 So, we will continue with the postulates first and then, so this is today's still an overview class before I bring you to the many electron theory. So, we completed the measurement part of the postulates and we said that if it is a stationary state, again I repeat, if it is a stationary state, then the measurement yields only one value, it always yields one value, so this is of course an interpretation, if it is not a stationary state, then there is only a probability of getting a value, but the value is always one of the eigenvalues of the operator, that is important to note, you will always get an eigenvalue of the operator, so I specifically mentioned that if I have eigenvalues of the operator which are ai and the eigenfunctions are phi i, but if psi is a non-stationary state, then of course I will get, I won't get a specific ai, but I will get a scatter of eigenvalues as a measurement, with a probability, with a probability which can actually be defined, so that probability can be defined once I know the psi, so that I will tell you how to define. So, we get a scatter of eigenvalues means we can get a1, we can get a2, we can get a3, so this is basically an interpretation and in fact the idea of the Strodinger Cat, the experiment that was conceived is basically part of this, that there is a half chance that the cat is alive half, depending on who does the experiment, so that is where I said there is a lot of subjectivity which has been brought into this Copenhagen interpretation, this is called the Copenhagen interpretation of quantum mechanics, but this probability is a defined probability. The definition comes by the fact that this psi, although it is non-stationary, can always be expanded in terms of the eigenfunctions of A, because the eigenfunctions of the Hermitian operator form a complete orthonormal set, so I can expand this in terms of a complete orthonormal set, where a particular ck can be written as phi k psi, inner product of phi k psi, which is basically again in full notation integral phi k star psi d tau, that is very easy to show, you expand psi as ci phi i, so if I write it again direct notation, expand psi i as ci phi i, then project with phi k, because all phi i's will be orthogonal to phi k, the only one that will survive is when this sum over i will include phi k and then the result is ck, okay, so again we have done this before in the previous class I am repeating, and then we say that the probability of getting ak, which is the kth eigenvalue, is nothing but mod ck square, mod ck square is the probability which is ck star ck, that's the probability of getting a particular eigenvalue ak, alright, and of course the sum of this probability, that is if I sum over ak, probability of ak, what should be the value 1, right, that is of course understood that the total probability of getting each of the eigenvalues is 1, okay, that brings us to a very interesting relationship which I must write now, so you have a sum over k, what is probability of ak mod ck square, so sum over k mod ck square is equal to 1, alright, so let me write down this ck square now, it is ck star ck equal to 1, sum over k, now let's write ck star, ck is nothing but phi k psi, so ck star will be psi phi k, correct, psi phi k, alright, and then again ck is phi k psi, note that in this interpretation that the probability is mod ck square, ck square we have also assumed that the psi is a normalized function, otherwise the normalization constant of psi has to be simply subtracted, so if you do this then we have found, we have got a relation which says that psi phi k, okay, phi k psi is equal to 1, if you sum over all k, right, so k comes here, so that's an interesting relation, since psi psi equal to 1, since psi psi equal to 1 we can now write from here that the sum over k, this k phi k, the conjugate phi k is equal to an identity operator, note this is again something that I have done before in the last class that this is the projection operator, each of them is a projection operator and the sum of all this projection operator is an identity operator and you have used it in those who have done the 4 to 5 in many examples what is called the resolution of identity and I again repeat that we just show that the resolution of identity that if you sum over all the eigenfunctions of any operator, any Hermitian operator with the ket and the conjugate, it is an operator and this is equal to 1, note again that there is a difference between the conjugate first in Dirac notation and the ket later and the reverse, so for example if I had phi k phi k that is a number, this is an integral whereas phi k phi k is an operator, this is an operator, again sum of these things that we have done in the previous class, I again want to repeat, this is an operator that is a number in Dirac notation, so this is phi k star, phi k integral, here there is no integration, it is just the phi k into phi k star, so if I sum these operators and each of this is called the projection operator and I will tell you why it is a projection operator, the sum is equal to an identity operator because you can very clearly see then this becomes psi identity psi which is equal to 1, correct, so this will actually become psi identity psi because now no longer the summation is required, the summation is included in the identity and obviously this is equal to 1, so that is how you get the sum over probability, so I have actually shown the sum over probability is 1 obviously and I can justify this by showing from the resolution identity, derived from