 In classical physics, if we want to measure the position and the momentum of a moving particle at any given instant in time simultaneously, then in principle, we can do that. However, in quantum mechanics, we cannot do that. In quantum mechanics, there is a fundamental limit in the accuracies of the measurement of the position and the momentum of a particle at a given instant in time simultaneously. This is known as the uncertainty principle and this represents one of the very important cornerstones of quantum mechanics. In fact, there are many important conclusions in the quantum mechanical world that arise directly from the uncertainty principle. In this particular video, I am going to talk about such five very important, very interesting, very peculiar and very non-common sensical conclusions that directly arise from the uncertainty principle. I am going to discuss all of them in detail and in fact, I am going to mathematically prove each one of them straight from the uncertainty principle. So, that is all what this video is going to be about. So, let's start. So, the Heisenberg's uncertainty relation simply says that we cannot measure the position and the momentum of a particle at a given instant in time simultaneously with an absolute degree of arbitrary accuracy. There is a fundamental limit to the accuracies of the measurement of the position and the momentum of a particle that we can measure at a given point in time. In fact, the product of the uncertainty in the position and the uncertainty in the momentum of a particle is always theoretically in principle greater than a constant h cut upon 2 where h cut is simply equal to Planck's constant upon 2 pi. Now, this particular relationship is not just for position and momentum. In fact, in quantum mechanics, there are pairs of physical quantities for which we have a corresponding uncertainty relation and there are pairs of physical quantities whose uncertainties are inversely correlated. In fact, another important uncertainty relation involves energy and time. So, if we are talking about measurements of energy of a given system over a period of time, then the uncertainties associated with the energy and the time also follows a similar kind of a relationship and there are other similar relations. I have spoken in detail as to the origins of this kind of an uncertainty relationship in quantum mechanics in one of my previous lectures where we have discussed how the concepts of wave particle duality concept of wave packet and the concepts of Fourier transform between the wave packet in position space and momentum space leads to the uncertainty relation. So, if you are interested in all that discussion as to how this equation comes up in the first place then you can check that lecture first. So, in this particular video I am interested in looking at some serious conclusions about the nature of the universe that we can derive directly from this particular principle. So, let's talk about the very first consequence. The very first consequence has to do with the non-existence of an electron inside a nucleus. Now, we know that in the atomic structure we have something called a nucleus and around the nucleus an electron goes around it. Now, almost 100 years ago when we were still figuring out the nature of such an atomic structure there were ideas floating around and one idea was that electrons can survive within the nucleus. Now, the origin of such an idea has to do with the radioactive decay processes. You see there are certain decay processes for example, the beta decay process in which we find that electrons are spontaneously emitted from the nucleus of an atom. So, for example, if you have some kind of a element like X which has a mass number of A and an atomic number of Z then we find that in certain beta decay processes then the element undergoes transformation. It becomes a new element where the atomic number increases by one, mass number remains the same and an electron is also emitted and an electron anti-neutrino is also emitted in the process. What essentially happens is that a neutron gets converted to a proton and electron is emitted in that particular process. So, there was an idea floating around that electrons exist within the nucleus and during certain mechanisms the electron is thrown out of the nucleus. So, now what I am going to show you is that that that idea is not possible. We can show from the Heisenberg's uncertainty relation that if an electron is confined to a very small region in space which is the dimension of the nucleus itself. Then it is going to have so much high amount of energy that we do not really see that in nature. So, how we can do that is we can assume that let us suppose we have a nucleus alright. And nucleus is usually of the size of the few femtometers ok. So, the nuclear dimensions are of the order of a few femtometers. So, let us suppose that if in case there is a nuclear electron which is confined within a nucleus then the uncertainty in the position of that nuclear electron would be somewhat similar to the size of a nucleus itself. So, nuclear sizes of the are of the order of few femtometers 10 to the power minus 15 meters. So, I am going to take a typical nucleus. So, let us