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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I apologize if I sat a little bit today, I’m just getting over a cold.
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If I have some sniffles and things like that, I hope you will forgive me.
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Today, we are going to continue on with our example problems.
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We already did one set and then we talked a little bit more about the quantum mechanics, the formal hypotheses of the quantum mechanics.
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Now, we are just going to do several lessons of problems.
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Let us just jump right on in.
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Before we start the example problems, I did want to go over just some of the high points.
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Just recall some of the equations because there was a lot going on mathematically with quantum,
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as there is with thermal, and anything else in physical chemistry.
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Sometimes, you have to pull back and just make a listing of some of things that are important that we remember.
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We solve the Schrӧdinger equation and we find this wave function ψ.
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That is a wave function and it represents the particle that we are interested in a particular quantum mechanical system.
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Instead of looking at the particle like a particle, we look at it like a wave.
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What we do is we play with this wave function to extract information from it.
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That is all that is actually happening in quantum mechanics.
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The ψ conjugate × ψ, we said was the probability of finding the particle whose wave function is ψ
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in a differential volume element called the DV at the point XYZ.
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You have this wave function which is going to be a function of XYZ.
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At some random XYZ, if you actually multiply, it is going to end up being the probability of
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finding the particle in that little differential volume element.
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Now we have the equation this ψ DV = 1.
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Actually, I should say this is not the probability, the ψ* × ψ is the probability density.
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But for all purposes, we can think of it as the probability.
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The actual probability is the ψ × ψ* × the differential volume element so that you actually have the probability.
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When we integrate all of the probabilities, we are going to get 1.
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This is the normalization condition.
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This was very important normalization condition.
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Again, one of the frustrating things about quantum mechanics is wrapping your mind about around things conceptually.
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But what is nice about it is, because it is so purely mathematical, even if you do not completely understand what is going on,
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as long as you have a certain set of equations at your disposal,
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You will at least get the right answer.
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Eventually, if you become more comfortable and solve for problems, conceptually it will start to make more sense.
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Now every observable in classical mechanics, corresponds to a linear hermitian operator in quantum mechanics.
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If we observe a linear momentum in classical mechanics, we have a linear momentum operator in quantum.
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If we observe angular momentum in classical mechanics,
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for some particle moving in a circular path or curved path we have a angular momentum operator in quantum mechanics.
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That operator, we apply it to wave function to give this information.
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This is what I mean by we extract information from the wave function by operating on the wave function.
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Doing something to it mathematically.
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An operator applied to some wave function in a particular state, is equal to A sub N ψ sub N.
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It is an Eigen value problem.
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Remember, we can express the Schrӧdinger equation as an Eigen value problem.
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Again, we are just going over some highlights of what is that we covered so that we have them
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in a one quick place before we start the example problems.
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That the ψ sub N or called the Eigen functions of the operator A.
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The A sub N are called Eigen values of A corresponding to the Eigen function, corresponding to the ψ sub N Eigen function.
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When a particular function is in a given state, let us say ψ₃, it is in that Eigen state for the operator.
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We speak of Eigen states, we speak of Eigen functions, we speak of Eigen values.
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Let us talk about what hermitian means.
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Hermitian also has a mathematical definition.
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Hermitian operator means it has to satisfy this.
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F* AG= the integral of GA* F*.
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If you have 2 wave functions F and G, if you do the left integral and if you do the right integral, those equal each other,
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then this operator is something that we call hermitian.
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If it is hermitian, if it satisfies this property.
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The hermitian operator implies that the Eigen values are real numbers.
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It is very important and I will actually do a lot to demonstrate why this hermitian property implies reality
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and implies orthogonality in some of the example problems.
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One of the first things that hermitian implies is the fact that the Eigen values are real.
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The other thing, that this hermitian property of the operator implies, double arrow for application.
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It implies that the Eigen functions are actually orthogonal.
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The integral of ψsub N conjugate × ψ sub P is equal to 0 for N not equal to P.
