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 continue our discussion of the particle in a box.
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Let us jump right on in.
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We have from the previous lesson, for a free particle in a one dimensional box, we have the following.
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I’m going to stop using the letter ψ because I personally do not like the letters.
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I’m going to use a different letter for the wave function.
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I’m going to use W.
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For a free particle and a one dimensional box, the particles just moving back and forth.
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It can only move one dimension between 0 and A.
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We have the wave function W sub N of X = 2/ A¹/2 × the sin of N π / A × X.
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The energy for that particular state of N was equal to H² N²/ A² × 8M.
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Let me make M a little bit clear here.
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We are going to say X is between 0 and A, and N of course takes on integer values 123 and so on.
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This was the solution to our particle in a box problem.
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Recall from the lesson on the mathematical interlude on probability and statistics, we have the following.
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We had the average value of X was equal to the integral of X × the probability of X DX.
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We have the average value of X² that was something that we call the second moment.
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That was going to be integral of X² × the probability of X DX.
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And we have something called the variance.
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The variance which we symbolized as that, that was equal to the integral of the real value of X - the average value of X² × the probability of X DX.
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And again, just treat these mathematically.
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A lot of what we are doing in quantum mechanics.
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If we do not entirely understand what is happening, it is okay.
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Just treat it mathematically, become accustomed to the mechanics and eventually the understanding will emerge as you do more and more problems.
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As you and your teacher and your friends and colleagues discuss things more.
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Just deal mathematically if all of this does not entirely make sense.
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That was also equal to the averages value of the second moment which is X² – the square of the average value.
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There are 2 ways, you can find this and find that, and define the variance that way.
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Or you can just go ahead and integrate this function.
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In all these cases, this piece of X , piece of X ,piece of X, the PX DX including the DX that is the probability.
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I will write where P of X DX is the probability of a given state.
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For a particle in a box, our probability, our PX DX = W* × W × DX.
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W* W DX = this × itself DX.
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I will go ahead and just write it, that is not a problem.
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2/ A¹/2 × sin N π AX × 2/ A¹/2 sin of N π / AX DX.
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This thing is our probability.
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Again, 0 less than or equal to X, less than or equal to A and it is equal to 0 otherwise outside of the interval.
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Let us go ahead and write that here just in case we forgot.
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The left of 0 to the right of A, the probability is 0.
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In other words, the particle will never be found there.
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That is all that means.
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We are going to use this probability to put it into these equations to calculate some average values and some standard deviations and variances.
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We use this W* W DX to calculate some average values and standard deviations.
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And if you remember, the standard deviation is the square root of the variance.
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The variance is the σ² X.
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Let us go ahead and calculate the average value of X.
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The average value of X, in other words the average value of the position of the particle.
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That is what X represents.
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X is just where the particle is.
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I take a measurement, I take another measurement, each measurement that I take, the particles are going to be somewhere.
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It is just going to be somewhere.
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Sometimes it is here, sometimes it is there.
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We take 10 measurements, 50 measurements, 100, 10,000, 100,000, a 1,000,000 measurements.
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Once I have those million measurements, I want to find the average value.
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It is going to equal this.
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It is going to equal by definition the integral of X PX DX, that is the definition of the average value.
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In this particular case, it is going to be the integral from 0 to A of X W* W DX.
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X × the probability, X × in this particular case the probability of a particle in a box is this.
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And we integrate it from 0 to A.
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What we end up having is the integral from 0 to A of 2/ A.
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I'm sorry X × 2/ A sin² N π/ A X DX.
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If you look at this particular integral, let me pull the constant out.
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2/ A, 0 to A of X sin² N π A/ X DX.
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If you look up this integral in a table, you will find it.
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Or have your software do it, you are going to end up with the following.
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It turns out that the average value of X is going to equal 2/ A × when you do this integral here, you are going to end up with A²/ 4 A/ 2.
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It turns out that the average value of the position of the particle is A/ 2.
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What is that mean?
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Our interval is 0 to A.
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If I take a million measurements on average, sometimes it is going to be here, sometimes here.
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If I average it out over many, it is going to be A or 2.
