WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today we are going to begin our discussion of Quantum Mechanics.
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Let us just jump right on in.
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We are going to be discussing the Schrӧdinger equation and something called operators.
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Einstein has demonstrated that the relation between a photon’s of momentum and its wavelength is this right here.
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A wavelength of a photon is able to planks constant divided by the photons momentum.
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A photon is a particle of light and this is planks constant.
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Louis de Broglie argued that matter also obeys this relation.
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That a particle of mass M and velocity V will have a wavelength of this.
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Momentum is just mass × velocity.
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When we are talking about a specific particle with a definite mass and a definite velocity, it is the same relation.
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This is the Broglie relation.
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The Broglie waves have been experimentally confirmed.
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In other words, particles do exhibit waves like behavior.
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If matter then behaves like waves then theoretically at least, there should be some wave equation that describes the particles behavior.
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Notice that I put these enclosed.
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There is an equation, this is the Schrӧdinger equation and it looks like this.
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I have written 2 versions of it.
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There are actually several different ways that you can write this.
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This is the important thing right here, let me go ahead and do this in red.
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This red thing that you see right here.
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This is the relationship that exists among between the different elements of this way function.
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You have got to see what is the function that we are looking for and we have this different equation.
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It just says that if I take the second derivative of this function, if I multiply it by some variation of planks constant divide by twice its mass, negate it.
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If I add to that the function itself multiplied by the potential energy, I end up getting the function multiplied by the total energy of the system.
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This is a different way of writing it and what I have done is basically taken this function and I have put it out here.
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It will make more sense a little bit later in the lesson when I talk about this thing called operators.
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Now the solutions of this differential equation, these right here, this particular function that we are looking for,
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they describe how a particle of mass M moose in its particular potential field.
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It is this Schrӧdinger equation that we are interested in,
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In any given particular system that we are dealing with, we are going to come up with a Schrӧdinger equation for it.
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We are going to solve the equation and then we are going to get the Z,
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these different functions that describe how the particle is behaving at a given time, at a given speed, at a given whatever.
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That is the whole idea.
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What we want, the whole idea of Quantum Mechanics is to find this wave function from
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the wave equation that we write down from the given set of data.
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Z sub x are called the wave functions of the particle and they will end up telling us everything
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we want to know about how the particle is behaving, that is the whole idea here.
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This wave function contains all the information about the particle.
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Whatever I need to know about it, its position, its momentum, its energy, is angle, whatever it is.
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It is part of this function and I extract information from this function.
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Z sub x is a measure of the amplitude of the matter wave.
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Again, we are looking at matter as if it is a wave.
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As if it is displaying wavelike properties.
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Since that is the case, there is a wave function that describes its behavior.
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Z sub x is a measure of the amplitude of that matter waves.
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We are saying more about this later.
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If you are interested in seeing how the Schrӧdinger equation can be obtained from the classical wave equation,
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that is an argument for the plausibility of the Schrӧdinger equation or why the wave equation is actually called an amplitude, please see appendix 1.
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These appendices that I'm going to be doing throughout this course, they are extra information.
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They are not necessary, as far as the continuity of the course is concerned.
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It is not like you have to necessarily watch them or do anything with them to continue on with the course.
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They are just extra information for those of you that are interested in going a little deeper,
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whether it is deeper conceptually, whether it is deeper mathematically, and things like that.
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Quantum Mechanics is entirely mathematical.
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At this level, my best advice is to accept and perform the mathematics without worrying too much about what the individual concepts mean.
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When I talk about the mathematics, particular technique that we may be using whether it is differentiation, integration, something else,
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we are going to be introducing some new mathematics that many of you may not have seen before.
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It is not that I’m not going to explain what this physical significance is,
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but in a lot of ways understanding in Quantum Mechanics is an emerging process.
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Like it is in most sciences, in all sciences but it is a lot more so with Quantum Mechanics that it is with classical sciences that you are accustomed to.
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It is really just a different way of thinking.
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Quantum mechanics has this reputation of being very esoteric and really hard to wrap your mind around.
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That is actually not true at all.
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What you have to do is pull yourself away from trying to wrap your mind around it conceptually and
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just developing a certain mathematical facility, just doing the math as is.
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As you do it, as you become more comfortable with it, it will start to make sense why the math is actually taking the form that is taking.
