Physical Chemistry Example Problems II

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

I apologize if I sat a little bit today, I’m just getting over a cold.0004

If I have some sniffles and things like that, I hope you will forgive me.0007

Today, we are going to continue on with our example problems.0011

We already did one set and then we talked a little bit more about the quantum mechanics, the formal hypotheses of the quantum mechanics.0014

Now, we are just going to do several lessons of problems.0022

Let us just jump right on in.0024

Before we start the example problems, I did want to go over just some of the high points.0027

Just recall some of the equations because there was a lot going on mathematically with quantum,0032

as there is with thermal, and anything else in physical chemistry.0038

Sometimes, you have to pull back and just make a listing of some of things that are important that we remember.0041

We solve the Schrӧdinger equation and we find this wave function ψ.0049

That is a wave function and it represents the particle that we are interested in a particular quantum mechanical system.0057

Instead of looking at the particle like a particle, we look at it like a wave.0067

What we do is we play with this wave function to extract information from it.0070

That is all that is actually happening in quantum mechanics.0076

The ψ conjugate × ψ, we said was the probability of finding the particle whose wave function is ψ0080

in a differential volume element called the DV at the point XYZ.0112

You have this wave function which is going to be a function of XYZ.0131

At some random XYZ, if you actually multiply, it is going to end up being the probability of0135

finding the particle in that little differential volume element.0143

Now we have the equation this ψ DV = 1.0149

Actually, I should say this is not the probability, the ψ* × ψ is the probability density.0163

But for all purposes, we can think of it as the probability.0168

The actual probability is the ψ × ψ* × the differential volume element so that you actually have the probability.0171

When we integrate all of the probabilities, we are going to get 1.0178

This is the normalization condition.0182

This was very important normalization condition.0184

Again, one of the frustrating things about quantum mechanics is wrapping your mind about around things conceptually.0192

But what is nice about it is, because it is so purely mathematical, even if you do not completely understand what is going on,0198

as long as you have a certain set of equations at your disposal,0206

You will at least get the right answer.0210

Eventually, if you become more comfortable and solve for problems, conceptually it will start to make more sense.0212

Now every observable in classical mechanics, corresponds to a linear hermitian operator in quantum mechanics.0218

If we observe a linear momentum in classical mechanics, we have a linear momentum operator in quantum.0254

If we observe angular momentum in classical mechanics,0261

for some particle moving in a circular path or curved path we have a angular momentum operator in quantum mechanics.0266

That operator, we apply it to wave function to give this information.0273

This is what I mean by we extract information from the wave function by operating on the wave function.0276

Doing something to it mathematically.0282

An operator applied to some wave function in a particular state, is equal to A sub N ψ sub N.0287

It is an Eigen value problem.0296

Remember, we can express the Schrӧdinger equation as an Eigen value problem.0300

Again, we are just going over some highlights of what is that we covered so that we have them0308

in a one quick place before we start the example problems.0311

That the ψ sub N or called the Eigen functions of the operator A.0315

The A sub N are called Eigen values of A corresponding to the Eigen function, corresponding to the ψ sub N Eigen function.0329

When a particular function is in a given state, let us say ψ sub 3, it is in that Eigen state for the operator.0357

We speak of Eigen states, we speak of Eigen functions, we speak of Eigen values.0366

Let us talk about what hermitian means.0372

Hermitian also has a mathematical definition.0379

Hermitian operator means it has to satisfy this.0385

F* AG= the integral of GA* F*.0394

If you have 2 wave functions F and G, if you do the left integral and if you do the right integral, those equal each other,0405

then this operator is something that we call hermitian.0419

If it is hermitian, if it satisfies this property.0423

The hermitian operator implies that the Eigen values are real numbers.0427

It is very important and I will actually do a lot to demonstrate why this hermitian property implies reality0434

and implies orthogonality in some of the example problems.0442

One of the first things that hermitian implies is the fact that the Eigen values are real.0447

The other thing, that this hermitian property of the operator implies, double arrow for application.0451

It implies that the Eigen functions are actually orthogonal.0457

The integral of ψsub N conjugate × ψ sub P is equal to 0 for N not equal to P.0463

If I have one Eigen function ψ sub 1 and I have another Eigen function ψ sub 15,0472