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In other words, on average I'm going to find it right in the middle.
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That is all what average is.
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It is a mean value.
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On average, I'm going to find the particle right there, that is all this is saying.
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This make sense, I mean the particles can sometimes be here, sometimes here,
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Over a bunch of measurements, it is going to average out to right down in the middle.
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If you flip a coin and get heads, if you flip a coin you get tails.
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If you keep flipping 100, 200, 300, or 1000 flips, you are going to end up getting just as many heads as you get tails on average.
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Let us go ahead and find the average value of X² which is something called the second moment.
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The average value of X² that is equal to the integral of X² × PX DX, that is the definition.
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It is going to equal the integral from 0 to A of X² × the wave function × itself DX, that is going to equal,
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I’m going to pull the constant out, the integral from 0 to A of X² sin² N π/ A DX.
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When you solve this integral, you are going to end up with A²/ 3 - A²/ 2 N² π².
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The variance is equal to the average value of X² - the average value of X².
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When I do that, I get A²/ 3 - A²/ 2 N² π² - A/ 2²,
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Because the average value of X was A/ 2.
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It was A/ 2².
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That is fine, I will just do it.
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It is going to be A²/3 – A²/ 2 N² π² – A²/ 4.
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I find myself a common denominator with the 4 and 3.
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Let us go ahead and do this A²/ 12.
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4A² – 3A²/ 12.
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A²/ 12 – A²/ 2 N² π².
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This is our σ² X.
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When I take the square root of that, I get the actual standard deviation S sub X.
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I get σ sub X is going to equal A²/ 12 - A²/ 2 N² π² all raise to the ½.
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I found the average value of X.
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I found the average value of X².
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And I use these two to find this one.
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That is that right there.
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What happens when I want to calculate the average value of the energy or the average value of the momentum?
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Now, we want to calculate the average value of the energy or the average value of the momentum.
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I recall that things like energy and momentum they are represented by operators.
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In this particular case, they are represented by differential operators so it creates a little bit of a problem how do we actually do that?
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Recall that energy and momentum are represented in quantum mechanics by differential operators.
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Remember, the energy operator which was the Hamiltonian operator which was – H ̅²/ 2 MD² DX² + VX that was the operator.
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Of course, we had the momentum operator which was -I H ̅ DDX.
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Now the question is how do I find the average value of the momentum?
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When the momentum is represented by this differential operator, on what function do I actually operate?
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We have a wave function, that is not a problem.
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We have our wave function, the 2/ A ⁺square root.
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√2/ A × the sin N π/ AX.
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We have the wave function and we know that if we want to extract some information,
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like something about momentum, we operate on that function.
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We want to find the average value so are we operating on the complex conjugate W?
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Are we are going to operate on W*?
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Are we going to operate on W? Are we going to operate on W* × W?
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What is it that we do?
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The question is, on which function does the operator operate?
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Is it the conjugate?
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Is it the function itself?
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Is it the probability density? Is it the square of the wave function?
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Which one is it?
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Let us go ahead and see if we can find out.
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Let us go ahead and recall how our operator version, our Eigen value problem.
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I have got the Hamiltonian operator operating on the wave function WN.
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It is going to equal the energy, the wave function.
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This was our Eigen value problem.
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This was the Schrӧdinger equation expressed.
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This is the Schrӧdinger equation in operator form or Eigen function, Eigen value form.
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WN is the Eigen function, E sub N is the Eigen value.
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Here is what I’m going to do.
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I’m going to fiddle around with this a little bit.
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I’m going to multiply on the left by the conjugate of the wave function and I’m going to integrate.
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I'm going to get the following.
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I'm going to get the integral of W* HW = the integral of W* E sub NW.
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I can pull the Z sub N, it is just a number.
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It is a scalar so I can pull it out.
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That is equal to E sub N × the integral of W* W.
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The wave function is normalized so the integral of the square of the wave function, this is just going to end up being 1.
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We end up with that.
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I found the energy simply by multiplying on the left by operating on the function and
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then multiplying on left by the complex conjugate, and then integrating over the particular interval.
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That ends up giving me my energy.
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This is extraordinary.