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For those of you that go on into higher science and particularly those of you
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that want to take other courses in mathematics like Fourier series, a theoretical algebra, things like that.
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All of this will actually come together.
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For right now, we want you to develop a good mathematical facility with what is going on.
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Do not worry too much if it does not entirely make sense to you.
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Treating it that way, it is going to be a lot easier than you will expect it, I promise.
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Let us go ahead and see what we can do.
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Let us talk about operators.
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I’m going to go ahead and rewrite the equation again.
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Let me go ahead and write it in blue.
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We have -H ̅² / 2 M.
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We have D² / DX² MC.
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I will go ahead and write that + the potential energy V × this C function = total energy of the system × this function.
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Again, it is this C that we are looking for, that is what we want to find.
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When you are doing algebra, you have something like 3x + 6 = 9 then solve for x.
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A differential equation is the same sort of thing.
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Now, instead of solving for a number x = 5, we want to get an actual function.
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We are looking for a function, it is just another variable.
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In this equation, this is the variable, this is your x.
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Except x happens to be a function.
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Let us go ahead and talk about operators.
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An operator is a symbol that tells you to perform a task.
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That tells you to perform not just a task, it could be one or more operations on a function, thus, producing a new function.
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In fact, what we are really doing is we are giving a name to something that you have been doing for years and years.
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We are producing a new function.
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For example, when you take the derivative of a function you get a new function back.
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The derivative of x² is 2x, the differential DDX is an operator, it is the differential operator.
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That is what you are doing, we are just giving a name to it.
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In other words, we start with some function f of x and we operate on it.
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Let us do A with a little caret symbol over and we spit out a new function.
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The symbol for the operator is symbolized by a capital letter with a caret over it.
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It is symbolized by a capital letter with a caret symbol.
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You have to define what the operator is.
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We might say A is due this, B is due this, and we will see a little bit of that in just a moment.
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Operators and operations are best described just by doing examples.
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I’m just going to launch into the examples rather than try to explain it and it will make complete sense.
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They are actually very easy to deal with.
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Example 1, we will let A, this operator A equal to D² Dx².
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In other words, the operator A means take the second derivative of some function.
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Now what we want you to do is to find A of the sin of X.
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Also written as A sin X.
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You do not necessarily need to put parentheses around the function that you are operating on.
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This says perform the operation A on the function sin X.
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Well nice and simple.
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You already know this, you have been doing this for ages.
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Sometimes I will write the parentheses, sometimes I will not.
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A = D² DX² of sin X.
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I like to do things pictorially, so sin X when I take a derivative of sin of 5x, I end up with Φ cos 5x.
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That is the first derivative.
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The operators take the second derivative also.
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When I take another derivative of that, I end up with -25 × the sin of 5x.
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Operator, here is my definition of the operator.
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The symbol A means do this and I have a function that I'm going to do that to, and I do it.
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I end up with a function, it is that simple.
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You have been doing it all along, you are just given it.
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The only difference is that some of our operators tend to go a little bit more complex.
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But you can handle it very easily.
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Example 2, we will the operator B, we will define it as this D² DX² + multiplication by this thing called V, whatever V is.
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V is a function of X.
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We want you to find B of sin of 5X, the same function.
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But now we want you to perform a different operation on it.
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The operator is defined by this.
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Although operators are only symbols, they can be treated just as though they were regular polynomials.
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This whole thing is the operator.
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I can just treat this, this way.
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Here is what happens, B of the sin 5X = this is the operator, it was going to be D²/ DX² +
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this function V of X × I will put sin of X here to perform this operation on sin X.
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I can just treat this even though it is a symbol, operators are just like polynomial.
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You can distribute them.
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This says take the second derivative of the sin 5X and then add to it this V of X × the sin of 5X.
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It is this and operate this way.
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The operations, you can distribute the operations the same way you would distribute any number.
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That is what makes these operators very powerful.
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The second derivative of the sin 5X.
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We already found that this before.
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This first part is going to be -25 sin 5X and here we have whatever V happens to be × the sin 5X.
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That is our new function from this operator, it is that simple.
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You just do exactly what it says and you treat the operator whether it is 2 things, a binomial operator, a trinomial operator,
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a quadranomial operator, you just distribute the way you do anything else.
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Let us see example number 3, I hope we are not elaborating the point too much but I think it is always good to see a lot of examples.
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Example 3, this time we will call the operator C.
