Good morning. We have been discussing, the behaviour of microscopic particles like electrons.

And we considered a situation where we had an electron beam incident on 2 slits and we

were looking at the pattern of the arrival of the electrons at a distance screen. And

what we found was that the arrival pattern of the electrons on the screen, look like

an interference patterns. And based on this to in order to explain this it was necessary

to invoke a wave associative wave with every electron. And what happened was when this

wave encountered the 2 slits each slit now act like a secondary source. And if you wish

to calculate the resultant wave at any point on the screen and it was the superposition

of 2 contributions. So, what we found was that, it was not possible

to say that the electron arrives at the screen through either slit 1 or 2 slit 2. And you

have to admit the possibility that the electron basically goes through both the slits at the

same time, because the wave corresponding to the electron, passes through both the slits.

So, the bottom line of all these discussion was that you have to give up the picture where

you can think of these microscopic particles. As particles which we are familiar with particles

in the familiar world can be thought of when they move from one position to another. You

can associate a trajectory with them. You write down the Newton’s equations of motion.

And you can solve them and it finally, at the end of the day you have the particles

position at every instant of time. As it travels from one point to another, but we found that

for microscopic particles like electrons you have to think of it in terms of a wave propagation.

And then as a consequence you have all the phenomena like interference diffraction associated

with waves also occurring for such particles. I also told you that this wave which you associate

with the particle you interpret as being the probability amplitude the modulus square of

this wave. So, this wave are the wave function.

Psi the mod square of this gives the probability density of finding the particle somewhere

this is, what I told you in the last class. So, if you ask the question what is the probability

of finding the particle, along the interval dx in the interval dx cantered at the point

x on the screen this will be given by. The probability density into the interval dx and

the probability density can be calculated from the wave psi by taking its modulus and

then squaring it. So, this psi is the probability amplitude.

And this is the probability density the modulus square of it gives a probability density,

whose interpretation we have discussed in the last class. Now, in today is lecture I

am going to tell you the basic rules of quantum mechanics. The previous lectures have all

been motivation and interpretation of the wave function and motivation why we need to

think of particles as having as in terms of waves. So, today we are going to go into the

mechanics of these waves. So, this is what is called quantum mechanics.

Or wave mechanics

and the first postulate of quantum mechanics as I am going to present them to you states

that for every possible state of a particle corresponding to every possible state of a

particle. There is a different wave functions psi associated with it. So, an electron for

example, could be in one of different possible states it could be many possible states and

at any time it could be in 1 of them. And or there could be many possible states in

which you may find the electron. So, for corresponding to each of these states there is a different

wave function psi. So, corresponding to every possible state of the of a microscopic particle

there is wave function psi. And this wave function is governed by a wave equation know

as the Schrodinger wave equation. So, let me also write down the Schrodinger wave equation.

So, the Schrodinger wave equation is the equation is the wave equation governing the evolution

of this psi. So, let me remind you once more the first postulate is that corresponding

to every state of your system. Say you are dealing with electron corresponding to every

possible state of the electron there is a different wave function psi.

And the evolution of these wave functions is governed by the Schrodinger wave equation,

and let me write down the Schrodinger wave equation. So, for a particle moving in which

can move in all 3 dimensions. The wave equation is i h cross del by del t psi now psi could

be a function of r and t this is equal to minus h cross square by 2 m, where m is the

mass of the particle. So, if it is an electron it is the mass of the electron Laplacian minus

h Cross Square by 2 m Laplacian of psi plus v which again could be a function of r and

t into sigma. So, this is the Schrodinger equation. So, this term over here is the partial

derivative with respect to time. This term has spatial derivatives the Laplacian which

we have already encountered earlier has spatial derivatives in it.

Now, when you have a particle under the influence of an external force which is the situation

quite often for example, if you have an electron the electron could be an external potential.