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Find the average value of a physical quantity like energy or like momentum that is associated with a quantum mechanical operator...
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Whatever, we have to find the physical quantity that is associated with an operator.
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If I have the momentum operator and if I want to find the average momentum, I take the wave function,
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I operate on the wave function, I multiply it on the left by the conjugate of the wave function, and then I integrate over the entire interval.
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That gives me the average value.
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This is the definition of finding the average value of a physical quantity that is associated with an operator.
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Here, L is the operator and L is the average value of that operator.
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Is the average value, I should say of the quantity for the particle in the state described by W sub N.
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Let us go ahead and calculate the average momentum of the particle in a box.
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We have the average momentum of the particle in a box.
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That is equal to the integral of W conjugate × the momentum operator W DX.
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That is going to equal the integral of 2/ A ^½.
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You literally just put everything in.
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It looks really complicated, but it is not.
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Sin of N π/ A × X.
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This momentum operator you have – I H ̅² DDX.
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We are going to operate on 2/ A ^½ sin N π/ A × X.
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And you are going to integrate all of that from 0 to A.
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Here, let us pull some things out.
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I pull this out, I pull this out, I can pull this out, and when I differentiate this,
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I’m not going to go through all the steps, here is what I end up with.
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-I H ̅ that takes care of that.
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-I H this is not squared, this is a momentum operator.
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-IH and then this and this, gives me 2/ A.
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All I’m left with is, take the derivative of this function, that is what we are doing.
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You are going to apply this operator to this function and then multiply it by that.
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When I take the derivative of sin of N π A/ N π/ A of X, I end up with N π/ A × cos of N π A/ X.
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That constant also comes out N π/ A × integral from 0 to A of the sin of N π/ AX × the cos of N π/ A × X × DX.
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This integral = 0.
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Therefore, my average momentum is equal to 0.
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Again, this makes sense and here is why.
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If I have 0 to A, this is an average value.
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This is that if I take 10,000 measurements, there going to be times that the particle is moving in this direction.
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There are going to be times that the particle is moving in that direction.
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This direction, that direction, that direction.
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When I average it out, these directions are going to cancel out.
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The average momentum of the particle is going to end up being 0, that is what this means.
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The average value that you get from taking thousands and thousands of measurements.
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Not even thousands, maybe just hundreds of measurements, maybe 50.
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On average, this is what is going to happen and it makes sense physically.
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In a way of looking at is your equally likely to find a particle moving to the left as it is moving to the right.
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On average, it is not moving at all.
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Let us go ahead and talk about something called the uncertainty principle.
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This is very important.
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It is fine, I will stick with blue.
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The uncertainty principle or the Heisenberg uncertainty principle.
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We have calculated the average value of X and we also found the σ of X.
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We found the standard deviation and we also found the average momentum.
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Let us go ahead and let us find.
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We found the average value of X and the standard deviation of X.
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We found the average momentum, the average value of P.
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Now, let us find the standard deviation of the momentum.
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Let us find σ sub P.
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OK so we know that σ² of P that is going to equal the average value of the P² - the average value for P quantity².
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I need to find this value now and subtract in order to find this, and take the square root of it, in order to get that right there.
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The first thing we need to do is find the second moment of the momentum.
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The average value of P² that is going to equal, what you got is the integral of the conjugate, the operator² that.
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Remember, an operator² is the same as just doing the operator and doing it again.
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That just means do it twice, that is all the squared means, do it twice.
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What we are going to have is the following.
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This is equal to.
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Let me write the whole thing.
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0 to A, 2/ A × sin N π/ A × X × -I H ̅ DDX, that is one operation.
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We have – I H ̅ DDX is the second operation and we are operating on that function which is sin of N π/ A × X.
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We are integrating from 0 to A.
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This is what we are integrating.
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This means take the derivative of this and then take the derivative of it again.
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That is a constant, that is a constant, I forgot the 2A over here.
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It is the hardest part of quantum mechanics, just keeping all of that straight.
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It is not that it is conceptually difficult.
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Let us go back to red.
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This is a constant, that is a constant.
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When you pull all of that out, you end up with the following.