suppose I take some average sized nucleus 5 times 10 to the power minus 15 meters or 5 femtometers. So, del x the uncertainty associated with the position of the electron confined to that nuclear dimension is 5 into 10 to the power minus 15 meter. What this does is that because there is an uncertainty associated with the position there is directly corresponding uncertainty associated with the momentum of the electron. So, according to the Heisenberg's uncertainty principle we can calculate that there is an uncertainty associated with the momentum which is equal to h cut upon 2 del x. Which is equal to h cut is h upon 2 pi h is equal to 6.626 into 10 to the power minus 34 joule seconds upon 2 pi is 2 into 3.14 times 2 into 5 into 10 to the power minus 15 meters. Now I have already calculated this particular value in my notebook. So, the value of this particular calculation comes out to be around 1.1 times 10 to the power minus 20 kg meter per second. So, this is the uncertainty associated with the momentum of a nuclear electron if the electron is confined to the nuclear size. Now if the uncertainty associated with the momentum is this much then the momentum itself is also going to be of a similar order. You see if we have uncertainty is associated with the measurement of some physical quantities then usually the physical quantity value itself is either equal to or greater than its particular uncertainty. So, I am going to say that if the uncertainty is this much then we can at least say that the momentum of that electron or at least the minimum value of the momentum ok. We are just looking at the limit here what is the minimum value of the momentum is almost equal to the uncertainty itself. So, this is equal to 1.1 times 10 to the power minus 20 kg meter per second. Now if the electron has this much momentum as a result of being confined within the nuclear size then it has some energy. So, let us calculate that energy ok. So, for a very general relativistic scenario the energy of any relativistic particle having some momentum is given by what? It is equal to E is equal to P square C square plus M naught square C to the power 4. This is something that you must be familiar with if you have studied a relativity. Now if we look at these quantities what is P square C square? So, I will let me plug in the values 1.1 square times 10 to the power minus 40 C is 3 to the power 8. So, 9 into 10 to the power 16 and this comes out to be some value that I have calculated in my notebook. So, let me just plug the value here it comes out to be 1. something or rather 10.89 into 10 to the power minus 24 joule square. Similarly, you have M naught square C to the power 4 which comes out to be what is the mass of a electron 9.1 into 10 to the power minus 31. So, this square times again 3 into 10 to the power 8 is the speed of light to the power 4 and this comes out to be I have again calculated this for you. You can check it for yourself 6.7 into 10 to the power minus 27 joule square. Now as you can see this particular term P square C square term is much larger compared to the M naught square C to the power 4 square. So, let's suppose that I just want to get an approximate idea about the nature of the energy and because the P square C square term is much higher. Almost you can say orders of magnitude of 1000 times higher than M naught square C to the power 4 to make my life easy to make your life easy to make the calculations easy. So, I am going to say that the energy because I just want to get an approximate idea of the energy is almost equal to root over P square C square. So, this is almost equal to root over this particular number. So, we have here 10.89 times 10 to the power minus 24 to the power half and because I want to convert it from joules to electron volt. So, one electron volt is equal to 1.6 into 10 to the power minus 19 joules. So, I divided by this number and finally, this actually comes out to be 20.6. You can do these calculations yourself in your notebook. I am just plugging in the values because I want to save some time in that process times 10 to the power 6 electron volt which comes out to be around approximately 20.6 mega electron volt. So, this is the order of the magnitude of the energy of a nuclear electron which is confined to nuclear sizes of the order of 5 femtometers. Experimentally speaking, when we look at the beta decay spectrum that means the energies of the particles emitted in the beta decay process. We find that the energies are of the order of 0 to around 4 mega electron volt which is very very less. In fact, majority of the cases the energies of the beta particle is somewhat around 1 or even less than 1 mega electron volt. So, usually even though we see electrons coming out of a nucleus experimental study of the energies of those electrons usually falls in the range of 0 to 4 mega electron volt. As we can see from the Heisenberg's uncertainty relation that if an electron actually exists inside a nuclear size then it should have an energy greater than 20 mega electron volt which is much larger compared to what we see in experiment. Which simply means that the electrons that are emitted in the radioactive decay processes are not electrons that exist