon m naught c. If m is the mass of electron is what is called Compton wavelength of electron. And if you just put the numbers, these are all standard numbers. The value of this turns out to be fairly small which is 2.43 into 10 to the power minus 3 nanometers. It is a fairly small value. All right? This is what you get. You also realize that delta lambda is equal to 0 for theta is equal to 0. If you put theta is equal to 0, then cos theta is equal to 1. So, there is no change. So, if the photon is going directly, it means essentially it does not get scattered. If it does not get scattered, obviously there cannot be a change in wavelength which is obvious. And is equal to 2 lambda times c for theta is equal to 180 degree. In fact, that is the largest change that you can have. The largest change can happen when this cos theta becomes equal to minus 1 which means theta is equal to 180 degrees. And at that time, the total change in the wavelength that you will be getting will be twice the Compton wavelength. Now, problem is that our wavelength is so small that if you use for example, Compton effect using the optical light which has typically the wavelengths of the order of 5000 angstroms to measure a change of 0.0504 angstrom, 0.2 angstrom in that 5000 angstrom to devise an experiment where you can measure such a small change in such a large wavelength is going to be extremely difficult. Therefore, we normally would though in principle even that particular photon will get a small Compton shift in the wavelength if it gets scattered, but that will be very very difficult to measure. So, if I want to experimentally measure, it is better to use a very small lambda, it means much higher energetic photons, then only there is a possibility that I will be able to detect the change in the wavelength. So, therefore gamma rays are used whenever we perform actually an experiment with Compton effect which are comparatively much higher energies. And therefore, to measure small changes in lambda becomes possible because you are never going to get change more than 0.04 angstrom. So, better to use one angstrom to angstrom then probably you can detect that it is a 4 percent change which is still small, but then you can measure or go to still lower lambdas, but in 5000 to measure change of 0.04 is going to be extremely difficult. The other thing which I wanted to also tell that this is for electron H upon M naught C, if I would have used heavier particle instead of for example, neutron or an atom, the shift would have been still smaller. Though there would not have been any change in expression because in principle in derivation I have never used electron, I have called it electron because most of the time we perform this experiment with electron, but on the other hand instead of electron it could as well be any other particle, no other property of the electron has been used other than its mass. Okay, so instead if I change the mass of the particular particle to any other particle the expression will still be valid. So, it means if this particular electron gets scattered by let us say a neutron or by an atom or by nucleus of an atom, there will still be a shift, but that will be much, much smaller. So, remember proton is 2000 times approximately heavier than electron. Okay, so shift will be 2000 times lower. Okay, so essentially if we hardly you will notice any change because electron happens to be one of the very light particles. Therefore, it becomes easier for to perform this experiment shift will be the largest and that also is not so large. Now, we compare experimental setup as I said there is always a gap between the way theoreticians thing and the way experiment has to be done. It is very easy to say theoretically that you know let us put an electron at rest and let the photon come and hit it. But where do I get in a laboratory electron which is at rest you know it is not very easy thing to be done and where do I get an electron which is sort of free you know which is not bound. Now, generally it is known that in metals there are large number of electrons which are approximately free I am using the word approximately free. If we discuss free electron theory you will realize that you know there are short commons with free electron theories. So, electrons cannot be 100% free even in the metal. Okay, therefore they are bound by forces. Okay, but to a good approximation remember in physics we always like to make approximation you know ideal situation can never be reproduced you know let us try to do whatever this we can do. So, we choose a metallic material in which there will be large number of approximately free electron and let these electrons scattered the photons. So, what we do we have a metallic target as I have mentioned here and there is incident gamma ray beam which is coming and hitting this particular target. Alright, then you put your detector at an angle theta from the incident direction of gamma rays and you start counting. Okay, there are lot of detectors you know from which you can count the number of photons since I do not remember but you know those people who are working in nuclear physics probably will know that many type of detectors from which you can actually count the number of photons which are coming and therefore you can find out in a given time how many number of photons have been coming at that particular angle of theta and not only that you can also find out their wavelength or their energies. Similarly, you can keep on changing your theta slowly and slowly and in principle find out at each value of theta what are the type of energies that you are getting and what are the numbers that you are getting. It is a very very simple experiment otherwise. I mean this experiment we do also in MSC and you know I mean in fact lot of people also do a research type of experiment there. Now this graph I have plotted roughly myself so this may not be very accurate graph. So, please excuse me for that if you want to look at an actual experimental graph you have to look at the textbook. Okay, if you plot the total number of photons coming as a function of lambda for a particular theta. So, I have chosen a particular theta fix my detector at that value of theta then I am calculating the total number of photons as a function of lambda prime. Okay, the different wavelength. What you find surprisingly you find two peaks one peak is completely sharper which is here and another followed by another peak which is larger lambda which is comparatively broader. All right. Now if you perform this particular experiment at different values of thetas what you find that the separation of these two peaks keep on changing. However, this particular lower lambda peak essentially remains more or less at the same position. All that is happening if you increase theta you will find