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 continue on with our example problems.
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Let us jump right on in.
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Our first example is, is the function E ⁺to the power – X²/ 2 / the interval for - infinity to infinity is a good candidate for a wave function?
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Recall that for a wave function to be viable, not any function can be a wave function, it has to satisfy certain properties.
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In other words, it needs to be what we call well behaved.
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The wave function to be viable must be well behaved.
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Well behaved means, one, that the integral of ψ* ψ must converge.
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By converge, we mean it must reach some finite number.
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You must get actual number not infinity.
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You must get 5, 7, 8, 6.2, π, 11 π, whatever it is, that is what we mean by converge.
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It must converge to a number.
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In other words, it must be finite.
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I can get some value, it cannot be improper.
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Two, that both the wave function and the derivative of the wave function must be finite.
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The functions can go off into infinity over their particular interval of definition, over their domain.
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Three, ψ and ψ prime must be continuous and must be smooth, no breaks in the graph.
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Four, both the wave function ψ and the derivative of the wave function ψ prime, they must be single valued.
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In other words, just a nice well defined function.
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It have to be a function.
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Let us go ahead and see if this particular function, E ⁻X²/ 2 over this interval, actually satisfies these properties.
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Let us go ahead and talk about the first one.
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We are going to have to integrate from -infinity to infinity.
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We have to do this, ψ * × ψ.
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In this particular case, ψ * is ψ because there is nothing complex in here.
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There is no I involved.
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It is E ⁻X²/ 2 × E ⁻X²/ 2 DX.
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This is a symmetric interval and the function itself is an even function.
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It is an even function, therefore, I can go ahead and write this as the integral from to infinity.
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It is symmetric about the Y axis so it is not a problem.
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I can just take half of it and just multiply by 2 to get the other half.
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I do not have to do this -infinity to infinity thing.
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E ⁻X² DX.
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This definitely does converge.
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Let us go ahead and write it out.
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I will do this interval and I recommend you let your software do it for you.
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It is going to be 2 × π/ 4 ^½.
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The integral converges.
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Again, we were able to do this thing because E ⁻X ^/ 2 is an even function.
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Remember what an even function is.
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It is where F of –X = F of X.
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It is something that is going to be symmetric about the Y axis.
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The same on the left it is on the right.
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That takes care of one, that is good.
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Now, we need to check to see whether it is finite.
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Ψ is equal to E ⁻X²/ 2 and ψ prime = -X × E ⁻X²/ 2.
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I hope you are checking my derivatives.
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Yes, both of these are absolutely finite on their particular interval as X goes to + infinity or –infinity.
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Here both are finite.
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So far so good.
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Now, we need to check continuity.
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This is continuous and this is continuous, both are continuous.
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There are no discontinuities, both are continuous, our domain of definition.
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And last, both are single valued.
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In other words, when you put 1X in there you will get 1Y value.
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You are not going to get 2 different Y values.
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Both are single valued.
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Yes, this is a good candidate for a wave function.
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This is not just any random function that can be a wave function.
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It has to be well behaved.
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It has to satisfy these properties.
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I apologize for a little bit of sniffling and my voice being a little different.
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I’m just getting over a bit of cold.
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Example number 2, look at the solutions to the problem of a free particle in a 2 dimensional box,
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These are the solutions to the Eigen value problem, is the Schrodinger equation expressed as an Eigen value problem.
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H is the hamiltonian operator.
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It is just a shorthand notation and E is the energy Eigen value.
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It shows that whenever the system is in one of its energy Eigen states,
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In other words whenever the system is one of the wave functions,
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One of the Eigen functions, that the variance of the energy is equal to 0.
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The standard deviation is equal to 0.
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In other words, that every time you measure the energy, you are going to get the same value.
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You are going to get some of that particular value and you will never get anything else.
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In other words, there is no going to be deviation.
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You are going to get the same point every time.
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55555555 a million × and the average of that is 5.
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The deviation is 0 because there is no other number except 5.
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That is what this means.
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Let us go ahead and find out what ψ is, look up the solutions.
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For a 2 dimensional box, ψ N sub X N sub Y, we have 2 quantum numbers,
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Is going to equal 2/ AB¹/2 × the sin of B sub X π/ A × X × the sin of N sub Y π/ B × Y.
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That is the wave function.