within the nucleus but they get created during some other process. We now know that these are processes which involves transformations between quarks and leptons and during the transformation between a neutron and a proton we see the creation of such lepton particles but which simply means that electrons cannot exist inside the nucleus. So, this was the first consequence of the uncertainty relationship. Let's look at the second consequence of the uncertainty relationship. Alright, the second consequence of the uncertainty relation is something which is very very interesting. It is called the zero point energy. It is something very peculiar about the nature of our universe that systems can never really go to absolute zero energy. To understand this let's imagine a classical system. A classical system like a simple pendulum which is at rest or a spring mass system which is an equilibrium they technically both have zero energy. In classical physics there is no restriction that a system cannot have zero energy. Systems are capable of having zero energy. If they are at rest, if there is no motion, there is no such kind of oscillation in these particular systems per se. But the same consequence does not apply to microscopic systems. Systems like atoms and molecules which are vibrating in space even vacuum does not have zero energy. So, the concept of zero point energy is simply what is the minimum possible value of energy of any given quantum mechanical system and the answer is not zero. The minimum possible value of energy of any given quantum mechanical system is actually a non-zero and a finite number and that non-zero energy is called zero point energy. So, the zero point energy is actually a non-zero finite minimum value of energy that any quantum mechanical system possesses as a result of the uncertainty relationship. To demonstrate that I am going to take the example of a system whose potential is demonstrated by a harmonic oscillator potential. You can equally demonstrate this for other potentials like an infinite square well or some other variation of a potential. You can in principle show that, but because the harmonic oscillator is a very common example for various kinds of bound systems as well as we can approximate various bound systems as harmonic oscillator. So, therefore, this is a very good example. So, the harmonic oscillator is a system in which the potential has this particular variation that vx is equal to half m omega square x square m being the mass of the particle omega being the angular frequency associated with that particular potential which is a characteristic of the particular potential and the force field. So, what I want to calculate is I want to calculate the minimum possible value of energy of a particle which is in this kind of a potential. So, if we want to calculate the energy, the energy is essentially kinetic energy plus potential energy. So, if you have kinetic energy plus potential energy that comes out to be p square upon 2m plus half m omega square x square. Now, let us suppose that there is a particle in this particular harmonic oscillator system and it has some position and it has some momentum. Now, the uncertainty principle says that the accuracy associated with the position can never go beyond a particular limit. So, if I try to make the position of the particle absolute, that means if I say, okay, let us suppose the particle is at x is equal to 0, it simply means that there will be an infinite amount of uncertainty associated with its momentum, right. So, there is always going to be a very small amount of uncertainty associated with the position and the momentum of the particle. So, let us suppose at a minimum energy configuration, the particle is at some sort of a position x and momentum p. Now, as I just now said, whatever is the minimum limit to the accuracy is a position and momentum. Let us suppose the uncertainty associated with the position is del x and the uncertainty associated with the momentum is let us suppose del p. I am going to say that the actual position and the actual momentum are sort of similar because usually whenever we measure a physical quantity, the value of the physical quantity is actually equal or greater than the uncertainty associated with its measurement. So, let us suppose x is almost equal to del x and therefore p is almost equal to del p. Now, from the Heisenberg's uncertainty principle, what does the Heisenberg's uncertainty principle says? It says that del x and del p is equal to h cut by 2. By the way, I am just taking the minimum limit. It is supposed to be greater than equal to h cut by 2 for all situations but I am just looking at the limit because we are trying to find out the minimum possible energy theoretically speaking. So, this simply means that del p is equal to h cut upon 2 del x. So, let us suppose that a particle is in a given location x, then in that situation it will have some momentum which is equal to the uncertainty associated with it which is given by this. So, if I plug in this particular value in this total energy expression, then the energy essentially ends up becoming 1 upon 2 m p square is equal to h cut upon 2 x square plus half m omega square x square. Now, this is