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It is equal to - i × H, I will do H bad DDX.
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By the way, H ̅ is just planks constant divided by 2 π.
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It is just some shorthand notation for it and we will see it again.
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It is just a constant, that is all it is.
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This says if I perform C on a given function, I’m going to take the first derivative of the function then I’m going to multiply it by H and multiply by -1.
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Clearly, operator can be complex, imaginary as well as real.
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It just do something to a function and get a new function.
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This time, we want you to find this of the function e ⁺INX, where n is just some number.
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The operator of A ⁺INX = this – I × H ̅ DDX of e ⁺INX.
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When I go ahead and take the derivative of this, I'm going to get.
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This is going to be - i × this H ̅, the derivative of e ⁺INX is IN e ⁺INX derivative of the exponential function in e ⁺INX.
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We have - I² H ̅ and A ⁺INX.
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I² = -1, - -1 that becomes +1.
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You are left with just H ̅ Ne ⁺INX.
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If you have noticed with the previous samples or example number 1,
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notice, the original function was e ⁺INX.
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Operating on that gave me back something × the original function.
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This is going to be very important in a little while.
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I probably noticed it with the first example, it is another example.
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You may notice it in a few more examples before we actually talk about other things.
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I just want to bring that your attention.
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Interesting enough, sometimes the operator will change and become a bit of a completely different function.
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Sometimes what the operator only does is multiply the original function by some constant, that is very important.
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Let us go back to blue here for our examples.
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Let us do example 4.
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Example 4, this time our operator D = DDX.
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This is the partial differential operator.
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For those of you who have not done partial differentiation, there is actually nothing to learn.
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If you have a function of 2 variables, let us say X² Y.
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All you are doing when you are taking the partial, just take the derivative only with respect to X.
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It means hold every other variable constant, that is all you are doing.
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You already know what to do here.
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Find D of XY² Z³.
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In this particular case, we have a function of 3 variables X, Y, and Z.
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It happens to be XY² Z³.
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This operator is asking you to take the partial derivative of this whole function with respect to X.
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All that means is that Y² is a constant, Z² is a constant.
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They do not exist, you just leave them alone.
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Let us see what we have got.
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I’m sorry this is DDZ not DDX.
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Sorry about that.
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We are going to hold X constant, we are going to hold Y² constant, and we are going to differentiate just the Z³.
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D of XY² Z³ = DDZ of XY² Z³.
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This is a constant so it stays XY² and the derivative with respect to Z is 3Z².
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I will just write it like this.
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If I want to put a number in front, I can, not a problem.
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You can write it anyway you want.
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You can leave like this or you can write it this way.
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There you go, that is it.
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In this particular case, we start with a function and we end up with a different function.
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This particular operator does not just multiply the original function by a constant where is the one before did.
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Sometimes it does, sometimes it does not.
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Again, that is going to be very important the differentiation between the two.
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Let us go ahead and do another example.
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This is going to be example 5, and this time we will go ahead and call our operator L.
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The operators is going to be D² DX² + 2 DDX -3.
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All this operator says is that if you are given some function, take the second derivative, add to it 2 ×
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the first derivative of it and then subtract the number 3.
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That is it, it just as a symbol, it is an operator.
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It is saying do this.
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We have got CL and this time we want you to do is see what is that we are going to find.
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We want you to find L of X³.
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L caret of X³ well that is equal to D² DX² + 2DDX - 3 of this function X³.
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We just distribute, this one, this one, and this one had.
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Adding and subtracting, very simple.
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We get D² DX² of X³ + 2 × DDX of X³ – 3.
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We have got the second derivative.
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We have got 3X² and we got 2 × 3X so we are going to end up with 6X over here.
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And this one we are going to have the derivative of the 3X³ is going to be 3X².
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It is going to be 2 × 3X² this is going to be - 3X³.
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We end up with 6X + 6X² -3X³.
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Nothing strange, nice and normal.
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Let us go ahead and do one last example.
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This time we are going to perform operators sequentially.
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Here we will let the operator A =- I H ̅ DDX.
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We will let the operator B = X³.
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When you see an operator equal to some function, that means multiply the function that you get by this.
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In other words, when I'm operating B for example, if I do B of X² it is going to be X³ × X².
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When some operator is just some function, it means multiply the function that you are supposed to operate on by this.
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It is just multiply by, that is all it is.