the resolution identity, so let me come back to this particular operator, there would be many, many different operators which will encounter in quantum mechanics, so this kind of operator is known as a projection operator, so this is of course sum of all projection operators, so projection operator by the very meaning of English, it means it projects something, so let us see what it projects, for example if this operator acts on any arbitrary function psi, so this is an operator which is acting on an arbitrary function psi, so by the Dirac notation, remember the Dirac notation is very, very convenient to write this projection operator, moment I write a ket vector in the next to the conjugate vector, this becomes a number, now it is an integral, what is this value, this value is nothing but ck correct, so this is nothing but ck times phi k, so what is this is an operator, so let us say this is pk, so I have written this pk acting on psi equal to, so pk is operator acting on psi, what does it give you from the psi, it gives you the component of phi k in psi, only the phi k, ck phi k, you have c1, phi 1, c2, phi 2, c3, phi 3, entire component of phi k it brings out from a function which consists of all phi k, all phi y, phi 1, phi 2, phi 3, so if this is p1 then only phi 1 will come out, if p2 this is only phi 2 will come out with this coefficient, so that is the reason these are called the projection operators, projection operators have lot of interesting properties, so one of the properties is called the idempotency, we have again done this before I remind you, projection operators are idempotent operator, what is idempotent operator that if you take a square of the operator, so for example if you take pk square, this is equal to pk, a single projection operator if you square it, it is equal to pk, again it is trivial to show this, write again by direct notation phi k, phi k that is your pk, then you write pk again, so then you come phi k, k, moment the k and conjugate come together they become a number and then you have the conjugate phi k, so this becomes a number which is 1 because all eigenfunctions can be normalized, so this is of course phi k, phi k, note that everything that we are writing is in a space of normalized functions, psi is normalized, eigenfunctions are normalized, if it is not normalized it is not a big problem you can always re-normalize, but everything we are talking about, so you can see that pk, pk is equal to pk again back and you can keep doing it, so all pk to the power n is equal to pk and we actually did an exercise in 4 to 5, again those who have done it that the eigenvalues of this pk are either 0 or 1, I hope you remember I had given an exercise actually somewhere that the eigenvalues of this projection operators are 0 and 1 which is very again very easy to prove, you can say that pk, phi k equal to lambda k, phi k if this is an eigenvalue equation then we said pk square, phi k equal to lambda k square, phi k by trivially doing it twice, however pk square is nothing but pk, so obviously this is same as lambda k, phi k, hence lambda k square equal to lambda k and hence lambda k equal to 0 or 1, I hope all of you remember the proof, it is again a trivial proof I am again reminding you what all we have done before because some of these things will again be used, is it clear the proof? Because pk square gives you with the same eigenfunction gives you lambda k square, however pk square is pk, so lambda k square must be equal to lambda k, which means lambda k can only be 0 and 1, right? If lambda square equal to lambda that means lambda square minus lambda equal to 0, take lambda as a common lambda into lambda minus 1 equal to 0, I mean all of you have done this algebra, right? Lambda equal to 0 or 1, so the eigenfuncts, eigenvalues of a projection operator or actually you can say eigenvalues of any idempotent operator are always 0 and 1, always any idempotent operator, so because this comes from this definition of idempotent operator, then all idempotent operators, remember again this is a Hermitian operator, I think that is a trivial to show, right? If you take a conjugate of this, this becomes ket, this becomes bra, so again it is a equal to itself, so it is a Hermitian operator, Hermitian operator eigenvalues must be real, those values are 0 and 1, so you cannot have any complex plane because later on you have a problem, then you can argue that pk to the power 3 is pk, by the same argument you can say lambda to the power 3 equal to lambda and then you can get a complex numbers, alright? But that is not allowed simply because Hermitian operators are real eigenvalues, I am just completing the entire chain of argument because somebody can confuse you, alright? So first tell the person that is a Hermitian operator, so no complex eigenvalues allowed, okay? And if they are real there is no other solution, there is no other solution, okay? 0 and 1, alright? So this is something that we discussed, however if it is a stationary state then everything is very fine, you get only one eigenvalue and that is a trivial extension of this, that you get only one eigenvalue and everything is fine and that is what we are actually despite doing everything non-stationary such a great detail, I may remind you we are interested only in the stationary states and that too of Hamiltonian, okay? However for all other observables this is need not be a stationary state, hence what we do is what we said last time that we calculate other states by after we get a stationary state simply take an