the general relationship for the lowest possible energy that a particle in this kind of a system can have. So, let us try to minimize it. Let us try to minimize this energy and see what we get. So, if we minimize the energy, what we can do is we can minimize it with respect to x here. So, I can say D e upon D x is equal to 0. So, this is a mathematical process of minimizing a function that is related to another variable. Here the variable is x. So, if I minimize this by doing D e upon D x is equal to 0, then what kind of a answer are we looking at here? So, D e upon D x here I am going to have h cut square upon 2 into to 4 into 8 m x square plus D upon D x of half m omega square x square is equal to 0. So, let us solve this and see what we get. So, here the constants will be out h cut square upon 8 m x square. So, here I will have minus 2 x cube plus again I have 2 upon 2 m omega square x is equal to 0. So, here 2 I will end up getting 4, the 2 gets cancelled. If I take this to the right hand side, I end up getting h cut square upon 4 m x cube is equal to m omega square x or I multiply this, it becomes x to the power 4 is equal to h cut square upon 4 m square omega square. So, h cut square upon 4 m square omega square and finally, we have x is equal to root over h cut upon 2 m omega. So, when we are trying to minimize the energy, we find that the value of x at which we will get the minimum possible value of energy is actually this particular expression. So, now let us plug this back into our equation. Let us suppose this is point number 1. So, using equation number 1, if I plug it back in, what do I get? So, E is equal to what? E is equal to h cut square upon 8 m. Now, here we have x square right. So, this can be written as h cut upon 2 m omega plus again here we have half m omega square times h cut upon 2 m omega. Let us see if we can simplify this, h cut gets cancelled, 2 gets cancelled with 8 and I get 4 in return, m m gets cancelled, here m gets cancelled and omega gets cancelled. So, this becomes h cut upon 4 times omega. So, 1 upon 4 h cut omega plus 1 upon 4 h cut omega. So, this finally comes out to be half h cut omega. So, the minimum possible value of energy. So, the minimum possible value of energy, let me write E minimum here. So, the E minimum for a particle in a harmonic oscillator potential is half h cut omega. So, this is a non-zero value. H cut is a constant. Omega is basically the angular frequency. It is a characteristic of the nature of the potential or the strength of the force field. It is a non-zero value. The minimum possible energy in a quantum mechanical harmonic oscillator is never equal to zero. It is a non-zero value. Therefore, whether it is a harmonic oscillator and we can demonstrate these for any other kind of a bound potential as well through a similar sort of a procedure, the minimum possible energy that is allowed in a quantum mechanical system is never zero. Particles in quantum mechanical potentials are never really allowed to settle down to rest configurations. They always have some minimum finite value of energy. So, whether you look at atomic or molecular vibrations, even at zero temperature, even at absolute zero temperature, they still will have some sort of a minimum value of energy. In fact, in quantum mechanics, even vacuum at absolute zero temperature has some minimum value of energy due to the Heisenberg's uncertainty relationship. In fact, later on when we study Schrodinger's equation and different kinds of potentials, we will see that the ground state energy of this kind of a potential actually comes out to be half H cut omega. So, this is the zero point energy concept. That means, in a quantum mechanical system, if we have a particle in some sort of a potential, it can never have zero energy unlike in classical physics where configurations can have zero energy. But in quantum mechanics, that is not allowed. That is the concept of zero point energy. The third consequence of the uncertainty principle is a restriction as to the size of an atom. So, it gives us an idea about atomic size. You see, usually atoms we know are basically configurations in which you have a positively charged nucleus and electron going around it. The uncertainty principle puts a limit as to how close the electron can be to the nucleus. You see, it puts a finite limit as to how close the electron can be towards the nucleus because the nucleus is positively charged and the electrons are negatively charged. So, they attract each other. The electrons should be able to come as close to the nucleus as possible, but the uncertainty relationship puts a limitation to that. There is a minimum limit as to the size of an atomic structure and that is what I am going to prove to you and thereby obtain the Bohr radius for a very simple configuration. So, let's make some couple of assumptions. So, let's say that if an electron is in an atom which is the smallest atom that is possible, then the uncertainty in its position is of the same order of magnitude as the radius of its orbit. And then we will have a momentum associated with it and the momentum is also going to come out to be somewhat equal or greater than the uncertainty associated with that particular