out that this particular peak which is a broader peak that keeps on going to still a larger and larger lambda. Now if you take lambda prime and you know what is your original lambda, lambda prime minus lambda corresponding to this high lambda peak actually satisfies the Compton effect experiment expression. Okay, however you there are two interesting observations which we realize here. First of all we find out that the width of this particular peak is much larger in comparison to the original peak and the second is that you find two peaks and the lower lambda peak essentially matches with the original lambda. And the way we explain this particular thing is that after all you are using a metallic target and electrons are coming and hitting that. There is no reason that electrons will get scattered only by electrons. There are also heavier particles like nuclei okay and electron has also a chance that they may get scattered at that particular value of theta after getting scattered from the nucleus. As we have just now discussed that if there are heavier particles the Compton shift is going to be extremely exceedingly small. So, essentially they will give you the same value of lambda because the change in lambda will be so small that it will be very difficult to measure. So, we explained that this particular low lambda peak that I am getting in the Compton effect experiment is because of those electrons which have got scattered from the heavier particle because almost impossible for us to isolate only electrons and perform this experiment only with electron. This second observation is that the width of the peak is much higher of the Compton scattered electrons which have been actually scattered from electrons, Compton scattered photons which have been actually scattered from electrons. And the reason for that is that these electrons which we have talked about are not really addressed. In original expression of Compton effect we have assumed that these electrons are addressed. Actually these electrons will be moving and they will not be moving only with a single velocity but they will be moving with large number of velocities. There is always the distribution of velocities of the electrons inside a metal ok. And when you look and take their velocities into consideration you will find that there is small changes in their lambdas of the scattered photon. So, this particular enlargement of peak width that I am getting in the Compton scattered photons because of the electrons is because of the fact that they are a large number of electrons with a large distribution of velocities and that is why this particular peak gets widened. But anyway what is the most important conclusion of that this particular shift that I am getting is actually matching the expression which we have theoretically derived. So, this was one of the experiment which conclusively proved that photons can be do behave like particles which have energy h nu and momentum h nu by c. So, it is a very strong proof of the fact that these particular photons do carry a momentum which is actually given by h nu by c ok. So, what we basically we have discussed was the dual nature of light. There are certain experiments like photoglyctal effect and Compton effect. We seem to show that light does behave like particles. Then we have another set of experiments like interference diffraction where the wave nature of the light comes into play. One of the biggest mystery is that how these particles have both these effects common. This is something on which we are going to spend little bit more time which basically realizes somewhat I will call it a somewhat philosophical aspect. Yes. Is it now okay to take a photoelectric effect as the proof of particle nature because the threshold kind of thing today with large intensity when you have multi proton absorption those things. So, those threshold will not be sharp threshold. So, is it not better to talk about black body radiation instead of photoelectric effect when that will give more equal to h nu thing. See in black body radiation in principle the way you originally derived it you never used any particle nature of the light.  That is the way originally black body radiation expression. See after all what Planck did the density of state he took exactly same as Rayleigh energies. Only thing what was different was the average energy. All right. So, in principle you do not require in the standard black body radiation thing a particle nature of light. But what I want to emphasize is not just the particle nature of the light the fact that this particular energy quantum is proportional to frequency that is what much more important okay and which directly is evidence only or was directly evidence for the first time only from photoelectric effect experiment. Because that was the experiment which clearly demonstrated that there were stomach potential depends on frequency which was essentially considered as the most direct evidence that these photons energy is proportional to nu. Well I do not think you get that direct evidence. See now that we know about these things no we can always find out arguments that okay this thing would not have happened if particles were not photon okay. But if you especially go to the classical time this was considered as the most direct evidence and as probably all of you know that this particular I mean most of these experiments in this Einstein actually had proposed this particular thing in 1905 and most of these experiments were done much later by Millikan. In fact Millikan also got Nobel Prize for that who could actually prove that these things whatever Einstein had predicted had come out to be true. So I will consider still that photoelectric effect experiment was much more direct evidence that energy of the photon depends on frequency let me put like that. Okay if that quantization comes in many way about quantization I will talk a little bit more. Yeah. This study of direct interaction brings fields because see basically photon is a boson and electron is a pharma. Yeah. When we are talking about the interaction of electron and photon basically these two effects are occurring okay. So now is there any study between that? Yeah I am not very sure I mean I am pretty sure the people who are doing particle physics they have discussed these things in details and you know how the actual interaction takes place in principle they are always scattering cross-section you know when the scattering takes place you know what are the probability of a particular type of a scattering happen. I mean I am pretty sure of course that is not personally not my area so I will not be able to tell you much about it. I have come across that spin electronics. Spin electronics is a very different thing. In that what we are doing is that we are creating spin polarized. Current. They are optically spin polarized the electron spin. See if you are talking about spin electronics this is very different if you mean you know I am not sure whether Professor Suresh will talk about spin electronics because he is giving one lecture on magnetism. That is a part of magnetism okay. Basically normally in the normal electrons we talk only of the current of the electrons okay. We do not bother about spin okay.  