Under the influence of an external electric field static electric field could be represented

in terms of a potential. So, if I have a uniform electric field I am sure we all know that

there is a corresponding electrostatic potential. So, any motion any force. So, when we do classical

mechanics we the force forces can be represented as gradients of potential under some conditions.

So, this external force action of the external force on the particle comes in here through

the potential v. So, if I have a particle in an external potential.

So, then I have to I have I have to also include this term this term is the potential the gradient

of the potential minus the gradient of the potential gives me the force acting on the

particle. So, when I thinking of this say a macroscopic particle and I think of it in

the usual classical picture. I would write down the equation of motion for a particle

moving in an external potential. And for a macroscopic particle I would write down the

equation let me just remind you what we are talking about.

So, for a macroscopic particle say a cricket ball or something like that. So, if I had

a microscopic particle under the influence of an external force. Then I would write down

an equation which could be that the force if the force could be represented in terms

of a potential if it for a potential force. Then usually the force is the minus the gradient

of the potential, for potential forces that is how you introduce the potential.

So, then I would write down the equation of motion for the macroscopic particle in terms

of the potential m into the acceleration is the force which is minus gradient of the potential.

And I would solve this equation of motion which would give me the trajectory of the

particle whereby I would know the position of the particle at every instant of time for

this under the influence of this external force. Now, if you are dealing with microscopic

particles we have to abundant this picture where the particle moves in a trajectory.

And we have to describe it in terms of a wave.

And the evolution of the wave is going to to be governed under the in the presence of

the same potential. The evolution of the wave is now going to be governed by this wave equation

which tells me how the wave is going to evolve under the influence of that external force.

Remember the force is the minus the gradient of this potential. So, I have to now replace

the trajectory I cannot I can no longer thing of the particle in terms of a trajectory.

I have to think of it in terms of a wave the evolution of the wave. And the mod square

of this wave gives the probability amplitude. If I can calculate this wave at different

positions the modulus square of that gives me the probability of amplitude this is some

which we have already discussed.

The modulus square of the wave function tells me the probability amplitude. So, once I know

the wave function I can tell what is the probability of finding the particles?

Somewhere, I cannot tell exactly where the particle is like I can do for macroscopic

particles, for which I can calculate trajectory this has to be abundant.

When we are dealing with microscopic particles which we have to think in terms of waves and

the evolution of the wave is governed by this wave equation. So, this is a time derivative

term which occur in the wave equation this has spatial derivatives and this is the term

that arises due to any external influences which we can represent through a potential.

So, this is the Schrodinger wave equation.

Now, we are going to restrict in this course, we are going to restrict our attention to

particles which are free to move only in 1 dimension

along the x axis. So, this simplifies the discussion that is all, but from this simplified

discussion we can get a clear picture of what would happen if we were to deal with particle

in 3 dimensions. It simplifies the mathematic analysis that is why we are restricting our

attention to particles which can move only in 1 1 direction as are on the x axis under

this simplification. The Laplacian now is just partial derivatives with respect to x

that is the only change which occurs psi is a function of x alone and the potential v

is also a function of just x. So, the equation now becomes i h cross del by del t psi which

is now function of x and t is equal to minus h cross by 2 m partial derivative with respect

to x second partial derivative. Second derivative with respect to x plus v, which could now

be a function of x and t into psi. We will also make another simplifying assumption we

will assume that v.

V is time independent. So, v is a function of x alone that is it is static it is a static

static potential. So, for example, if I have an a charge particle in an external electric

field which is static for example, the uniform electric field we know the potential is proportional

to the distance x. We have a static potential, but suppose I have a charge particle let us

say an electron bound in an atom and we shine light on this atom. Now, we have discuss that

light is essentially an oscillating electric and magnetic field. So, we have a time dependent

external force the time dependent external electric field. So, this a situation that

is not static the analysis of such situation is more complicated. We are not going to deal

with such situations in this course, we are going to deal with situations where the potential

is static it is just a function of x and it is not dependent on time. So, under such conditions

we will now proceed to look for solution under such conditions.