quantity. So, from the Heisenberg's uncertainty relation, again here I am going to use del x del p is equal to h-cut. So, you see that the Heisenberg's uncertainty relation is actually an inequality, right? It gives you a limit beyond which the uncertainties are manifested in real life. So, it basically gives you a limit that we cannot really cross. So, it's an inequality. So, sometimes we use h-cut, sometimes we use h-cut by 2. So, in this case I just want to obtain a much more better result. So, I am taking h-cut here. So, here the momentum of an electron confined to that particular region should come out to be h-cut upon let's suppose r here, ok? So, if I plug this into the total energy of an electron. So, if an electron is interacting with a positively charged nucleus, then the electron has kinetic energy. So, it has p square upon 2m and it is interacting with the potential because of the electrostatic interaction of the nucleus and the electron, right? Which is essentially equal to minus e square upon 4 pi epsilon naught r here. So, this is the electrostatic potential. So, if I plug in this particular value here, then what should I get? This comes out to be 1 upon 2m. Here it is h-cut upon r square minus e square upon 4 pi epsilon naught r. So, this comes out to be h-cut square upon 2m r square minus e square upon 4 pi epsilon naught r. So, this is the energy expression for an electron which is in an atom that looks like this. Now, I want to obtain the minimum size. So, I want to obtain the r corresponding to this particular minimum energy configuration. So, if I again follow the same procedure that we followed in the last sort of a consequence at point number 2. So, again I am going to minimize the energy here, ok? And let's suppose we are interested in the minimum possible value of energy. So, minimizing the energy, because ultimately we are talking about the minimum possible value of momentum here, right? So, again I am going to say, ok, dE upon dr this time is equal to 0. So, if I perform the calculations d upon dr, I have h-cut square upon 2m r square minus e square upon 4 pi epsilon naught r is equal to 0. So, d by dr this gives us the constants are out 2m and then you have r cube and minus 2. And here you have e square upon 4 pi epsilon naught r square and a plus minus 1. So, this becomes 0. So, if I take this to the right hand side, I should get e square upon 4 pi epsilon naught r square is equal to 2h-cut square upon 2m r cube. So, here 2 gets cancelled, right? And all the other terms are constants. So, again r square I end up only having r left. So, what is the r? So, this is the r corresponding to the minimum energy configuration. In fact, this is the minimum possible value of the radius that is allowed by the uncertainty relationship. This comes out to be 4 pi epsilon naught h-cut square upon m e square. You see these are all constants here. Now, I can plug in all the values. So, let us say if I want to plug in all the values. So, this is 4 into 3.14 times epsilon naught is the permittivity of free space. I can say, ok, 8. I think it is 8.85 into 10 to the power minus 12. And then we have the Planck's constant here 6.6 into 10 to the power minus 34 again square. And in the denominator, we have mass which is 9.1 into 10 to the power minus 31 and electronic charge 1.6 into 10 to the power minus 19 square. And then h-cut we will get a 2 pi in a denominator. So, this is 2 into 3.14 square. So, if I multiply all these terms, then I have actually calculated the value for this. And the value for this actually comes out to be 0.52 into 10 to the power minus 10 meters which is almost equal to 0.52 angstroms or 0.53 angstroms which is what? You must be familiar with this number. 0.52 angstrom is what? It is a Bohr radius. We will see later on when we talk about the Bohr model of the atom in a hydrogen atom, you have a proton and you have an electron going around it. Then the Bohr radius, the distance at which the electron is supposed to orbit the nucleus is actually this particular number. In fact, later on when we study Schrodinger's equation and we find out the nature of the probability distribution of the electron cloud around the hydrogen atom, then the distance at which the probability distribution or the probability density is most probable or has a peak is actually this particular number. And we have obtained the same number for the atomic size using the Heisenberg's uncertainty principle. So, even though the electron is negatively charged and the nucleus is positively charged and the electron is supposed to be nearer to the nucleus as much as possible, but the uncertainty principle provides a limitation to that particular size and it comes out to be somewhat of the order of a few angstroms. The fourth consequence of the uncertainty principle has to do with the lifespan of atomic transitions and how that affects spectral width. So, we are going to discuss this later on in one of my videos on atomic structure and atomic transitions, but effectively what happens is that in atoms there are different levels of energy. So, there are different energy levels and the atom can make a transition from a higher excited state to a lower excited state. So, essentially what