WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to talk about the Schrӧdinger equation as an Eigen value problem.
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Let us get started, let us go ahead and work in black today.
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Recall example 3 from the previous lesson.
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We had an operator which was I believe C ̂ and we said that this operator was defined by -I × H ̅ DDX of something, whatever F was.
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We said that F was equal to E ⁺I × n × X.
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We operated on this and we found that this C ̂ of our particular F was equal to n × H e ⁺INX.
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Now, notice that operating on F is the same as multiplying it by we just ended up, the original function was e ⁺INX.
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We operated on function was nh × e ⁺INX.
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We ended up just multiplying the original function by some scalar, by some number.
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Positive or negative actually does not matter.
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This is the general arrangement.
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Operating on F is equal to some scalar × F.
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The only thing this particular operator did was multiply F by some constant.
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In other words, if my function is X², I may end up with 17X².
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That is it, I just multiplied the X² by a factor 17, that is all.
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Recall example 4, in example 4 we had that the operator D was actually equal to partial derivative of the function with respect to Z.
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In that particular case, RF was equal to XY² and Z³.
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In this particular case, the operation on F ended up equaling 3XY² Z².
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Here operating on F did not just multiply it by some scalar factor, the D of F was 3X² Y² Z².
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Make sure this is clearly to.
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The original function was XY² Z³.
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In this case, it did not do that.
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When an operator operates on a function and gives back the original function multiplied by some scalar, this is profoundly important.
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Here is what we are getting into some very deep and important mathematics.
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The general expression is this, operating F is equal to some A × F.
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I will actually go ahead and do here.
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Let me go ahead and put the A of X = some constant × F of X.
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Personally, I do not like the X, they tend to distract me.
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I will just go ahead and do it this way.
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A of F is equal to some constant A × F of some function that we are operating on.
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Here the F of X or the F, it is called an Eigen function of the operator.
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A is called an Eigen value for the Eigen function.
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I actually called V, the Eigen value associated with that particular Eigen function.
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Operating on F in a situation where when you operate on F, you just end up getting F back multiplied by some scalar.
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The F of X is called the Eigen function of the operator.
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A is called the Eigen value that is associated with that particular Eigen function, when this thing is satisfied.
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Now, given a particular operator A ̂, finding F of X and its corresponding A is called an Eigen value problem.
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You are going to be presented with some operator.
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It is your task to find all of the functions and all of the values, the scalars that satisfy this equation.
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That is what we are saying.
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Which function when you operate using this particular operator that we give you,
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gives you back the original function multiplied by a scalar, that is the problem.
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It turns out that the Schrӧdinger equation is just an Eigen value problem.
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We will show it in just a minute.
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For example 3, our C of F = A of F.
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We have 3 things, we have the operator, we have the Eigen function, and we have the Eigen value.
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In this particular case, the operator was n-I H ̅ DDX, that was the operator.
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The Eigen function was E ⁺INX.
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Any function E ⁺INX satisfies this relation.
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Of course, the Eigen value.
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You can write it as two words, I will write it as one word.
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The Eigen value that was NH, that is the whole idea.
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Operator, Eigen function, Eigen value, profoundly important.
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Those of you who have taken linear algebra, chances are you already have seen this when we are talking about matrices.
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This is this and that is that, that is what is going on here.
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Let us go ahead and talk about what this has to do with the Schrӧdinger equation.
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Let us look back at the Schrӧdinger equation and let us write it out like this.
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We had - H ̅²/ 2M × D² DX² of our particular C.
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I’m not going to go ahead and put the X there.
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It is a function of X that we are looking for + this potential energy function × our Z or ψ if you like.
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Equals E × Z this was Schrӧdinger equation, that is the function.
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Notice something here, notice I have the Z here × something, the Z here × something.
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I can factor out the Z and write this left hand side as an operator.
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Here is what it looks like.
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It ends up looking like this.
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Let me go back to a black.
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I’m going to factor out the, so it was going to be –H ̅²/ 2N D² DX² + V of X.
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I’m going to write the Z out here.
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Z equals E × C.
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Notice, we know we can do this because operators they distribute the way that polynomials do.
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Notice what we got.
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Once again, here is my Z and here is Z, this as an operator.
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This right here, this is just some number.
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The energy of the system is just a constant.