And we are looking solutions to this equations where x is v is now just a function of x not

of time.

So, we going to use the method of separation of variables which is familiar. Remember we

have introduce this method when we were discussing standing waves. And what you do is we have

to we can we will take a trail solution of the form where psi x t is a function of is

product of 2 functions capital X which is a function of the position alone and capital

T which is a function of time alone. So, we will take a trial solution like this this

recollect that this is the method of separation of variables. So, we are going to put a trial

solution like this into our wave equation.

Now the partial derivative of this psi with a partial derivative with respect to time

is going to act only on this function and it is going to be now a total derivative.

So, the equation the Schrodinger equation in one dimension now reads as follows. So,

I have to replace this with x into t and the x will go out I will have the time the function

of time left over here.

So, what I have is I h cross x d by dt of capital T that is the first term again I replace

the this psi as a product of capital x and capital T. Capital T will now go outside because

this is the derivative with respect to x. And what it gives me is…

This equal to minus h cross square by 2 m capital T now comes out d square dx square

of capital X and I will do the same replacement is here where is going to be no change.

V which is just a function of x now into capital X into capital T. So, this is the Schrodinger

equation with the separation of variables put in. Now, what we do if you remember we

have to divide this equation complete equation by psi which is X into T. So, when I divide

this equation with X into T. If I divide this term with X into T X cancels out and what

I have is i h cross I am dividing it by X and T. So, 1 by T remains the d by dt of capital

T if i divide this by X into T capital X into capital T what I have is minus h cross square

by 2 m. I am dividing by capital X and capital T. So, capital T will cancel out I will have

a capital X left. This is and this term will give me this v which is a function of x alone

I am dividing by capital X and capital T. Now, notice that the left hand side is a function

of time alone the right hand side is a function of x alone. I am free to change time without

changing x I am free to change x without changing a time the fact that this equality must hold.

Basically tell us that these 2 terms must be equal to a constant and I am writing that

constant as E. So, we have used the method of separation of variables and it tells us

that these terms should separately be equal to a constant we have to now solve these equations.

Let us first look at the time part of this equation. So, the time part means that we

have to look at this term which is equal to a constant E. What this gives us [Vocalized

Noise] is this differential equation.

i h cross this equal to T times E where E is a constant this capital T is a function

of time. Now, the solution to this is easy to write down it the solution is T as a function

of time. Capital T is the function time is equal to there could be a constant which I

am not writing down and overall constant into e to the power minus i E t by h cross. And

you can easily check that if you differentiate this once with respect to time you will pull

out a factor of minus i E by h cross minus i into plus i will give me 1. You have a one

by h cross will cancel out this h cross here and i will get E. So, what this if i differentiate

this i will get exactly the right and side. And there could be a overall constant which

I have not written down this is the time part of the solution. Now, let me write down the

the the spatial part of the solution. So, when dealing with the spatial part of the

solution it is convenient to introduce a constant another constant which is define like this.

So, we are going to introduce another constant P square which is 2 m into E. So, we are going

into introduce another constant P square which is 2 m into E. We are now looking at the spatial

part ho before that. we are now going to focus our attention on a particular situation where

we have a free particle.

Let me first make this point. So, the first situation that we are going to consider is

where we are dealing with the free particle. So, for a free particle the potential v…

Let me put this elsewhere this need not come here. So, we are going to deal with the free

particle and see what kind of solution. This Schrodinger equation gives us remember a free

particle refers to a particle on which there is no external force. So, this kind of a particle

has no potential is as no potential acting on it. So, this potential v is going to be

set to 0 if i set the potential v to zero the spatial part of this equation now becomes.

D square capital X dx square is equal to minus 2 m E by h cross square into X. Now, it is

convenient to introduce a new variable let me put it here. So, it is convenient to define

a new variable. So, define this numerator to m E as a new variable P square equal to

2 m into E. So, with this new in terms of this new variable P the equation governing

capital X. Now, becomes d square X dx square is equal to minus P square by h cross square

into X, and this has to 2 solutions basically this we know that this is the simple harmonic

oscillator equation. And the solution to this is quite straight forward to write down.