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We want to show that this is equal to 0.
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This is equal to the average value of E² – the average value of E².
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Whenever that system is in one of its energy Eigen states, that means this is satisfied.
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Its Eigen value is satisfied.
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When the system is in one of its energy Eigen states, remember the Hamiltonian operator is the operator for the total energy of the system.
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The E and H, when we talk about the total energy of the system, we are talking about the Hamiltonian operator.
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It is one of its energy Eigen states, that means that the H of ψ N sub X N sub Y is equal to the energy N sub X N sub Y of ψ N sub X N sub Y.
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We are just applying this, we are just working symbolically and formally.
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N² ψ, just apply it twice.
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N sub X N sub Y= E², N sub X N sub Y ψ, N sub X N sub Y.
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It is just an application of this.
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This is true.
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The expectation value of the energy is going to equal the double integral.
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Double integral because we are talking about 2 variables.
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We have X and Y of ψ * N sub X N sub Y × the energy Hamiltonian operator are the same.
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Ψ N sub X N sub Y.
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This is the definition of the average value.
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This is a double integral because it is a two variable problem.
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This is going to equal the double integral.
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This is this, this part this part is this part, I’m just going to substitute in.
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It is going to be ψ* N sub X N sub Y × E N sub X N sub Y ψ N sub X N sub Y.
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This is the definition, this thing is just this thing and I just substituted this in for here to get this.
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E now is a scalar, it comes out of the integral.
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This equals E N sub X N sub Y × the double integral of ψ*.
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N sub X N sub Y × ψ N sub X N sub Y which is equal to E, because this integral =1.
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This is a normalized wave function and this is the normalization condition.
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Because the integral equals 1.
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D² that is equal to the integral of ψ* N sub X N sub Y × the hamiltonian² ψ N sub X N sub Y.
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I just substitute in, equals the double integral of ψ sub * N sub X N sub Y E N sub X N sub Y ψ N sub X N sub Y.
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This comes out because it is a scalar.
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This is², N sub X N sub Y × the double integral.
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We are back to N sub X N sub Y ψ N sub X N sub Y is equal to E² N sub X N sub Y.
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Our σ² of E which is equal to the expectation value of E² - the expectation value of E that is² is equal to E² N sub X N sub Y - E N sub X N sub Y² is equal to 0.
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We did what we were supposed to do just by manipulating the basic definition of the Eigen value problem.
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That and that were the case of the hamiltonian operator and hamiltonian operator of ψ is equal to the energy × ψ.
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The Schrodinger equation and we just use this and the definition of the expectation value.
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Evaluate L sub X L sub Y.
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Very simple statement in the problem.
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Let us see what we can do.
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This is the angular momentum operator in the X direction.
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Angular momentum in the Y direction.
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The X component of the angular momentum, the Y component of the angular momentum, the commutator of those two.
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Let us go ahead and write out what these are.
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Let me go back to blue here.
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You know what, let me go back to black, I think.
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I have L sub X = - I H ̅ Y DDZ - Z DDY and the Y component of the angular momentum operator equals - I H ̅ Z DDX - X DDZ.
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Trying to keep all of these symbolism straight is the hardest part of quantum mechanics, I promise you.
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I really do not like these brackets.
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I do not like physically actually drawing them out so I hope you will forgive me, I tend to use a slightly different notation.
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I just tend to write comma, L sub X, L sub Y, commutators are not a big deal.
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You can use whatever symbolism you want as long as you know what is being said.
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I’m just making the brackets for some odd reason just that bothers me.
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That is equal to L sub X L sub Y of F – L sub Y L sub X of F.
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Basic definition of commutator.
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Let us go ahead and do this first one.
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I’m going to blue, let us go ahead and do this first one.
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L sub X L sub Y of F = - IH Y DDZ - Z DDY of - I H ̅ Z DF DX - X DF DZ.
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We have to multiply this out.
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Let me go ahead and take care of the constants first.
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This and this, - IH and - IH ends up becoming a -H ̅².
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An operator, you can treat it just like a binomial or a polynomial.
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You have a binomial operator.
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A binomial operator is when I do this × this, this × that, this × this, this × that.
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We are just multiplying, we are operating so we have to be careful here.
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This one here, let us do this first.
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This and this, this is the DDZ differential operator.
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Here we have Z × a function of XYZ so this is going to be a product rule.