happens is that if you have a ground state which is E0 and a first excited state or a higher excited state which is E1, let's suppose, which is basically an excited state. So, if the atom goes into the excited state, then after sometime the electron comes down from the excited state to the ground state. What happens is that the energy difference is sort of radiated in the form of a photon having some frequency new or wavelength. Now, if we perform a sort of a spectroscopy analysis of this particular atomic transition, so basically we want to look at the wavelengths coming in and we want to look at the intensity of the wavelengths coming in a graph with respect to the different kinds of wavelengths and frequencies. We will find that if this is the intensity then there is a spot or there is a particular line corresponding to let's suppose some wavelength or some frequency. So, this corresponds to what we call the emission spectra of that sort of an atom whatever we are studying. So, this is the emission spectra. Now, this kind of an atomic transition doesn't really happen instantaneously. It takes a certain amount of time to happen and there the uncertainty relationship comes in. So, the uncertainty relationship between energy and time is given by del E del T is almost equal to h cut where del T is the uncertainty associated with the time period corresponding to this kind of a de-excitation and del E is the energy associated with the or the uncertainty in the energy associated with the energy difference. Now, if we apply the uncertainty relationship to this kind of an atomic transition then what sort of a conclusion can we derive from here? The first conclusion is that usually many of the atoms that experience this kind of a de-excitation process their lifespan is of the order of 10 to the power minus 8 seconds. So, if del T is of the order of 10 to the power minus 8 seconds then that means there is some sort of a non-zero uncertainty in the energy associated with it. So, del h cut or h cut upon del T. So, this comes out to be of 6.626 into 10 to the power minus 34 joule seconds upon 2 into 3.14 times 10 to the power minus 8. I have calculated this. The value of this comes out to be 1.05 into 10 to the power minus 25 joules or 10 to the power minus 26 joules which in terms of electron volt comes out to be 0.66 into 10 to the power minus 7 electron volts. Now, what does this mean? You see whenever a system is in the ground state then an atomic system can be in the ground state indefinitely. So, when we are talking about the ground state the atomic system can be in the ground state indefinitely and because the atomic system can be in the ground state indefinitely. So, del T tends to infinity. So, del E tends to 0 that means we can measure the ground state energy with a very high amount of accuracy. However, the excited state only lasts for 10 to the power minus 8 seconds and because of this reason the uncertainty in the measurement of the excited state there is a certain amount of broadening to it. So, the moment we make measurements if you look at this diagram. So, let me redraw this particular diagram here. Then if you look at this particular transition then we cannot really measure this excited state energy level to that high accuracy. In fact, there is a certain element of what we can say broadening of the energy level that takes place. So, there is a certain element of broadening of the energy level that takes place. You see that there is a broadening of the energy level that has taken place by this particular value and as a result of the broadening of these energy levels there is also a degree of uncertainty associated with the wavelength that is emitted during this kind of a de-excitation process. So, if we look at the emission spectra then you also see a certain element of broadening of this particular spectra around it something like this. You see that you do not get a sharp line. In fact, you see a broadening of the spectral line here by a certain amount. Let us suppose we call this as del lambda. So, this broadening of both the energy level e1 the excited energy level as well as the spectral line is a result of the uncertainty relationship. So, essentially you do not get an infinitely narrow spectral line or emission spectra that you would otherwise get in the absence of uncertainty relationship. But instead you get a sort of a natural spectral bandwidth that arises as a result of the uncertainty relationship. In fact, we can calculate what del lambda is. So, if I want to calculate what del lambda is then we know that the energy that is emitted as a result of the de-excitation process is equal to what? It is equal to h nu. So, nu is the frequency corresponding to the energy difference. So, that is this can be written as h upon 2 pi let us suppose and 2 pi times nu is equal to c upon lambda. So, this is equal to 2 pi and then we have h cut c upon lambda. So, if I want to say from here what is del e? So, del e if e is equal to 2 pi h cut c upon lambda. So, here 2 pi h cut c they are all constants and lambda is the one that changes it is in the denominator. So, 2 pi h cut c and then here I should get lambda square here and del lambda here. But what is del e? Del