And the solution is X as a function of x is a constant again I could have a constant which

I am not writing explicitly into e to the power plus minus i into P into x by h cross.

If I differentiate this twice I will pick up minus P square by h cross square which

is the equation that I have over here.

So, it is very easy to check that this is just the simple harmonic oscillator equation

and it is quite straight forward.

To realize that this is the solution there could be either plus here plus P or minus

P because we have to take the square root when we are dealing with that equation. So,

this could plus or minus we can choose 1 of them and go ahead.

So, combining this with the temporal solution over here what we find is that the resultant

is Psi x t can be written as some constant. I am now putting in the constant remember

both the this and the temporal part had constant in front and I can take the product of those

2 and get a another constant E to the power minus i by h cross. So, I am writing the temporal

part first.

The part is minus i E by h cross into t.

So, which I have written as minus i by h cross E into t minus P into x which is the spatial

part. So, the spatial part has a solution which is given over here plus minus i P x

by h cross I have taken only the plus the solution with the plus sign which is the 1

I have taken here. You could also have a solution with the minus sign which will introduce a

plus sign over here. So, bear in mind that there are 2 possibilities I have just taken

1 of them over here. So, this is solution of the of of the Schrodinger equation for

a free particle where there could be an overall constant which is still on determined. There

is also a an arbitrary constant E which appears which I have not told what it signifies and

there is another constant P which is related to E which is not independent, but related

to E as follows.

P square is 2 m E.

So, there is basically 1 constant either you can think of it has either P or E which I

have to tell and E. If I tell you P then E can be E is fixed over if I fix then P is

fixed and there is an overall normalization constant over here. And the E and P are related

as follows. E is equal to P square by 2 m. So, this is a solution of the Schrodinger

wave equation and solution has an overall constant over here. And it has 2 constants

over here which i are not independent which are related like this. So, I can fix any 1

of them and I can get the other constant if I fix P I will E from that and that is it.

So, there are this is a solution of the Schrodinger equation we have still not discussed the significant

of this constant E. Which appears when I integrate this equation the Schrodinger wave equation

or the constant P which appears, when we have when we integrate the Schrodinger differential

equation. So, there are till now unknown constants whose

significance I have not discussed. Now, if you remember de Broglie hypothesis then you

can straight away say what the significance of this E and P are. But we shall come back

to the significance of this E and P in later on as we go long in this lecture or possibly

in the next lecture. Another point which I should make here let us ask the question,

what is the angular frequency of this wave? Now, you can straight away read the angular

frequency of the wave if you think of this as e to the power of minus i omega the so,

omega you see is e by h cross. If you ask what is the wave number of this wave for this

wave the wave number if you write this as e to the power of i k x. Then you can straight

away read the fact that P by h cross is the wave number.

So, this solution of the Schrodinger equation has arbitrary constants which come about.

There is only one of them that is independent in the exponent there is only one of them

which is independent. And it is this constant that also determines the angular frequency

of the wave. And the wave number of the wave. So, this is a plane pure sinusoidal wave solution

for the Schrodinger wave equation. If I had chosen the minus sign I would have this is

the wave propagating to the right, if I chose the minus sign I would have got the wave propagating

to the left. Now, the super position of two such solutions is also a solution let me also

write that down. So, I can have two different you see this constant e is arbitrary or in

other words the constant P is arbitrary. Suppose I chose two different values for E

and P if I so, I will refer to them as E 1 P 1 and E 2 P 2. Then i E 1 P 1 the set of

values E 1 and P 1 will give me particular solution thus of the Schrodinger equation

which as I have told you right in the start corresponds to a particular state of the particle.

