WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to close out our discussion of the postulates and principles of Quantum Mechanics.
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Let us dive right on in.
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We have already seen that operators can be applied sequentially.
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Let us go ahead and write that down.
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We have already seen that operators can be applied sequentially.
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In other words, if I do A, B of some function this just means working from right to left.
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Apply B first then apply it to what you got.
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It means apply B to F first, then apply A to the result.
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In general, AB of F does not equal BA of F.
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In general, if you apply B first and apply A to what you got, it is not the same as applying A and applying B to what you got.
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When it does happen that way, when AB of F, when it does equal BA of F, we say the operators AB commute.
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In other words, A and B are commutative just like the numbers 2 × 4 = 4 × 2.
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When numbers is fine with the operator, sometimes it is, sometimes it is not.
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When it does happen that way, we call it commutative.
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Let us go ahead and calculate a couple of things.
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Let us calculate, pick a couple of operators.
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Let us calculate the linear momentum that of F.
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P ̂ X ̂ of F and then let us calculate X ̂ P ̂ of F.
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Let us see what happens.
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Well P ̂ X ̂ of F = - I H ̅ DDX of XF.
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The X operator just means multiply the function by X.
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When we do that, we get - I H ̅.
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I will go ahead and use it.
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DDX, this is X and this is a function of X so we have to use the product rule.
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This × the derivative of that + that × the derivative of this.
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This is going to be X DF DX, this × the derivative of this + that.
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That is that one, that is P ̂ X ̂.
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Let us go ahead and do X have P ̂.
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X ̂ P ̂ of F is going to equal X ̂ apply to - I H ̅ DDX of F.
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I will go ahead and just do it as a single.
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This is going to be DF DX = - I H ̅ X DF DX, that is that one.
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Notice, this and this are not the same.
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P ̂ X ̂ of F does not equal X ̂ P had of F.
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They are not commutative.
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These two operators, in other words the linear momentum operator, and the position operator do not commute.
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Let us calculate another pair.
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Let us calculate the kinetic energy and linear momentum.
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This is sub X because we are just doing in the X direction.
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Let us do K ̂ P ̂ and let us do P ̂ K ̂.
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I think I'm going to actually go ahead and drop the X.
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I do not think it is going to be very confusing.
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For right now, I’m going to stick with working in one dimension.
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I’m realizing that it is actually too many symbols floating around and it is a little confusing.
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K ̂ P ̂ and P ̂ K ̂, let us see what we get.
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The K ̂ P ̂ of F is going to be – H ̅²/ 2 MA² DX² apply to - I H ̅ DF DX.
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What is in the brackets is this one, P apply to F first and then I'm going to apply K to what I got.
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When I do that, I end up with -I H ̅³/ 2M D³ F DX³.
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That takes care of K ̂ P ̂ of F.
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Let us do the other.
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Let us do P ̂ K ̂ of F.
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That is going to equal - I H ̅ DDX of –H ̅²/ 2 M D² F DX².
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We apply K first to F, that is what is in the brackets and then we are going to apply P.
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When we do this, we end up with I H ̅³/ 2 M D³ F DX³.
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Notice, this and this are the same.
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The kinetic energy operator and the linear momentum operator they do commute.
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K ̂ P ̂ = P ̂ K ̂, this is commutative.
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The kinetic energy operator and the linear momentum operator, they do commute.
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We just solved the linear momentum operator and the position operators do not commute.
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When A and B, when the two operators commute then we know that AB of F = BA of F.
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I'm going to move this over to the left hand side.
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This is going to be A ̂ B ̂ of F - B ̂ A ̂ of F.
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I’m just working symbolically.
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An operator is just the thing that you can treat like a number, = 0.
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Let me go ahead and factor out the F.
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I get A ̂ B ̂ - B ̂ and A ̂ of F = 0.
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What I’m going to do is I’m going to introduce symbol here, 0 ̂ of F,
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Where this 0 ̂ is just the 0 operator and it means multiply by 0.
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Again, we just want to be mathematically consistent.
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We have an operator operating on F, an operator operating on F.
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We know it equal 0 but we want to symbolically be consistent.
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Is the 0 operator meaning multiply by 0, multiply by the number 0.
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This expression right here, let me go back to red.
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This expression right here is called the commutator of A and B.