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I will write that here.
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Sometimes you just need to do that, you need to multiply by X³.
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Our task in this example is to find A caret, B caret of sin X.
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We also want you to find B caret A caret of sin X.
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It will be both.
00:24:01.600 --> 00:24:08.900
Operators and sequence, when they are written like this, you start from the right most operator and work with your left.
00:24:08.900 --> 00:24:12.500
In this particular case, we will do this first one.
00:24:12.500 --> 00:24:17.700
A caret B caret of sin X.
00:24:17.700 --> 00:24:21.400
Let us make B a little bit more clear here, sorry about that.
00:24:21.400 --> 00:24:30.500
A caret B caret of sin X that is equal to A caret of B caret sin X.
00:24:30.500 --> 00:24:36.800
I’m going to perform B first and I’m going to perform A on what it is that I got.
00:24:36.800 --> 00:24:48.200
This is going to be A caret, now B sin X, B sin X was multiplied by X³ so I’m going to get X³ sin X.
00:24:48.200 --> 00:25:04.100
A, perform the operation A on X³ sin X that is going to equal – I H ̅ DDX of X³ sin X.
00:25:04.100 --> 00:25:05.500
We have a product rule here.
00:25:05.500 --> 00:25:10.500
X³ and sin X are both functions of X, we will go ahead and leave that one out.
00:25:10.500 --> 00:25:21.700
We end up with is - I H ̅ this × the derivative of that is s going to be X³ × the cos X + that × the derivative of this.
00:25:21.700 --> 00:25:25.800
It was going to be 3X² sin X, there you go.
00:25:25.800 --> 00:25:42.200
This is A caret B caret, we perform the B first then we performed A.
00:25:42.200 --> 00:25:44.600
Let us go ahead and do the other one.
00:25:44.600 --> 00:25:49.500
Let us do B caret A caret of sin X.
00:25:49.500 --> 00:25:53.700
That is equal to B caret, we will do A first.
00:25:53.700 --> 00:26:09.000
A caret of sin X is going to equal B caret of - I × H ̅ DDX of sin X = B caret,
00:26:09.000 --> 00:26:16.100
The derivative of sin X is cos X, we have – I H ̅.
00:26:16.100 --> 00:26:34.100
This is going to be cos X and then B means multiply by X³ so we end up with –I H ̅ X³ × the cos X.
00:26:34.100 --> 00:26:43.400
This one, we perform the operation A first and then we apply the operator B.
00:26:43.400 --> 00:26:51.200
Notice, in general, in this particular case AB does not equal BA.
00:26:51.200 --> 00:26:53.300
Operators do not commute.
00:26:53.300 --> 00:26:59.800
In other words, you know the 2 × 4 is 4 × 2, that is the property of the real number system, that is commutability.
00:26:59.800 --> 00:27:02.500
Operators do not commute in general.
00:27:02.500 --> 00:27:05.900
This like matrix multiplication, they do not commute in general.
00:27:05.900 --> 00:27:21.000
In general, AB performed on some function F does not equal BA performed on some function F.
00:27:21.000 --> 00:27:26.900
Operators do not commute, in other words.
00:27:26.900 --> 00:27:38.900
Operators do not commute, in general that has profound consequences for quantum mechanics.
00:27:38.900 --> 00:27:45.900
There are going to be times when the operators do commute, that has profound consequences for quantum mechanics, not commute in general.
00:27:45.900 --> 00:27:48.600
We will be seeing this again.
00:27:48.600 --> 00:27:55.600
In general, operators do not commute.
00:27:55.600 --> 00:27:59.500
Let us go ahead and talk about our next topic here.
00:27:59.500 --> 00:28:07.000
Back to operators, we have defined what operators are and done some examples, now operators can be linear or nonlinear.
00:28:07.000 --> 00:28:12.400
Now we are going to give a very specific mathematical definition of what linear is.
00:28:12.400 --> 00:28:16.200
Those of you who studied linear algebra, you already know this definition or you have seen it.
00:28:16.200 --> 00:28:23.200
Those of you who have not done linear algebra, this is going to be they real mathematical definition of what linear means.
00:28:23.200 --> 00:28:26.800
Linear does not just mean that the exponent on a variable is 1.
00:28:26.800 --> 00:28:30.200
You have treated it like that for years now, ever since middle school
00:28:30.200 --> 00:28:37.200
but now we are going to give you what the mathematical definition is, the criterion for linearity.