If I chose a different set of values E 2 and P 2 I will get a different solution which

corresponds to a different state of the particle and I can now superpose these two solutions

that also is guaranteed to be a solution of the Schrodinger equation. Because it is an

linear equation. So, the superposition of two solutions is also a solution and I can

write down this super position solutions. So, let we write it down here.

So, this psi is a super position of two different solutions. This solution has the constants

in for the in this solution the constants have value P 1 and E 1 the overall amplitude

is A 1 the amplitude remember could be complex. This solution the constant has values P 2

and E 2 and there is an overall amplitude A 2 which is again different from this. This

corresponds to a particular state of the particle in this case the electron if you are dealing

with an electrons wave function this corresponds to a different state from this for the particle.

The superposition of these 2 is again at a different solutions. So, it corresponds to

a third state of the particle. And if I change these constants and I will get different each

of them will give me a different state of the electron. I can generalize this further.

So, I can write down a solution.

That looks like this. So, here the constant P which appears in a in our solutions can

take any value in the range minus infinity to infinity. I am superposing all such values

with different amplitudes.

Here I have only 2 such values I am superposing them with different amplitudes so, corresponding

to P 1. There is a E 1 and I have a particular amplitude for that wave corresponding to P

2 there is an E 2 which is the function of P 2 and for that. For this particular wave

I have a different amplitude and the superposition of this also gives me a solution.

Now, what I am doing is I am superposing infinite some of infinitely many momenta. So, I represented

as an integral. So, P will change and when P changes the particular wave corresponding

wave is going to be added with the different amplitude. So, this amplitude also changes

with P and I am superposing all of these solutions to produce another solutions psi of x of psi

which is a function of x and t. So, different set of amplitudes will give me different psi

wave functions and 1 sign no psi I can calculate the probability amplitude of finding the particle

somewhere. So, if I know if I ask the question what happens if I make a measurement of the

position of the particle if I measure

the particles position I know how to what I can I can predict with the wave function

what I can predict with the wave function is as follows.

If I know for a particular solution psi the mod square of this gives the probability density

rho x and the probability of finding the particle and an interval dx. So, if this is the region

space I am dealing with this 1 dimensional if I ask the question what is the probability

of finding the particle in an interval dx around this point x that is given. That probability

can be calculated if I know the wave function. And it is it has the value the rho x. So,

rho x into the interval dx gives me the probability of finding the particle in this small interval

around the point x. So, we know we can make some kind of a prediction

of what we expect. If I make the measurement of the position of the particle what we can

do with the wave function is that, we can predict the probability of finding it at some

at any position. So, once we know the wave function we can tell, what is the probability

of finding the particle at different positions? But for a particle free to move in 1 dimension

position is not the only physical observable attribute that you can measure physical observation

that you can do. You can measure various other things for example, you could measure the

particles momentum or you could measure the particles energy.

For a particle which can move in 3 dimensions all 3 dimensions you can also talk about the

particles angular momentum. The question is in quantum mechanics where we represent the

state of the particle as waves. How do we deal with other such observable quantities?

Physical observable quantities how do we deal with a situation where I measure the momentum

of the particle? So, I have represented the state of the particle by a wave the question

is, what will happen when I make a measurement of the momentum? So, here we have to introduce

another postulates. So, I have already told you that the state of the particle is represented.

By a wave function psi, the second thing is physical quantities physical observable quantities

are represented by Hermitian operators. So, corresponding to every physical observable

quantity like position momentum energy etcetera for a particle there is an operator there

is a Hermitian operator. So, we have now introduced to new words I have to explain what they mean

the first thing is the Hermitian and the second thing is operator or you may say the other

way around I have introduced this Hermitian and operator. We going to take a operators

first and then I am going to tell that what you mean by a Hermitian operator.