average value, alright? If it is non-stationary state, sorry if it is non-stationary state this is always the best description to take an average value, so if I have a wave function which is good for energy it may not be good for dipole moment, so then how I will calculate dipole moment I will take an average value, so just I thought I would tell you that this is how we go, obviously it will be very nice if a state is an eigenstate of Hamiltonian as well as eigenstates of other operators, that would be very nice because then everything is very beautiful and that is what symmetry achieves, so if you remember in group theory I do not know how many of you have done group theory course yet, has any of you done, okay? So some of you have done, some of you will have an opportunity to do next semester, essentially group theory achieves this to find out operators which commute to the Hamiltonian because I already told you yesterday they have two operators commute, they have simultaneous eigenfunctions, so if the operators commute with each other then Hamiltonian eigenfunctions will also be the eigenfunctions of them but that is a very nice thing, it is not always true, when it is not true that is how you have to calculate, alright? So with that as preamble for the postulates let me write down the postulates for the measurement, let me write down the final postulate which actually relates to the time, so this is postulate 6, I said postulate 5 as the average value postulate, so the time dependence Schrodinger equation which says that if H psi equal to ih cross del psi identity which means we can also calculate the time dependence of any function, remember this is a postulate because this is for any arbitrary function psi, stationary or non-stationary, it does not matter that is the reason it is postulate, if it is stationary then of course your H psi, so if it is stationary again when I mean by stationary I mean energy stationary of H, eigenfunction of H that is a conventional thing then your E psi becomes ih cross del psi identity because your H psi now becomes a number times psi then this differential operator can be very easily solved and you can, you have psi as an exponential minus IET over H cross into the space part, so this space or spin part whatever, space spin whatever, so this is the time part which simply becomes a phase factor, so that can be easily taken out and all observables remain time independent, all observables remain time independent in the stationary state of Hamiltonian because even if I take an average value the phase factor will always cancel out, so there is nothing will remain time dependent as a time independent, so real time dependence will come actually when this psi is not an eigenfunction of H then the non-trivial time dependence will start to come otherwise it is a trivial time dependent, in fact that is also a very important significance of a non-stationary state that the time really starts to come only when the state is a non-stationary state or not an eigenfunction of the Hamiltonian. In this context we also showed a very important theorem called the Ehrenfest theorem which actually can be derived from this, I do not know if you remember the Ehrenfest theorem, who remembers, okay good good, so derivative of an average of any operator, so let me at least state the Ehrenfest theorem which can be actually proved from the time dependent Schrodinger equation, so if I have an average value of an operator in any state, remember in any state that is very important, not necessarily in stationary state, the Ehrenfest theorem is much more powerful then what happens now tell me, you are asking did it time, so one is of course derivative of del A del T, so whenever please note whenever I am using this symbol it means there is a psi on the left, there is a psi on the right, so that is a standard symbol that is used, so any operator A anything A means it is psi A psi, okay and I am again assuming this psi psi of course psi is normalize you do not need to write, but otherwise you have to write, so now I am doing this total derivative of this average value, so that is what we are doing, so it is del A del T average, so del A del T average then plus yes commutator of what, A with H, A with H or H with A, okay, yeah A with H good, okay, so I again do not write psi psi, then anything else is there, there is a factor here, 1 by H cross, okay, please see this, make sure that this is right, there is only one error which could be there that either if it is HA or AH, there is only one error that is possible, the rest everything is okay and depends on how you write your IH, that you can also kind of manage by write minus IH cross, IH cross, so I by H cross, so it depends on how you want to write, anybody thinks this is wrong, it is very easy to check, take the derivative del psi, right, so you want to do del psi del T, it is 1 by IH cross H, psi, psi H, so why do not you do that if you have doubt, so let us not worry about this term, let us worry about the second and the third term, let us just go through the simple calculation, so first is del psi del T, right, H psi plus psi H del psi del T, correct, there are two terms which will survive, so what is this conjugate of this, remember conjugate of this, so del psi del T is 1 by H cross H psi, so del psi del T, so let me write the second term first, H 1 by H cross, we are writing the operator A, so A, so now it will be H psi, right, so del psi del T is nothing but 1 by IH cross H psi, so that is easy to write, then you write this last