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If I have one Eigen function ψ₁ and I have another Eigen function ψ sub 15,
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If the operator is hermitian, the operator that gave rise to the Eigen functions, the Eigen functions are going to be orthogonal.
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That is analogous to two vectors being perpendicular.
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Two vectors are orthogonal when their dot product is equal to 0.
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Two Eigen functions are orthogonal when their integral of their product is equal to 0.
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It is completely analogous, that is all that is happening here.
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When measuring an observable in quantum mechanics, we only get the Eigen value of
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the operator corresponding to the observable when the wave function is an Eigen function of the operator.
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In other words, when a quantum mechanical system happens to be in a state that is represented
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by a wave function that happens to be an Eigen function of the operator,
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then what we observe when we take a measurement is going to be one of the Eigen values.
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When the quantum mechanical system is in a state that is represented by the Eigen function of
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the operator of interest then what we observe is going to be one of the Eigen values of the operator.
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If the wave function ψ is equal to ψ 1 ψ 1 + ψ 2 ψ 2 + so so, is written.
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Ψ is the wave function of the quantum mechanical system.
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If this wave function happens to be written as a linear combination also called the super position.
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I do not like the word super position but that is fine.
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It is written as a linear combination of Eigen functions of the operator of interest, whatever operator we happen to be dealing with.
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Then what we observe are the Eigen values A₁, A₂, A₃, and so on.
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Let me go to the next page.
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With probabilities C₁² C₂² C₃² and so on.
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This is for normalized wave functions.
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For the most part, all of our wave functions are going to be normalized.
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If they are not normalized, we are going to normalize them.
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That is not a problem.
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Basically, what we are saying is if we have some wave function ψ of a quantum mechanical system and
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let us say it is represented by 1/2 I × ψ₁ – 1/5 ψ₃ + 2/7 ψ sub 14.
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Let us say it is represented as a linear combination of Eigen functions of the operator of interest.
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Then what I'm going to observe are the Eigen values A₁, A₃, A sub 14, every time I make a measurement,
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I’m going to see one of these gets one of these three numbers.
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The extent to which I get one number over the other is going to be square of that, the square of that, the square of that.
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Those are the probabilities.
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1/5² is going to be 1/25.
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1/ 25 of the time, at every 25 measurements, one of those measurements I'm going to get an A3.
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That is all this is saying.
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That is all this represents.
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This probably will not play a bigger role in what we do.
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These are one of the hypotheses that we discussed.
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Let us go ahead and say a little bit more.
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After many measurements, the average value also called the expectation value.
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The average value is symbolized like that and it is going to be the integral of ψ sub * the operator and ψ.
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And this is for normalized.
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We will go ahead and put the one for un normalized.
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This definition right here, it applies when the ψ is written as a linear combination or not.
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If it is if this thing, then this thing goes in here and here.
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The definition is universal.
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The average value of a particular observable is this.
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The general definition for an un normalized wave function, it is just good to see it.
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We have the average value of A is going to equal the integral of ψ sub *.
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These are just integrals, all you are doing is literally plugging the functions in.
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Operating on this, multiplying it by to ψ conjugate.
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Putting it in the integrand and integrating it with respect to the variables.
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If it is a one dimensional system, it is a single integral.
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If it is a 2 dimensional system, it is a double integral.
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If it is a 3 dimensional system, XYZ, it is a triple integral.
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You have your software to do the integral for you.
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The integral of ψ A ψ ÷ the integral of the normalization condition, this thing.
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Remember, when it is normalize, this thing is equal to 1 which is why it is equal this, just the numerator.
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This is the definition for an un normalized wave function.
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If ψ is a linear combination is written as a linear combination.
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In other words, ψ = C1 C1 + C2 C2 +… ,
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Then the average value is really simple.
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It is actually equal to C 1² × A1, the Eigen value + C 2² × the Eigen value + …,
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It is equal to the sum I, C sub I² A sub I.
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There is another way of actually finding it when it is written as a linear combination.
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The final thing you want to review is something called the commutator.
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We have operator AB, the symbol this means this is called the commutator of the 2 operators.
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And it is defined as AB – BA.
