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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In this particular lesson, we are going to discuss the plausibility of the Schrӧdinger equation.
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This is one of those appendix lessons where it is not necessary for your understanding.
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This is just a little bit of extra information if you happen to be curious, if you want to dig a little bit deeper, things like that.
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I'm going to be presenting these every so often throughout the lesson just to give you a little bit extra.
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You are absolutely welcome to skip this, it is not a problem at all.
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It is called the plausibility of the Schrӧdinger equation and what I'm going to do is I'm going to cool to derive the Schrӧdinger equation.
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The Schrӧdinger equation is not really derived.
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The way we are presenting Quantum Mechanics to you and the way it is present in general now a days,
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is as a series of patio, as a series of axiom's, things that we accept that are true.
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And then from those handful of axioms 345, however they maybe, we develop the entire theory of Quantum Mechanics.
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The Schrӧdinger equation is actually one of those axioms but just to show you where it may have come from,
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we are going to present something like this.
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Just so it seems to make a little bit more sense to you if you need to wrap your mind around it.
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Let us go ahead and get started.
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Let us go ahead and start with the classical wave equation.
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There is a differential equation, there is a partial differential equation, there is a second order equation that relates.
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Let us just go ahead and write it down, the wave of equation.
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The classical wave equation is the equation that when you solve it, it actually gives you the equation for how a wave behaves.
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This is the following so D² of F DX² = 1/ V² × D² of F DT².
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This says the following, if I have some function which is a function of both position and time, X and T the function of 2 variables,
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This wave equation is a function of X and T.
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If I take the second partial derivative with respect to X it actually happens to equal 1/ the velocity of the wave × the second derivative with respect to T.
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This equation is what gives us our wave equation for how the wave actually behaves at a given X and at a given T.
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The data that we collect gives rise to the equation.
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We can solve the equation to actually get a function.
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That is the function that we use for the rest of our problems.
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In this particular case, V is just the velocity of the wave.
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A particular solution to this equation is the following.
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The solution is the function of X and T happens to be some function of X which we will call the C × the cos of ω T.
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This is not a W, this is ω, this is the Greek letter ω and happens to stand for the angular velocity of this particular wave.
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This thing you know from your work back in high school, this is the amplitude of the wave.
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This function, this is a wave equation.
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It is a periodic or harmonic function like that and this thing in front represents the amplitude.
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You are used to seeing things like this, A cos X where this is the amplitude.
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When an amplitude is actually fixed at some number 1, 2, 3, 15, 34 things like that.
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It represents the height, at fixed height.
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In this particular case, the wave equation, the amplitude is actually a function of X, the function of position.
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Different amplitudes depending on where you are.
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This is a more sophisticated version of the wave equation that you are accustomed to.
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This is called an amplitude and because this represents a solution to the wave equation, which is why we called C.
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This is the function we are going to be concerned with Quantum Mechanics, the wave function for the particular particle.
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This is why it is called the amplitude of the wave function because it comes from the fact that
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it is an amplitude of the classical wave function.
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That is why it is called the amplitude.
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Amplitude of the wave and it is a function only of X.
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It is very interesting, this function of 2 variables actually comes from multiplying a function of one variable by a function entirely of the other variable.
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It is actually very cool.
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It is called the spatial amplitude because it happens to deal with where you are in space.
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In this particular case, one dimension just X where you are in space.
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We will go ahead, ω once again is the angular frequency.
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I apologize it is angular frequency.
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That is fine, angular velocity and angular frequency.
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And ω also happens to equal to π × μ which is the normal frequency.
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Μ × λ happens to equal the velocity.
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In this particular case, μ and V are the same thing.
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V is velocity of the wave, μ is the frequency of the wave.
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In other words, how many cycles it goes through in 1 sec.
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Let us go ahead and play with this.
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Let us go ahead and take the second derivative with respect to X, the second with respect to T,
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set them equal to each other, multiply by 1/ V² and let us see what we get.
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DF DX = ψ prime of X × cos ω T.
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Again, this partial derivative notation, when you take the derivative with respect to X it means treat T as a constant.
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If you treat T as a constant then cos of ω T is just a constant.
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We ignore it for all practical purposes.
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And we take the D² of F DX² and we end up with the ψ double prime of X × cos ω T.
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We will take DF DT.
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We will differentiate with respect to T so we are going to hold the X constant.
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It is just equal to – ψ of X × ω × sin T because the derivative of cos ω T = -ω sin ω T.
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Just make sure I get everything in here, ω T.
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These symbols are making me crazy.
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= -ψ of X × ω² × cos of ω T.
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I’m going to put each of these into the classical equation and let us see what it is that we actually end up getting.
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Let me go ahead and work in red.
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I’m going to go ahead and take this one and this one.
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I'm going to get ψ double prime of X × cos ω T = 1/ V².
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This is going to be - and then I'm going to put the ω² in front.
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I’m just going to put on top here – ω.
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It is probably better if we see it this way.
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It is going to equal 1/ V² × -ψ of X × ω² × cos ω T.
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We end up getting ψ double prime of X × cos ω T = -ω²/ V² × ψ of X × cos ω T.
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We will go ahead and cancel that and we will go ahead and move this part to the left and we are left with the following.
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We are left with the ψ double prime of X + ω² / V² × ψ of X = 0.
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Now, we have a differential equation that only deals with this spatial amplitude.
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Our task is to find that ψ, that spatial amplitude what does that represent.
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Now ω = 2π × the frequency, and the frequency = the velocity/ λ.