Essentially people realize that if you tried if you could manipulate the spin of the electrons or produce a current which you call as a spin polarized current which has current with electrons only one spin that can give rise to many more effect especially in the memory in fact why it caught so much of attention was because of you know the memories of the hardest you know went up you know drastically only because of that particular spin tonic effect. But that spin tonics and what you are talking according to me these are two different things because when you are talking of electron photon interaction that is something different because in a spin tonic device I am not normally talking about the photon. I am only talking to the electronic current okay. So according to me these two are somewhat different yeah. There is magnetic current scattering is a very well-established area what we are talking about and these experiments are done by using the synchrotron radiations at spring at Japan and the one at European ESRF. It is a very well-established area and in that particular experiment using the spin of the electron complex scattering is studied. That is what I said see actually there is different area unfortunately physics is divided into multiple areas. People will do these scattering experiments especially like using synchrotron yeah okay those are different and generally they are working in the area of particle physics and other things you know because I am not particle physics so I cannot give you direct information. But I know that yeah so I mean I am pretty sure people will be doing that particular type of experiments I am pretty sure about that. Excellent sir. Yeah. Regarding this Compton effect can we measure this zero Compton for theta equal to zero experimentally? See the thing is that anyway I am measuring even at theta is equal to zero you see only the original pic that you can always measure there is no difficulty. But all I am trying to say that even if you are at a different value of theta okay you are always getting some photons which essentially appear to be unshifted in their wavelength all right. So the two peaks that I am talking here this particular peak is essentially more or less unshifted energy of the photon which I am explaining that actually we have seen that they have scattered because we are measuring them at a different angle okay but the change in the energy is so small that you are not able to measure them effectively. Obviously if you put a direct beam you will still get this particular type of photon so you can always measure. But see when you are talking how small we can measure then becomes very very difficult because then you know I mean for example when we take a monochromatic beam of photons as we will be discussing now not today about uncertainty principle you cannot have a perfectly monochromatic beam at all okay. They will always raise distribution of energy they are always the distribution there always has to be natural line width otherwise this violates uncertainty principle. But here we are not changing the theta theta is kept constant. See in this particular figure theta is kept constant as I said if I measure theta if I change theta for example if I increase theta what will happen this peak more or less remains unshifted but this peak goes to higher lambda values. Can I interrupt sir yes it is there if you put 20 detectors 20 10 whatever you on different angles you will get the peaks as sir is saying it is seen in that particular experiment you can measure it. If you can vary vary the alignment of the detector you can get only a single peak that is called elastic peak and another one is in an elastic peak. So it is better to use one detector and scan it along the angular position that will be much but it is very difficult because the detector is very huge and it is costing in Indian currency around 25 to 50 lakhs. So you have to put it on a particular angle and that angle is to be optimized and then you have to do it. Okay I think let's go ahead you know because otherwise you know we can discuss it outside you know some of these things which are yeah. What is the effect and interference difference and all this and I want to ask you something will you give me examples at which point a particle behaves as a wave and you also give some example when a wave behave as a particle it will be better for me if you think that I am not very sure whether we are aware in this situation what way a particular particle behave I mean at least I am not aware of it. So these are things which are predecided by nature let me put it like that you know I mean depending on situation the particles behave but you know see examples more difficult point about which I will talk is how do these people what type of problem that will face if these two things have to coexist which eventually leaves to uncertainty principle that's what I would like to mention. Okay I am not getting the answer properly. See the thing is that if you want me to answer this particular question that if a particular particle is going and I want to perform an experiment whether I can know a priori whether this is going to show a wave behavior or it is going to show a particle behavior I do not know the answer that's what I am trying to say. Okay once you perform the experiment or once we have performed the experiment then we know that in such a situation the particle will behave like a either a wave or a particle but to predetermine I am not very sure. Okay sir. Thank you. It is the simultaneous property of the particle. If you say it is larger it shows the wave and particle here it only for one thing like it's according to. That's what is believed in that was the famous complementary principle that you know in an experiment both the properties cannot be simultaneously seen there people have been trying to do some experiment whether this complementary principle is violated I am not sure whether anybody has done but you know there have been people who have been doing this. Being possessed we say that thing it is simultaneous property only one can be identified. That's what it's believed in. Sir this can be better explained by the taking the consideration of group velocity when we just plot a envelope or group velocity. I am going to talk about group velocity today. Then we can mention that if we just checking the group velocity then there will