If the operator is hermitian, the operator that gave rise to the Eigen functions, the Eigen functions are going to be orthogonal.0478

That is analogous to two vectors being perpendicular.0484

Two vectors are orthogonal when their dot product is equal to 0.0487

Two Eigen functions are orthogonal when their integral of their product is equal to 0.0491

It is completely analogous, that is all that is happening here.0497

When measuring an observable in quantum mechanics, we only get the Eigen value of0505

the operator corresponding to the observable when the wave function is an Eigen function of the operator.0538

In other words, when a quantum mechanical system happens to be in a state that is represented0572

by a wave function that happens to be an Eigen function of the operator,0581

then what we observe when we take a measurement is going to be one of the Eigen values.0585

When the quantum mechanical system is in a state that is represented by the Eigen function of0592

the operator of interest then what we observe is going to be one of the Eigen values of the operator.0599

If the wave function ψ is equal to ψ 1 ψ 1 + ψ 2 ψ 2 + so so, is written.0609

Ψ is the wave function of the quantum mechanical system.0631

If this wave function happens to be written as a linear combination also called the super position.0634

I do not like the word super position but that is fine.0650

It is written as a linear combination of Eigen functions of the operator of interest, whatever operator we happen to be dealing with.0653

Then what we observe are the Eigen values A sub 1, A sub 2, A sub 3, and so on.0680

Let me go to the next page.0702

With probabilities C sub 1² C sub 2² C sub 3² and so on.0704

This is for normalized wave functions.0725

For the most part, all of our wave functions are going to be normalized.0734

If they are not normalized, we are going to normalize them.0738

That is not a problem.0739

Basically, what we are saying is if we have some wave function ψ of a quantum mechanical system and0740

let us say it is represented by 1/2 I × ψ sub 1 – 1/5 ψ sub 3 + 2/7 ψ sub 14.0745

Let us say it is represented as a linear combination of Eigen functions of the operator of interest.0767

Then what I'm going to observe are the Eigen values A sub 1, A sub 3, A sub 14, every time I make a measurement,0775

I’m going to see one of these gets one of these three numbers.0783

The extent to which I get one number over the other is going to be square of that, the square of that, the square of that.0787

Those are the probabilities.0796

1/5² is going to be 1/25.0798

1/ 25 of the time, at every 25 measurements, one of those measurements I'm going to get an A3.0802

That is all this is saying.0809

That is all this represents.0814

This probably will not play a bigger role in what we do.0816

These are one of the hypotheses that we discussed.0818

Let us go ahead and say a little bit more.0823

After many measurements, the average value also called the expectation value.0828

The average value is symbolized like that and it is going to be the integral of ψ sub * the operator and ψ.0845

And this is for normalized.0860

We will go ahead and put the one for un normalized.0863

This definition right here, it applies when the ψ is written as a linear combination or not.0868

If it is if this thing, then this thing goes in here and here.0873

The definition is universal.0877

The average value of a particular observable is this.0879

The general definition for an un normalized wave function, it is just good to see it.0884

We have the average value of A is going to equal the integral of ψ sub *.0894

These are just integrals, all you are doing is literally plugging the functions in.0899

Operating on this, multiplying it by to ψ conjugate.0905

Putting it in the integrand and integrating it with respect to the variables.0908

If it is a one dimensional system, it is a single integral.0911

If it is a 2 dimensional system, it is a double integral.0914

If it is a 3 dimensional system, XYZ, it is a triple integral.0916

You have your software to do the integral for you.0918

The integral of ψ A ψ ÷ the integral of the normalization condition, this thing.0924

Remember, when it is normalize, this thing is equal to 1 which is why it is equal this, just the numerator.0934

This is the definition for an un normalized wave function.0938

If ψ is a linear combination is written as a linear combination.0946

In other words, ψ = C1 C1 + C2 C2 +… ,0960

Then the average value is really simple.0972

It is actually equal to C 1² × A1, the Eigen value + C 2² × the Eigen value + …,0978

It is equal to the sum I, C sub I² A sub I.0992

There is another way of actually finding it when it is written as a linear combination.1001

The final thing you want to review is something called the commutator.1005

We have operator AB, the symbol this means this is called the commutator of the 2 operators.1010

