WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, and welcome to www.educator.com, welcome back to Physical Chemistry.
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We have talked about the harmonic oscillator.
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We talked about the rigid rotator and now we are going to do some example problems.
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Let us jump right on in.
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Before we jump on in, I always like to recall what it is that we learned.
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Take a look at the major equations so we have a certain set of tools so that we know what we are going to be dealing with.
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Let us recall what we have.
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Let us recall what we have.
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I think I’m going to work in blue this time.
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The energy of the harmonic oscillator.
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The energy was E sub R is the state, R is a quantum number.
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It is equal to H ̅ × K/ ν¹/2 × R + ½, where R is equal to 0, 1, 2, and so on.
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Other versions of this are H ̅ O R + ½ or this ω here is going to be this thing,
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this is the angular velocity, if you want think about it that way.
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That is that.
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Another version of that is H × the frequency R + ½.
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We have the ω is equal to this K / ν¹/2.
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This ν is equal to 1/2 π K/ μ¹/2 is equal to ω/ 2 π.
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Just so we have everything on the table here.
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K is the force constant of the bond of the harmonic oscillator.
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The μ is the reduced mass.
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The definition was 1/ μ is equal to 1/ mass 1/ mass 1 + 1/ mass 2.
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Or if you prefer, μ = M1 × M2/ M1 + M2.
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I will go ahead and do on the next page.
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The selection rule for the harmonic oscillator was that δ R = + or -1.
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The selection rule for the harmonic oscillator, when the harmonic oscillator
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is moving from one energy state to another, it is not going to jump from the first state to the fifth state.
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It is going to go one at a time.
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It is going to go up by 1 or down by 1, that is all that means.
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The observe frequency of radiation, we hit it.
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When we hit, some system was rotating, some quantum system is rotating with a certain amount of energy.
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It is going to absorb radiation as certain frequency.
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What frequency is that?
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That is what this is.
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The observed is equal to 1/ 2 π K/ μ¹/2.
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We often represent this actually in terms of wave numbers, which the unit is inverse centimeters.
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And that is going to equal 1/ 2 π C K/ μ¹/2.
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In spectroscopy, we tend to use these units, this wave numbers.
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The unit is inverse cm.
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This thing is just that divided by C.
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Anytime you see a tilde over something, just find the corresponding variable and just divide by the speed of light.
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Adjacent states are the same, δ E apart.
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State 1, state 2, state 3, state 4, are the same δ E apart.
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When we take a look at spectrum for the harmonic oscillator, it is going to be a single line.
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The spectrum is a single line at this.
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Let us take a look at the harmonic oscillator wave functions.
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We are just collecting our little bits of information, the most important bits of information that
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we know to solve these problems.
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The harmonic oscillator wave functions.
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We have ψ sub R is equal to N sub R H sub R, the argument is Α¹/2 X E ^- Α X²/ 2.
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Where Α is equal to K × μ/ H ̅²¹/2.
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The normalization constant, it actually changes with R.
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It is equal to 1/ 2 ⁺R × R!¹/2 × Α/ π¹/4.
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These are the Hermite polynomials.
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This H sub R are the Hermite polynomials for our rigid rotator.
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The harmonic oscillator, now rigid rotator.
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The energy of the rigid rotator, we have energy sub J is equal to H ̅/ 2I J × J + 1 K,
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or J is going to equal to 0, 1, 2, 3, and so on.
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I here is the rotational inertia, it happens to equal the reduced mass × the radius²,
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where R is the distance between the masses.
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Of course, this μ is the reduced mass which we saw earlier.
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We also have D sub J is equal to 2 J + 1.
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This the degeneracy, this is the number of states that actually have this energy 2 J + 1.
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If you are at the J = 2 state, you are going to have 2 × 2 + 4, 2 × 2 is 4 + 1 is equal to 5.
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You have 5 different states that have the energy.
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The selection rule for the rigid rotator is δ J = + or -1.
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You can only move to different states, passing through adjacent states.
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The frequency of absorption transition is the frequency is equal to H/ 4 π² I × J + 1.