be either we can find the exact location then the. I will be discussing that in more detail today so again I want to interrupt. There was an experiment and one report on March 15 this year in which electron behaved as a wave as well as particle. As I say there are a lot of people who are trying to do experiment but see things that we always keep on publishing even I am not very sure whether there is a final word on this particular thing. Actually it's not to my knowledge but you know it's possible as I say I am aware that there are large number of people who are working on this particular aspect having whether they know this particular complementary principle can be violated or not. Okay if you have finished now let's go a little faster now. So what we are talking about the second cloud see in earlier time we have talked about my first lecture we talked about that there are two clouds on the classical physics. One was about the motion of earth in ether about which you have discussed quite a bit in detail. The second was about the equipartition law. I will not go into the details of the equipartition law but basically this particular equipartition law was used to explain variety of phenomena in classical physics and three phenomena which I have listed is specific heat of gases, the specific heat of solids and black body radiation. Earlier what at that particular time whatever experiments were done at least in the first two aspects of this particular thing there was a fair amount of agreement. So it appeared that the equipartition law of energy did work for specific heat of gases and specific heat of solids in order to explain these things. However as time progressed and we had much better data on specific heat especially the specific heat data as a function of temperature we realize there are discrepancies from what we got from equipartition law of energy. But in older times one of the biggest problem which was faced by people was to explain the black body radiation curve. In fact there is a curve which is called universal black body radiation curve which everybody is supposed to show that you know if you plot something as a function of something you plot msc power divided by t2 power 5 as a function of lamp battery or whatever it is. And this particular curve was something which one could not explain there are lot of attempts and one could not explain until blank came. It was the first explanation of the black body radiation curve and which was based on two very very important assumptions which of course at that time could not be proven it was just a hypothesis just a sort of phenomenology you can call it or you know the two assumptions is first that the whenever there is exchange of energy takes place in harmonic oscillator that take takes place in quantum and this quantum of energy is proportional to the frequency that is the second one. So, the two basic assumptions the first assumption is that the energies are quantized so an oscillator can only have energy h nu 2 h nu 3 h nu 4 h nu and this quantum is actually proportional to the frequency these are the two important assumptions only after these assumptions you could find the first explanation of the black body radiation. All right of course as lot of people believe that himself never believed in his experiments in his theory he was always trying to find another way but that is most beyond the thing. Now, this particular idea of quantization was later picked by many other experiments again phenomenologically there was no real way of saying there was no theory from which this quantization was being proven ok for every system you evolve your different quantization laws. Specific heat of solids in fact Einstein again published another paper where he showed that you know in place of now the standard k t which we normally use for equipartition law instead of that if you use the Planck's formula you can at least explain why the specific heat of the solid 10 to 0 as temperature tends to 0 ok. So, this was one of the positive step in that particular direction one of the very very important positive step was the Bohr's model because we knew hydrogen spectrum very well for the ages ok. It was fitted we would know about reverse constant we would knew that you know this follows 1 upon n square plus 1 minus 1 upon n you know which is not by the way trivial thing because this expression is not all that simple but still people knew that this particular type of expression is valid people could differentiate between these series Lyman series Barmer series and all those series but without any explanation. And at that time Bohr was the first person who proposed this particular thing again introduced quantization ok of course he introduced quantization of angular momentum but other than using quantization use everything classically for him the electron was really moving in a circle ok. So, instead angular momentum quantization energy also automatically gets quantized the radii also get automatically quantized ok but because the quantization was comparatively simpler in comparison to angular momentum he started with angular momentum quantization that angular momentum of the electron must be equal to n h cross and from that particular thing he derived various type of energy expressions and they provided a very good explanation of all these hydrogen line spectrum which is considered as a wonderful success. Of course the basic problem still remains why the oscillators or why the angular momentum gets quantized as n h cross and what is interesting as probably all of you know that that quantization of angular momentum is not correct the energy by which you obtained is still correct that still can be quantum mechanically derived but the expression of angular momentum is not correct because angular momentum of electron now depends on a different quantum number as all of you know alright. Now, let us just try to talk about quantization per se let us just try to discuss as I say little more synopsis aspect. If you go philosophically I think there is nothing very surprising about quantization there are many things which are quantized in the word as we say that you know for example, monetary system is quantized you know we can never make a transaction which is you know smaller than a paisa it has to be a multiple of paisa ok you cannot give somebody 0.1 paisa or 0.2 paisa that is not possible ok every monetary system is quantized physically also there are many other systems which are quantized so we never use this word quantization earlier ok I will give you some examples. For example, when we did a black border radiation curve we said only certain frequencies which form a standing wave can propagate inside the cavity ok in a way that was also quantization because you are allowing only certain frequencies to be traveling inside