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You will end up with - H ̅² × 2/ A the integral from 0 to A.
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I’m just going to go ahead.
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That is fine, I will write it all out.
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Sin of N π/ A × X.
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Now we have D² DX² of the sin N π/ A × X DX.
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What you will end up with is, we have 2 H ̅² N² π²/ A × A² × the integral from 0 to A of sin² N π/ A X DX.
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The derivative of sin is cos, the derivative of cos is negative sin.
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The negative and negative cancel to give me a positive.
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The derivative of sin of this thing, that constant comes out once, the constant comes out twice.
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You are going to get N² π²/ A².
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Here is the N² π²/ A².
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We have H ̅ 2/ A, H ̅ 2/ A.
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That is all where this comes from.
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Now when I do this, I’m going to get the following.
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I'm going to get 2 H ̅² N² π²/ A × A² × A / 2.
00:30:15.200 --> 00:30:22.300
In other words, when I solve this integral I'm going to end up with A/ 2.
00:30:22.300 --> 00:30:27.900
To again, just use the table, use mathematical software whatever it is that you need to do in order to integrate it.
00:30:27.900 --> 00:30:52.800
A cancels with A, 2 cancels with 2, and what I am left with is the average value of the momentum² is going to equal H ̅² N² π²/ A².
00:30:52.800 --> 00:30:57.700
There we have that.
00:30:57.700 --> 00:31:03.800
We know that the average value of the momentum = 0 and we know that
00:31:03.800 --> 00:31:14.900
the average value of the second moment of the momentum = H ̅² N² π²/ A².
00:31:14.900 --> 00:31:24.900
Now, the variance = this² - that².
00:31:24.900 --> 00:31:27.600
This is just 0 so this goes to 0.
00:31:27.600 --> 00:31:39.800
I'm left with P = H ̅² N² π²/ A²,
00:31:39.800 --> 00:31:48.500
Which implies that the σ of the P = H ̅ N π/ A.
00:31:48.500 --> 00:31:52.000
There we have it.
00:31:52.000 --> 00:32:00.100
We have σ P, we have the σ X, let us see what we can do.
00:32:00.100 --> 00:32:06.600
Let us go back to blue here.
00:32:06.600 --> 00:32:43.000
Both σ P standard deviation and σ P² variance, are measures of the extent of deviation from the mean value.
00:32:43.000 --> 00:32:44.800
That is what they represent.
00:32:44.800 --> 00:32:46.400
You have a certain set of data.
00:32:46.400 --> 00:32:48.600
That certain set of data has an average value.
00:32:48.600 --> 00:33:01.300
The standard deviation is a numerical measure of the extent to which all of the data as a whole deviate from that mean value.
00:33:01.300 --> 00:33:05.100
The standard deviation of the mean value.
00:33:05.100 --> 00:33:52.600
Therefore, we can interpret σ P as a measure of the uncertainty involved in the measurement.
00:33:52.600 --> 00:33:59.300
You have a set of data, that set of data has an average value.
00:33:59.300 --> 00:34:09.300
If I take any particular measurement that I have made and if I subtract from it the average value, if I take the absolute value,
00:34:09.300 --> 00:34:15.200
Basically, it is the difference between any one measurement and the mean value of all the measurements,
00:34:15.200 --> 00:34:17.100
there is going to be some sort of a gap there.
00:34:17.100 --> 00:34:20.400
That gap is what the variance is.
00:34:20.400 --> 00:34:27.700
It is a numerical measure of the actual deviation from the mean value.
00:34:27.700 --> 00:34:36.100
For example, if I had a bunch of values and a mean value happens to be 5 and if I take some random data point 5.3.
00:34:36.100 --> 00:34:41.200
That 5.3 and the 5, there is a difference of 0.3.
00:34:41.200 --> 00:34:46.400
There is some sort of an error if you will, in that measurement.
00:34:46.400 --> 00:34:54.500
The standard deviation, it is a measure of the extent to which any given measurement actually deviates from the average value.
00:34:54.500 --> 00:34:59.600
We are going to interpret it as the uncertainty in any given measurement.