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Which is more commonly written as frequency is equal to 2 B × J + 1, where this B is equal to this H / 8 π² I.
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B is called the rotational constant, very important.
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In terms of wave numbers, we have that it is equal to 2 B J + 1.
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The only difference is like we said, any time you see a tilde over, you just divide everything by the speed of light.
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We have H/ 8 π² CI.
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Let us go ahead and start doing our example problems.
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The first one is going to be very simple.
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Just computation of a reduced mass.
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Calculate the reduced mass of the hydrogen atom.
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Use the full decimal values listed in your book for the masses of the proton and the electron.
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We said that the reduced mass is going to equal mass 1 × mass 2/ mass 1 + mass 2.
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Very simple.
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We have 9.109390 × 10⁻³¹ kg × 1.672623 × 10⁻²⁷ kg and divided by this + that.
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9.109390 × 10⁻³¹ + 1.672623 × 10⁻²⁷, all of that is kilograms.
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When we do the math, we end up with 9. 104432 × 10⁻³¹ kg.
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Kg × kg is kg² on top.
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When we add the numbers, this many kilograms × this many kilograms, the units are multiplied so it will be squared.
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When we are adding, it is just 1 unit of kilogram, that cancels that is why we end up with kg here.
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Nice and simple.
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Example 2, the force constant K of the nitrogen 14 nitrogen 14 bond is 2243 N/ m.
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Calculate the fundamental vibration frequency and the 0 point energy of this molecule.
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The fundamental vibration frequency is the frequency that we actually see at that single spike,
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that absorption spike on the spectrum.
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That is called the fundamental vibration frequency.
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We are also going to be finding the 0 point energy.
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The energy where we are at the 0 state, where R = 0.
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Let us see, we have ν ̃ is equal to 1/ 2 π C × K/ μ¹/2 and this is in units of inverse CM.
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We are going to be using wave numbers.
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Let us go ahead and calculate the reduced mass.
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The reduced mass is going to be, we need that first.
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We have K, we have reduced mass, that is the only one that we have to find here.
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Like we said, it is going to be M1 M2/ M1 + M2.
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It is going to equal, this is nitrogen 14 14 so we have 14 atomic mass units × 14 atomic mass units/ 14
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+ 14 atomic mass units × 1.66 × 10⁻²⁷ kilograms per atomic mass unit.
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That and that, they all cancel.
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What I’m left with is 1.162 × 10⁻²⁶.
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When we are given the molecule this way, these are not g or kg, they are atomic mass units.
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To work in mass units then you have to convert.
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1 atomic mass unit is equal to that many kilograms.
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We want the mass in kg.
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This is equal to 1/ 2 π × 3.0 × 10¹⁰.
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We have to use the speed of light in cm/ s, not m/ s.
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K is 22 43 N/ m, 1.162 × 10⁻²⁶ kg¹/2.
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You end up with 2331 inverse cm.
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The spectrum would look, I would see a single spike at 2331.
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Let us go ahead and do our 0 point energy.
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Do I have another page here? Yes I do.
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I think I can go ahead and do the 0 point energy.
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We said that the energy is going to equal H μ × R + ½.
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We want it to be for R = 0, the 0 point energy.
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Our energy₀ is going to equal H μ/ 2.
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Let us see what we have got here.
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We know that μ/ C is equal to ν ̃ which means, therefore, that ν itself is equal to ν ̃ × C
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because ν ̃ is actually the value that they gave us, or the value that we just found because the formula is actually ν.
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Let us go ahead and calculate ν.
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Ν is equal to this 2331 inverse cm × the speed of light 3.0 × 10¹⁰ cm / s.
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The centimeter and centimeter go away.
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And I'm left with a frequency of 6.993 × 10¹³ inverse seconds which is Hz.
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Therefore, our 0 point energy is going to equal ½ × planks constant 6.626 × 10⁻³⁴ J-s × 6.993 × 10¹³.
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/ s, that goes away and we will do our calculation.
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If I have done my arithmetic correctly which I probably have not, so please check it for me.