the system. One more important thing is the vibration of a string see if you have a musical instrument you know that there is a natural frequency with which you tune it depends on tension and mass per unit length this is a standard expression and frequency is equal to n upon pool under root t by m ok. There also you realize that only selected frequencies can travel which satisfy this boundary condition that displacement of the spring must be 0 at this end and must be 0 at this particular end. In that way also this was also quantization because you are allowing only certain frequencies to travel inside this particular string and how does this quantization though people never use the word quantization at that time but how does this quantization come because we applied certain boundary conditions and the boundary conditions is that the displacement here is 0 displacement here is 0 ok which forces that this particular string must travel must have frequencies which are satisfying this particular equation. So, they get quantized not every frequency can travel only this frequency n 1, n 2, n 3, n 4 or n 1, n 2, n 3 or whatever it is only those frequencies can travel inside this particular system. So, in a way this particular application of boundary condition led to a quantization alright. So, in that sense quantization by its own self existed earlier though we never talked about it in that clear manner. In a way frequencies that can be excited in a string are quantized the boundary condition has limited the frequencies in the string to certain number of discrete values alright. Now, let us assemble our thought let us imagine that we are at that particular time where things seem to be essentially in dark. So, let us just try to assemble our thought. The electromagnetic wave showed dual nature we have discussed that light seems to showing dual nature sometimes it shows a particle nature sometimes it shows a wave nature. We discussed many experiments where particle nature is clear. Classically experiments involving interference diffraction are explained purely on wave nature of electromagnetic wave. Second thought nature loves symmetry. In relativity massive particle and photons are governed by the same equation. Photons can behave like particle or wave depending upon the experiment that is what we just now discussed. Isn't it logical to assume that massive particle may also have some wave property alright. If there is a symmetry in the nature if massive particles and massless particles over same equation relativity is it possible that the massive particles also have wave nature after all photons seem to have dual nature. So, is it not possible that the particles also have a dual nature. If the particles could have wave property if at all then these waves could also be subjected to some boundary conditions. We have just now seen after see light was subjected to some boundary condition we got some selected frequencies in a cavity. Normal wave traditional sound wave we clamped the two ends of the string it allowed certain quantization of frequencies. Okay. So, if the particles could have wave properties then these waves could also be subjected to some boundary condition that is a possibility that they could they could subject them to some boundary condition depending upon the problem whatever problem you are describing. These boundary conditions may then limit the wavelength like we have seen in our earlier cases alright. It is possible you know we are just assembling our thoughts and in case this wavelength depends on energy then energy will also get quantized. Okay. So, this is the way we are putting our thought process that may be particles behave like waves and if this wavelength depends on energy and if this they are the whole system subjected to certain boundary conditions then eventually you may limit the frequencies or wavelengths of these particular waves which are particle waves and in that case the energy may automatically get quantized. Essentially using this idea De Broglie in fact recently French the pronunciation is De Broglie know most of the people call it De Broglie. So, in 1924 he proposed that probably particles also behave like waves and he used this particular expression lambda is equal to h upon p. See remember the expression h nu upon c is valid for photons. Okay. So, lambda is equal to h upon p will be valid both for photons and also for massive particle while h nu upon c will be valid only for photons because a massive particle does not terminate the speed of light. Okay. So, this particular expression lambda is equal to h upon p is valid both for photons and particles while h nu upon c is valid only for photons. This is what he proposed. Later it was experimentally verified of course these are some certain things which I think let skip probably you all know that actually he was trying to fit a stationary wave into the Bohr's orbit realizing why what is so special about the Bohr's orbit and that is how he landed on to this particular expression. And also just to give you the order of magnitude of the wavelength realizing that for massive particle the wavelength I mean for normal bigger particle macroscopic particle if you talk of cricket ball the wavelengths are going to be so small that you never able to measure it only for a very lighter particles like fundamental particles the wavelength will be of that particular order which can be typically measured. For example, if you take for electron a thermal electron means the electron which has typically the energy of 3 by 2 kT. T is the room temperature. In that case its wavelength will be of the order of an angstrom it means these electrons in principle could be diffracted. Okay because the typical entero atomic spacing in a solid is also of the order of an angstrom. So, if you have a wave of wavelength of the order of an angstrom this particular wavelengths can be catered. So, then there is subsequent experiment by Daveson and Germer and later by GP Thompson who showed that electrons actually can be diffracted. Of course, as you are probably all aware that electrons are actually diffracted electron diffraction x-ray diffraction neutron diffraction these are very well known techniques experimental techniques which are available. So, all the experimental cases in fact I generally always show in a typical electron diffraction diagram to the students just to show this is the diagram you can see rings here which are basically because of electron diffraction which obey exactly the same condition 2D sin theta is equal to n lambda which actually were originally derived for the case of x-rays. So, electrons do behave like waves. Now, let us discuss little bit more of a wave. If there is a wave which is associated going to be associated with a particle what property it could happen? Normally when we