00:28:37.200 --> 00:28:53.600
Operators can be linear or nonlinear.
00:28:53.600 --> 00:28:55.900
We deal only with linear operators.
00:28:55.900 --> 00:29:00.600
In quantum mechanics, we are only concerned with linear operators.
00:29:00.600 --> 00:29:14.200
We deal only with linear operators which is very convenient because non linear operators are quite difficult.
00:29:14.200 --> 00:29:23.000
Here is the definition of linear.
00:29:23.000 --> 00:29:31.300
Here is what it say, they are 2 things that you have to check when you are given some operator to check whether is linear.
00:29:31.300 --> 00:29:40.500
The definition is A of F + G.
00:29:40.500 --> 00:29:49.600
I’m not going to use the X, these are functions of X or functions of Y, or function of whatever.
00:29:49.600 --> 00:29:50.200
I’m not going to put the variable.
00:29:50.200 --> 00:29:53.400
I’m just going to put F+ G.
00:29:53.400 --> 00:29:59.300
It is equal to A of F + A of G.
00:29:59.300 --> 00:30:01.900
What this says is the following.
00:30:01.900 --> 00:30:07.700
We know that operators are things that you do to functions, you operate on a function.
00:30:07.700 --> 00:30:15.000
A linear operator has to satisfy this, it says that if I’m given a function F and I’m given a function G,
00:30:15.000 --> 00:30:21.200
if I add those two functions first and then operate on what I get when I add them,
00:30:21.200 --> 00:30:28.500
that I will get the same thing if I operate on F separately, operate on G separately and then add them.
00:30:28.500 --> 00:30:32.400
That is what linear means, it means I can switch the order of addition and operation.
00:30:32.400 --> 00:30:41.300
Add first then operate, or operate first then add, that is what linear means.
00:30:41.300 --> 00:30:44.100
That was the first thing you has to satisfy.
00:30:44.100 --> 00:30:47.000
The second thing that you saw was the following.
00:30:47.000 --> 00:30:53.200
A of CF = CA of F.
00:30:53.200 --> 00:31:01.100
If I'm given some function and if I multiply the function by some constant C and operate it, I should get,
00:31:01.100 --> 00:31:08.200
If the operator is linear, it means I can go ahead and take the function, operate on it first and then multiply by the constant.
00:31:08.200 --> 00:31:14.400
Here linearity implies that I can switch the order of operation and multiplication by a constant.
00:31:14.400 --> 00:31:22.800
Also, I can switch the order of addition of two functions and operation or operation than addition.
00:31:22.800 --> 00:31:30.600
These two things have to be satisfied when for an operator to be called linear.
00:31:30.600 --> 00:31:39.700
When you are presented with an operator, in order to check linearity you have to check these two things.
00:31:39.700 --> 00:31:41.800
Let us go ahead and write that down.
00:31:41.800 --> 00:32:41.600
Confirm linearity we have to verify that for a given operator, then 1 and 2 are satisfied for a given operator.
00:32:41.600 --> 00:32:42.200
Let us go ahead and do some examples.
00:32:42.200 --> 00:32:49.500
This is the only way this is going to make sense.
00:32:49.500 --> 00:32:55.500
Determine whether the operator defined by A of F = S² is linear or nonlinear.
00:32:55.500 --> 00:32:58.600
This is a different way of defining it.
00:32:58.600 --> 00:33:01.000
Notice, in the previous examples I gave you the operator and I set it this.
00:33:01.000 --> 00:33:03.300
Here it actually specifies it explicitly.
00:33:03.300 --> 00:33:06.200
A of F is the same is just S².
00:33:06.200 --> 00:33:12.400
Operating on F means just taking a function S and squaring it, that is what the operation is.
00:33:12.400 --> 00:33:15.800
The operation square, that is what it is.
00:33:15.800 --> 00:33:18.900
We have to show whether this is linear or not.
00:33:18.900 --> 00:33:22.600
Here is what we have to verify, the definition of linearity.
00:33:22.600 --> 00:33:25.700
Let me go ahead and work in red here for these examples.
00:33:25.700 --> 00:33:41.100
I have to show that A of F + G= A of F + A of G.
00:33:41.100 --> 00:33:43.200
I’m given two functions F and G.
