 We continue our discussion of quantum harmonic oscillators today. What we will do in this module is that we will talk about the energies of harmonic oscillator and we will talk about actually one wave function. We will perform a more detailed discussion of the wave functions and how they are interrelated in the next module. Today we will talk about the lowest energy wave function. Let me start with a spoiler what you see here well we have seen this in the last module as well is a the potential energy surface of a quantum harmonic oscillator in which we have shown the different energy levels that are allowed. Remember energy is quantized here and quantization arises out of the boundary condition that the wave functions must become 0 at x equal to plus infinity and minus infinity. And as you can see the wave functions do not exactly terminate at the potential energy surface they go a little bit beyond and it was your homework to find out why this is the case how this is justified please read about this and post on the forum. But if you look at the wave functions they look somewhat similar to the particle in a box wave functions the lowest one is pretty much a Gaussian function without any node. There is one node in the second one, two in the third one and as you go higher up number of nodes in the wave function increases as a rule of thumb energy and number of nodes go hand in hand a more energetic wave function would have a greater number of nodes. This is perhaps something that you have studied axiomatically at well in high school as well. Today let us go ahead and see how one can arrive at the expression of the energies and at least one of the wave functions. But before that let us do a brief recap of what we have studied in the last couple of modules. We have said that harmonic oscillators are those which obey Hooke's law. So in the classical domain solution for x the displacement from mean position would be a sin omega t plus b cos omega t where omega is the angular frequency of oscillation. Please note the term angular frequency of oscillation. By the time we close the discussion in this module we will have something to say about that. So this angular frequency of oscillation is popular because it is simply root over k by m where k is the force constant of the spring and m is the mass. So for a harmonic oscillator from classical mechanics itself we know from classical mechanics itself we know that the potential energy is parabolic. So what we do is we take this parabolic potential and we plug it into the Schrodinger equation in place of v. So the first term remains the same the kinetic energy term. The potential energy term becomes half m omega square x square psi equal to e psi. As we have said earlier the beauty of Schrodinger treatment of quantum mechanics is that you start with classical quantities and then when you try to interpret the wave functions the unique features of the quantum world manifest themselves automatically. That is what will happen here as well. So this is your Schrodinger equation and we said that there are two ways of solving it. First is the power series method which we will come to later on and what we are doing right now is the algebraic method and the most important thing about the algebraic method is the ladder operators to which we have been introduced in the last module. So what the way we got to the ladder operators is that we wrote Schrodinger equation and remembering the operator for linear momentum h cross by i d dx we plugged it in and we wrote the kinetic energy term as p square psi by 2m remember we are not writing the hat here but p is really an operator the momentum operator it is not a number. Second term since there is a 1 by 2m here we take 1 by 2m common from the second term as well so that you get square of m omega x operating on psi. So the Schrodinger equation becomes like this 1 by 2m then p square plus square of m omega x the whole operator this is the Hamiltonian this Hamiltonian operates on psi to give you e psi. So what we have done essentially is that we have rewritten the Hamiltonian in terms of the linear momentum operator and then what we said is that this Hamiltonian here is sort of u square plus v square kind of term. So if it was a number you could have written it as i u plus v whole thing multiplied by minus i u plus v but then since these are operators that will not be the case but let us see what happens when we take this i u plus v u here is p so i p plus v is m omega x i p plus m omega x and minus i p plus m omega x i u plus v minus i u plus v. What we said is let us take these two quantities anyway work out their products and see how far we are from the Hamiltonian. And as we said this factor that is there that is because we already know what the solution is going to be so solutions look better if you have this 1 by root over 2 h cross m omega. A minus and a plus these two are operators in themselves these are our ladder operators. And we also asked you to work out that the commutator of a minus a plus is equal to plus 1 which means commutator of a plus a minus is minus 1. And since the commutator is not equal to 0 what happens is the Hamiltonian is not just a minus a plus or a plus a minus what we get is that the Hamiltonian is written as h cross omega multiplied by a plus a minus plus half or you can write it as h cross omega a minus a plus minus half. Both are valid forms we use the one that is more convenient given the