talk of a wave an ideal wave is something like this which is given by A sin kx minus omega t. Sir, you mentioned that for a particle if we assume that energy depends in the wave and if there is a proper boundary condition where we can put the boundary condition in the wave the energy will be quantized because energy depends on the frequency of the wave. If I extend the same article argument to photon then suppose a photon is free there is no boundary condition here. So, from that argument I should say that energy is not quantized. No, no see I am not sure that this argument how far you can take you know the as I said these are only basically the thoughts which we are trying to incorporate. Photons unfortunately cannot be normally derived with the normal quantum mechanics because these particular particles you know see typical Schrodinger equation is only non relativistic and photon because it travels speed of light is not there and for that you require a totally different type of dynamical principle. So, therefore which is definitely not in the scope of this particular thing. So, I will not be able to tell but the thing is that you know it is true that if you are looking at the let me put it like that if you are looking at the symmetry of the argument which we normally always give ok. This is probably more to realize go from the historical aspects ok, but they cannot always be carried too far. But the fact is that I mean if you talk about the wave quantum mechanics the wave mechanics happens to be only one part of the quantum mechanics. I mean in principle it is not really necessary for many of the system to at all talk about the wave nature and still get into the quantization of the energy because you know say for example, harmonic oscillator can be solved purely by using operators ok. So, in that sense you know I mean many of these things they are a good starting point especially if you put like that they are good convincing point to a pressures ok. But many of these arguments cannot be 100 percent carried forward to ultimate in that sense you know. So, this is a sort of a normally a sine wave. We normally define what we call as a wave vector which probably all of you know, but you know normally we tell students because most of the time when they come out of high school they are generally used to talk in terms of frequency and lambda when they have talked about the wave. They are while in the quantum mechanics or generally in most of the time physics we use k and omega and not nu and lambda. So, I take some time to define this particular wave vector k which is 2 pi by lambda and omega is equal to 2 pi into nu which makes the wave equation a little simpler a sine kx minus omega t. So, that is what we normally use and then we define what we call as a freeze velocity. It means if you find out how much time it takes for a particular wave to travel. In fact, I define what I mean by velocity. So, I said you know at a later time if the wave moves to the dotted line then I can calculate how much is the distance ok. And this distance divided by the time it takes to move from here to there will be actually called the what we call as a phase velocity and that is given by omega by k. I also try to sort of insist and try to tell them that can you tell me what is the direction in this particular wave will be moving. And I always say that if you have basically depends on the sine of x and t term because if you take sine kx minus omega t then what happens in order to have the same value of phi the same value of displacement ok. If time increases x has also to increase therefore, this particular wave travels in a positive x direction. While if you go to this particular case where it is kx plus omega t and if I want to retain same psi value then as time increases x has to decrease. Therefore, this particular wave will actually be traveling in minus x direction. This is generally useful when we talk about you know when we are taking step potential and we try to describe the direction of velocity to the particles. It is basically the time difference of the x term and the t term which will determine in which direction a wave moves. So, this is a particular thing which no people know, but the students generally are not very well aware. So, I try to emphasize on this particular wave equation. Yeah. Usually represent using sine cosine. Yeah. But wave equation we represent using the real part there is cosine of kx minus omega t or kx plus omega t. Yes. And we do not represent by sine because the imaginary part we just No, no, no, no. Here I have not even introduced complex numbers. See, think is that as far as sine wave is concerned and cosine wave is concerned, they are just differed by phase. Nothing else happens. Okay. If I am talking of sine kx, I prefer sine kx because that x is equal to 0, you know, this shows displacement 0. If I am taking cosine wave that x is equal to 0, this will show the maximum amplitude. So, that is only the difference. See, I have not introduced the imaginary numbers at all at this particular moment. I am talking of a very, very traditional classical wave. Of course, I mean, see if we use imaginary, we have to use the imaginary, we are rather forced to use the imaginary numbers when we are talking wave mechanics. Okay. But many times we also use the imaginary number for the convenience. For example, when we use electrical circuits and we use AC waves, okay, AC passing through that, we represent it by complex numbers and then we talk about real part and imaginary part. Here I am not talking of any of those things. Here I am talking of only a real sine wave. Okay. Now this can be represented by and if you are not very happy, you can always talk about cosine wave. I mean, nothing of physics will change or nothing of what arguments I am giving will change if sine is replaced by cosine. It depends on this one's own likeliness. Okay. So, I mean, also in the electrical circuits, instead of real parts, if you start talking about imaginary parts, nothing will change. So, I mean, we always talk in terms of real parts, but we could also talk in terms of imaginary parts. Okay. Similarly, in quantum mechanics, also another way I am choosing, I can always choose the opposites and I will get the same results. Nothing of that will change. I mean, we use a convention to define that way. That is a different question. Thank you. Now then we sort of introduce that actually none of the wave is a real, none of the wave that we have come across and real are actually a wave of the type which we have described earlier in which the displacement is supposed to be true for all values of x and for all values of time. All waves are limited in space as well as limited in time. Any realistic wave cannot be of this particular form which will vary for all values of x and for