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You apply AB to the function then you apply BA to the function and you subtract one from the other.
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This is called the commutator.
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And we also have sigma A² = A² - A², there is that one.
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The uncertainty in the measurement, the variance, if you take the square root of that you get the standard deviation.
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And the sigma of B² is equal to squared.
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Of course, the final relation which is the general expression for the Heisenberg uncertainty principle is the following.
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The sigma of A, sigma of B is greater than or equal to ½ the absolute value of the integral of ψ sub *.
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The commutator of AB applied to ψ.
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That is the general expression for the uncertainty principle and it is based on this commutator.
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If you do AB of the function of the BA of the function.
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If you subtract one from the other you get 0 and those operators commute.
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If they commute then you can measure any of those 2 things to an arbitrary degree of precision.
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If they do not commute like for example the position of the momentum,
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the position of the momentum operator do not commute.
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Based on the original thing that we saw, the original version of the Heisenberg uncertainty principle that we saw,
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we know that we cannot measure the momentum and
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the position of a particle to an arbitrary degree of an accuracy or precision simultaneously.
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We have to sacrifice one for the other and we have to find the balance.
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Whatever it is that we happen to want depending on the situation.
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With that, let us go ahead and start some example problems.
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I do not know it that helped or not but that was nice to see.
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Let ψ sub θ = E ⁺I θ for θ greater than or equal to 0 and less than or equal to 2 π.
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We want to normalize this wave function.
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Quite nice and easy.
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Normalize the wave function.
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Let me go ahead and do this in blue, just to change the color a little bit.
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Normalized means we have some constant that we have multiply the wave function by, to make the normalization condition satisfied.
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Normalized is ψ of θ is equal to some normalization constant × the function.
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The normalization condition is this.
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It is that equal to 1.
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We need to solve this integral and find N, the normalization constant.
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That is what we do.
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If we take the integral of ψ sub *.
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In this particular case, ψ*= E ⁻I θ because it is a conjugate and ψ is equal to E ⁺I θ.
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We do not have to watch out for it.
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Sometimes the conjugate is not the same as the real number.
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This become N × E ⁻I θ × ψ which is NE ⁺I θ.
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It is going to be E θ and we are going to set it equal to 1.
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We are going to get N² × the integral of E ⁻I θ × E ⁺I θ E θ.
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This is going to equal 1.
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We are going to get N² of E ⁺I E ⁻I θ × E ⁺I θ is E⁰ which is 1.
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It is going to be D θ.
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We are integrating from 0 to 2 π.
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D θ is equal to 1.
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This is going to be N² × 2 π is equal to 1.
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N² is equal to 1/ 2 π which implies that N is equal to 1/ 2 π ^½, or if you like 1/ √2 π if you prefer older notation.
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I should do it down here.
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Ψ sub θ of the normalize wave function is equal to 1/, 2 π ^½ E ⁺I θ.
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That is your normalize wave function.
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You want to normalize a wave function, apply the normalization condition.
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Give me that extra page here.
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There is a little one missing here.
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The wave function in example 1 is that a particle moving in a circle, what is the probability that the particle will be found between π/ 6 and π/ 3?
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The probability density we said is ψ * ψ which is also equal to the modulus of that.
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This was equal to the probability density.
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Ψ is equal to 1/ radical 2 π × E ⁺I θ.
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Ψ* is equal to 1/ radical 2 π × E ⁻I θ.
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So far so good, let us go ahead and find the probability density.
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We will just multiply these 2 together.
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Ψ* × ψ is going to equal 1/ 2 π × E ⁺I θ × E ⁻I θ which is going to equal 1/ 2 π.
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The probability is equal to the probability density × the differential element.
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D θ in this case because we are working with θ.
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Therefore, our probability is going to equal 1/ 2 π which is equal to this part, D θ.
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Now, we want to find the total probability of finding it within a particular region and we said π/ 6 and π/ 3.
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We are going to integrate from π/ 6 to π/ 3.
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Therefore, the probability of finding the particle when θ is between π/ 6 and π/ 3 is equal to the integral π/ 6 to π/ 3 of the probability.
