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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;