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Therefore, ω = 2π × the velocity/ λ which implies that ω² = 4π² V²/ λ².
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I will go ahead and divide by V² which means ω²/ V² = 4π²/ λ².
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I will go ahead and put 4π² λ² wherever I see ω²/ V².
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I end up with the ψ double prime of X + 4π²/ λ².
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I have this expressed in terms of the wavelength which you will see why I wanted this in just a moment.
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To see if X = 0, now I have this differential equation expressed in terms of λ.
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To this, I'm going to start applying this idea of De Broglie matter way.
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We said that De Broglie relation, matter if it is moving with a certain linear momentum, certain mass × its velocity, it is going to have a wavelength.
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There is going to be a wavelength associated with it.
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That is the wavelength that we are going to put in here.
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We are trying to find an equation for matter treated like a wave.
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We started with a wave equation now we brought it on the λ.
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We are going to put the De Broglie equation into this and when we find the ψ,
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that is going to give us the wave equation for the particle that is behaving like a wave.
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That is all we are doing here.
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Let us go ahead and just briefly, we have the energy = the kinetic energy + the potential energy.
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The energy = ½ the mass × the velocity² + the potential energy.
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The energy = linear momentum/ twice the mass + the potential energy.
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And recall that the potential energy is the energy of a particular particle or system based on its position.
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The potential energy is nothing more than the energy possessed by an object by virtue of its position, that is all potential energy means.
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By virtue of its position, in other words you know that if some object
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happens to be 8m above the surface of the earth, its potential energy is MGH.
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There is some function that depends on the position, that is why we say V of X.
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It depends on the position that is what the potential energy of that particular object is.
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I’m going to go ahead and solve this for P.
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When I do that, I end up with P = √2M × the total energy - the potential energy.
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De Broglie relation is the following.
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If you remember that is equal to.
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De Broglie relation said that the wavelength of a particular matter wave =
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planks constant/ its momentum or planks constant/ its mass × its velocity.
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I’m going to go ahead and put this over here and I'm going to find L and I'm going to put that into here, this λ.
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We end up with the following.
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We end up with λ = planks constant/ P which is 2M × total energy - potential energy.
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This V is the capital V, not the velocity.
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I will put this λ which is a function of the total and potential energies back into the original equation.
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Let us write that, we had ψ double prime of X + 4π²/ λ² × ψ of X = 0.
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I’m going to stick this into here and I get ψ double prime, I’m going to leave off the X here + 4π² ÷ H² / 2M × E - V × ψ = 0.
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I will just go ahead and leave the X in there.
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+ 4π² × 2M × E - V/ H² × ψ = 0.
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This is the Schrӧdinger equation right here.
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We can clean it up just a little bit.
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Let us go ahead and do that.
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If I can go ahead and turn the page here.
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Let us introduce a shorthand.
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Let us introduce this thing called H ̅ that is going to equal just H/ 2π.
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Therefore, H ̅² = H²/ 4π².
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Therefore, 1/ H ̅² = 4π²/ H.
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I have 4 π²/ H, I can replace that.
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Therefore, I have ψ double prime of X + 2M × E - V ÷ H ̅² × ψ of X = 0.
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Let us go ahead and see if we can fiddle with this some more.
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Let us go ahead and move this over to the other side.
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We have ψ double prime of X = -2M × E – V/ H ̅² × ψ of X.
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We move that over to the right.
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I'm going to go ahead and multiply by H ̅² ÷ -2M.
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I'm going to be left with the following.
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I get – H ̅²/ 2 M × ψ double prime of X is equal to,
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I'm going to go ahead and write it.
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Then I have E - V × ψ of X.
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I’m going to multiply, these are just more things I can multiply.
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I’m going to distribute so I get -H ̅²/ 2M × ψ double prime of X = E × ψ of X - V × ψ of X.
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I’m going to go ahead and bring this V over to the other side and I'm left with –H ̅²/ 2M × ψ double prime of X + V of X.
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V is a function of X × ψ of X = the total energy × ψ of X.
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There you go, now let me go ahead and express this double prime as something a little bit different.
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The notation that you are used to, the DDX stuff.
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-H ̅²/ 2M D² DX² of ψ of X.
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This is this + V of X × ψ of X = E × ψ of X.
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And of course, if I want to I can go ahead.
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This is fine and I can go ahead and do it like, I can leave it like this.
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But since we introduce the notion of the Eigen value problem, this is ψ of X.
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E is just a scalar, this is a function, these are all operators.
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I’m going to go ahead and write my final one like this.
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I have –H ̅²/ 2M D² DX² + the potential energy × ψ of X = energy × ψ of X this is the Hamiltonian operator.
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I have H of ψ of X = E of ψ of X.
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Here we have the Schrӧdinger equation expressed in operator notation, the Eigen value version of the Schrӧdinger equation.
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Again, we just thought we started with the wave equation, the classical wave equation.
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We know a solution to the wave equation we just took the spatial factor, the amplitude.
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And then we took the derivatives, we put it back to the wave equation and
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we are able to derive another equation just for the amplitude, just for the ψ of X.
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We introduced the notion of De Broglie wavelength into that equation because there is a λ.
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And then we managed to get an equation that expresses a relationship between the potential energy and the wave function.
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The kinetic energy, the wave function, and the total energy, the wave function, express this is an operator.
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Kinetic energy operator, potential energy operator, this is just a scalar.
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The total energy of the system multiplied by the function and we express the Schrӧdinger equation as an Eigen value problem.
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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.