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I actually prefer to write it differently.
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I prefer my differential element to be separate.
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I do not like to write it on top.
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This is going to equal 1/ 2 π × θ as it goes from π/ 6.
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2 π/ 3 which is equal to 1/ 2 π × π/ 3 - π/ 6, which is going to equal 1/ 2 π.
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Π/ 3 – π/ 6, 2 π/ 6 – π/ 6 is π/ 6.
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The π cancels, leaving you with the probability of 1/ 12.
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The probability density is ψ* ψ.
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The probability ψ* ψ D θ.
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If you want the probability between two certain points, in this case two certain angles, use integrate from the point to the other point.
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We will see later that the general wave equation for a particle moving in a circle is ψ sub θ = E ⁺I × M sub L θ.
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Where M sub L is a quantum number like the N in the equation for the particle in a box.
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Just another quantum number for a circular motion.
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Shows that ψ₂ and ψ₃ are orthogonal.
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In order to show orthogonality, we need to show the following.
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We need to show that the integral of ψ* of 2, ψ of 3 is equal to 0.
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We need to show that they are perpendicular.
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We need to show that they are orthogonal.
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Orthogonal was the general definition.
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We need to show that the integral of their product is equal to 0.
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Let us go ahead and do it.
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The integral of ψ* to ψ 3, it does not matter which order you do it.
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You can do ψ* ψ 3, it really does not matter.
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That is going to equal the integral from 0 to 2 π, that is our space from 0 to 2 π.
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We are talking about circular motion.
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Ψ₂ is equal to, that is the 2 and 3, that is the NL.
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We have 1/ radical 2 π × E ^- I 2 θ × 1/ radical 2 π.
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All I’m doing is just putting in the equation, plugging them into the equations that I have developed already.
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That is the nice thing about quantum mechanics.
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There is a lot going on but at least it is reasonably handle able because you have the equations.
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In fact, you just plugged them in.
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As far as the integration is concerned, sometimes you are going to have something that you can integrate really easily like these.
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Sometimes you are going to have to use your software, not a big deal.
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If you have long integration problems, please do not do the integration yourself.
00:28:13.300 --> 00:28:14.600
If you want to use tables, that is fine.
00:28:14.600 --> 00:28:19.500
I think it is nice but at this level you want to concentrate more on what is going on underneath.
00:28:19.500 --> 00:28:21.400
You want to leave the mechanics to machines.
00:28:21.400 --> 00:28:22.900
Let the machines do it for us.
00:28:22.900 --> 00:28:27.200
That is what they are for.
00:28:27.200 --> 00:28:30.800
We have E ⁺I × 3 θ D θ.
00:28:30.800 --> 00:28:34.500
This is the integral that we have to solve.
00:28:34.500 --> 00:28:37.000
It turns out to be really nice integral.
00:28:37.000 --> 00:28:41.300
We have 1/ 2 π, let us pull that out.
00:28:41.300 --> 00:28:49.200
0 to 2 π E ^- I 2 θ or 2 I θ × E³ I θ.
00:28:49.200 --> 00:28:55.100
Just add them up and you are going to end up with E ⁺I θ D θ.
00:28:55.100 --> 00:29:06.500
That is going to equal 1/ 2 π × when I integrate this, I'm going to get 1/ IE ⁺I θ.
00:29:06.500 --> 00:29:11.600
I'm going to take it from 0 to 2 π.
00:29:11.600 --> 00:29:16.500
I will do all of this in one page.
00:29:16.500 --> 00:29:36.200
= 1/ 2 π I × E² π I – E⁰ which is equal to 1/ 2 π I.
00:29:36.200 --> 00:29:41.900
Remember, E² π is cos of 2 π + I × sin of 2 π.
00:29:41.900 --> 00:30:10.800
The Euler’s relation, cos of 2 π + I × the sin of 2 π -1 is equal to 1/ 2 π I × cos of 2 π is 1 + sin of 2 π 0 -1 = 0.