And it is defined as AB – BA.1019

You apply AB to the function then you apply BA to the function and you subtract one from the other.1024

This is called the commutator.1030

And we also have sigma A² = A² - A², there is that one.1036

The uncertainty in the measurement, the variance, if you take the square root of that you get the standard deviation.1049

And the sigma of B² is equal to squared.1056

Of course, the final relation which is the general expression for the Heisenberg uncertainty principle is the following.1067

The sigma of A, sigma of B is greater than or equal to ½ the absolute value of the integral of ψ sub *.1074

The commutator of AB applied to ψ.1088

That is the general expression for the uncertainty principle and it is based on this commutator.1094

If you do AB of the function of the BA of the function.1101

If you subtract one from the other you get 0 and those operators commute.1105

If they commute then you can measure any of those 2 things to an arbitrary degree of precision.1110

If they do not commute like for example the position of the momentum,1119

the position of the momentum operator do not commute.1122

Based on the original thing that we saw, the original version of the Heisenberg uncertainty principle that we saw,1126

we know that we cannot measure the momentum and1132

the position of a particle to an arbitrary degree of an accuracy or precision simultaneously.1137

We have to sacrifice one for the other and we have to find the balance.1143

Whatever it is that we happen to want depending on the situation.1148

With that, let us go ahead and start some example problems.1153

I do not know it that helped or not but that was nice to see.1156

Let ψ sub θ = E ⁺I θ for θ greater than or equal to 0 and less than or equal to 2 π.1160

We want to normalize this wave function.1167

Quite nice and easy.1168

Normalize the wave function.1171

Let me go ahead and do this in blue, just to change the color a little bit.1172

Normalized means we have some constant that we have multiply the wave function by, to make the normalization condition satisfied.1176

Normalized is ψ of θ is equal to some normalization constant × the function.1188

The normalization condition is this.1200

It is that equal to 1.1202

We need to solve this integral and find N, the normalization constant.1206

That is what we do.1211

If we take the integral of ψ sub *.1214

In this particular case, ψ*= E ⁻I θ because it is a conjugate and ψ is equal to E ⁺I θ.1222

We do not have to watch out for it.1235

Sometimes the conjugate is not the same as the real number.1236

This become N × E ⁻I θ × ψ which is NE ⁺I θ.1241

It is going to be E θ and we are going to set it equal to 1.1252

We are going to get N² × the integral of E ⁻I θ × E ⁺I θ E θ.1256

This is going to equal 1.1265

We are going to get N² of E ⁺I E ⁻I θ × E ⁺I θ is E⁰ which is 1.1267

It is going to be D θ.1275

We are integrating from 0 to 2 π.1276

D θ is equal to 1.1280

This is going to be N² × 2 π is equal to 1.1287

N² is equal to 1/ 2 π which implies that N is equal to 1/ 2 π ^½, or if you like 1/ √2 π if you prefer older notation.1296

I should do it down here.1319

Ψ sub θ of the normalize wave function is equal to 1/, 2 π ^½ E ⁺I θ.1322

That is your normalize wave function.1336

You want to normalize a wave function, apply the normalization condition.1339

Give me that extra page here.1347

There is a little one missing here.1349

The wave function in example 1 is that a particle moving in a circle, what is the probability that the particle will be found between π/ 6 and π/ 3?1354

The probability density we said is ψ * ψ which is also equal to the modulus of that.1365

This was equal to the probability density.1377

Ψ is equal to 1/ radical 2 π × E ⁺I θ.1383

Ψ* is equal to 1/ radical 2 π × E ⁻I θ.1394

So far so good, let us go ahead and find the probability density.1402

We will just multiply these 2 together.1406

Ψ* × ψ is going to equal 1/ 2 π × E ⁺I θ × E ⁻I θ which is going to equal 1/ 2 π.1409

The probability is equal to the probability density × the differential element.1425

D θ in this case because we are working with θ.1436

Therefore, our probability is going to equal 1/ 2 π which is equal to this part, D θ.1439

Now, we want to find the total probability of finding it within a particular region and we said π/ 6 and π/ 3.1452