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This one is going to end up being , let us see here.
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YZ is going to be this Y and then the derivative of this is going to be this × the derivative of that.
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It is going to be D² F DZ DX + that × the derivative of this.
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The Y and Z part, we just multiply that.
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Y the derivative of this thing is this time the derivative of that which is why I get the Y × Z.
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D² of DZ DX.
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It is going to be the derivative of this.
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This × the derivative of this which is 1.
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We are going to get Y DF DX.
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I hope that made sense because this is Z and this is a function of XY and Z.
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This is the product rule so when I apply this differential operator to this term, I have to do the product rule.
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The rest are easy.
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This × this, we will do this × this.
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This is going to be - XY D² F DZ².
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We will do this × this, not a problem these are just straight because there is no function of that variable.
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I'm not taking the derivative with respect to a variable of, you know something that includes that variable like this one.
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They are the normal product rules.
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This we get - Z² D² F DY DX +,
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I will do it this right here.
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It is going to be XZ D² F DY DZ.
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Let us go ahead and do L of Y, L of X of F.
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That is going to be - I H ̅ Z DDX - X DDZ that is the operator of LY, × - I H ̅ of Y DF DX - X DF DY.
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Again, you notice I use an F, I did not just work with the operators.
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For this one, I included the F.
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For this part here, I include the F.
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I have working on some function, I do not want to lose my way.
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This is going to equal to – H ̅².
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It is going to be this in this, it is going to be a YZ D² X.
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Let me see if I got this here.
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DF DZ this is XYZ.
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We have to be careful here.
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Notice, the LX operator, X is going to be YZ ZY.
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This is the Y operator which is going to be ZX XZ.
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There we go, now we are good.
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We have YZ D² F DX DZ.
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I will do this one, this is going to be –Z² D² F DX DY.
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Let me have this one which is going to be –XY D² F DZ².
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I have Z and Z, I’m going to get two terms.
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This is going to be product rule, again.
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It is going to be + XZ D² F DZ DY.
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In other words, this × the derivative of this and it is going to be + X.
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This × the derivative of this DF DY.
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Everything actually ends up canceling out.
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YZ D² of DZ DX YZ D² F DX DZ.
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Remember, mixed partials are equal so that cancels with that.
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That stays and that stays.
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XY D² of DZ² XY D² of DZ².
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This - - becomes +, so that cancels.
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Z² D² of DY DX D² of DX DY cancels.
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X of Z D² of DY DZ D² of DZ DY cancels.
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What I’m left with is the following.
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I’m left with LX LY F – LY LX of F = -H ̅² × Y DF DX – X DF DY.
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I'm going to go ahead and switch this around.
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I’m going to go ahead and write this as H ̅², flip that.
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Basically pull out a -1 from here I get X DF DY - Y DF DX.
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I’m going to write this as - I H ̅ × - I H ̅.
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- and – is a + here, X D FDY - Y DF DX.
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If you notice this thing right here, this part right here happens to be L of Z.
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We get I H ̅ L of Z of F and we can drop the F.
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Our final operator notation I H ̅ L of Z.
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I will just go ahead and use the XLY, it equals that.
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Notice that this does not equal to 0 operator.
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It is not possible.
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These operators the LX and LY do not commute.
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It is not possible to measure any two components of the angular momentum simultaneously to an arbitrary degree of precision.
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In other words, this is the same as the linear momentum position operator.
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If you measure one really well, you have to lose the measurement of the other.
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You cannot measure both to any degree of accuracy or precision that you want.
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In the case of the angular momentum, any 2 components of the angular momentum cannot be measured simultaneously to any degree of precision.
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The same holds for LX LZ and LY LZ.
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LX LY and LY LZ, LX LZ, you are always going to get something like this.
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A little messy but nice.
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This is the stuff that you have to do.
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You just have to do it through it.
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Example 4, this is going to be a not so much of a problem, this is going to be a demonstration.
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It is a demonstration of why the necessity of a real Eigen values imposes the hermitian property on operators.
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Remember what we said about when we take our measurements in quantum mechanics,
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What we are measuring, we need the measurements to be real numbers.
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In other words, we need the Eigen values to be real numbers.
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Operators and Eigen functions can be complex.
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They can be real or they can be complex.