00:33:43.200 --> 00:33:45.900
I’m going to add them and then operate them and square it.
00:33:45.900 --> 00:33:49.300
What I’m going to do is square the square of G and I’m going to add them.
00:33:49.300 --> 00:33:51.000
I’m going to see if the left side and the right side are the same.
00:33:51.000 --> 00:33:54.600
If they are, it is linear, it is a linear operator.
00:33:54.600 --> 00:33:56.100
If not, it is not a linear operator.
00:33:56.100 --> 00:33:58.200
It is that simple.
00:33:58.200 --> 00:34:00.800
Let us go ahead and do, I will go ahead and write the second one too.
00:34:00.800 --> 00:34:07.300
I have to show that A of C of F = C A of F.
00:34:07.300 --> 00:34:13.800
In other words, I'm going to take F and I’m going to square it and I’m going to multiply by a constant.
00:34:13.800 --> 00:34:15.700
And then I'm going to take F and multiply by a constant and I will square it.
00:34:15.700 --> 00:34:17.800
If those two ends up being the same, it is linear.
00:34:17.800 --> 00:34:19.800
If they end up not being the same, it is non linear.
00:34:19.800 --> 00:34:21.900
Both have to be satisfied.
00:34:21.900 --> 00:34:25.300
One might, the other might not, that does not count.
00:34:25.300 --> 00:34:27.900
Both have to be satisfied.
00:34:27.900 --> 00:34:31.800
Let us go ahead and check number1.
00:34:31.800 --> 00:34:38.600
Let us go ahead and do A of F + G.
00:34:38.600 --> 00:34:42.800
A of F + G, A of F is squaring.
00:34:42.800 --> 00:34:50.600
If I take F + G and I square it, that is going to equal F + G².
00:34:50.600 --> 00:35:00.700
F + G² I just multiply that out, that is equal to S² + 2 FG + G².
00:35:00.700 --> 00:35:04.900
This is my left side, this side right here.
00:35:04.900 --> 00:35:17.700
Now the question is, does it equal A of F + A of G?
00:35:17.700 --> 00:35:21.400
A of F = F².
00:35:21.400 --> 00:35:25.000
A of G = G².
00:35:25.000 --> 00:35:28.900
Does F² + 2 FG + G²= F² + G²?
00:35:28.900 --> 00:35:31.000
No, it does not.
00:35:31.000 --> 00:35:33.100
This is not a linear operator.
00:35:33.100 --> 00:35:39.300
It is that simple, you just have to perform the operations on the left, operations on the right, and see if they are equal.
00:35:39.300 --> 00:35:45.600
At this point I can stop, number 1 is dissatisfied.
00:35:45.600 --> 00:35:47.700
Therefore, it is not linear.
00:35:47.700 --> 00:35:48.400
I do not have to check 2.
00:35:48.400 --> 00:35:53.800
However, if you want to go ahead and check 2, that is not a bad idea.
00:35:53.800 --> 00:35:58.300
A of CF that is the left side.
00:35:58.300 --> 00:36:02.000
A of CF or A of whatever = the whatever².
00:36:02.000 --> 00:36:08.700
This is equal to CF² which is equal to C² F².
00:36:08.700 --> 00:36:17.800
Our question is if this equal to C of A of F?
00:36:17.800 --> 00:36:22.000
A of F is F², this is C of F².
00:36:22.000 --> 00:36:24.600
Does C of F² = CF²?
00:36:24.600 --> 00:36:28.000
It does not, not linear.
00:36:28.000 --> 00:36:30.900
You do not have to do 1 before 2, you can do 2 before 1.
00:36:30.900 --> 00:36:31.200
It does not matter.
00:36:31.200 --> 00:36:35.100
This is not a linear operator.
00:36:35.100 --> 00:36:41.800
The operation that tells you to square something, whatever it is that you are given is not a linear operator.
00:36:41.800 --> 00:36:44.200
But you knew that already.
00:36:44.200 --> 00:36:50.200
You knew that from the fact that back in high school and calculus, it is not linear, it is quadratic.
00:36:50.200 --> 00:36:55.400
Quadratic functions are not linear.
00:36:55.400 --> 00:36:57.700
Let us go ahead and do this one.
00:36:57.700 --> 00:37:04.800
Determine when the operator defined by A of F = D² DX².
00:37:04.800 --> 00:37:09.700
F is the operation of taking the second derivative of something a linear operator.
