 Okay, so now we are on week 5 and we will be discussing tutorial 5, okay. So let us do a quick recapitulation of what was done in week 5. We discussed about the commutator operators. Going from the classical picture where we have portion bracket from portion bracket how one talks about when you go to the quantum mechanical analogy, you obtain the commutator operators. So these commutator operators when you talk about them, if the two operators say A and B commute that is A and B commutator is 0, then one can do the simultaneous measurement of these two operators. And if A and B do not commute, then the simultaneous measurement of A and B is not possible. That is what resulted into what you know now as uncertainty principle. And then once you talk about the commutators, you talk about compact and non-compact observables. And this tutorial will cover all these topics which will include commutators that is compact operators, non-compact operators, determination of eigenvalues and eigenvectors. Is it possible to do a simultaneous measurement when two operators in the form of matrix are given to you? Then we will talk about the time evolution operator and then how the eigenstate is evolved in time. So when you go from time t is equal to 0 and when you find the eigenstate or eigenvector in the evolved state using the unitary operator. So this choice of the basis we will talk about that, then we will talk about from the commutator relation how one obtains various observables or various terms like for example in problem number 2 you can see or in problem number 4 and 5. So these 3 problems out of 5 you will be using commutator relation obtaining certain observables or for example in the fourth problem you can just see here in the fourth problem you have to obtain the mean momentum of the state using the commutator operator. So I hope you will enjoy this tutorial. So let us just start with the first problem which talks about a two state quantum system as the basis state plus ket and minus ket and these basis state are not the eigenstate of the Hamiltonian of the system. So at time t is equal to 0 the system is either in state so to start with it is in plus ket state. Then as it evolves in time the state is going from t is equal to 0 to t to t etc then the system is in plus state and then when you are in time t by 2 sorry t by 2, 3t by 2 etc you are in minus ket state. So you have to construct a Hamiltonian which will describe the above physics plus and minus ket are not the eigenstate the Hamiltonian this is what is given to you because the system is either in state plus or minus at different time. So when it is evolving in time so what is given to you is that at in this state plus ket for time t is equal to 0 t to t that is to say that nt that time the system is in plus ket state and the system is in minus ket state for t by 2, 3t by 2 that is you can write this as 2n plus 1t by 2 so the system is in minus ket state. So now there are two ways to go about this what I can do is I can start with the basis which will be the eigenstate of the Hamiltonian and use a special case that is I can use a special case where in I know the eigenstate of the Hamiltonian and using that I will obtain the Hamiltonian. So first thing would be we choose a basis state or the eigenstate choose the eigenstate of Hamiltonian that is to say that I would choose these states 0 and 1. Now these are so called you can say that as conjugate basis state. So let me rewrite this conjugate basis state. So that is to say that we already know that these definitions we already know right 1, 0 and this is nothing but this state. So you can define these in this form. So what you have is you have these eigenstate which you can actually write in terms of these plus and minus kets. So how will you do that? In this one more assumption we make that the eigenvalues thus in this case what you have is the eigenvalues will be 0 and E. So one of the eigenvalue I choose it to be 0 and that would not affect because the global phase will not change. So the choice I make here is 0 and E that is the eigenvalue one of the eigenvalues 0 and the other is E. So the Hamiltonian will look like so my Hamiltonian will take the form. So Hamiltonian will take the form this will be the form of the Hamiltonian. So we know this okay. Now the thing is the question was that we have to obtain a relation between these time evolution I mean this time with respect to the energy. So how will you do that? First step would be I would again just a reminder that this plus and minus ket are not the eigenstates of the Hamiltonian but it is possible to have a linear combination of plus and minus ket such that they are the eigenstate of the Hamiltonian. So what I will do is I will define plus state. So this is how I define plus state it is a linear combination I will write again it is a linear combination of plus and minus state and ket 1 is a linear combination. So this is the two possible combination I can have. So one has to invert it. So when I invert this would give me the definition of plus and minus ket in terms of the defined what you call the defined 0 and 1 state this is what we have right. So this would be I will write this as after inversion I will have 0 plus 1 this you can do it easily. So I have ket plus and ket minus in terms of the kets 1 and 0 and 1 I need to change this okay. So the next step would be now I have these kets in terms of plus and minus eigen ket. Next I would do is so this was step 1 next step what I will do is I will have a so since you can see this is a evolution in time. So now the unitary time evolution operator take me from time t is equal to 0 to time t is equal to some time t. So here we know that initially to start with at time t is equal to 0 the particle is in or the system is in plus ket state. So you have to go from so let me write a general expression first. General expression for time evolution operator is nothing but if you have n such eigen states or eigen basis then you can write a general expression as this is nothing but a complete set of state. So let me write here this is nothing but complete set of eigen state. So in this case we have 0 and 1 as the ket. So this would get modified as so basically when you write t, 0 it will mean that I am going from t is equal to 0 to some time t. So that will be nothing but so here we actually do not need this summation sign because I am writing term by term. So this would result into one of first eigen state is 0 I have eigen value is 0 which I have chosen and the second eigen value let me write it values. So second eigen value is E. So that would result into state 0 plus I will have e raise to i minus i t upon h cross 1 1. So this is my time evolution operator in terms of my complete set of state that is ket 0 and ket 1. Now using this time evolution operator I can evolve the wave function or the eigen state, state vector from time t is equal to 0 to another time t and this we need to do to find out the Hamiltonian. So the next step would be next step would be I will write the state as the unitary operator that is t, 0 which I had defined in terms of so at time t is equal to 0 you will evolve this from time t is equal to 0 to t. So this will be the time evolution operator or the time evolution equation for the eigen vector. So at time t is equal to 0 the system is in plus state. So let me rewrite this as t, 0 this is plus state again now we have this this is nothing but you had written let me go back. So here we had 0 and 1 as the ket 0 and 1 as the ket. So coming back here let me write this as psi of t now will be I will write in terms of the plus state again the plus state or the plus ket can be written in terms of 0s and 1s. So this in turn this would be so this is what I have now the wave function or the eigen state Schrodinger time dependent equation. So let us go to the Schrodinger time dependent equation. Let me write here Schrodinger time independent equation using Schrodinger time dependent equation what we do is we obtain. So this is the special case we have considered for obtaining the relation between the time and the energy eigen state. We will also discuss a general case after this. So this will be dou by dou t. So this is the e part and that is nothing but h psi of t that is. So this will be a time evolution of the eigen state and using this we obtain the Hamiltonian. So now we have to calculate the left hand part this can be easily done and you can do it as diy one step. I am leaving this space for you you can just obtain the Hamiltonian from here which will be of this which will be of this form if and only if your energy e is holding this relation to pi h cross upon t. So let me quickly go back to the question once again. So what one need is to construct a one parameter Hamiltonian which produces this above physics that is system is in plus minus state plus state and then minus state and then plus state at a regular periodic time of t by 2 and what we have done is we have constructed that Hamiltonian which depends on t. So coming back here this is what we obtain. Now I also want to discuss a general scenario. So apart from this what we have done I will discuss a general scenario in which so let me write this here as a general case. So as I already said you all that this plus and minus state are not the eigen state but it is a possibility that a linear combination would give us the eigen state. So let me just write general expression general expression eigen vector what I do is I start by writing a general eigen vector which is nothing but the general form that is psi of some t. So previously we tried to evolve this eigen vector from psi is equal to 0 I mean psi at t is equal to 0 to psi of t. Now here what I am doing is I am starting with a general expression for the eigen vector. So the eigen vector of the system the system which is the system the system we are talking about which is periodically is either in state plus ket or in state minus ket. So let me write a general expression. So let me define some alpha of t plus beta of t minus. So this general case I am starting with this general case wherein the wave function or the eigen state vector is either in plus state or minus state. So how do we understand this? So actually my alpha of t is nothing but cos pi t by t. So you can easily see this because when I have alpha of t as cos pi t by t that means I am for the case if I say psi of 0 what will sorry this has to be. So psi of 0 will be alpha of t that is I am talking about this cos pi t the system will be in the plus state with some constant. There will be some constant. Similarly I have this beta of t. I am defining this beta of t as sin. So this will satisfy this relation. So I will check for this. This will be again a constant times I get the eigen. So this C constant C which I am writing here can be plus or minus 1. How? So let us just see an example wherein when t is equal to 0 for the first case when t is equal to 0 or pi also t to t etc. For these values of t you will see that alpha of t becomes plus or minus 1. So always the system is in plus ket state and beta of t becomes 0 and when I take the second condition I mean the second step. So when t is equal to t by 2 3 t by 2 etc. So this goes up to n t and this goes up to 2n plus 1 t by 2 which we have already seen in the previous general case when we talked about the general case. So I am sticking to the same color and then here what you will see is alpha of t is always 0 for this condition and beta of t takes the value plus or minus 1. So from there you can see that C will always take value plus or minus 1. So now what I do is let me go to the next step. Now it is easier, now you just have to use the Schrodinger time dependent equation and obtain the result. So the wave function or the eigenvector which we have defined let us write in the matrix form. So when I am writing it in the matrix form how will I write it? I will write this as psi of t will be I am trying to write in the matrix form. So the plus plus component will be alpha of t and this will be beta of t. So this is the eigenvector or the eigen basis which we have used. So this will be nothing but cos sign. So once you define the general eigen basis or eigen state the next step what you will do is use the Schrodinger time dependent equation time dependent. So that is nothing but ih cross dou psi of t that is equal to h or when I am writing it in this representation wherein I am talking about just let me write it in this form that is what I want to this state. So from this to be very quick I will just put this terms and you must verify this by yourself. This will be I will have a factor of ih cross pi by t. So I have a i coming from the exponent. So that would give me just h cross square upon t and you can just go back and just scan through. So this will be sin pi t upon t cos that will be h of cos pi t upon t sin. So I have sorry there is a ih cross coming from here this pi t when you take the derivative and the signs will change. So this would imply that my h is nothing but after you obtain I mean using this you just substitute for h as a 2 cross 2 vector and you obtain h as 0 minus ih cross pi by t and 0. So basically when you try to write in terms of E the energy will be the same as what we obtain in the previous expression 2 pi h cross upon t. So E what we obtain here will be similar to you can see the relation. So if you want to write h in terms of E so basically you will obtain the energy as 0 minus E E and 0. So we have obtained Hamiltonian in terms of a single parameter t. So you have I think already done some relations like x. So there are commutators like x, p. So famous commutators like x, p and then you have you can check whether a Hamiltonian operator commutes with the momentum operator, the position vector. So let me put a cap here. So x cap, p cap where x and p are the operators and operators are represented by a hat or a cap. So this is an operator representation. And this is called as a commutator brackets. 2 operators commute operator a and b commute that means that a and b when you put in the commutator bracket it is 0. You will see in coming lectures that why do we check whether 2 operators commute or not. So when 2 operators commute they form a simultaneous eigen ket. So the measurement of a to not affect the measurement of p that is what it would mean. So when you operate an operator a, b on a eigen vector this will be equal to, so interesting part of these 2 operator is that the measurement of a and b are independent. And hence you can form a simultaneous eigen ket for this vectors. And this will be useful for coming lectures. This is a slight background. And as we know that operator x and p gives ih cross. This is the relation we obtain. And I think you might have proved some time or if you have not proved this please try your hands on prove this. I will not discuss the proof in this tutorial but you must try out x, pn. You can also try this is 1. You can also try what do you obtain n operator p. So x raise to n on p. So x on pn this I will give you this result which you can use in the problem. But it would be I would suggest that you must prove this identity. So this we are going to use in this tutorial too. So in second question you are given 2 operators position operator x hat and the momentum operator p hat. And you have to write this operator evaluate the commutator of these 2 operator. So let us try evaluating the commutator bracket or the commutation relation of x and alpha e raise to alpha p. e raise to alpha p cap I can write this as 1 plus alpha p hat plus alpha square upon 2 factorial p hat square and so on. You can write the exponential series. Now when I do this would be commutation of the position operator with these terms of the exponents. This is a simple way to do plus dot dot dot. So the first part this is the first term. Second term would be this third term would be x hat p square plus all the terms. So remember the operator times a number will always give you 0. First term is 0 plus alpha i h cross. Now second term you will use this here it will be square this square n is 2. So you have 2 i h cross p. So this is 2 i h cross p plus alpha square 2 i h cross you have a p hat upon 2 factorial. So what will be the result of this alpha i plus you have the other terms alpha i h cross plus you have 2, 2 will get cancelled. You have alpha square h cross p hat I have not missed anything okay. I have a factor of i also. Plus n terms. So in general form can I write it as i. So this I can take i h cross alpha and I have 1 plus second term would be alpha p hat plus so on. So this is nothing but again you get an exponential series. So this would be x e raise to alpha p hat is nothing but i h cross e raise to alpha p hat. So this is what I get after performing the and I have missed the alpha which I put here. So this is the expression I obtain for commutator of x e raise to alpha p. Now if I replace alpha by my question is i a upon h cross. So let me replace alpha by i a upon h cross. e raise to i a p cap upon h cross will be this would be h cross this will be a i h cross e raise to i a p cap upon h cross that is nothing but minus a e raise to i a p cap upon h cross. So this is what I have obtained after finding the commutator of these 2 operators. Remember a is a constant and it is having the dimension of length dimensionally it is the dimension of length you can see from here it is very simple to see from here. Now the second part of the question is show that x is the eigenstate of this operator. The eigenstate of this operator I have a exponent e raise to i a p hat upon h cross I operate on x. So when I operate x hat on x I obtain x times x. So x is some eigenstate of a positive operator you have to evaluate that. So now if I operate this on e raise to i h cross so now we have the commutation relation. So if I write x hat as this operator I define as commutator of x i a operated on x minus or plus let me see this is x e raise to i then I have a minus so I will have a plus e raise to i a p hat upon h cross on x. So this is what I have this would give me this relation we have seen gives this commutator gives minus a e raise to i h cross and operating x on the vector will give me x. So I have x minus a e raise to minus e raise to i a p hat upon h cross. So the final step is e raise to i p hat upon h cross x will be x minus a e raise to i a p hat upon h cross x. This is what I had correct. So what do we observe this is some another vector x. So what do we obtain here is that e raise to i p upon h cross x is an eigenstate this operator let me call it as some capital X. So this capital X is an eigenstate of operator x with eigenvalue x minus a. So this operator is the eigenstate of this ket is an eigenstate of the operator small x with eigenvalue x minus a that implies that e raise to i p cap upon h cross is a unitary operator is a unitary operator that what does this unitary operator does it translates x to x minus a particle to minus a. So if this operator is operated on x it should give me same x and then get x capital X back but here it does what it does is it gives me x minus a that means it translates the particle to minus a. So this is a very good exercise wherein you understand the operator how the unitary operator comes into picture for this particular problem in the previous problem also you have seen a different flavor. So with this we stop here.