00:30:10.800 --> 00:30:20.000
They are orthogonal, nice and simple.
00:30:20.000 --> 00:30:23.800
Let us see what we got here.
00:30:23.800 --> 00:30:28.700
Let ψ be a wave function for a particle in a 1 dimensional box.
00:30:28.700 --> 00:30:37.600
Calculate the expectation value, average value of the kinetic energy operator for this function.
00:30:37.600 --> 00:30:48.500
The average value of the kinetic energy operator is equal to the integral of ψ* × the operator and apply to ψ.
00:30:48.500 --> 00:30:50.400
That is the integral that we have to solve.
00:30:50.400 --> 00:30:55.000
We just plug everything in.
00:30:55.000 --> 00:31:01.200
The particular ψ, I had written here.
00:31:01.200 --> 00:31:02.600
Sometimes the problem will give you the equation.
00:31:02.600 --> 00:31:04.300
Sometimes it will not give you the equation.
00:31:04.300 --> 00:31:09.900
You have to be able to go to the tables or places in your book where you are going to find the equations you need.
00:31:09.900 --> 00:31:14.200
Much of the work that you actually do will knowing where to get the information you need.
00:31:14.200 --> 00:31:19.100
You do not necessarily have to keep the information in your head, you just have to know where to get it.
00:31:19.100 --> 00:31:35.300
If we recall or if we can look it up, the equation for a particle in a 1 dimensional box is equal to 2/ A¹/2 × sin of N Π/ A × X.
00:31:35.300 --> 00:31:39.100
The length of the box is from 0 to A.
00:31:39.100 --> 00:31:43.500
That is the equation that we want to work with, that is ψ.
00:31:43.500 --> 00:31:45.600
In this particular case, this is a real.
00:31:45.600 --> 00:31:49.700
Ψ* is equal to ψ so we can go ahead and write that down.
00:31:49.700 --> 00:31:55.900
Ψ* is equal to ψ, it is not a problem.
00:31:55.900 --> 00:31:58.900
The kinetic energy operator, let us go ahead and write down what that is.
00:31:58.900 --> 00:32:08.600
The kinetic energy operator is –H ̅/ 2 M D² DX².
00:32:08.600 --> 00:32:09.900
We are going to apply that.
00:32:09.900 --> 00:32:11.300
We are going to do this part first.
00:32:11.300 --> 00:32:15.900
We are going to apply the kinetic energy operator to ψ.
00:32:15.900 --> 00:32:21.500
K apply to ψ is equal to –H ̅.
00:32:21.500 --> 00:32:27.800
I would recommend you actually write everything out during the entire course.
00:32:27.800 --> 00:32:31.100
If you want to get in the habit of writing everything out, do not do anything in your head.
00:32:31.100 --> 00:32:32.200
There is too much going on.
00:32:32.200 --> 00:32:34.000
I do not do anything in my head.
00:32:34.000 --> 00:32:36.500
I write everything out.
00:32:36.500 --> 00:32:44.100
D² DX² of ψ which is 2/ A.
00:32:44.100 --> 00:32:45.700
Do not let the notation intimidate you.
00:32:45.700 --> 00:32:50.500
Most of it is just constants that go away.
00:32:50.500 --> 00:32:56.800
Sin N π/ A × X.
00:32:56.800 --> 00:32:58.900
Like I said, most of it is just constant.
00:32:58.900 --> 00:33:03.900
When I take the derivative of the sin N π of A twice, the derivative of sin is cos.
00:33:03.900 --> 00:33:08.000
The derivative of cos is –sin.
00:33:08.000 --> 00:33:12.600
The - and – go away and I'm left with a +.
00:33:12.600 --> 00:33:23.700
Let me write everything out here.
00:33:23.700 --> 00:33:31.000
We are going to get the H ̅²/ 2 M.
00:33:31.000 --> 00:33:35.800
We are going to pull this one out 2/ A¹/2.
00:33:35.800 --> 00:33:42.000
Again, we have the sin when we differentiate twice but because of this N π/ A × X,
00:33:42.000 --> 00:33:49.600
that is going to come out twice and it is going to be N² π²/ A².