00:34:59.600 --> 00:35:02.400
That is what we are going to do.
00:35:02.400 --> 00:35:21.300
Let us go ahead and take the variance of our position σ sub X.
00:35:21.300 --> 00:35:32.100
We said that was equal to A²/ 12 - A²/ 2 N² π².
00:35:32.100 --> 00:35:34.400
I’m going to write this in a way that makes it a little bit more convenient.
00:35:34.400 --> 00:35:47.100
I’m going to write this is A²/ 4 π² N² × N² π²/ 3 – 2.
00:35:47.100 --> 00:35:57.600
I also have the variance of the momentum which is H ̅² N² π²/ A².
00:35:57.600 --> 00:36:04.100
Notice, as far as the variance with a measure of the standard deviation,
00:36:04.100 --> 00:36:13.000
measure of the uncertainty and as far as the position is concerned, everything else here is a constant.
00:36:13.000 --> 00:36:17.700
It is a function of A but it is A in the numerator.
00:36:17.700 --> 00:36:24.400
For the uncertainty and the momentum, A is in the denominator.
00:36:24.400 --> 00:36:26.600
This is important.
00:36:26.600 --> 00:36:29.000
Watch what happens here.
00:36:29.000 --> 00:36:35.000
Let me go ahead and write it actually on the next page again.
00:36:35.000 --> 00:36:47.900
Σ² of X = A²/ 4 π² N² × N² π²/ 3 – 2.
00:36:47.900 --> 00:36:58.800
And then I have over here, I have the variance of the momentum which is equal to H ̅² N² π²/ A².
00:36:58.800 --> 00:37:03.100
Here is A is in the numerator, here A is in the denominator.
00:37:03.100 --> 00:37:11.800
Here is what happens.
00:37:11.800 --> 00:37:19.900
As A increases, the σ sub X also increases.
00:37:19.900 --> 00:37:28.500
Σ sub X², I just took the square root of this, it also increases.
00:37:28.500 --> 00:37:37.700
The σ sub P, as A increases the σ sub P decreases.
00:37:37.700 --> 00:37:49.100
As A decreases, the uncertainty in the position decreases but the uncertainty on the momentum increases.
00:37:49.100 --> 00:37:54.500
Basically, if I have some interval from 0 to A, if I now make a bigger.
00:37:54.500 --> 00:38:04.000
In other words, if I give more room for the particle to be my uncertainty in where the particle is, goes up.
00:38:04.000 --> 00:38:07.200
Now, it is very delocalize.
00:38:07.200 --> 00:38:09.400
It could be anywhere from 0 to A.
00:38:09.400 --> 00:38:15.100
It is a huge area but mathematically, as A gets bigger, the uncertainty and the momentum drops.
00:38:15.100 --> 00:38:18.600
Now, I can be very certain about what the momentum is.
00:38:18.600 --> 00:38:22.600
If I make A smaller, I’m actually localizing the particle.
00:38:22.600 --> 00:38:25.300
I'm saying the particle is there.
00:38:25.300 --> 00:38:32.600
If I’m making A smaller and smaller, my uncertainty in where the particle actually is become smaller
00:38:32.600 --> 00:38:38.200
but the problem is as A gets smaller, this whole quantity gets bigger.
00:38:38.200 --> 00:38:42.200
The momentum of the particle now I can say anything about the momentum.
00:38:42.200 --> 00:38:47.900
This is the relationship and it is based on the mathematics like that.
00:38:47.900 --> 00:38:50.900
Let us go ahead and take the product of the two.
00:38:50.900 --> 00:39:20.400
Σ X σ sub P = this is going to be A/ 2 π N × N² π²/ 3 – 2 ^½ × H ̅ N π/ A.
00:39:20.400 --> 00:39:28.500
When I multiply the two, the A cancels with the A.
00:39:28.500 --> 00:39:34.200
N cancels the N, the π cancels the π.
00:39:34.200 --> 00:39:36.600
And I'm left with the following.
00:39:36.600 --> 00:39:55.300
The σ X σ P = H ̅/ 2 × N² π²/ 3 - 2 all to the ½.
