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;

let us go back to the harmonic oscillator and let me calculate i h cross d x by d t,

x is an operator. In the Schrodinger picture x operator is a constant. Please note x operator

is a constant, P operator is a constant. In the Heisenberg picture x operator becomes

a function of time okay. So, let me put x H, so this is equal to x H comma H.

So, let me substitute from the values of x, if you go back we had written this h cross

by 2 m omega a dagger plus a comma H operator is h cross omega a dagger a plus half this

is the x operator. All these operators they suppose be function of time, a dagger is a

function of time, now a is a function of time everything is seems to be function of time,

but let me substitute the x operator and the H operator here. So, if I expand this, I will

get h cross by 2 m omega comes out and h cross omega comes out h cross omega comes out what

will happen to this factor half. That will not contribute because, a dagger plus a commutator

with a dagger a, and a dagger plus a commutator with half. I can open the commutator into

two commutators a dagger plus a commutator with a dagger a and a dagger a commutate with

half and because, half is a number a dagger a commutate with half just disappears. So,

I will get a dagger plus a commutator with a dagger a and I will leave it you to show you can use

the commutation relations between a dagger and a and what you will get is essentially

sorry into i h cross okay. So, let me give on this what you get is actually

h cross omega under root h cross by 2 m omega into a minus a dagger. This is what you get

and this final leads to the fact that x H by d t is equal to P h. This is the same as

a classical equation of motion d x by d t is equal to P by m. Now it is an operator