00:37:09.700 --> 00:37:11.800
Let us find out.
00:37:11.800 --> 00:37:15.700
I think I may go to black, I’m sorry.
00:37:15.700 --> 00:37:18.300
Again, let me go back to black, sorry.
00:37:18.300 --> 00:37:26.900
We need to show that A of F + G = A of F + A of G.
00:37:26.900 --> 00:37:30.000
Add first then operate, operate on each then add.
00:37:30.000 --> 00:37:33.700
They have to be equal to each other.
00:37:33.700 --> 00:37:35.700
Let us go ahead and do the left side first.
00:37:35.700 --> 00:37:45.900
A of F + G = D² DX² of F + G.
00:37:45.900 --> 00:37:51.100
I know that when I do differentiation, I know that the differential operator is linear.
00:37:51.100 --> 00:37:57.600
I know it from calculus, we do not call it an operator but the process of differentiation is linear.
00:37:57.600 --> 00:38:11.400
Therefore, this is going to equal D² F DX² or if you like F double prime + D² G DX² or G double prime.
00:38:11.400 --> 00:38:17.900
Let us do A of F + A of G.
00:38:17.900 --> 00:38:38.500
This is going to equal D² D A² of F is this + A of G is going to be D² G DX².
00:38:38.500 --> 00:38:40.600
They are equal.
00:38:40.600 --> 00:38:44.200
Let us go ahead and do 2.
00:38:44.200 --> 00:38:51.000
We operate A of CF, that is going to equal the second derivative of this thing C of F.
00:38:51.000 --> 00:39:01.900
I know I can pull constants out, that is equal to C D² F DX² or CF double prime.
00:39:01.900 --> 00:39:14.400
And if I have C of A of F that is equal to C × D² F DX².
00:39:14.400 --> 00:39:16.500
This and this are equal.
00:39:16.500 --> 00:39:21.400
Yes, this is a linear operator.
00:39:21.400 --> 00:39:27.700
You just have to check 1 and check 2.
00:39:27.700 --> 00:39:36.800
Let us see the next one, determine whether the operator defined by A of F = LN of F is linear or nonlinear.
00:39:36.800 --> 00:39:44.600
I’m given some function and I take the log of that function, that is my operator, taking the log of whatever is that I’m given.
00:39:44.600 --> 00:39:46.700
Is this linear or nonlinear?
00:39:46.700 --> 00:39:53.900
I think you are accustomed to this, it is good to write out what the criterion is.
00:39:53.900 --> 00:40:11.900
We need to show that A of F + G, the definition in other words = A of F + A of G and we need to show that A of C of F = C × A of F.
00:40:11.900 --> 00:40:23.300
Let us go ahead and do A of F + G = log of F + G.
00:40:23.300 --> 00:40:38.700
A of F + A of G = the log of F + log of G.
00:40:38.700 --> 00:40:45.400
This and this do not equal each other.
00:40:45.400 --> 00:40:48.300
I will just go ahead, at this point we can stop, it is nonlinear.
00:40:48.300 --> 00:40:50.900
However, let us go ahead and do the other one.
00:40:50.900 --> 00:41:02.100
We have A of C of F = the log of C of F.
00:41:02.100 --> 00:41:04.400
I will go ahead and put our little caret there.
00:41:04.400 --> 00:41:17.200
C of A of F, I can probably do a little bit more with this one.
00:41:17.200 --> 00:41:20.500
The log of something × something, let us go ahead and expand it.
00:41:20.500 --> 00:41:27.500
This is going to be equal to the log of C + the log of F.
00:41:27.500 --> 00:41:29.800
We will go ahead and leave that one.
00:41:29.800 --> 00:41:37.600
The C of F of A = C × the log of F.
00:41:37.600 --> 00:41:41.000
This and this, they are not equal.
00:41:41.000 --> 00:41:52.200
This is not a linear operator.
00:41:52.200 --> 00:41:54.600
We have introduced the Schrӧdinger equation.
00:41:54.600 --> 00:42:00.000
We have introduced the notion of operators which is profoundly important in Quantum Mechanics.
00:42:00.000 --> 00:42:02.100
We will go ahead and close this lesson off like this.
00:42:02.100 --> 00:42:04.500
Thank you so much for joining us here at www.educator.com.
00:42:04.500 --> 00:42:05.000
We will see you next time.