00:33:49.600 --> 00:33:53.000
And you are going to get sin of N π A/X.
00:33:53.000 --> 00:33:55.400
This is just basic differential from first year calculus.
00:33:55.400 --> 00:33:57.600
Nothing going on here.
00:33:57.600 --> 00:34:30.500
This is the K of ψ, the ψ* × K ψ is going to equal 2/ A¹/2 sin of N π/ A × X × H ̅² N² π²/ 2 MA² × 2/ A¹/2.
00:34:30.500 --> 00:34:32.900
I’m just putting things together.
00:34:32.900 --> 00:34:49.100
Sin of N π/ A × X and that is going to equal 2/ A.
00:34:49.100 --> 00:34:52.100
Let me write everything.
00:34:52.100 --> 00:34:56.200
2/ A ^½ and 2/ A¹/2, I’m going to do it like this.
00:34:56.200 --> 00:35:03.500
It is going to be 2 on top, there is going to be A × A ^½, that is A on the bottom.
00:35:03.500 --> 00:35:05.600
A and A² becomes A³.
00:35:05.600 --> 00:35:21.600
We get H ̅² N² π²/ 2MA³ and we get sin² N π/ A × X.
00:35:21.600 --> 00:35:24.000
The 2 and 2 cancel.
00:35:24.000 --> 00:35:29.700
Now, we need to integrate this thing so we are going to have.
00:35:29.700 --> 00:35:32.400
I hope I have not forgotten any of my symbols here.
00:35:32.400 --> 00:35:41.100
H ̅², I should have an N², I should have a π², I should have an M and I should have an A³ ×
00:35:41.100 --> 00:35:50.300
the integral from 0 to A of sin² N π/ A × X.
00:35:50.300 --> 00:35:59.000
This is going to equal H ̅² N² π²/ MA³.
00:35:59.000 --> 00:36:04.500
This is going to be, when I look this up in a table or in this particular case I will use the table entry.
00:36:04.500 --> 00:36:09.200
You can have the software do it for you.
00:36:09.200 --> 00:36:12.000
This integral is going to end up being A/ 2.
00:36:12.000 --> 00:36:14.100
I will go ahead and write it out.
00:36:14.100 --> 00:36:26.300
-A × sin of N π/ A × X/ 4 N π from 0 to A.
00:36:26.300 --> 00:36:35.300
And it is going to equal H ̅² N² π²/ MA³ × A/ 2.
00:36:35.300 --> 00:36:50.700
This is A/ 2, A cancels one of these and turns it into A² and we are left with H ̅² N² π²/ 2 MA².
00:36:50.700 --> 00:36:53.400
That is correct, yes.
00:36:53.400 --> 00:36:55.000
That was what we wanted.
00:36:55.000 --> 00:36:56.300
Let me see, do I have an extra page here?
00:36:56.300 --> 00:36:59.000
Yes, I do.
00:36:59.000 --> 00:37:03.300
The expectation value of the kinetic energy operator.
00:37:03.300 --> 00:37:10.900
When I measure the kinetic energy, this is what I'm going to get.
00:37:10.900 --> 00:37:13.300
Let us do another approach to this problem.
00:37:13.300 --> 00:37:15.500
We are going to do that to the next page.
00:37:15.500 --> 00:37:18.500
Another approach to this problem.
00:37:18.500 --> 00:37:24.400
It was nice to revisit momentum every so often because momentum and
00:37:24.400 --> 00:37:29.800
angular momentum are huge in quantum mechanics, in all physics actually.
00:37:29.800 --> 00:37:38.000
Another approach to the problem.
00:37:38.000 --> 00:37:42.800
We know that K is equal to P²/ 2 M.
00:37:42.800 --> 00:37:46.400
That is just another way of writing the kinetic energy, ½ mass × velocity²
00:37:46.400 --> 00:37:55.800
is actually equal to the mass × the velocity which is the momentum²/ 2M.
00:37:55.800 --> 00:38:01.800
The average value of K is equal to the average value of P²/ 2.