situation we are in. So this is how you rewrite the Hamiltonian in terms of the ladder operators a plus and a minus. Then we came to this question why is a plus called a plus why is a minus called a minus because we proved that when a plus operates on psi it produces a wave function which has an eigenvalue of E plus h cross omega E is the energy of psi the original wave function. So when a plus operates on that wave function energy of the resultant wave function is increased by h cross omega which is one vibrational quantum for this harmonic oscillator. That is why a plus is called a raising operator it takes us up by one step in the energy ladder of the quantum harmonic oscillator. Similarly when a minus operates on psi the eigenvalue that we get for the new wave function a minus psi again unfortunately I have repeated the error of last module sorry about that this will be a minus of course but the wave function the eigenvalue is E minus h cross omega a minus psi this is a minus please correct it in your notes a minus. So this is called a lowering operator and the good thing about this is that in principle if you know one wave function we should be able to work out the others by going up and going down the ladder using the appropriate ladder operator. If you know the energy of one level we can work out all others and this is something we will do in the next 15 20 minutes right. So this is what we had said and now with this background let us to try to find out the wave function of the lowest energy state of course the question that should arise here is that is there a lowest energy state or is it a bottomless bit I mean can you go down further and further can you go down to minus infinity energy can you go down to plus infinity on the higher side can you do you have to stop at 0 well minus infinity and all might be stretching things a little bit what is the meaning of negative vibrational energy but then if you think of the classical harmonic oscillator then we can expect that the lowest energy state should have energy equal to 0 as we touched upon briefly in the module before last for classical harmonic oscillator it can come to rest right it does not have to vibrate all the time does not have to oscillate all the time what happens here. So to do that we will use a clever trick and once again I am using the treatment used in Griffiths quantum mechanics book let us say psi 0 is the wave function for the state with lowest energy if there is one well there will be one. So psi 0 is the wave function of the state with lowest energy see if I apply a plus on psi 0 what happens I go up higher up the ladder yeah I get the wave function or I get a function that is proportional to the wave function that has energy which is higher than this lowest energy wave function by one step one quantum h cross omega what happens when I apply a minus on this supposedly lowest energy wave function there is nothing below it. So when a minus operates on this wave function let us say what have I call it I call it psi 0 when a minus operates on psi 0 will you agree with me that I should get 0 not because the energy is 0 not because the eigen value is 0 but because the resultant wave function itself should be 0 everywhere or in other words it should not exist because if psi 0 is the lowest energy wave function you operate a minus on it where do you go there is nowhere to go that state does not exist. So wave function cannot be there for it so that wave function has to be 0 at all values of x in other words we can think that probability of finding that state is 0 everywhere so psi psi star probability density is 0 everywhere therefore psi also has to be 0 everywhere psi has to be 0 everywhere. So a minus psi 0 produces a wave function that is 0 everywhere that is why we write a minus psi 0 equal to 0 I hope you understand this this is a very crucial step see mathematics and all will work out but it is important to get the logic and the logic is quite subtle sometimes if you miss it then we might not get it in a lifetime all right so a minus psi 0 equal to 0 now we know what a minus is right this is a minus 1 by root over 2 h cross m omega multiplied by i p plus m omega x when this operates on psi 0 I get 0 we know what x operator is is just multiplication by the value of the position x what is p operator this is p operator h cross by i d dx so what we will do now is that what are we trying to do we have somehow trying to find what psi 0 is to do that we need to set up some equation in psi 0 so that becomes easy when we take this expression for the momentum operator h cross by i d dx plug it into the expression for a minus and we get 1 by root over 2 h cross m omega multiplied by h cross d dx plus m omega x psi 0 equal to 0 it is a differential equation right first order differential equation first thing I can do is I can get rid of this factor 1 by root over 2 h cross m omega because that will be equal to that is of course not equal to 0 so rest of it has to be equal to 0 so what I get is h cross d psi 0 dx plus m omega x psi 0 equal to 0 right simple first order equation I think most of you would be able to solve it without me doing it but we will do it anyway now let us rearrange in which I have everything in psi 0 on one side and everything in x on one side so this is what I get h cross b psi 0 by psi 0 is equal to minus m omega x dx what would the next step be very obvious next step would be to integrate it yeah h cross