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If we call this operator, if we get a symbol, if you call this operator on the left, operator H ̂, we can rewrite this whole thing as H ̂.
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Let me go back to black.
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H ̂ of Z is equal to E × Z.
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We just expressed the Schrӧdinger equation as an Eigen value problem.
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If you given some wave function and if you operate on it with this thing, this we have not given a name to it yet, it is actually called a Hamiltonian operator.
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If we operate on this wave function, we actually end up getting the wave function back multiplied by some constant.
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The constant happens to be the total energy of the system.
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This is profoundly beautiful.
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We expressed the Schrӧdinger equation as an Eigen value problem.
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That is exactly what it is, as an Eigen value problem.
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We have presented with a particular Schrӧdinger equation for a particular system.
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Our task is to find the function Z, the wave function and the associated Eigen values that happened to coincide with these operators.
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The operator is the same, it is the Hamiltonian operator.
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We are going to take this operator and see if we can find a function and the Eigen values that are associated with it.
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For each function, there is some energy for the system.
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When the system is in this particular energy state, the particle is behaving this way according to the wave function, that is all we are doing.
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The rest of it is just math, it really is.
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When we have solved the equation, we will not only have found a wave function and Z of X but also the total energy of the particle.
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The total energy of the particle in the state is represented by the wave function.
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In the state that is represented by the particular wave function the Z of X.
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Once again, the operator is called the Hamiltonian.
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HR is called the Hamiltonian operator.
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Let us go ahead.
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Our Hamiltonian operator happens to equal - H ̅²/ 2 × the mass, the second derivative +
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this potential energy function that is the Hamiltonian operator right there.
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Notice, if we have - H ̅²/ 2M D² DX² + V of X × some C is equal to the energy × the Z.
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Notice, this is potential energy, this part right here.
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This is the total energy is equal to the potential energy + the kinetic energy.
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Therefore, this makes the kinetic energy operator.
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I will use E kinetic.
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Operators are quantum mechanical.
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The operators, this and this, or this thing together, are quantum mechanical.
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The energy is classical mechanical, that is the relationship.
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There is an association.
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Classical quantities, things like energy, momentum, angular momentum, position, things like that, are represented in quantum mechanics,
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But operators that is the whole idea.
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Operators are quantum mechanical.
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Scalars like the energy are classical mechanical.
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When there is some quantity in classical mechanics like momentum, position, energy, angular momentum, whatever it is,
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In quantum mechanics those are represented by the kinetic energy operator.
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The position operator, the momentum operator, they are represented by operators.
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The reason they are is because all the information is contained about a particular system of quantum mechanical system is contained in the wave function.
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In order to extract information from that function, we have to operate on it.
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If I want to find the angular momentum of a particle, I'm going to take the angular momentum operator of the wave function.
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If I want to find the position, I'm going to take the position operator of that function
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and it is going to give me some certain information that I can do something with.
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Operators are quantum mechanical.
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Scalars are classical mechanical.
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Such associations are the very heart of quantum mechanics.
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Now given what is above,
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Let me do this on the next page, that is fine I can do it here.
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Given what is above, in another words this thing that we just did.
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Let us define the kinetic energy operator that is equal to -HR²/ 2N DDX².
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This is the kinetic energy operator.
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If I have a wave function and I want to know what the kinetic energy is at a given moment,
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I will go ahead and apply the kinetic energy operator to that wave function and it tells me something.
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When I say apply it to that wave function, I will be more specific about that when we actually talk about
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how we are going to extract information that we can actually measure and see from the wave function.
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But that is what we are doing.
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We defined V which is just V of X which means multiply the function by this function V of X.
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This is the potential energy operator.
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In other words, if I want to know what the potential energy of a particular particle is,
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of an electron in this particular system whatever happens to be, I take the wave function and multiply it by the potential energy function.
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This is the potential energy operator.
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In this particular case, the operation is just multiply by.
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It is fine, let me go ahead and in the next page.
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The Hamiltonian operator that is equal to the sum of the kinetic and potential energy operators, that is equal to.
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I think I can go ahead and put on the next page.
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Let us do this.
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H ̅ that is equal to –H ̅²/ 2M D² DX² + VX this is the total energy operator.
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Or just the energy operator.
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These are operators.
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Let us go ahead and fiddle with this a little bit.