00:38:01.800 --> 00:38:08.500
2M is just a constant so it ends up being the average value of P²/ 2M.
00:38:08.500 --> 00:38:14.500
From our previous lesson, we have already calculated this PM.
00:38:14.500 --> 00:38:25.900
It was H ̅² M² π²/ A².
00:38:25.900 --> 00:38:35.900
We have H ̅² M² π²/ A² / 2M.
00:38:35.900 --> 00:38:38.300
Just put this over the 2 M.
00:38:38.300 --> 00:38:43.400
We will put the 2M down here and we get the same answer as before.
00:38:43.400 --> 00:38:50.700
You can do it with the definition of expectation value or you can do it with something else based on something that you already done.
00:38:50.700 --> 00:38:53.200
This is a really great relation to remember.
00:38:53.200 --> 00:39:00.500
Kinetic energy is the momentum² or twice the mass.
00:39:00.500 --> 00:39:04.100
Where are we now?
00:39:04.100 --> 00:39:10.600
Evaluate the commutator of P sub X P sub Y and the commutator X² P of X.
00:39:10.600 --> 00:39:13.500
Let us go ahead and do this first one.
00:39:13.500 --> 00:39:18.100
When we evaluate these commutator relations, use a generic function F.
00:39:18.100 --> 00:39:22.700
Just use F, do not try to do these symbolically without a function.
00:39:22.700 --> 00:39:25.700
At least until you become very comfortable with this.
00:39:25.700 --> 00:39:27.900
I, myself is not comfortable with it.
00:39:27.900 --> 00:39:32.500
I like to put my function in there because I know I’m operating on a function and the end just drop the function
00:39:32.500 --> 00:39:38.400
and you are left with your operator symbol.
00:39:38.400 --> 00:39:39.400
You write that down here.
00:39:39.400 --> 00:40:01.600
When doing these, use a generic F and by generic F I mean just the symbol F.
00:40:01.600 --> 00:40:14.100
Do not use just the operators until you become much more proficient and familiar with operators.
00:40:14.100 --> 00:40:21.100
This P sub X P sub Y, the most exhausting part of quantum mechanics is writing everything down.
00:40:21.100 --> 00:40:25.400
This is the symbolism, this is just so tedious.
00:40:25.400 --> 00:40:27.900
Applied to some generic function F.
00:40:27.900 --> 00:40:40.300
That is equal to P sub X P sub Y of F – P sub Y P sub X of F.
00:40:40.300 --> 00:40:43.000
We know what we are doing here.
00:40:43.000 --> 00:40:53.800
P sub X - IH DDX that is the P sub X operator and the P sub Y.
00:40:53.800 --> 00:41:00.000
This is P sub Y applied to F, then do P of X applied what you got.
00:41:00.000 --> 00:41:01.700
We are working from right to left.
00:41:01.700 --> 00:41:04.100
Remember, sequential operators.
00:41:04.100 --> 00:41:10.900
This is going to be - I H ̅ DF DY.
00:41:10.900 --> 00:41:15.200
Notice, I put the F in there so operate on a function.
00:41:15.200 --> 00:41:28.200
- I H ̅ DDY - I H ̅ DF DX.
00:41:28.200 --> 00:42:00.900
Here we get – H ̅² D² F DX DY - H ̅² D ⁺F DY DX is actually = 0.
00:42:00.900 --> 00:42:23.900
And the reason it is equal 0 because for all of the functions that you are going to be dealing with, use mixed partial derivatives.
00:42:23.900 --> 00:42:26.400
And again, we saw this in thermodynamics.
00:42:26.400 --> 00:42:33.300
Mixed partial derivatives, by mix partial we mean the partial with respect to X first then the partial with
00:42:33.300 --> 00:42:39.000
the respect to Y is equal to the partial with respect to Y first and then the partial with respect to X.
00:42:39.000 --> 00:42:46.200
The order in which you operate, the order in which you take the derivative, it does not matter for all well behaved functions.
00:42:46.200 --> 00:42:50.600
By well behaved, it just means to satisfy certain continuity conditions.