all values of time. So, therefore, the actual wave or physical wave is something or what we call as a realistic wave is something like this, which we call as a wave group or a wave packet in which displacement of the particles last, does not last forever and does not last for all values of x is limited both in space as there is in time. Now, if we create a wave like this, basically question that we want to answer is that what will be the wave on this particular wave packet, whether it could be moving with the same wave omega by k as we have been talking earlier or a different wave, a different speed. And the second question is that how can we form such a wave packet? How can we form such a realistic wave group? So, this is where I spend a little bit of time. So, as most of the textbooks do, we start with a standard beats problem where we take two waves, two sine waves, you can again take two cosine waves if you want with slightly different values of k and slightly different values of omega, a well-known beat phenomenon which all of the students know in high schools. And when you superimpose these two waves, essentially you get a wave of this particular type where you have a cosine term which becomes a part of the amplitude because its delta k is a small, so its wavelength is very, very small. So, this will essentially modulate the amplitude of the waves. And basically your carrier wave which is sine represented by sine term will still be present. So, this is something a wave which like to move something like this. And at that particular time, I start describing of two different type of velocities, one which is the velocity of the original wave and the second the velocity with which this amplitude or the envelope is travelling. So, I have just a small sort of a clip which I will show. So, it is exactly the same thing which I am allowing to move as a function of time, ok. There are two situations which I have written there, one on the top, another on the bottom. Let us focus our first attention on the bottom wave. If we focus our attention on the bottom wave, see I have put two black spots. If you look one is here, another is here, ok. This particular black spot here tells with what speed this envelope is moving.  Now, we concentrate our attention to the top one. If we concentrate our attention to the top one, you realize that these two black spots are moving with the different velocity. For example, this black spot which the envelope velocity is going only this much, while you remember the original velocities are going much, much larger. So, this particular point is moving much faster than this particular black point which is moving with much smaller velocity, while in this particular case, you find that this and this both are moving with the same velocity. So, essentially you tell the idea that it is possible to create a situation in which this envelope is moving with a different velocity than the original carrier waves, alright. Is my point clear? Whatever I am trying to say. Now, question is that with what speed the envelope moves. If I have to look at the speed with which the overall envelope moves, then I have to look at this particular term which is in the square bracket which is actually amplitude modulating, which is modulating its amplitude which is given by this particular term. And the wave vector of this particular envelope is given by delta K. The omega of this particular envelope is given by delta omega. So, if I have to look at the envelope velocity, see like the velocity of the carrier wave depended on omega and gamma, the speed of this particular envelope will depend on delta omega by delta K. So, it is basically the group velocity, this is what we call as group velocity will depend on d omega dK. Now, it so happens that in high school we come across most of the waves in which omega turns out to be proportional to K, it means the speed of wave does not depend on the wavelength, then group speed and the phase speed can be same. It means the envelope speed and the speed of the carrier waves may be same, but in many of the waves also with the particle waves and for that matter even elastic waves inside a solid this does not happen. Omega actually is a function of K, the speed does depend on the wavelength and in that case the group speed and the phase speed can be different. Now, why it is important, so this is the value of the group speed which we call as d omega dK, of course it has to be evaluated at a particular value of K. Is it the same as phase speed? Yes, if omega is equal to vK, it means with v is constant, in other word the speed is not wavelength dependent. Now, in order to say what is the significance of this envelope speed, why we at all talk about the speed of this particular envelope, I come to the concept of how I can localize a particular wave. If I have to localize a particular wave, because people normally at the first year level would not know Fourier transform, so I essentially want to introduce them in a very very indirect fashion. I said consider two mechanical waves of wavelength 10 millimeter and another of 11 millimeter and let us say p is equal to 0, so I mean I am not talking of the time dependent term in the wave. If we superimpose, then we realize that wave will be given by this psi 1 and this psi 2, let us superimpose, if we superimpose this is what we will be getting. Now, my question is that value of x is equal to 0, this particular wave would be giving you maximum, here I have taken cos wave because this is little easier for me to put my argument. So, if I put x is equal to 0, my amplitude will be equal to a1 plus a2. Now, at what other value of x, I will get my amplitude as a1 plus a2. It is very easy to see that I will get amplitude a1 a plus a2 after 110 millimeter, because first one has a 10 millimeter another has a wavelength of 11 millimeter, both of these cosines have to become 1 at the same time at the same value of x. Then only it is possible that amplitude will again become a1 plus a2, that will happen only after 110 mm. Now, instead of two waves, if I add three waves, I add one more wave which has a wavelength of let us say 13 millimeter, ok. In that case, if I have the amplitude, no a1 plus a2 plus a3 at x is equal to 0, next time my amplitude will become a1 plus a2 plus a3 only after a distance of 10 multiplied by 11 multiplied by 13, which essentially means that if I could really impose infinite number of waves which are slightly varying in their wavelength, then in that case, if my amplitude has become 1, the second time when it becomes some, it will essentially be shifted to infinity, and therefore, I will be able to completely localize the wave packet. So, essentially a localized wave packet is a combination of sine waves with different values of omega that case. So, this is the point which