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There has to be a certain property of the operator that guarantees that the Eigen value is always real because what we measure has to be a real number.
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It cannot be a complex, we cannot measure complex numbers.
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We need a real number.
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Here we are going to demonstrate why this is the case, before we just threw it out there and say that it is the case.
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We are going to demonstrate why the necessity of having a real Eigen value forces the operator to be hermitian.
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Let us go ahead and do that.
00:29:49.300 --> 00:29:54.000
Should I work in red or black?
00:29:54.000 --> 00:29:57.500
Let us go ahead and work with red.
00:29:57.500 --> 00:30:09.400
We have A ψ = λ ψ, where λ is a real number.
00:30:09.400 --> 00:30:11.600
It is in a real number system.
00:30:11.600 --> 00:30:14.000
This is the general Eigen value problem.
00:30:14.000 --> 00:30:18.900
The operator × ψ = ψ × some constant.
00:30:18.900 --> 00:30:28.100
You are not getting ψ back when you operate on it but you end up getting it back multiplied by some real value.
00:30:28.100 --> 00:30:37.000
What I'm going to do is I'm going to go ahead and take this and multiply on the left by ψ* both sides.
00:30:37.000 --> 00:30:41.400
Whatever you do to both sides, you retain the equality.
00:30:41.400 --> 00:30:49.200
Ψ * A ψ = ψ * λ ψ.
00:30:49.200 --> 00:30:53.500
All of them have taken the basic definition and I have done this to it.
00:30:53.500 --> 00:30:57.100
I’m going to integrate both sides.
00:30:57.100 --> 00:31:01.900
The integral of that side is equal to the integral of this side.
00:31:01.900 --> 00:31:08.900
This integral over here on the right, this is a real number so I could pull it out of the integral side.
00:31:08.900 --> 00:31:13.800
This is going to be λ × the integral of ψ * ψ.
00:31:13.800 --> 00:31:17.300
Ψ * ψ is the normalization condition and it equals 1.
00:31:17.300 --> 00:31:23.300
This equal to λ × 1 which equals λ.
00:31:23.300 --> 00:31:26.500
That is it, that is the first part.
00:31:26.500 --> 00:31:36.900
Let us take the complex conjugate of this thing, start with that and see what we can do.
00:31:36.900 --> 00:31:43.400
Let us start again with A ψ = λ × ψ.
00:31:43.400 --> 00:31:45.500
Let us go ahead and take the complex conjugate.
00:31:45.500 --> 00:31:54.200
This is going to be this conjugate, ψ conjugate = λ conjugate ψ conjugate.
00:31:54.200 --> 00:31:55.100
We are going to manipulate this.
00:31:55.100 --> 00:31:59.500
I’m going to multiply on the left by ψ.
00:31:59.500 --> 00:32:02.400
I have ψ, I’m just manipulating.
00:32:02.400 --> 00:32:06.200
All I’m doing is a mathematical manipulation and see where it takes me.
00:32:06.200 --> 00:32:09.400
This is ho we did it.
00:32:09.400 --> 00:32:18.300
Ψ λ * ψ *.
00:32:18.300 --> 00:32:21.300
I’m going to integrate both sides.
00:32:21.300 --> 00:32:27.300
When I integrate the side, I integrate this side.
00:32:27.300 --> 00:32:37.500
Λ * is just a scalar so it comes out as λ * × ψ ψ *.
00:32:37.500 --> 00:32:48.400
This is just the normalization condition so this is equal to λ *,but Λ is a real number.
00:32:48.400 --> 00:32:52.100
Therefore, the conjugate of a real number is the number itself.
00:32:52.100 --> 00:32:55.700
This equals λ.
00:32:55.700 --> 00:32:57.800
Both integrals equal λ.
00:32:57.800 --> 00:33:00.300
Let me go to blue.
00:33:00.300 --> 00:33:14.000
This integral equals λ, this integral equals λ.
00:33:14.000 --> 00:33:21.900
If they are both equal λ, then they are both equal to each other.
00:33:21.900 --> 00:33:55.000
Therefore, I have the integral of ψ* A of ψ equals the integral of ψ A *ψ * which is the definition of the hermitian property.
00:33:55.000 --> 00:34:02.800
In other words, we arrived at this λ being real forces these integrals to be equal.
00:34:02.800 --> 00:34:08.500
We use this final step in our derivation as the definition of hermitian property.