We are going to integrate from π/ 6 to π/ 3.1458

Therefore, the probability of finding the particle when θ is between π/ 6 and π/ 3 is equal to the integral π/ 6 to π/ 3 of the probability.1462

I actually prefer to write it differently.1487

I prefer my differential element to be separate.1489

I do not like to write it on top.1492

This is going to equal 1/ 2 π × θ as it goes from π/ 6.1495

2 π/ 3 which is equal to 1/ 2 π × π/ 3 - π/ 6, which is going to equal 1/ 2 π.1509

Π/ 3 – π/ 6, 2 π/ 6 – π/ 6 is π/ 6.1525

The π cancels, leaving you with the probability of 1/ 12.1531

The probability density is ψ* ψ.1537

The probability ψ* ψ D θ.1539

If you want the probability between two certain points, in this case two certain angles, use integrate from the point to the other point.1543

We will see later that the general wave equation for a particle moving in a circle is ψ sub θ = E ⁺I × M sub L θ.1563

Where M sub L is a quantum number like the N in the equation for the particle in a box.1573

Just another quantum number for a circular motion.1577

Shows that ψ sub 2 and ψ sub 3 are orthogonal.1580

In order to show orthogonality, we need to show the following.1588

We need to show that the integral of ψ* of 2, ψ of 3 is equal to 0.1592

We need to show that they are perpendicular.1607

We need to show that they are orthogonal.1609

Orthogonal was the general definition.1610

We need to show that the integral of their product is equal to 0.1613

Let us go ahead and do it.1617

The integral of ψ* to ψ 3, it does not matter which order you do it.1621

You can do ψ* ψ 3, it really does not matter.1628

That is going to equal the integral from 0 to 2 π, that is our space from 0 to 2 π.1632

We are talking about circular motion.1638

Ψ sub 2 is equal to, that is the 2 and 3, that is the NL.1642

We have 1/ radical 2 π × E ^- I 2 θ × 1/ radical 2 π.1650

All I’m doing is just putting in the equation, plugging them into the equations that I have developed already.1663

That is the nice thing about quantum mechanics.1668

There is a lot going on but at least it is reasonably handle able because you have the equations.1671

In fact, you just plugged them in.1679

As far as the integration is concerned, sometimes you are going to have something that you can integrate really easily like these.1681

Sometimes you are going to have to use your software, not a big deal.1686

If you have long integration problems, please do not do the integration yourself.1689

If you want to use tables, that is fine.1693

I think it is nice but at this level you want to concentrate more on what is going on underneath.1694

You want to leave the mechanics to machines.1699

Let the machines do it for us.1701

That is what they are for.1703

We have E ⁺I × 3 θ D θ.1707

This is the integral that we have to solve.1711

It turns out to be really nice integral.1714

We have 1/ 2 π, let us pull that out.1717

0 to 2 π E ^- I 2 θ or 2 I θ × E³ I θ.1721

Just add them up and you are going to end up with E ⁺I θ D θ.1729

That is going to equal 1/ 2 π × when I integrate this, I'm going to get 1/ IE ⁺I θ.1735

I'm going to take it from 0 to 2 π.1746

I will do all of this in one page.1751

= 1/ 2 π I × E² π I – E⁰ which is equal to 1/ 2 π I.1756

Remember, E² π is cos of 2 π + I × sin of 2 π.1776

The Euler’s relation, cos of 2 π + I × the sin of 2 π -1 is equal to 1/ 2 π I × cos of 2 π is 1 + sin of 2 π 0 -1 = 0.1782

They are orthogonal, nice and simple.1811

Let us see what we got here.1820

Let ψ be a wave function for a particle in a 1 dimensional box.1824

Calculate the expectation value, average value of the kinetic energy operator for this function.1829

The average value of the kinetic energy operator is equal to the integral of ψ* × the operator and apply to ψ.1837

That is the integral that we have to solve.1848

We just plug everything in.1850

The particular ψ, I had written here.1855

Sometimes the problem will give you the equation.1861

Sometimes it will not give you the equation.1862

You have to be able to go to the tables or places in your book where you are going to find the equations you need.1864

Much of the work that you actually do will knowing where to get the information you need.1870

You do not necessarily have to keep the information in your head, you just have to know where to get it.1874