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Commutator of A and B symbolizes as follows.
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It is A, B and we put in brackets.
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This symbol means AB – BA.
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It is called the commutator of A and B.
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We saw that the commutator of the kinetic energy and the linear momentum operator is equal to 0.
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Because PP - PK was equal to 0, KP = PK.
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We also saw that the linear momentum and the position operator does not equal that
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because the commutator does not equal 0.
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The linear momentum position operator of F - the position operator linear momentum of F equals the following.
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That was - I H ̅ X DF DX - I H ̅ F -,
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When I did this, I got some value.
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When I did that, I got another value.
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When I subtract these two, which is the commutator of that, I'm going to just put those values and then subtract.
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- and - I H ̅ X DF DX, this is a + this is a -, these cancel.
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I'm just left with this = - I H ̅ F.
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I hope that makes sense.
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I have P X bar of F - X ̂ P ̂ of F = this.
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We can write this as, let me factor out the F.
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P ̂ X ̂ - X ̂ P ̂ of F = - I H ̅ I ̂ of F.
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Where I ̂ was being symbolically consistent.
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I ̂ is called the identity operator and it just means multiply by 1.
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It is the identity operator and means multiply by 1.
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In other words, take this function and multiply by 1.
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It is just the way of operator, you have some value.
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Operator operating on a function, operator operating on function.
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We are trying to establish some consistency, multiply by 1.
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This is the commutator.
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P ̂ X ̂ is equal to - I H ̅ identity operator.
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All I have done is move the X out of the way, now I’m just dealing in operator notation.
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The function itself is not irrelevant but from the mathematical perspective, you do not have to deal with it because any function will do.
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It is the operator that is actually important.
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Let me go back to black here.
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We already know that σ, the standard deviation of the position of a particle × the standard deviation of the momentum of a particle is ≥ H ̅/ 2.
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This is the Heisenberg uncertainty principle.
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Standard deviation, uncertainty, that is what uncertainty is.
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A standard deviation gives me a degree of uncertainty as far as quantum mechanics is concerned.
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It represents the uncertainty in the measurement that I make.
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If my standard deviation is 0 that means that every measurement that I make is the same number over and over and over again.
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If I take 1000 measurements, not all the measurements are going to be the same, it is going to be some leeway.
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We have already seen that the uncertainty in a position × the uncertainty in the momentum is going to be ≥ H ̅/ 2.
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In other words, I cannot simultaneously measure the position and
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the momentum of a particle to an arbitrary degree of the precision or accuracy.
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I’m limited.
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As if I'm better if I measure that one really well, then the other I’m not going to know.
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If I measure the other one really well, the other one I'm not going to know.
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That is what this means.
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That is the position and momentum cannot simultaneously be measured to arbitrary degree of accuracy.
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The non commutativity of two operators like the position of linear momentum operator, we saw that they are non commutative.
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Those operators do not commute.
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The non commutativity of two operators and we also saw that the uncertainty in the linear momentum
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and the uncertainty in the position is ≥ some number.
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The non commutativity of two operators and the uncertainty relationship between them is not coincidental.
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We already knew that the uncertainty in the position × the uncertainty in the linear momentum is going to be≥ H ̅/ 2.
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We also saw that those two linear operators, the linear momentum operator and the position operator do not commute.
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That is not coincidental, it is always going to be like that.
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I'm going to write down the general uncertainty relationship, the broadest one.
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This is the uncertainty principle.
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What we found before was actually a special case and we are going to go through it and derive it in a second from the general principle.
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The general uncertainty principle, given two operators A and B, the uncertainty in the measurement of A × the uncertainty
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in the measurement of B is going to be ≥ ½ × the absolute value of the integral of C sub *.
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I cannot draw brackets anymore.
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A B C the absolute value.
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Where σ A and σ B are the standard deviations uncertainties in the measure quantities, corresponding to their respective operators.
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This is the general uncertainty principle.
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It says that if I have a particular wave function and let us say I go ahead and measure some property of that wave function A,
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and then I measure some property of that wave function B,
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I make multiple measurements and end up getting uncertainties.
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I’m going to end up getting standard deviations for those particular measurements.
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If I apply the commutator, in other words if I apply AB of ψ - BA of ψ and I multiply that by the conjugate of ψ.