00:42:50.600 --> 00:42:55.400
For our purposes, you will never run across a function that does not satisfy this.
00:42:55.400 --> 00:42:58.400
We will always be dealing with functions that satisfy this property.
00:42:58.400 --> 00:43:02.700
This and this, even though the orders are different, they are actually equal.
00:43:02.700 --> 00:43:06.000
- + you end up with 0.
00:43:06.000 --> 00:43:10.800
Mixed partial derivatives are equal.
00:43:10.800 --> 00:43:21.100
In other words, the D ⁺2F DX DY is absolutely equal to D² F DY DX.
00:43:21.100 --> 00:43:23.800
That is a fundamental theorem in multivariable calculus.
00:43:23.800 --> 00:43:28.700
The order of differentiation does not matter.
00:43:28.700 --> 00:43:32.200
Let us try our next commutator.
00:43:32.200 --> 00:43:43.300
We want to do the X² PX was going to equal X².
00:43:43.300 --> 00:43:53.100
If you remember the X operator, the position operator just means multiply by X and the PX operator is - I H ̅.
00:43:53.100 --> 00:43:58.700
Again, we are going to do DF DX -, now we are going to switch them.
00:43:58.700 --> 00:44:08.700
We are going to do PX X² - IH DDX.
00:44:08.700 --> 00:44:13.800
We are going to do X² F.
00:44:13.800 --> 00:44:21.900
It is this × this - this × this and this × this order.
00:44:21.900 --> 00:44:25.200
Be very careful here.
00:44:25.200 --> 00:44:35.700
This is going to be - I H ̅ X² DF DX, I just change the order here.
00:44:35.700 --> 00:44:37.600
Nothing strange happening.
00:44:37.600 --> 00:44:43.500
And then this one is going to be +.
00:44:43.500 --> 00:44:49.900
Notice, now I have an X² F.
00:44:49.900 --> 00:44:53.700
Let me go ahead and write this up.
00:44:53.700 --> 00:45:02.100
This is , X² PX - PX X².
00:45:02.100 --> 00:45:06.200
We have X² F, this is a function × a function and differentiating that.
00:45:06.200 --> 00:45:12.900
I have to use the product rule so it is going to be this × the derivative of that + that × the derivative of this.
00:45:12.900 --> 00:45:25.400
It is going to be, the negative cancels so I get + I H ̅ this × the derivative of that is going to be X² DF DX
00:45:25.400 --> 00:45:41.100
+ that × the derivative of this + I H ̅ 2 XF – I X² DF DX + IHX² DF DX.
00:45:41.100 --> 00:45:50.000
These go away, I'm left with I H ̅ 2 X F.
00:45:50.000 --> 00:45:55.400
I know I can go ahead and drop that F in terms of we know it is not equal 0.
00:45:55.400 --> 00:46:05.100
What is happening now, I can go ahead and drop the F part and just deal with the operator part.
00:46:05.100 --> 00:46:14.300
It is equal I H ̅ 2 X which definitely does not equal the 0 operator.
00:46:14.300 --> 00:46:23.200
This is the operator, this is our answer.
00:46:23.200 --> 00:46:28.600
The a commutator of this is equal to that.
00:46:28.600 --> 00:46:32.800
We include F in order to keep track of our differentiation properly.
00:46:32.800 --> 00:46:37.000
If we did not include the F, we would not have F here, we would not have the F here.
00:46:37.000 --> 00:46:42.200
It might cause some confusion as far as where is the product rule.
00:46:42.200 --> 00:46:43.400
That is why we are putting it in there.
00:46:43.400 --> 00:46:47.700
It is very important to put it in there until you become very accustomed to operators.
00:46:47.700 --> 00:46:51.000
I, myself, do not, I use F.
00:46:51.000 --> 00:46:54.200
That is it, thank you so much for joining us here at www.educator.com.
00:46:54.200 --> 00:46:56.900
We will see you next time for a continuation of example problems.
00:46:56.900 --> 00:46:58.000
Take good care, bye.