If we recall or if we can look it up, the equation for a particle in a 1 dimensional box is equal to 2/ A¹/2 × sin of N Π/ A × X.1879

The length of the box is from 0 to A.1895

That is the equation that we want to work with, that is ψ.1899

In this particular case, this is a real.1903

Ψ* is equal to ψ so we can go ahead and write that down.1905

Ψ* is equal to ψ, it is not a problem.1910

The kinetic energy operator, let us go ahead and write down what that is.1916

The kinetic energy operator is –H ̅/ 2 M D² DX².1919

We are going to apply that.1928

We are going to do this part first.1930

We are going to apply the kinetic energy operator to ψ.1931

K apply to ψ is equal to –H ̅.1936

I would recommend you actually write everything out during the entire course.1941

If you want to get in the habit of writing everything out, do not do anything in your head.1948

There is too much going on.1951

I do not do anything in my head.1952

I write everything out.1954

D² DX² of ψ which is 2/ A.1956

Do not let the notation intimidate you.1964

Most of it is just constants that go away.1966

Sin N π/ A × X.1970

Like I said, most of it is just constant.1977

When I take the derivative of the sin N π of A twice, the derivative of sin is cos.1979

The derivative of cos is –sin.1984

The - and – go away and I'm left with a +.1988

Let me write everything out here.1992

We are going to get the H ̅²/ 2 M.2004

We are going to pull this one out 2/ A¹/2.2011

Again, we have the sin when we differentiate twice but because of this N π/ A × X,2016

that is going to come out twice and it is going to be N² π²/ A².2022

And you are going to get sin of N π A/X.2029

This is just basic differential from first year calculus.2033

Nothing going on here.2035

This is the K of ψ, the ψ* × K ψ is going to equal 2/ A¹/2 sin of N π/ A × X × H ̅² N² π²/ 2 MA² × 2/ A¹/2.2037

I’m just putting things together.2070

Sin of N π/ A × X and that is going to equal 2/ A.2073

Let me write everything.2089

2/ A ^½ and 2/ A¹/2, I’m going to do it like this.2092

It is going to be 2 on top, there is going to be A × A ^½, that is A on the bottom.2096

A and A² becomes A³.2103

We get H ̅² N² π²/ 2MA³ and we get sin² N π/ A × X.2105

The 2 and 2 cancel.2121

Now, we need to integrate this thing so we are going to have.2124

I hope I have not forgotten any of my symbols here.2130

H ̅², I should have an N², I should have a π², I should have an M and I should have an A³ ×2132

the integral from 0 to A of sin² N π/ A × X.2141

This is going to equal H ̅² N² π²/ MA³.2150

This is going to be, when I look this up in a table or in this particular case I will use the table entry.2159

You can have the software do it for you.2164

This integral is going to end up being A/ 2.2169

I will go ahead and write it out.2172

-A × sin of N π/ A × X/ 4 N π from 0 to A.2174

And it is going to equal H ̅² N² π²/ MA³ × A/ 2.2186

This is A/ 2, A cancels one of these and turns it into A² and we are left with H ̅² N² π²/ 2 MA².2195

That is correct, yes.2211

That was what we wanted.2213

Let me see, do I have an extra page here?2215

Yes, I do.2216

The expectation value of the kinetic energy operator.2219

When I measure the kinetic energy, this is what I'm going to get.2223

Let us do another approach to this problem.2231

We are going to do that to the next page.2233

Another approach to this problem.2235

It was nice to revisit momentum every so often because momentum and2238

angular momentum are huge in quantum mechanics, in all physics actually.2244

Another approach to the problem.2250

We know that K is equal to P²/ 2 M.2258

That is just another way of writing the kinetic energy, ½ mass × velocity²2263

is actually equal to the mass × the velocity which is the momentum²/ 2M.2266

The average value of K is equal to the average value of P²/ 2.2276

2M is just a constant so it ends up being the average value of P²/ 2M.2282

From our previous lesson, we have already calculated this PM.2288

It was H ̅² M² π²/ A².2294

We have H ̅² M² π²/ A² / 2M.2306

Just put this over the 2 M.2316

We will put the 2M down here and we get the same answer as before.2318

You can do it with the definition of expectation value or you can do it with something else based on something that you already done.2323