I want to emphasize at this particular point, and then I just give this particular expression that there is a standard thing which is called Fourier transform from which you can actually determine these amplitudes of these particular waves knowing a particular type of wave packet. Now, I try to introduce these two things saying that let us suppose we have a classical wave packet which is associated with the particle. Now, when I want to try to mix these two ideas, what type of problems I have? Classically, we assume that a particle is a localized entity, while again a classical wave is an extended entity. If you are talking of a particle, if a particle is kept here, I have to send some signal there in order to detect the property of the particle. Either I have to send a light signal or any other type of signal, something localized here, particle is lying only here, it is not here, it is not here, it is only localized there. But if you look at let us say an ocean wave, the ocean wave could be very, very long wave, a person could be standing at that particular point in ocean, and another person could be standing at this particular point in ocean. Both will simultaneously see waves in front of coming here. So, wave will be observed there also, wave will be observed there also. But as far as a localized particle is concerned, it is observed only at a particular point. So, classically, a particle is always believed, believed to be localized entity, while wave is supposed to be extended entity. Now, if I have to mix these two ideas, how do I mix together? So, that is what I have written. Classically, a particle is localized in a certain region of space, while an ideal wave is not. In order to detect the property of a particle, some signal must be received from the point of localization. For example, color, if I want to know what is the color of this particular point, I have to send a light signal, when it gets reflected, comes to our eye, then only I can find out what is the color of the particle. An ideal wave is infinite. People sitting in different places far away from each other can simultaneously detect the properties of the wave. How can these contradicting properties coexist in a quantum particle? If a particle has to behave like a wave, how can these things coexist? We had earlier said that a realistic wave is like a wave packet and has somewhat localized. We even said that the wave associated with a particle is expected to be form of wave packet. This is what we have already said, that any wave which is realistic has to be some sort of wave packet. So, it has to be localized. Nevertheless, I cannot localize it completely. It has to have some finite extent. And if it has to have some finite extent, wherever this particular finite extent is lying, in principle, the property of the particle can be detected there. If this is a wave packet which is representing a particle, somebody may perform an interference experiment here and detect the property of the particle, somebody may perform an interference experiment here and can detect the particle property there. So, so long this particular wave represents this particle and this wave, whatever might be the finite extension, whatever is the extension, particle property can be detected there. I can find the presence of a particle here. Another person sitting here performing an experiment here can find a particle property particle. Nevertheless, whatever is the extent of the wave packet, the particle property can be detected there. So, the position of the particle is uncertain to the order of the length of the wave packet. In fact, therefore, I can never be very precisely sure where my particle is lying. A particular person sitting here will perform an experiment here and say my particle, I have found particle here. Another person sitting here will say, I found particle here like a wave. Person sitting here will say that, okay, I saw the wave here. Person standing there also will say that I found the wave here. Both are correct. So, it means wave is uncertain. Similarly, the particle property also becomes uncertain. So, this is what situation I have depicted. You have an ideal wave which has infinite extension.  I have localized it further. Now, this particular person, it does not detect the particle property of the wave here. I have localized it further. Now, these two particles, persons do not see particle, only this person sees. More and more I localize, okay, position of the particle becomes more and more certain, all right. But in order to localize, I have to use more and more wavelengths to localize more and more, okay. Now, if I do that, what will happen? If I take a perfect ideal wave, it is lambda is precise. If I have two waves, lambda of 10 millimeter, 11 millimeter, okay. It is uncertain. Wavelength is uncertain. It is 10 millimeters, 11 millimeters, as well as 11 millimeters. And when I have infinite number of waves to create an entire wave packet, okay. Essentially, all the wavelength is uncertain. You cannot define a unique wavelength for a wave packet. And what is wavelength related to? My de Broglie wavelength momentum. It means more I make it compact, more my momentum becomes uncertain. I can make position more and more precise, only at the cost of making momentum more and more uncertain. On the other hand, if I want my momentum to be very certain, I have to use only one wave, then my position becomes completely uncertain. So, this is basically what is uncertainty. Shorten the wave packet. You can shorten wave packet to any amount by superposing ideal waves of a larger and larger range of wavelengths. But a unique value of wavelength can be assigned only to an ideal wave. Larger range of wavelengths mean their wavelength becomes more uncertain. As wavelength is related to the momentum by de Broglie relationship, the momentum becomes more and more uncertain for shorter waves. This is the basis of uncertainty principle. An ideal wave has a precise wavelength, and hence its momentum is very precise. But when this wave extends from plus infinity to minus infinity, hence there is an infinite uncertainty in position. More we want to localize the wave packet, we make the position of the particle more precise, but the loose control on the precision of its momentum. Uncertainty principle protects wave particle duality. This is something which I will discuss a little bit more. After the superposition, as you have said, the two peaks will be separated to infinity, but that will again mean that the wave packet is spreaded till the infinity. No, I mean, if you put really infinite numbers, it will be 100% localized. There is nothing like infinity in the real thing. See, that particular example was given only to just prove without actually talking about the Fourier transform. Okay, I think I will stop here.