integral d psi 0 by psi 0 is equal to minus m omega integral x dx simple what is d psi 0 by psi 0 everybody knows it is ln psi 0 what is integral x dx it is x square by 2 and let us keep it indefinite for the time being so we will get h cross ln psi 0 minus m omega x square by 2 plus some constant of integration okay pretty simple now for the sake of convenience what I will do is I will write the constant in logarithmic form as well because if I can somehow bring it here inside this log then I can just take antilog on both sides raise both sides to the power of e and I will get a nice expression so what I will do is I will write this constant as h cross ln a then it becomes simple you just take it to the other side it will become h cross ln psi 0 by a then raise both sides to e we get psi 0 equal to a multiplied by e to the power minus m omega x square by 2 h cross right simple so we have got psi 0 haven't we we have already found what this lowest energy wave function is well almost but not quite because we still have to find out a and if you remember the properties of wave function the wave function has to be normalized so we can treat a as a normalization constant and when we consider normalization what it means is that integral for values of x from minus infinity to plus infinity of psi 0 square dx has to be equal to 1 so we write like that a square ideally I should have written mod a square but I know in this case we are going to get real value of a so I have been a little lazy and we have written just a not mod so a square is a constant comes out of the integral sign integral minus infinity to plus infinity e to the power minus m omega x square by h cross see here we had e to the power minus m omega x square by 2 h cross take square of that this 2 in the denominator would go simple so e to the power minus m omega x square by h cross integrated for x equal to minus infinity to x equal to plus infinity that should be equal to 1 now this minus infinity to plus infinity integral this is a standard integral the value is known you can look up a compendium of standard integrals and you can find out what the value is generally if you are asked to work this out in an exam the value of this integral has to be given right so we will do that and believe me when I say that this value is root over pi h cross divided by m omega so a square multiplied by root over pi h cross by m omega is equal to 1 rest is simple square root of m omega by pi h cross is equal to a square so a is equal to m omega pi pi h cross raise to the power 1 by 4 of course you can ask why have I not taken the negative root I could have does not matter for wave function actually you have to define one wave function and work with respect to it so if you want to write minus I will be my guest but generally wherever possible we keep the positive value there is a convention so a is there now we can say that we have found the wave function psi 0 and it is 1 well m omega to the power pi h cross raise to the power 1 by 4 multiplied by e to the power minus m omega by 2 h cross x square what kind of function is this Gaussian function so if you remember the diagram I had shown you at the beginning of this module the lowest energy wave function was Gaussian right this is it so we have been able to work out the wave function of the lowest energy state right what is the next step the next step is to find the energy okay since you know the wave function what we will do is we will take that wave function and we will make the Hamiltonian operate on it so energy should come as the eigen value of this wave function that will be the lowest energy so what we will do is we will use a convenient form we could have used the differential form also but since we are using ladder operators and since ladder operators are so convenient we will use the form of Hamiltonian with the ladder operator you could have used the other one as well this is actually more suitable in this case you will see in a moment why it is so so h equal to h cross omega multiplied by a plus a minus plus half this is the form of Hamiltonian we will use so h psi 0 then becomes h cross omega multiplied by a plus a minus psi 0 plus half psi 0 that should be equal to e psi 0 right now one thing to remember from the last slide is that we know very well that a minus psi 0 is equal to 0 right we started with that consideration the psi 0 is the lowest energy operate lowest energy wave function so if you try to lower it there is nothing to go nowhere to go so the wave function that you will get is actually 0 everywhere so this is very convenient a minus psi 0 is 0 so even if you apply a plus to that a plus 0 is also 0 it is 0 everywhere you do not know what to do with it so conveniently the first term becomes 0 only the second term survives so what you get then is that the second term is half h cross omega multiplied by psi 0 so Schrodinger equation reduces to h psi 0 equal to h cross omega by 2 psi 0 what is the eigenvalue here h cross omega by 2 and that then is e 0 the energy of the lowest energy vibration state okay psi 0 has an energy of half h cross omega this if you remember from our introduction is called 0 point energy this is the smallest energy a quantum harmonic oscillator can have it can never be 0 because if it was 0 then it would have violated uncertainty principle so even if you take the oscillator down to 0 Kelvin it will oscillate with the frequency of half h cross omega this