00:34:08.500 --> 00:34:12.800
In other words, their operator has to satisfy this condition.
00:34:12.800 --> 00:34:16.600
That integral has to equal this integral, that is the definition.
00:34:16.600 --> 00:34:20.400
The necessity of then being real forces to operator to have this property.
00:34:20.400 --> 00:34:23.000
This property we call hermitian.
00:34:23.000 --> 00:34:27.600
There you go.
00:34:27.600 --> 00:34:51.700
If some operator satisfies the hermitian property regardless of whether
00:34:51.700 --> 00:35:16.300
The operator or the wave function is complex or not, the Eigen values λ sub N are always real.
00:35:16.300 --> 00:35:19.800
This is very profound on so many levels.
00:35:19.800 --> 00:35:24.400
Those of you who are going to continue on with some higher mathematics or go further with quantum mechanics,
00:35:24.400 --> 00:35:32.200
It is quite extraordinary what is going on here mathematically.
00:35:32.200 --> 00:35:35.800
Let us see something else.
00:35:35.800 --> 00:35:39.600
We said that the hermitian property guarantees that the Eigen values are real.
00:35:39.600 --> 00:35:40.800
We say why, that is the case.
00:35:40.800 --> 00:35:45.200
We also said that it guarantees that the Eigen functions are orthogonal.
00:35:45.200 --> 00:35:50.900
In other words, the integral is equal to 0 but they are perpendicular to each other in some sense.
00:35:50.900 --> 00:35:59.500
In this problem, it is going to be a demonstration of why the Eigen functions of hermitian operators are orthogonal.
00:35:59.500 --> 00:36:15.800
Let us go ahead and start with A ψ sub M = λ sub M ψ sub M.
00:36:15.800 --> 00:36:26.900
Let us start with A of ψ sub M= λ sub M ψ sub M.
00:36:26.900 --> 00:36:30.100
I’m going to go ahead and do the same thing that I did before.
00:36:30.100 --> 00:36:35.000
We are going to multiply on the left by the conjugate and take the integral.
00:36:35.000 --> 00:36:53.100
I got ψ conjugate AN, ψ sub N = ψ N conjugate × λ sub N × ψ of N.
00:36:53.100 --> 00:36:56.900
I’m going to integrate those two.
00:36:56.900 --> 00:37:05.300
I’m going to call this integral 1 and I’m going to call this integral 1 prime.
00:37:05.300 --> 00:37:09.600
Over here, I’m going to get the conjugate first.
00:37:09.600 --> 00:37:22.300
I’m going to take A conjugate ψ sub M conjugate = λ sub M conjugate ψ sub M conjugate.
00:37:22.300 --> 00:37:30.100
I’m going to multiply both sides on the left.
00:37:30.100 --> 00:37:35.200
I’m sorry this is going to be ψ sub M not ψ sub N.
00:37:35.200 --> 00:37:44.100
Ψ sub M and over here I’m going to multiply on the left by ψ sub N.
00:37:44.100 --> 00:38:05.900
I got ψ sub N × A conjugate ψ sub M conjugate = ψ sub N × λ sub M conjugate ψ sub M conjugate.
00:38:05.900 --> 00:38:09.800
Let me see if I got this right.
00:38:09.800 --> 00:38:18.200
I’m going to call this integral 2 and I’m going to call this integral 2 prime.
00:38:18.200 --> 00:38:26.000
I’m going to take 1 prime, this integral – 2 prime.
00:38:26.000 --> 00:38:30.100
I’m going to go to red.
00:38:30.100 --> 00:38:47.400
I got 1 prime – 2 prime = the integral ψ sub M conjugate × λ sub N ψ sub N – 2 prime which is.
00:38:47.400 --> 00:38:55.000
I forgot to once I get to this point I’m going to integrate both sides.
00:38:55.000 --> 00:39:10.400
-2 prime which is ψ sub N λ sub M conjugate ψ sub M conjugate.
00:39:10.400 --> 00:39:14.400
That is going to equal, we are going to pull the λ out.
00:39:14.400 --> 00:39:25.000
Λ N – λ M conjugate × the integral of ψ sub M conjugate ψ sub N.
00:39:25.000 --> 00:39:29.300
Ψ sub N ψ sub N conjugate is the same thing.