This is a really great relation to remember.2331

Kinetic energy is the momentum² or twice the mass.2333

Where are we now?2340

Evaluate the commutator of P sub X P sub Y and the commutator X² P of X.2344

Let us go ahead and do this first one.2350

When we evaluate these commutator relations, use a generic function F.2353

Just use F, do not try to do these symbolically without a function.2358

At least until you become very comfortable with this.2363

I, myself is not comfortable with it.2366

I like to put my function in there because I know I’m operating on a function and the end just drop the function2368

and you are left with your operator symbol.2372

You write that down here.2378

When doing these, use a generic F and by generic F I mean just the symbol F.2379

Do not use just the operators until you become much more proficient and familiar with operators.2401

This P sub X P sub Y, the most exhausting part of quantum mechanics is writing everything down.2414

This is the symbolism, this is just so tedious.2421

Applied to some generic function F.2425

That is equal to P sub X P sub Y of F – P sub Y P sub X of F.2428

We know what we are doing here.2440

P sub X - IH DDX that is the P sub X operator and the P sub Y.2443

This is P sub Y applied to F, then do P of X applied what you got.2454

We are working from right to left.2460

Remember, sequential operators.2462

This is going to be - I H ̅ DF DY.2464

Notice, I put the F in there so operate on a function.2471

- I H ̅ DDY - I H ̅ DF DX.2475

Here we get – H ̅² D² F DX DY - H ̅² D ⁺F DY DX is actually = 0.2488

And the reason it is equal 0 because for all of the functions that you are going to be dealing with, use mixed partial derivatives.2521

And again, we saw this in thermodynamics.2544

Mixed partial derivatives, by mix partial we mean the partial with respect to X first then the partial with2546

the respect to Y is equal to the partial with respect to Y first and then the partial with respect to X.2553

The order in which you operate, the order in which you take the derivative, it does not matter for all well behaved functions.2559

By well behaved, it just means to satisfy certain continuity conditions.2566

For our purposes, you will never run across a function that does not satisfy this.2570

We will always be dealing with functions that satisfy this property.2575

This and this, even though the orders are different, they are actually equal.2578

- + you end up with 0.2583

Mixed partial derivatives are equal.2586

In other words, the D ⁺2F DX DY is absolutely equal to D² F DY DX.2591

That is a fundamental theorem in multivariable calculus.2601

The order of differentiation does not matter.2604

Let us try our next commutator.2609

We want to do the X² PX was going to equal X².2612

If you remember the X operator, the position operator just means multiply by X and the PX operator is - I H ̅.2623

Again, we are going to do DF DX -, now we are going to switch them.2633

We are going to do PX X² - IH DDX.2639

We are going to do X² F.2649

It is this × this - this × this and this × this order.2654

Be very careful here.2662

This is going to be - I H ̅ X² DF DX, I just change the order here.2665

Nothing strange happening.2676

And then this one is going to be +.2677

Notice, now I have an X² F.2683

Let me go ahead and write this up.2690

This is , X² PX - PX X².2694

We have X² F, this is a function × a function and differentiating that.2702

I have to use the product rule so it is going to be this × the derivative of that + that × the derivative of this.2706

It is going to be, the negative cancels so I get + I H ̅ this × the derivative of that is going to be X² DF DX2713

+ that × the derivative of this + I H ̅ 2 XF – I X² DF DX + IHX² DF DX.2725

These go away, I'm left with I H ̅ 2 X F.2741

I know I can go ahead and drop that F in terms of we know it is not equal 0.2750

What is happening now, I can go ahead and drop the F part and just deal with the operator part.2755

It is equal I H ̅ 2 X which definitely does not equal the 0 operator.2765

This is the operator, this is our answer.2774

The a commutator of this is equal to that.2783

We include F in order to keep track of our differentiation properly.2788

If we did not include the F, we would not have F here, we would not have the F here.2793

It might cause some confusion as far as where is the product rule.2797

That is why we are putting it in there.2802

It is very important to put it in there until you become very accustomed to operators.2803

I, myself, do not, I use F.2808

That is it, thank you so much for joining us here at www.educator.com.2811

We will see you next time for a continuation of example problems.2814

Take good care, bye.2817