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Schrӧdinger Equation as an Eigenvalue Problem
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- Intro
- Schrӧdinger Equation as an Eigenvalue Problem
- Operator: Multiplying the Original Function by Some Scalar
- Operator, Eigenfunction, & Eigenvalue
- Example: Eigenvalue Problem
- Schrӧdinger Equation as an Eigenvalue Problem
- Hamiltonian Operator
- Quantum Mechanical Operators
- Kinetic Energy Operator
- Potential Energy Operator
- Total Energy Operator
- Classical Point of View
- Linear Momentum Operator
- Example I
- Intro 0:00
- Schrӧdinger Equation as an Eigenvalue Problem 0:10
- Operator: Multiplying the Original Function by Some Scalar
- Operator, Eigenfunction, & Eigenvalue
- Example: Eigenvalue Problem
- Schrӧdinger Equation as an Eigenvalue Problem
- Hamiltonian Operator
- Quantum Mechanical Operators 16:46
- Kinetic Energy Operator
- Potential Energy Operator
- Total Energy Operator
- Classical Point of View
- Linear Momentum Operator
- Example I 26:01
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Transcription: Schrӧdinger Equation as an Eigenvalue Problem
Hello, welcome back to www.educator.com and welcome back to Physical Chemistry.0000
Today, we are going to talk about the Schrӧdinger equation as an Eigen value problem.0004
Let us get started, let us go ahead and work in black today.0009
Recall example 3 from the previous lesson.0017
We had an operator which was I believe C ̂ and we said that this operator was defined by -I × H ̅ DDX of something, whatever F was.0023
We said that F was equal to E ⁺I × n × X.0039
We operated on this and we found that this C ̂ of our particular F was equal to n × H e ⁺INX.0045
Now, notice that operating on F is the same as multiplying it by we just ended up, the original function was e ⁺INX.0068
We operated on function was nh × e ⁺INX.0090
We ended up just multiplying the original function by some scalar, by some number.0095
Positive or negative actually does not matter.0099
This is the general arrangement.0101
Operating on F is equal to some scalar × F.0103
The only thing this particular operator did was multiply F by some constant.0124
In other words, if my function is X², I may end up with 17X².0149
That is it, I just multiplied the X² by a factor 17, that is all.0159
Recall example 4, in example 4 we had that the operator D was actually equal to partial derivative of the function with respect to Z.0162
In that particular case, RF was equal to XY² and Z³.0174
In this particular case, the operation on F ended up equaling 3XY² Z².0181
Here operating on F did not just multiply it by some scalar factor, the D of F was 3X² Y² Z².0192
Make sure this is clearly to.0207
The original function was XY² Z³.0209
In this case, it did not do that.0212
When an operator operates on a function and gives back the original function multiplied by some scalar, this is profoundly important.0220
Here is what we are getting into some very deep and important mathematics.0278
The general expression is this, operating F is equal to some A × F.0283
I will actually go ahead and do here.0292
Let me go ahead and put the A of X = some constant × F of X.0296
Personally, I do not like the X, they tend to distract me.0308
I will just go ahead and do it this way.0310
A of F is equal to some constant A × F of some function that we are operating on.0313
Here the F of X or the F, it is called an Eigen function of the operator.0321
A is called an Eigen value for the Eigen function.0345
I actually called V, the Eigen value associated with that particular Eigen function.0357
Operating on F in a situation where when you operate on F, you just end up getting F back multiplied by some scalar.0391
The F of X is called the Eigen function of the operator.0399
A is called the Eigen value that is associated with that particular Eigen function, when this thing is satisfied.0404
Now, given a particular operator A ̂, finding F of X and its corresponding A is called an Eigen value problem.0411
You are going to be presented with some operator.0452
It is your task to find all of the functions and all of the values, the scalars that satisfy this equation.0455
That is what we are saying.0464
Which function when you operate using this particular operator that we give you,0465
gives you back the original function multiplied by a scalar, that is the problem.0469
It turns out that the Schrӧdinger equation is just an Eigen value problem.0474
We will show it in just a minute.0477
For example 3, our C of F = A of F.0481
We have 3 things, we have the operator, we have the Eigen function, and we have the Eigen value.0496
In this particular case, the operator was n-I H ̅ DDX, that was the operator.0500
The Eigen function was E ⁺INX.0510
Any function E ⁺INX satisfies this relation.0521
Of course, the Eigen value.0525
You can write it as two words, I will write it as one word.0531
The Eigen value that was NH, that is the whole idea.0535
Operator, Eigen function, Eigen value, profoundly important.0545
Those of you who have taken linear algebra, chances are you already have seen this when we are talking about matrices.0550
This is this and that is that, that is what is going on here.0558
Let us go ahead and talk about what this has to do with the Schrӧdinger equation.0566
Let us look back at the Schrӧdinger equation and let us write it out like this.0572
We had - H ̅²/ 2M × D² DX² of our particular C.0579
I’m not going to go ahead and put the X there.0593
It is a function of X that we are looking for + this potential energy function × our Z or ψ if you like.0594
Equals E × Z this was Schrӧdinger equation, that is the function.0607
Notice something here, notice I have the Z here × something, the Z here × something.0621
I can factor out the Z and write this left hand side as an operator.0628
Here is what it looks like.0634
It ends up looking like this.0636
Let me go back to a black.0638
I’m going to factor out the, so it was going to be –H ̅²/ 2N D² DX² + V of X.0642
I’m going to write the Z out here.0658
Z equals E × C.0660
Notice, we know we can do this because operators they distribute the way that polynomials do.0665
Notice what we got.0672
Once again, here is my Z and here is Z, this as an operator.0676
This right here, this is just some number.0687
The energy of the system is just a constant.0692
If we call this operator, if we get a symbol, if you call this operator on the left, operator H ̂, we can rewrite this whole thing as H ̂.0700
Let me go back to black.0727
H ