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  • Today, I want to just continue with the review of quantum mechanics, introduce the Schrodinger

  • and the Heisenberg pictures before, we start quantize electromagnetic fields.

  • So, yesterday when we finished, we were looking at some problems for example; suppose I take

  • a harmonic oscillator in one of the eigenstates of the energy and ket. So, the energy of this

  • state is n plus half h cross omega. Can you calculate the expectation value of

  • x in this state. We have done this yesterday is it okay, so this is 0 expectation value

  • of x square, So, I have to calculate n x, x n. i substitute x cap in terms of a and

  • a dagger with this be 0, this will not be 0. Let me give you the value x square expectation

  • value is h cross by m omega into n plus half, expectation value of x is 0. Expectation value

  • of x square is so much similarly, expectation value of P is 0 and expectation value of P

  • square is m h cross omega into n plus half. So, as I mentioned yesterday, the energy eigenstates

  • have well-defined energy and the phase of the oscillator is completely undetermined

  • and that is why when you take an expectation value from an ensemble of systems, you will

  • get a value of 0. I will recall that we defined uncertainty

  • in the x as x square average minus x average square raise to the half. So, because, x average

  • is 0 this is simply h cross by m omega into n plus half raised power half. The uncertainty

  • in the position of the oscillator in the nth eigenstate is square root of h cross by m

  • omega n plus half raised power half. Similarly, I can calculate delta P, the uncertainty

  • in the momentum P square minus P square. That square root of this quantity because, expectation

  • value P is 0.

  • So, I defined the product of uncertainties delta x times delta P, I can calculate from

  • here and what I find is delta x times delta P is equal to h cross into n plus half. I

  • So, I substitute the values of delta x and delta p and I find the uncertainty product

  • in position momentum of the linear harmonic oscillator is h cross n plus half.

  • And the lowest uncertainty product appears for n is equal to 0 state and for this state

  • delta x delta P is equal to h cross by 2. All higher order higher excided states have

  • a larger value of uncertainty product. Now I need to understand what is the time evolution

  • of the system in one of the energy Eigen states for example; So, suppose I take a state which

  • happens to be in an energy eigenstate. So, let me take a state for example; H E is

  • equal to E E. So, for any system suppose I take ket E is the energy eigenstate, So H

  • E is equal to E times E, E is the energy Eigen energy eigenvalue. Remember the Schrodinger

  • equation is i h cross Del by Del, Del t of E of the state is equal to H times E. If I

  • want to understand how the system generated in Eigen ket E evolves with time this Eigen

  • ket must satisfy this equation.

  • For any state for any state psi will satisfy these equation, but if the state is an energy

  • eigenstate H times E is simply E times E and I can then integrate this equation and I get

  • E as function of time is a function of time equal to E at 0 into exponential minus i E

  • t by h cross. So, Energy eigenstate evolve with time according to this equation, there

  • is only a phase change. The phase of the oscillator change with time and the time dependence is

  • exponential minus i E t by h cross. Sir excuse me.

  • Yeah? Sir if e is not an eigenstate

  • Yes I have generated state of a system that is

  • supervision of different eigenstate. Yeah.

  • Can we equation into different parts corresponding to each eigenstate.

  • No, not this equation the equation is still this. So, I will have i h cross del by del

  • t of psi is equal to h psi .What I can do is psi at t is equal to 0, I can write as

  • sigma C n, E n where, E n's are the eigenkets. So, then I can substitute here and then integrate

  • but because, I know that each eigenstate energy eigenstate evolve with time as exponential

  • minus i E n t by h cross, so I can write this as is equal to sigma C n exponential minus

  • i, E n t by h cross into E n. Sir it is like essentially by taking this

  • equation the first equation you have written and saying that each eigenstate is an independent

  • variable in this and split. Yeah they are all independent.

  • Each eigenstates are all orthonormal to each other. Essentially, they do not mix among

  • themselves each one evolves independently and with a time dependence exponential minus

  • i, E n t by h cross okay. Now before I look at the time evolution of

  • superposition states under two different pictures. Let me introduce this two pictures one is

  • the Schrodinger picture and one is the Heisenberg picture.

  • As I mentioned to you yesterday, in the Schrodinger picture the Eigen the kets evolve with time

  • and the operators are independent of time. In the Heisenberg picture the kets are fixed

  • in time and the operators evolve with time. Let me look at let me start with an example;

  • I have the Schrodinger equation. So, this equation i h cross Del by Del t of psi is

  • equal to h psi this is a Schrodinger equation and this is an equation of how the ket evolves

  • with time and so this is a Schrodinger picture. In this picture the operators like position

  • operator and momentum operator they are all the independent of time. This is a Hamilton

  • in operator, so if I am in conservative systems where the total energy is conserved h is also

  • independent of time. So, we will only look at systems where the

  • Hamilton in his independent of time. So, this H is also independent of time, but psi evolves

  • with time. Now as I said what is the important is that, the expectation value of any operator

  • must be the same whether I look at in the Heisenberg picture or in the Schrodinger picture.

  • Now if this if H is independent of time, I can write a formal solution of this equation

  • like this psi as a function of time is equal to exponential minus i H t by h cross into

  • psi at t is equal to 0. If I have an exponential of an operator this is 1 plus A plus 1 by

  • 2 factorial A A plus 1 by 3 factorial A A A etcetera.

  • With single operator, there is no problem because, A and A commute but, if I had for

  • example; exponential a into exponential b, I have to be little careful, I will give you

  • a formula for that, but so that will be I defined this exponential of the operator.

  • You can verify that this is a solution by differentiating both sides and substituting

  • into this equation.

  • If you differentiate both sides for example; if I calculate i h cross del psi by del t

  • so del psi by del t this will be equal to i h cross del by del t of exponential minus

  • i H t by h cross psi at t is equal to 0. And if I had written this exponential operator

  • as a series in differentiate, I will get essentially minus i by h cross H exponential minus i H

  • t by h cross psi at t is equal to 0. You can expand this exponential in terms of

  • a series differentiate and you will find that differentiate at one which essentially gives

  • you the same as we have differentiating just like normal quality. So, this is equal to

  • i into minus i is 1 h cross cancels off and I get H into this remaining is nothing but,

  • psi of t. So, this solution which I wrote this formal

  • solution I wrote is the solution of the Schrodinger equation and this equation represents the

  • way the state evolves with time exponential minus i H t by h cross that is the Schrodinger

  • picture. Now for example; in this picture if I able to calculate the expectation value

  • of an operator a as a function of time, I will have this A. A is independent of time,

  • but this expectation value can change with time because, ket psi changes with time. Now

  • I want to another picture where I want to take away the time dependence from the ket

  • into the operator.

  • So, let me define the ket corresponding to the Heisenberg picture, I just put a subscript

  • h here as psi at t is equal to 0 that has not changed with time anymore and this equation

  • I can invert this formally and write this as exponential i H t by h cross into psi of

  • t. I have just taken this I have multiplied by an operator exponential i H t by h cross

  • on both sides and then because, the operator H is the same this becomes is equal to 1 identity

  • and I get this equation just gets reversed into psi of H, if psi of t is equal to 0 is

  • equal to this thing and by definition this is independent of time we need now this product

  • is a independent of time. This operator operating on this will always be at of psi at t is equal

  • to 0 okay.

  • Now the expectation value of an operator is remember psi s, so no subscript means a Schrodinger

  • picture into A into psi of t. So, now I have this equation for psi of t, which I use in

  • this equation. So, what is what is bra psi of t this is ket psi of t, exponential plus

  • i h t by h cross. Please remember H is a Hermitian operator.

  • So, this will be equal to psi at t is equal to 0, exponential i H t by h cross A, exponential

  • i H t by h cross psi at t is equal to 0. I have just substituted fresh and for the evolution

  • of the ket with time into this equation, I have just substituted for psi of t from this

  • equation essentially I am taking it back here and replacing in terms of psi at t is equal

  • to 0 minus here. No. There is a minus here.

  • So, this this is psi of t is equal to 0 is a function of psi of t. So, I am replacing

  • psi of t as a function psi of t is equal to 0 right. So, there is a minus sign here there

  • is a plus sign here. So, I define this as the operator in the Heisenberg

  • picture and this is nothing, but psi of H. So, the operator in the Heisenberg picture

  • is equal to exponential i H t by h cross operator in the Schrodinger picture minus i H t by

  • h cross. This is independent of time. What is a function

  • of time. And please note that H may not commute with A, so I cannot interchange A and exponential

  • i H t by h cross. If A commutes with x H then for example; what will be the Hamiltonian

  • in the Heisenberg picture. This will be H and H commutes with exponential i H t by h

  • cross, so this is simply be H, so this independent of time. In the Heisenberg picture the Hamiltonian

  • in the Heisenberg picture and the Hamiltonian in the Schrodinger picture are the same because,

  • if I replace A by H and if I expand the exponential, this H commutes with all the H's anywhere.

  • So, I can actually interchange this H and exponential and I get unity that means the

  • Hamiltonian is the same whether, you are looking at the Schrodinger picture or the Heisenberg

  • picture. Sir why does Hamiltonian commute with the

  • exponential terms. Because this is also H only this is an H operator

  • there is an H operator here, H operator always commutes with the H operator. So, I can if

  • you expand this exponential you will have H and that H and all these H will commute

  • anywhere. So, I can take this H out and reform back and the exponential then I will actually

  • that means I can interchange these two because, this and all this function commute with each

  • other okay. Now, I need to calculate what is the time

  • evolution, I need I can calculate an equation, disturbing the time of evolution of this operator

  • in the Heisenberg picture so for this I differentiate this equation.

  • So, I differentiate this with respect to time, and let me so let me calculate d a h by d

  • t. Let me assume in our analysis here that A has no time dependence in the Schrodinger

  • picture they could also be operators which are depending on time in the Schrodinger picture

  • itself, this is called an explicit dependence on time, but we will now look at that. Let

  • me assume in the Schrodinger picture, this operator i is independent of time. For example;

  • position operator, momentum operator they will all be independent of time, so when I

  • differentiate this I do not have to differentiate A with respect to time okay.

  • So, when I differentiate this equation what do I get. I differentiate first exponential,

  • so I get i H by h cross into exponential i H t by h cross A exponential minus i H t by

  • h cross plus exponential i H t by h cross A H okay. See when I differentiate the second

  • exponential, I will have a i H minus i H by h cross because, that commutes with this.

  • I can draw the exponential first and then the factor which comes out a differentiation

  • afterwards. So, I Just write the differential of this comes and the differential of this

  • comes and what is this, this is nothing but, A H of t this is also A H of t, so this is

  • nothing but, i by h cross A H minus A H. So, if I take the i h cross on the other side,

  • I get i h cross d a H by d t is equal to commutator A h H this, is called the Heisenberg equation

  • of motion. So, there is no i h cross Del psi by Del t is equal to H psi in the Heisenberg

  • picture because, psi is independent of time. I would have to solve this equation to get

  • how the operators vary with time. From the operator variation with time, I can always

  • calculate the expectation value because, psi does not change with time at all.

  • So, in the Schrodinger picture, I calculate how psi varies with time to calculate the

  • expectation value. In the Heisenberg picture, I calculate how the operators vary with time

  • to calculate expectation values and other quantities.

  • And actually in this particular picture is closed to classical mechanics for example;