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2.33 × 10⁻²⁰ J, I really dislike arithmetic.
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You are going to find very quickly if you have not found that already,
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that the problem of quantum mechanics is not the difficulty.
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The problem of quantum mechanics is the tedium of the calculations.
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Let us do example 3, show that the product of 2 even functions is even but
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the product of 2 odd functions is even, and that the product of an odd and even function is odd.
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What this has to do with quantum mechanics?
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You will see in a minute in our next problem, when we start actually doing integrals.
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We are going to be integrating odd functions, even functions, odd functions × odd, odd functions × even functions.
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It makes our integration easier, I just want to do this real quickly.
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Let us recall what even and odd mean.
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An even function means that if I do F of – X, that I end up just getting my original X back.
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In other words, if I have some function F of X, if every place there is an X, I stick a –X then and I simplify it,
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if I end up getting the original function back, that is an even function.
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Something like X² is an even function, something like cos of X is an even function.
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An odd function, it means that if I stick in a - X wherever I see an X in my function,
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I end up actually getting the negative of the original function back – F of X.
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X³ is a perfect example of an odd function.
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Sin X is a perfect example of an odd function.
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An even function is symmetric around the Y axis.
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An odd function is symmetric about the origin.
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This is the X² symmetric about this.
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X³ is symmetric about the origin, that is the difference between the two.
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That is the geometric version, this is the algebraic version.
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We want to show that the product of 2 even functions is even.
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The first one is both are even.
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The product function is going to be F of X × G of X.
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The P of – X, because we want to show that the product is odd or even, so we put in a - X wherever there is an X.
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The P of - X is just going to equal F of -X × G of –X, but they are both even functions.
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If they are even, F of – X is F of X.
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G of –X is G of X, so this is just F of X × G of X.
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That just = P of X.
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P of - X = P of X so it is even.
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P of -X = P of X which means we have an even function.
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The product of 2 even functions is an even function.
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Let us take a look at B.
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B says that if they are both odd, if you multiply an odd function and an odd function, you should get an even function.
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In this case, both odd.
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We need to find P of – X.
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P of - X is equal to F of - X × G of –X.
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F of X and G of X are both odd which means that F of -X is –F of X and G of -X is -G of X.
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When I multiply them together, the negative cancel so I still end up with F of X and G of X, which is equal to P of X.
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Therefore, P of -X = P of X.
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When both functions are odd, their product is even.
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C, we have one odd and we have one even function.
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We do P of –X, P of -X is equal to F of -X × G of –X.
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If one of them is odd, let us say the F is odd, that is going to be –F of X.
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G of –X, that one is even, that is just going to be G of X.
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We are going to end up with -F of X G of X, that is going to be –P of X.
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Therefore, P of -X is equal to –P of X, that is the definition of an odd function.
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If you have odd function and even function, their product is odd.
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Let us see what we have got.
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We can go ahead and apply this to a particular problem.
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For the harmonic oscillator, demonstrate that ψ 1 and ψ 2 are orthogonal, and that ψ 1 and ψ 3 are orthogonal.
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Remember what orthogonal means.
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Let us just go ahead and write the wave functions first.
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I think I would go back to black.
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Let us write out, we have to do this explicitly.
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Let us just write off the wave functions first.
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We have ψ 1, it is going to equal 4 α³/ π¹/4 × X E ^-α X²/ 2, that is the wave function for ψ 1.
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Ψ 2 is equal to Α/ 4 π¹/4 × 2 Α X² -1 × E ^-Α X²/ 2, that is ψ₂ for the harmonic oscillator.
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The wave function ψ 3 is equal to α³/ 9 π ¼ 2 Α X³ -3 X E ^-Α X²/ 2.
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These are the 3 wave functions.
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Let us recall what orthogonal means.
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Orthogonal for the harmonic oscillator means, the harmonic oscillator moving along the X axis.
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Theoretically, we have to integrate over all of space so it can go to infinity, positive infinity and negative infinity.
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Ψ * ψ MN, we need that integral to equal 0.
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If the integral = 0, it demonstrates that they are orthogonal.
