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 1 answerLast reply by: Professor HovasapianFri Feb 13, 2015 4:13 PMPost by matt kruk on February 11, 2015hi professor do we get the solution of just psi(x) by solving the 2nd order DE equation with some arbitrary conditions since the solution would be complex (sines and cosines)?is that where it comes from?

### The Plausibility of the Schrӧdinger Equation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• The Plausibility of the Schrӧdinger Equation 1:16
• The Plausibility of the Schrӧdinger Equation, Part 1
• The Plausibility of the Schrӧdinger Equation, Part 2
• The Plausibility of the Schrӧdinger Equation, Part 3

### Transcription: The Plausibility of the Schrӧdinger Equation

Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.0000

In this particular lesson, we are going to discuss the plausibility of the Schrӧdinger equation.0005

This is one of those appendix lessons where it is not necessary for your understanding.0011

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.0015

I'm going to be presenting these every so often throughout the lesson just to give you a little bit extra.0022

You are absolutely welcome to skip this, it is not a problem at all.0029

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.0033

The Schrӧdinger equation is not really derived.0039

The way we are presenting Quantum Mechanics to you and the way it is present in general now a days,0042

is as a series of patio, as a series of axiom's, things that we accept that are true.0047

And then from those handful of axioms 345, however they maybe, we develop the entire theory of Quantum Mechanics.0054

The Schrӧdinger equation is actually one of those axioms but just to show you where it may have come from, 0062

we are going to present something like this.0069

Just so it seems to make a little bit more sense to you if you need to wrap your mind around it.0071

Let us go ahead and get started.0076

Let us go ahead and start with the classical wave equation.0078

There is a differential equation, there is a partial differential equation, there is a second order equation that relates.0082

Let us just go ahead and write it down, the wave of equation.0090

The classical wave equation is the equation that when you solve it, it actually gives you the equation for how a wave behaves.0094

This is the following so D² of F DX² = 1/ V² × D² of F DT².0109

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,0123

This wave equation is a function of X and T.0132

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.0135

This equation is what gives us our wave equation for how the wave actually behaves at a given X and at a given T.0150

The data that we collect gives rise to the equation.0157

We can solve the equation to actually get a function.0161

That is the function that we use for the rest of our problems.0164

In this particular case, V is just the velocity of the wave.0170

A particular solution to this equation is the following.0182

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.0189

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.0203

This thing you know from your work back in high school, this is the amplitude of the wave.0212

This function, this is a wave equation.0218

It is a periodic or harmonic function like that and this thing in front represents the amplitude.0221

You are used to seeing things like this, A cos X where this is the amplitude.0230

When an amplitude is actually fixed at some number 1, 2, 3, 15, 34 things like that.0236

It represents the height, at fixed height.0241

In this particular case, the wave equation, the amplitude is actually a function of X, the function of position.0246

Different amplitudes depending on where you are.0251

This is a more sophisticated version of the wave equation that you are accustomed to.0254

This is called an amplitude and because this represents a solution to the wave equation, which is why we called C.0259

This is the function we are going to be concerned with Quantum Mechanics, the wave function for the particular particle.0270

This is why it is called the amplitude of the wave function because it comes from the fact that 0277

it is an amplitude of the classical wave function.0288

That is why it is called the amplitude.0289

Amplitude of the wave and it is a function only of X.0296

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.0306

It is actually very cool.0315

It is called the spatial amplitude because it happens to deal with where you are in space.0317

In this particular case, one dimension just X where you are in space.0323

We will go ahead, ω once again is the angular frequency.0336

I apologize it is angular frequency.0344

That is fine, angular velocity and angular frequency.0347

And ω also happens to equal to π × μ which is the normal frequency.0362

Μ × λ happens to equal the velocity.0375

In this particular case, μ and V are the same thing.0381

V is velocity of the wave, μ is the frequency of the wave.0384

In other words, how many cycles it goes through in 1 sec.0390

Let us go ahead and play with this.0395

Let us go ahead and take the second derivative with respect to X, the second with respect to T, 0397

set them equal to each other, multiply by 1/ V² and let us see what we get.0401

DF DX = ψ prime of X × cos ω T.0407

Again, this partial derivative notation, when you take the derivative with respect to X it means treat T as a constant.0417

If you treat T as a constant then cos of ω T is just a constant.0424

We ignore it for all practical purposes.0426

And we take the D² of F DX² and we end up with the ψ double prime of X × cos ω T.0432

We will take DF DT.0444

We will differentiate with respect to T so we are going to hold the X constant.0447

It is just equal to – ψ of X × ω × sin T because the derivative of cos ω T = -ω sin ω T.0451

Just make sure I get everything in here, ω T.0473

These symbols are making me crazy.0484

= -ψ of X × ω² × cos of ω T.0486

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.0497

Let me go ahead and work in red.0505

I’m going to go ahead and take this one and this one.0507

I'm going to get ψ double prime of X × cos ω T = 1/ V².0511

This is going to be - and then I'm going to put the ω² in front.0526

I’m just going to put on top here – ω.0544

It is probably better if we see it this way.0548

It is going to equal 1/ V² × -ψ of X × ω² × cos ω T.0550

We end up getting ψ double prime of X × cos ω T = -ω²/ V² × ψ of X × cos ω T.0563

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.0582

We are left with the ψ double prime of X + ω² / V² × ψ of X = 0.0589

Now, we have a differential equation that only deals with this spatial amplitude.0598