00:39:29.300 --> 00:39:38.800
I’m just going to write it as ψ sub M conjugate ψ sub N.
00:39:38.800 --> 00:39:42.300
That is going to be that one.
00:39:42.300 --> 00:39:45.800
I’m going to take 1-2, that integral.
00:39:45.800 --> 00:40:15.600
1-2 is going to equal the integral of ψ sub M conjugate A ψ sub N – the integral of ψ sub N A conjugate ψ sub M conjugate.
00:40:15.600 --> 00:40:18.300
This integral is going to equal 0.
00:40:18.300 --> 00:40:24.200
The reason it is equal to 0 is because the operator we said was hermitian.
00:40:24.200 --> 00:40:26.400
This is the definition of hermitian.
00:40:26.400 --> 00:40:29.900
This thing and this thing, they are hermitian.
00:40:29.900 --> 00:40:31.100
They are the same integral.
00:40:31.100 --> 00:40:36.000
Therefore, this integral is equal to this integral.
00:40:36.000 --> 00:40:38.200
This - that is equal to 0.
00:40:38.200 --> 00:40:46.000
This is the case because A is hermitian.
00:40:46.000 --> 00:41:04.100
This is the definition of hermitian ψ sub M * A ψ sub N is ψ sub N A * ψ sub M*.
00:41:04.100 --> 00:41:06.000
So far so good.
00:41:06.000 --> 00:41:13.800
I have got this thing and I got this thing.
00:41:13.800 --> 00:41:17.100
The 1 prime -2 prime is equal to this.
00:41:17.100 --> 00:41:21.700
1 -2 is equal to 0.
00:41:21.700 --> 00:41:31.700
1 is 1 prime, 2 is equal to 2 prime.
00:41:31.700 --> 00:41:38.200
1 prime -2 prime is equal to 1 -2.
00:41:38.200 --> 00:41:49.600
Therefore, my 1 prime -2 prime which we said was λ sub N – λ sub M conjugate ×
00:41:49.600 --> 00:42:01.500
The integral of ψ sub M conjugate ψ sub N is equal to 1 - 2 which we said is equal to 0.
00:42:01.500 --> 00:42:08.800
I have something × something is equal to 0.
00:42:08.800 --> 00:42:13.600
That means this is 0 or this is 0.
00:42:13.600 --> 00:42:30.900
If M does not equal to N, then this λ sub N – λ sub M conjugate do not equal each other.
00:42:30.900 --> 00:42:31.700
They are different Eigen values.
00:42:31.700 --> 00:42:34.700
A different Eigen values are not going to equal each other.
00:42:34.700 --> 00:42:37.900
These are not going to be equal to each other.
00:42:37.900 --> 00:42:44.700
If they are not equal to each other, this thing is not 0, which means that this has to be 0.
00:42:44.700 --> 00:42:45.600
Let me say it again.
00:42:45.600 --> 00:42:53.400
If M does not equal N, then λ sub N and λ sub M do not equal each other.
00:42:53.400 --> 00:42:57.500
Which means that λ sub N – λ sub M does not equal 0.
00:42:57.500 --> 00:43:02.100
This equation =0 so that means this one has to equal 0, this term.
00:43:02.100 --> 00:43:22.700
Therefore, the integral of ψ sub M conjugate × ψ sub N does equal 0 which we take as the definition orthogonality.
00:43:22.700 --> 00:43:31.100
We already use the definition and now the justification for why it works, for why it is, what it is.
00:43:31.100 --> 00:43:45.400
Which we take as the definition of orthogonality.
00:43:45.400 --> 00:43:49.500
A hermitian operator, all the operators in quantum mechanics are linear.
00:43:49.500 --> 00:43:52.000
They are hermitian, they guarantee that the Eigen values are going to be real and
00:43:52.000 --> 00:43:57.900
They guarantee that the Eigen functions are going to be orthogonal with respect to each other.
00:43:57.900 --> 00:44:01.700
The integral is going to equal 0.
00:44:01.700 --> 00:44:06.300
It is actually quite extraordinary.
00:44:06.300 --> 00:44:08.500
With that, we will go ahead and end it here.
00:44:08.500 --> 00:44:10.700
Thank you so much for joining us here at www.educator.com.
00:44:10.700 --> 00:44:11.000
We will see you next time, bye.