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I need to demonstrate that ψ 1 ψ 2, the integral of the product of those two is equal to 0 and
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the integral of ψ 1 ψ 3 is equal to 0.
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I need to do these integrals explicitly.
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Notice that, this E ^- Α X²/ 2, this is an even function.
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Even × you have this X here, this is an odd function here.
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Ψ₁ is odd, ψ₂ is even because of that X², and this is odd.
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The ψ₁ is odd, ψ₂ is even, ψ₃ is odd.
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Let us see what we have got.
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In general, ψ sub R is even when R is even.
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One is an odd number, it is an odd function.
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2 is an even number, it is an even function.
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3 is an odd number, it is an odd function.
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That is the general case for the harmonic oscillator.
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In general, ψ sub R is even when R is even and ψ sub R is odd when R is odd.
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Let me go back to black here.
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Ψ * 1 ψ * 2, is equal to ψ 1 ψ 2 because it is a real function.
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The complex conjugate of a real function is the function itself because ψ is real.
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Basically, what you have now is an odd function × an even function.
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You have an odd function × an even function.
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What do we say?
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An odd function × an even function is going to be an odd function.
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It is going to end up being odd.
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This ψ 1 ψ 2 is odd.
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Recall, when we integrate -A to A, when we integrate along a symmetric interval,
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some odd function the integral actually = 0.
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The integral from - infinity to infinity of ψ 1 ψ 2 is the integral of an odd function about a symmetric interval that is equal to 0.
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You do not have to actually have to solve the integral.
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Therefore, ψ 1 and ψ 2 are orthogonal.
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Let us do ψ 1 and ψ 3.
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Ψ 1 is odd and ψ 3 is odd.
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Ψ 1 * ψ 3 is equal to ψ 1 ψ 3, odd × odd, we are going to end up with an even function.
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In this case, ψ 1 ψ 3 is even.
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We do not have things simple like that so we would actually have to do this integral explicitly.
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We must integrate explicitly.
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Let us go ahead and move on to that.
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Our integral is going to look like this.
00:31:25.200 --> 00:31:36.000
The integral from -infinity to infinity of ψ 1 ψ 3 DX is going to equal,
00:31:36.000 --> 00:31:40.300
I’m going to pull out the normalization constants out of the integral.
00:31:40.300 --> 00:31:58.300
It is going to be 4 Α/ π¹/4 × Α/ 9 π.
00:31:58.300 --> 00:32:01.100
My numbers are getting crazy.
00:32:01.100 --> 00:32:13.800
To the ¼ × the integral from 0 to infinity, because it is a symmetric interval, I can just multiply the integral by 2.
00:32:13.800 --> 00:32:18.300
Instead, of going from –infinity to 0 is the same as 0 to infinity.
00:32:18.300 --> 00:32:41.100
Therefore, I will just go 2 × 0 to infinity of X × E ^- Α X²/ 2 × 2 Α X³ - 3X × E ^-α X²/ 2 DX.
00:32:41.100 --> 00:32:44.300
This is the integral, this is what I have to solve.
00:32:44.300 --> 00:32:49.300
Do not worry, I know it looks really scary but we have software.
00:32:49.300 --> 00:32:50.600
I’m just going to go ahead and call this as Z.
00:32:50.600 --> 00:32:54.100
I do not want to keep writing the normalization constant.
00:32:54.100 --> 00:33:01.900
This is actually going to equal to Z × the integral from 0 to infinity.
00:33:01.900 --> 00:33:05.900
I’m going to multiply the X × this thing and I’m going to multiply this and this.
00:33:05.900 --> 00:33:10.600
When I multiply this and this, the two goes away so I’m left with the following.
00:33:10.600 --> 00:33:22.600
I'm left with 2 Α X⁴ - 3X² × E ^-Α X² DX.
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You can use a table of integrals or you can just use your math software.
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When you do this integral, when you solve it, it is equal to 0.
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I’m not going to go ahead and solve it manually.
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I actually do so in the next lesson, but in this case, one of the nice things
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about having math software is you can do complicated integrals like this.
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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.