Our task is to find that ψ, that spatial amplitude what does that represent.0603

Now ω = 2π × the frequency, and the frequency = the velocity/ λ.0611

Therefore, ω = 2π × the velocity/ λ which implies that ω² = 4π² V²/ λ².0625

I will go ahead and divide by V² which means ω²/ V² = 4π²/ λ².0649

I will go ahead and put 4π² λ² wherever I see ω²/ V².0658

I end up with the ψ double prime of X + 4π²/ λ².0665

I have this expressed in terms of the wavelength which you will see why I wanted this in just a moment.0675

To see if X = 0, now I have this differential equation expressed in terms of λ.0681

To this, I'm going to start applying this idea of De Broglie matter way.0687

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.0693

There is going to be a wavelength associated with it.0705

That is the wavelength that we are going to put in here.0707

We are trying to find an equation for matter treated like a wave.0710

We started with a wave equation now we brought it on the λ.0716

We are going to put the De Broglie equation into this and when we find the ψ, 0719

that is going to give us the wave equation for the particle that is behaving like a wave.0724

That is all we are doing here.0731

Let us go ahead and just briefly, we have the energy = the kinetic energy + the potential energy.0734

The energy = ½ the mass × the velocity² + the potential energy.0742

The energy = linear momentum/ twice the mass + the potential energy.0750

And recall that the potential energy is the energy of a particular particle or system based on its position.0761

The potential energy is nothing more than the energy possessed by an object by virtue of its position, that is all potential energy means.0773

By virtue of its position, in other words you know that if some object 0803

happens to be 8m above the surface of the earth, its potential energy is MGH.0808

There is some function that depends on the position, that is why we say V of X.0816

It depends on the position that is what the potential energy of that particular object is.0820

I’m going to go ahead and solve this for P.0827

When I do that, I end up with P = √2M × the total energy - the potential energy.0832

De Broglie relation is the following.0845

If you remember that is equal to.0847

De Broglie relation said that the wavelength of a particular matter wave = 0853

planks constant/ its momentum or planks constant/ its mass × its velocity.0858

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.0867

We end up with the following.0878

We end up with λ = planks constant/ P which is 2M × total energy - potential energy.0881

This V is the capital V, not the velocity.0889

I will put this λ which is a function of the total and potential energies back into the original equation.0894

Let us write that, we had ψ double prime of X + 4π²/ λ² × ψ of X = 0.0902

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 = 0.0915

I will just go ahead and leave the X in there.0942

+ 4π² × 2M × E - V/ H² × ψ = 0.0946

This is the Schrӧdinger equation right here.0965

We can clean it up just a little bit.0967

Let us go ahead and do that.0969

If I can go ahead and turn the page here.0972

Let us introduce a shorthand.0976

Let us introduce this thing called H ̅ that is going to equal just H/ 2π.0980

Therefore, H ̅² = H²/ 4π².0987

Therefore, 1/ H ̅² = 4π²/ H.0999

I have 4 π²/ H, I can replace that.1006

Therefore, I have ψ double prime of X + 2M × E - V ÷ H ̅² × ψ of X = 0.1008

Let us go ahead and see if we can fiddle with this some more.1030

Let us go ahead and move this over to the other side.1034

We have ψ double prime of X = -2M × E – V/ H ̅² × ψ of X.1038

We move that over to the right.1052

I'm going to go ahead and multiply by H ̅² ÷ -2M.1054

I'm going to be left with the following.1063

I get – H ̅²/ 2 M × ψ double prime of X is equal to,1065

I'm going to go ahead and write it.1080

Then I have E - V × ψ of X.1080

I’m going to multiply, these are just more things I can multiply.1087

I’m going to distribute so I get -H ̅²/ 2M × ψ double prime of X = E × ψ of X - V × ψ of X.1092

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.1122

V is a function of X × ψ of X = the total energy × ψ of X.1123

There you go, now let me go ahead and express this double prime as something a little bit different.1134

The notation that you are used to, the DDX stuff.1140

-H ̅²/ 2M D² DX² of ψ of X.1144

This is this + V of X × ψ of X = E × ψ of X.1155

And of course, if I want to I can go ahead.1167

This is fine and I can go ahead and do it like, I can leave it like this.1170

But since we introduce the notion of the Eigen value problem, this is ψ of X.1176

E is just a scalar, this is a function, these are all operators.1184

I’m going to go ahead and write my final one like this.1189

I have –H ̅²/ 2M D² DX² + the potential energy × ψ of X = energy × ψ of X this is the Hamiltonian operator.1193

I have H of ψ of X = E of ψ of X.1214

Here we have the Schrӧdinger equation expressed in operator notation, the Eigen value version of the Schrӧdinger equation.1220

Again, we just thought we started with the wave equation, the classical wave equation.1232

We know a solution to the wave equation we just took the spatial factor, the amplitude.1240

And then we took the derivatives, we put it back to the wave equation and 1245

we are able to derive another equation just for the amplitude, just for the ψ of X.1251

We introduced the notion of De Broglie wavelength into that equation because there is a λ.1258

And then we managed to get an equation that expresses a relationship between the potential energy and the wave function.1264

The kinetic energy, the wave function, and the total energy, the wave function, express this is an operator.1271

Kinetic energy operator, potential energy operator, this is just a scalar.1278

The total energy of the system multiplied by the function and we express the Schrӧdinger equation as an Eigen value problem.1283

Thank you so much for joining us here at www.educator.com.1291

We will see you next time, bye. 1293