term, the first term, del psi del T, so what will be, so if you write del psi del T, you have to be very careful here, 1 by IH cross H psi, correct, so now what is the conjugate of this, you will have psi H but minus 1 by IH cross, right, so you will have psi H A psi but you will have a minus 1 by IH cross and that is the only way the commutator will come, remember, this is only way the commutator will come, otherwise of course it will become plus, anti commutator, so if you now put the commutator, it is A IH minus A IH, H IH, so what we wrote is correct, okay, but that just want to make sure that you understand, alright, so it is very easy to do this algebra, so this is actually the Arranfest theorem which all incidentally tells a lot of things because this Arranfest theorem is not just for stationary states, it is for any state because I have used only this equation, which is the postulate of quantum mechanics, I have not used this equation, however if this psi happens to be a stationary state then it is very easy to show that this is 0 and if A is explicitly time independent then all operators are time independent, of course that is understood that the operator itself is explicitly time independent, so this is all operators here will become 0 which I showed in a very different way by saying that the phase factor will cancel out, but from the Arranfest now you can very easily see because this expectation value will become 0 for any psi, remember which is an eigenfunction of either H or A, again that was the question that I gave in 4 to 5, if you have psi which is either a eigenfunction of A H or an eigenfunction of A it will be 0, probably in the quiz I had given as a multiple choice I do not remember where I gave, but so it will become 0, correct because it is very easy to show this because HA, H psi will become a number time psi so it will become A psi and then when I do minus HA that will cancel, so it is very easy to show this, please get yourself convinced, so that automatically comes, but there is something else that comes that even for any other state if H commutes with A again it is 0, correct, so both of them that they are part of the group theory that I said also actually follows from here, so let us assume that A is explicitly time independent and that is what we will of course if A is explicitly time dependent then this part will always remain, so that is the part that we are not bothered, so let us assume that A is explicitly time independent, then there are two important cases if psi is an eigenfunction of H or A then d dt of A is equal to 0, let us write a hat I think just to be sure, d dt of A is 0 because explicitly it is time independent to in the case B if A commutator H is equal to 0 then for any arbitrary psi even if it is non-stationary any arbitrary psi d dt of A equal to 0, no that is of course all operators that we are discussing Hermitian operator, these are all measure I think I have said that in the beginning that all physically observables are Hermitian operator everything is Hermitian, sure, so if this operator commutes with H then for any arbitrary psi not just for the stationary psi d dt of A is 0, so there are many ways the total average value of the total expectation and the time derivative of the average value of A would be 0 is either the psi is an eigenfunction already of H, of course this is more relevant but I must also say that even if it is eigenfunction of the operator it will be 0 or if the operator itself commutes with the Hamiltonian assuming of course that there is no explicit time dependence otherwise only this part will survive that anyway you cannot help, if the operator is explicitly time dependent then anyway this something has to be there but there is no contribution from the time dependence of psi that is the important thing. So this is a consequence of Ehrenfest theorem and the part of A where we said psi is an eigenfunction of H could be realized directly without doing anything it could be anywhere realized but the fact that psi is an eigenfunction of the A also would give you this cannot be realized so easily unless you do the Ehrenfest, it's interesting because you should be able to do this from here because Ehrenfest is nothing but derived from here, so think of it, Ehrenfest is also derived from here but when I write the Ehrenfest certain things become much more obvious and of course the fact that A commutes with H that is basically the symmetry operators so all those operators which commute with H are basically called the symmetry operators and that is what is the subject of the group theory in chemistry to find out all those operators. Again we have done this, done the Ehrenfest, we have also done another theorem which is a consequence of Ehrenfest which is actually called the Virial theorem, I will actually enumerate the Virial theorem for specific example with Virial theorem cannot be written I mean theorem can be written generally for a particular potential, for a general potential but I think we are more interested in particular potentials and we will write the Virial at that time but Virial is a very important part of the wave function and in fact it often tests the goodness of the wave function, the quality of the wave function is also tested by the Virial theorem, this is okay, again I have summarized many of the basic concepts note that after we did this we actually went to the solution of the stationary states without bothering about time because time is something that we are not interested for exactly solvable problem.