00:39:55.300 --> 00:40:02.600
This right here is greater than H ̅/ 2.
00:40:02.600 --> 00:40:16.600
The reason is because this term right here, because N² π²/ 3 -2¹/2 is always greater than 1.
00:40:16.600 --> 00:40:20.400
Because it is greater than 1, this is always to be going to be greater than this.
00:40:20.400 --> 00:40:22.800
We have it.
00:40:22.800 --> 00:40:28.000
Any uncertainty in the measurement of the position, multiplied by the uncertainty
00:40:28.000 --> 00:40:33.600
in the measurement of the momentum is always going to be greater than H ̅/ 2.
00:40:33.600 --> 00:40:44.700
In other words, if I become more certain of the position, I become less certain of the momentum.
00:40:44.700 --> 00:40:51.000
As I become less certain of the position, I become more certain of the momentum.
00:40:51.000 --> 00:40:55.000
Maximizing and minimizing the relationship between them is this.
00:40:55.000 --> 00:40:57.500
This is an expression of the uncertainty principle.
00:40:57.500 --> 00:41:02.700
When it comes to position an momentum, I can only maximize.
00:41:02.700 --> 00:41:05.200
If I maximize one, I minimize the other.
00:41:05.200 --> 00:41:07.100
If I minimize one, I maximize the other.
00:41:07.100 --> 00:41:10.300
There is a point, I have to come to some sort of compromise.
00:41:10.300 --> 00:41:12.500
I have to decide what is important to me.
00:41:12.500 --> 00:41:13.000
Do I want to know more about the position?
00:41:13.000 --> 00:41:15.200
Do I want to know more about the momentum?
00:41:15.200 --> 00:41:17.100
Or do I want to know a little bit about both?
00:41:17.100 --> 00:41:24.700
This expresses the relationship between the uncertainties in these measurements.
00:41:24.700 --> 00:41:30.100
Again, let me write final page here.
00:41:30.100 --> 00:42:26.600
As I increase the space over which the particle can move, the uncertainty in where the particle is rises.
00:42:26.600 --> 00:42:48.800
But the uncertainty of the particle’s momentum drops, vice versa.
00:42:48.800 --> 00:42:52.800
As I decrease the space over which particle can roam.
00:42:52.800 --> 00:43:18.800
In other words, as I can find the particle more and more, as I can find the particle to a smaller region,
00:43:18.800 --> 00:43:30.700
I have a better idea of where the particle is.
00:43:30.700 --> 00:43:54.500
In other words, my uncertainty of my particle’s position drops but I have a better idea of where the particle is.
00:43:54.500 --> 00:44:11.200
I have a worse idea of the particle’s momentum.
00:44:11.200 --> 00:44:16.100
The uncertainty in the position and the uncertainty of the momentum are inversely related.
00:44:16.100 --> 00:44:26.400
Once again, the uncertainty in the particle’s position × uncertainty in the particle’s momentum is going to be greater than H/2.
00:44:26.400 --> 00:44:32.900
This is one of the statements of the uncertainty principle.
00:44:32.900 --> 00:44:40.700
In the last couple of lessons we have been just been going over material and presenting theory, we have not done any problems.
00:44:40.700 --> 00:44:47.500
I want you to know that the problem sets are going to be in several lessons to come.
00:44:47.500 --> 00:44:49.300
I’m going to be doing them all at once.
00:44:49.300 --> 00:44:57.100
The nature of the material was such that with quantum mechanics, it is true that you can present a little bit of the topic and do a problem.
00:44:57.100 --> 00:45:07.400
I think it is better to just go ahead and present a certain amount of theory and then go back and then do a whole bunch of problems,
00:45:07.400 --> 00:45:12.000
Because I’m given a chance to actually review the material as we are doing the problems.
00:45:12.000 --> 00:45:18.300
If you are wondering where the problems are, do not worry we are going to be doing it and absolute ton of them and a variety of them.
00:45:18.300 --> 00:45:20.700
Do not worry about that.
00:45:20.700 --> 00:45:22.600
Thank you so much for joining us here at www.educator.com.
00:45:22.600 --> 00:45:24.000
We will see you next time, bye.