e according to the uncertainty relationship is this one this quantity. So, if I substitute this quantity here then this should become h cut upon del t which is equal to 2 pi h cut c upon lambda square del lambda. So, here h cut gets cancelled and del t is equal to 10 to the power minus 8 seconds or we end up getting del lambda is equal to lambda square upon 2 pi c del t. Now, we can just take a very simple example. Let us suppose that we are looking at some sort of a spectra where the wavelength that is emitted is 600 nanometer. If I take a very simple standard example and the time span or the life span is equal to 10 to the power minus 8 seconds and I plug in all these values. So, 600 into 10 to the power minus 9 meters square upon 2 into 3.14 into 3 into 10 to the power 8 meters per second into 10 to the power minus 8 seconds. So, in that situation del lambda comes out to be I have calculated the value here approximately 2 into 10 to the power minus 14 meters or 2 into 10 to the power minus 5 nanometers. That means, we are able to predict that if a de-excitation of an atom takes place for a wavelength 600 nanometers then the spectral width del lambda would be equal to somewhat around 2 into 10 to the power minus 5 nanometers. So, the uncertainty principle simply says that whenever this kind of an atomic transitions happen it does not happen instantaneously. It happens over a certain life span and because of this there is an uncertainty associated with the energy level of the excited state as well as the wavelength corresponding to that particular transition. So, we end up seeing these natural spectral bandwidth that arises from the uncertainty relationship here. Another very important very interesting and very peculiar prediction of the uncertainty principle is that of virtual particles. You see to understand virtual particles I have a very specific example here. So, when earlier we were trying to explain the nature of the forces inside the nucleus which is basically the nuclear force. Scientists were trying to figure out a mechanism through which nuclear force you know works between various nucleons. And one very popular theory was the Yucca was model of nuclear forces. So, in the Yucca was model of nuclear forces you have nucleons protons and neutrons which are experiencing an attractive force as a result of a continuous exchange of particles between them. Let me repeat. The nucleons inside the nucleus a proton and a neutron experiences an attractive residual nuclear force as a result of a continuous exchange of particles between them. And these exchange particles for this particular purpose is later on revealed to be what is called a pi meson. So, I here have a chart. So, you have a proton and a neutron the proton emits a pi meson undergoes a transformation. The pi meson after sometime reaches the other neutron gets absorbed this undergoes a transformation and this process happens continuously in a very fast manner. And the exchange of that particular particle results in a certain transfer of momentum that creates the attractive force which we know as the residual nuclear force. However, if we look at this particular process of this Yucca was model of nuclear forces we realize that pi mesons are not very light they have a certain amount of mass. And if such particles are indeed being exchanged by the nucleons then we should find the nucleons with different masses or masses changing with time but it doesn't really happen like that. We don't really see the masses of the nucleons inside the nucleus undergoing change with time. So, technically speaking this kind of a process should violate the conservation of energy principle. However, this violation of the conservation of the energy principle is allowed because of the uncertainty principle because the uncertainty principle essentially has to do with measuring physical quantities. And as long as there is a limitation to the accuracy of measurement we can always figure out a scenario in which the energy associated with some sort of a non-conserving process a process where energy is not conserved. The amount of energy or the violation of that particular energy is allowed as long as the exchange of that energy happens at a very tiny fraction of a time period which is given by the uncertainty principle. So, the uncertainty principle says that the uncertainty associated with energy and time is always greater than equal to h-cut by 2. I am going to take h-cut here just for simplicity. So, as long as the exchange of that particular energy happens in such a small time period so that whatever amount of energy or uncertainty that actually causes that particular violation of the energy conservation principle is exchanged in a time limit below this that way we are unable to actually measure it with instruments then that sort of a process is allowed or that sort of an exchange of particles having some sort of a mass energy is allowed. And those particles, those exchange particles are known as virtual particles. So, let me do a very quick calculation and sort of give you a little bit of an idea about this. So, let us suppose that we have these nucleons this is just a Feynman diagram corresponding to the same process where a certain