So, an operator

an operator which I will denote with a any symbol with a cap on top. So, this is the

operator o with the cap the cap fact that there is a cap on top or a hat on top tells

going to tell us at this is an operator not a number. And operator is something that acts

on a function to give me another function. So, let me give you the examples an example

of an operator is d by dx. So, the operator acting on the function let us say sig k x

is equal to d by dx of sin k x which is equal to Cos k into Cos kx. So, what we see is that

this operator acts on a function to give me an another function. In this case the function

is sin k x the function that I get when the operator o acts on this function or operate

at d by dx acts on this function is k into Cos kx. So, this is what is an operator let

me give you another example of an operator.

We have an operator o I am using the same symbol let us say who which is multiplication

by 2. So, o acting on let us say x square when where o is defined as multiplication

by 2. So, 2 times this is 2 into x square that is is the the operator here multiplication

by the number 2. So, when this operator acts on the function x square it gives me two into

x square. So, this is what we mean by an operator an operator acts on a function to give me

different function.

Now there are situations where an operator acts on a function to give me a different

function, but this different function which is the resultant of the operator acting on

psi is the same old function psi multiplied by a number. If this is true then this number

is said to be an Eigen value of the operator o. And this functions psi is said to be an

Eigen function. So, example an example of a of an Eigen function and an Eigen value

is given below I am going to give you 1 now. So, let us consider the operator d by dx and

this operator is going to act e to the power i k x this is going to give us i k into e

to the power i k x. So, what we see is that the function e to the power of i k x is Eigen

function of this operator d by dx. And it has a Eigen value i into k let me give you

another example the next example is when the same operator d by dx acts.

On sin k x it gives me k into Cos k x we see that this is not an Eigen function sin k x

is not an Eigen function of d by dx right. Because when d by d the operator d by dx acts

on sin k x it gives me a different function which is not the same function there is way,

i can write this as the same function multiplied by a number. So, this is not an Eigen function

and it has no Eigen value.

Therefore for this operator similarly this operator this Eigen this function is an Eigen

function of this operator. And it has an Eigen value i into k now I told you that the that

what we that in quantum mechanics corresponding to every physical observable.

There is a Hermitian operator, what do we mean by a Hermitian operator? I have told

you what we mean by an operator I have also told you what we mean by the Eigen value and

the Eigen function of an operator. Now, the Hermitian operator is is a little complicated

there is definition which I shall not go into the as far as we are concerned and we are

we are mainly interested in one property of Hermitian operators. And we I am going to

just tell you about this property. So, Hermitian operators are operators all of whose Eigen

values are real so.

Hermitian operators are type of operators have this special feature that its Eigen values

are all real. So, operators can have typically more than one can have many Eigen values.

For examples the operator d by dx which I have introduce as a example d by dx has many

Eigen values the function e to the power of i k x is an Eigen function for any value of

k this is an Eigen function. So, for each value of k I will have a different Eigen value

i into k. So, if k is 1 there will be a particular Eigen value i if k is 2 the Eigen value is

2 i, if k is 3 the Eigen value is 3 i. Any value of k is an Eigen is an Eigen value is

a different Eigen value of this operator.

So, operators typically have many Eigen values many and correspondingly many Eigen functions.

Now, Hermitian operators are a special class of operators all of whose Eigen values are

real.

And it is the postulate in of a quantum mechanics that corresponding to every physical observables,

there is a Hermitian operator associated with it. So, let me give me an example of a Hermitian

operator the operator. So, let me give you an example.

. So, let us consider the operator i d by dx and let us look at the function e to the

power let us i k x let us consider instead minus i d by dx. So, let us look at the function

e to the power i k x. And when this operator acts on this it give us what does it give

us. So, if I differentiate this with x I will pick up a sign of factor of i into k and i

will retain the original function i into minus i gives 1. So, I have a k. So, this gives

me k into e to the power i k x. So, what we see is that this operator o defined as minus

i d by dx this is Hermitian because its Eigen values are all real. Then you can check that

it has no imaginary Eigen values Eigen values are all real you could I could also give you

an example of an operator let us look at this operator d by dx.