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From the classical point of view, classically the kinetic energy of a particle you know was equal to ½ MV² that is also equal to this.
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If you take ½ MV² it is going to be equal to the momentum squared divided by twice the mass.
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We will go ahead and just start playing with this formally, mathematically.
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Formally just means we are working with this symbolically.
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We are going to define the kinetic energy operator is equal to this momentum operator squared/ 2M.
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Let us go ahead and multiply here, the momentum operator squared is going to equal 2M ×
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the kinetic energy operator and it is going to equal 2M × the kinetic energy operator which happens to be - H ̅²/ 2M D² DX².
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The 2M cancels and we get ourselves a square of the linear momentum operator which is going to equal - H²² DX².
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We are able to derive another operator.
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These exponent here on the operator, it is nothing more than sequential operation.
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Here is operator² is equivalent to doing P again.
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If you saw the linear momentum operator cubed, it will just be P again.
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The squared is just a symbol for sequential operation.
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This does not mean this is symbolic for the operation.
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It does not mean take the linear momentum operator, whatever you get square it.
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That is not what it means.
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When you see an operator squared, it means operate sequentially.
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Let us go ahead and break this down.
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Squared is equal to -H² D² DX².
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Let us go ahead and separate this out.
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- I H DDX in other words we are going to factor this out and - I H DDX.
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This linear momentum operator squared actually factors out.
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If I multiply these two, if I operate and operate, I end up getting this thing.
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Therefore, we can go ahead and define the straight linear momentum operator in the X direction which is one of these – I H ̅ DDX.
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This is the linear momentum operator,
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if I have some quantum mechanical system and if I have another wave function for it.
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If I want to know something about a linear momentum, whatever is that I want to know, I operate on the wave function with this operator and it gave me some information.
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And again, what I mean by operating on it is not just going to be a direct operation.
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We will see what it is, we are still going to play with this mathematically but is essentially what I’m doing.
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I will leave that for the time being.
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Let us see, now we have our linear momentum operator.
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We have our Hamiltonian which is the total energy operator.
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We have our kinetic energy operator.
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We have a potential energy operator.
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We have a majority of things that we need to actually get started.
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Let us go ahead and finish off with an example here.
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We will close this lesson out.
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One more, here we go.
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Let P = -I H ̅ DDX and in this particular case our functions are going to be sin X.
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We want to show that the operator squared of F does not equal the operation on F².
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In the last lesson we show that operators do not commute.
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We want to also make sure that you understand the difference between squaring operator and taking something and getting a function and squaring that.
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They are not the same thing.
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Let us go ahead and do this.
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The operator squared of F is equal to,
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Again, we set it as a sequential operation on F.
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It is going to equal this one, that is the far left one and it is going to be – I H ̅ DDX of sin X.
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We end up with this being - I H ̅, the derivative of sin X is cos of X.
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I think I’m going to it up here.
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We have the - I H ̅ of the –I H ̅ of cos of X and we end up with - -, we end up with I² H ̅².
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I’m sorry I forgot my derivative operator.
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Let me go back and erase this, it is – I H DDX of - i H ̅ cos X.
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When I take the derivative of cos X, I get - sin X - -is + I and I is I²,
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H ̅ and H ̅ is H ̅².
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I² is -1.
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Let us skip a few steps.
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- and - is going to be positive, I and I is going to be I².
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I end up I² is -1 and – and - is going to cancel and become a positive.
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We are just left with an I² H ̅² and sin X.
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Let us go ahead and do P of F and we will square that.
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The P of F is going to be – I H ̅ DDX sin X².
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It is going to be – I H ̅ cos X² is going to equal -1 × -1 is positive.
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I × I is going to be I².
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H ̅² cos² X is equal to – H ̅² cos² X.
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Clearly, this and this do not equal each other.
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In general, some operator raised to some exponent and then applied to F absolutely
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does not equal that operator applied to F and then raised to the exponent.
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This is a symbolic representation of how many times you are going to operate in sequence.
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This squared right here is an actual mathematical operation.
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These are not the same thing.
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We have to be very careful about our symbolism, about our mathematics.
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Profoundly important.
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Just take your time and work slowly.
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Now that we have introduce this notion of an operator, there are a lot of symbols floating around.
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In any case, we will go ahead and leave it that.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time for a continue discussion of operators in Quantum Mechanics, bye.