kind of a mass particle is exchanged between the nucleons. And let us suppose this mass particle has so we are going to call this as pion and it has a mass of m pi and it travels at a very near speed of light speed from one particle to the other. And let us suppose this R is of a few femtometers so it is of the range of in which nuclear forces actually exist in. So, this is of the order of 1.7 femtometers ok. So, if the R is 1.7 femtometers which is the range of nuclear forces then the time period it takes for the pion to go from one nucleon to the other nucleon is equal to distance upon speed. Let us suppose the speed is nearest velocity of light so this is equal to 1.7 femtometers upon the speed of light. So, if I plug this here ok in the Heisenberg's uncertainty relation right. If I plug this here what do we get we can actually calculate the discrepancy in the energy. What is the discrepancy in the energy? The discrepancy in the energy is actually the energy of the pion particle as a result of its mass. So, just for the sake of simplicity in the calculations I am going to say ok del e is equal to mass of the pion c square. So, mass of the pion c square times del t which is equal to R upon c is equal to h cut. So, if I plug in all the values m pi so cc gets cancelled into 3 point ok. So, let us suppose I take it to the other side ok what is m pi. So, m pi is equal to you have h cut upon R c which is equal to 6.626 into 10 to the power minus 34 joule seconds upon 2 into 3.14 into R is equal to 1.7 femtometers so 1.7 into 10 to the power minus 15 femtometers into 3 into 10 to the power 8 alright. So, I have already calculated the value of this particular mass. So, the value of this particular mass comes out to be around 2.06 2.06 into 10 to the power minus 28 kgs 10 to the power minus 28 kgs which if we compare with the mass of an electron is actually equal to 220 times the mass of the electron. So, essentially if our measurements are limited by the uncertainty principle in such a manner that there is an exchange of this amount of energy between the nucleons in such a short time period that we cannot really measure it experimentally then the mass exchange is equal to this much. And later on after 10 years this kind of a prediction of the Yukawa's model was sort of experimentally verified and it was found that experimentally speaking such particles M pi or the positive pi meson was actually found to have a mass of 273 times the mass of the electron and the mass of the negative pi on was found to have a value of 264 times the mass of the electron which is kind of very much similar to actually what has been predicted by the uncertainty relation. So, what the concept of virtual particles is that if this kind of an exchange of particles is happening continuously between particles, nucleon particles then the exchange particles can violate the conservation of energy principle technically as long as the amount of energy associated with that violation is exchange in a time period given by the uncertainty relation. As such that kind of a process would be allowed and therefore these particles are called virtual particles. Virtual particles in the sense that they are not really real particles because we can't really measure them in that process. We can measure it separately later on in some other circumstances but as the process is happening it's a theoretical process these particles are not really actual particles but transient particles which share the characteristics with ordinary particles but are limited in terms of being real by the uncertainty relationship. So, because they are so short lived we can have such kind of virtual particles existing almost everywhere in quantum mechanics. So, in quantum mechanics the interaction between two charged particles happens through exchange of virtual photons. In quantum mechanics I already told you that vacuum has a energy right a zero point energy then that sort of a energy can be seen or demonstrated in phenomena like chasmere effect. So, uncertainty principle therefore which is one of the cornerstones of quantum mechanics leads to some very interesting very peculiar and very non-common sensical predictions about how microscopic world works and these predictions the fact that electron cannot exist inside the nucleus the zero point energy of any kind of a quantum mechanical system the limitation to the size of any kind of an atomic structure the bandwidth of spectral sort of emission and the existence of virtual particles or virtual photons or virtual any kind of an exchange particle these all can be shown by the uncertainty principle. So, I hope that you have understood not just the uncertainty principle but also how it can be applied in these very specific examples that I have demonstrated in today's video. So, that is all for today. I will see you in the next video where I am going to discuss the atomic structure right. So, how these ideas of wave particle duality and uncertainty principle leads to concepts of the Rutherford model the Bohr model and then further we will talk about atomic transitions and spectroscopy etc. So, till then I am Divya Jyothidas this is for the love of physics. I will see you next time take care bye bye.