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And if I integrate that over my region of interest, whatever number I get, the absolute value of that divided by 2, this is the relationship.
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All certainties of those measurements multiplied by each other is going to be ≥ that.
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For operators that do not commute like the linear momentum and position operators, that happened to be H ̅/ 2.
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In other words, I cannot know the position really well and the linear momentum really well.
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I have to choose which one I know or I have to strike a balance between them.
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But for operators that do commute, we saw that the commutator is equal to 0,
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which means that this here, the σ A × the σ B is going to be equal to 0,
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which means that I can measure each to an arbitrary degree of accuracy.
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I can find out the kinetic energy and the linear momentum of a particle as closely as I want to.
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That is what is going on here, that is the uncertainty principle here.
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Once again, the uncertainty of a measurement for one operator × the uncertainty in the measurement for another operator
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is going to be ≥ ½ the absolute value of this integral, where this thing is the commutator.
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This commutator just says apply AB – BA to ψ, that is all that is going on here.
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Let us go back.
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In the case of this linear momentum and position operator.
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Let us take a look of this one, this is a special case.
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Let us use the definition that we just got.
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The σ of P, the σ of X, it says that is ≥ ½ the absolute value of the integral of ψ* ×,
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Apply the commutator 2 ψ DX.
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This is going to equal the absolute value, the integral of ψ sub *.
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We have already found what the commutator of this is.
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It is equal to - I H ̅ × the operator of ψ DX, we found this already.
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The commutator is what we just substituted in.
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It is ½, I forgot the ½ here.
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½ × the absolute value of - I H ̅ and I'm going to go ahead and pull the constant out.
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The integral of ψ sub * I of ψ DX.
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This right here, this identity operator just mean multiply by 1.
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This integral here ends up becoming the integral of ψ* ψ DX.
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The normalization condition guarantees that this is equal to 1.
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This goes to 1.
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What we have is ½ × the absolute value of - I H ̅ × 1.
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This is just the absolute value of –I × the absolute value of H ̅.
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The absolute value of –I is just 1.
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Remember, the complex plane, here is 1, here is I, here is -1, here is –I.
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The absolute value, the distance from 0 to -I is 1.
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Therefore, this is just equal to H ̅/ 2,
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Which means that σ of P × the σ of X is ≥ H ̅/ 2, but we knew that already.
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But we ended up deriving it from the general expression of the uncertainty principle based on the commutator.
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That is all we have done.
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Let me go ahead and close this out.
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Let us go back to black.
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When the commutator of two operators = the 0 operator, then this integral,
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I will go ahead and put the absolute values in.
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Ψ sub * of ABC will equal 0, the commutator = 0.
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It was a 0 operator which means this is just going to be the integral of ψ sub * × the 0 operator of ψ.
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The 0 operator × operator of ψ is just 0.
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This ends up being the integral of 0, it just equal 0.
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The σ of A × the σ of B is going to be ≥ 0 or = 0.
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Your σ A can be 0, your σ B can be 0.
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The two things multiply together it equal 0.
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One is 0 and the other is 0, or both can be 0.
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When you have a standard deviation which is 0, when you have an uncertainty that is 0,
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that means that you know exactly what the value is.
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There is no uncertainty, there is no deviation of the data.
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If you make 1000 measurements, those 1000 measurements are all going to be the same number.
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If they all end up with the same number, your standard deviation is going to be 0.
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If your standard deviation is 0, there is no uncertainty in that measurement.
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I'm not uncertain, I know exactly what it is, it is that number.
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In the case of operators that commute, I can measure each one to any degree of accuracy that I want.
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In the case of the kinetic energy and the linear momentum, I can know the kinetic energy very well.
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I can know the linear momentum very well.
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In the case of operators that do not commute, these uncertainties have to be a certain minimum value.
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They have to be bigger than that.
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In the case of the σ X σ P, they have to be bigger than that.
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In other words, I’m limited.
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If I know the momentum really well, I do not know the position.
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If I know the position really well and accurately, I do not know the momentum.
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They have to have a strike of balance, that is what is going on here.
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In this case, A and B, as observables, can be measured to any degree of accuracy.
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That is the general expression for the uncertainty principle and
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it is based on this thing called the commutator of those two operators.
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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, bye.