psi 0 will be the only state that will be occupied at that case in fact for most of the molecular harmonic oscillators this is the only state that is occupied in room temperature anyway to know further please look up our NPTEL course on molecular spectroscopy but let us proceed from here e 0 is half h cross omega great now what can we do if you remember if you use this step up operator a plus ladder operator on psi then in Schrodinger equation the wave function becomes e plus h cross omega okay so what I am saying is this you just keep on applying a plus let us apply a plus to psi 0 what will I get I should get psi 1 the wave function that is that has energy that that is just one step above that of psi 0 so that e 1 will be equal to e 0 energy of psi 0 that is half h cross omega plus h cross omega from here what will e 2 be e 2 will be well sorry so e 1 will be then I can take I can write this e 0 as half h cross omega and I can take h cross omega common outside the bracket so I will get 1 plus half h cross omega 3 by 2 h cross omega but let me write it in terms in the form 1 plus half because that will lead us to the general form of the energy anyway so e 2 what will e 2 be I can again get e 2 by making a plus operating on psi 1 then it should be e 1 plus h cross omega e 2 should be e 1 plus h cross omega what is e 1 e 1 is 1 plus half h cross omega add 1 to that what do you get simply you get 2 plus half whole thing multiplied by h cross omega and so on and so forth so what we have seen is that even though we have not really worked out the wave functions for the higher energy states we have been able to work out the energies so what happens for e v e v would be e v minus 1 plus h cross omega that would be v plus half h cross omega what is v here what is the range of the v ranges from 0 and it increases by an integer each time 0 1 2 3 4 right so this is the vibrational quantum number and what we see here is the most general form of the energy of a quantum harmonic oscillator e v equal to v plus half multiplied by h cross omega where v equal to 0 1 2 3 etc etc etc can go up to infinity these are the vibrational quantum numbers so what we have been able to do is we have been able to find out the expression of energy of not only the lowest energy state but all energy states of the quantum harmonic oscillators simply by using ladder function e v equal to v plus half h cross omega and v equal to 0 1 2 etc etc right now before going further I would like to say one thing since I am a spectroscopist I like to rewrite this equation somewhat remember omega is angular frequency um a spectroscopist usually likes frequency better than angular frequency no particular reason just convention so what I can do is I can write h crosses h by 2 pi and omega is 2 pi multiplied by nu right omega is angular frequency radian per second and nu is frequency per second right so this is a dimensionally consistent so this 2 pi and that 2 pi will cancel and you will be left with h nu so what I could do is what I like to do is I like to write e v equal to v plus half into h nu what is h nu one quantum of vibration okay and I like this because when there is a transition from one state to the other what you are left with is h nu and this frequency of vibration turns out to be the frequency of light that it absorbs as well if it undergoes a v to v plus 1 transition and as it happens for simple harmonic oscillators quantum harmonic oscillators only v to v plus 1 transitions are allowed there is something called selection rule just because you have energy levels in a system does not mean that one can absorb it can absorb light and go from any energy level to another only certain transitions are allowed others are disallowed I am well one selection rule that I think most of us are familiar with is Laporte selection rule remember DD transitions are not allowed right so similarly in quantum harmonic oscillators only v to v plus 1 transitions are allowed others are not for such transitions the energy difference simply turns out to be h nu where nu is the frequency of vibration of the oscillator and then that will equal to h nu where nu is a frequency of light also so interestingly a quantum harmonic oscillator absorbs light whose frequency matches its own vibrational frequency we have dealt with this a little in a little more detail in the molecular spectroscopy course okay so this is our expression where we wanted to get e v equal to v plus half h nu we can we know the expressions for the expression for the energy of all the vibration energy levels of the quantum harmonic oscillator but until now we know only one wave function psi 0 equal to m omega to the power pi h cross raise to the power 1 by 4 multiplied by e to the power minus m omega by 2 h cross x square what should the next goal be the next goal obviously should be to find psi 1 psi 2 psi 3 psi 4 okay how can we do it we can do it by using ladder operator but then we will also use this analytic technique which will allow us to arrive at some very beautiful relationship among the wave functions in a vibrational manifold and in fact that is what leads to the selection rule so what we will do in the next module is that we will try and understand how to find out the expressions for psi v the wave functions for the vibrational levels other than v equal to 0 equal to 0 you already know so that will require maybe two modules after this today for this module we close our discussion