So, d by dx acting on i k x gives me i k into e to the power of i k x now notice that this

e to the power i k x is an Eigen function of this operator and it has an imaginary Eigen

value such an operator is not Hermitian. So, I have given you an example of Hermitian operator

I have also give you an given you an example of an operator that is not Hermitian. And

in quantum mechanics corresponding to every physical observable there is a Hermitian operator

let me give you an example. Examples so, for a particle moving in one dimension we have…

What are the physical observable set you can think of

the position the corresponding to the position. We have an position operator x which is defined

as follows when x acts on a function psi x it gives me the variable x into psi x. So,

the function psi x is not an Eigen function of this operator x because x acting on psi

x the operator x acting an psi x is the variable x into psi x. If you change x this is under

constant if you change x the value of this variable also changes. So, this is not an

Eigen function of this operator. And corresponding to position we have this operator x which

is defined like this corresponding to momentum. We have the momentum operator P which is minus

i h cross del by del x. So, this is the momentum operator and the other quantity which you

can measure for a particle observable quantity for a particle free to move in 1 dimension

is the energy. Now, the operator corresponding to energy

or more rigorously strictly speaking there is an operator corresponding to the Hamiltonian

which as far as we are concerned in this course is the energy. And we shall denote this by

h and this is this is i h cross del by del t i h cross del by del t. So, what we see

that we have to postulate 2 things 2 postulates that we have encountered of quantum mechanics.

The first postulate the first assumption is that associated with every state of the particle

there is different wave function. And these wave functions are governed by the Schrodinger

equation I have told you what the Schrodinger equation is associated with every physical

variable which you can measure every physical observable quantity of the attribute of the

particle that you can measure. For example, for a particle in 3 dimension in moving only

in a single dimension you have its position. Momentum and energy corresponding to the position

there is a position of operator corresponding to momentum operator corresponding to energy

or strictly speaking the Hamiltonian there is Hamiltonian operator. Which is given over

here i h cross del by del t the momentum operator is minus i h cross del by del x. So, corresponding

to every physical quantity there is an operator. In the next lecture I shall tell you how to

what happens when we make the measurement what is the physical interpretation I mean

what is the what role do these operators play that I shall discuss in the next class. There

is a point which I should make before I close today’s lecture in today’s lecture we

worked out.

In the solution corresponding of the Schrodinger equation corresponding to a free particle

and in this solution I told you that there is this constant E which comes up when I solve

this equation there is this constant E which comes up. And related to E there is a constant

P which is related to E like this. So, P is essentially the square root of 2 m into E

with the plus plus or minus sign possible right. Now, the point which I forgot to tell

you is that the E the constant E has to necessarily be positive it cannot be negative and the

reason for this is as follows. The constant E is related to the constant

P over here. So, P is equal to plus minus 2 m E and till now our in our discussion E

and P are both are arbitrary constants which are related to one another. And they essentially

fix the dispersion relation for the wave. Now, E has to be real and positive because

if E is negative. The constant P becomes imaginary if I have an imaginary constant over here

and I multiplied with i then the x dependence now becomes of the form e to the power some

real constant into x let me make this point clear. Why does E have to be necessarily positive?

If E is negative then P becomes imaginary and if I have E to the power minus i plus

i P x that the x dependence if this P is imaginary. Let us say P is i alpha then this becomes

minus alpha x because i and i gives minus 1 and the range of x is from minus infinity

to plus infinity. Now, notice if the x dependence of the wave function is of this form. It blows

up when x goes to minus infinity the wave function should not blow up because the wave

function gives the probability and the modulus square of the wave function integrated should

be 1. So, if you want this to be satisfied the wave

function should not blow up which basically tells us that P cannot be imaginary, and if

P cannot be imaginary E has to be real. So, in this arbitrary solution in this solution

that we have worked out E could be arbitrary an arbitrary constant as long as it is real.

That is the point which I have forgotten to mention. So, let me bring today is lecture

to an end over here we shall resume over discussion of how to interpret these operators how to

manipulate with them in the next lecture.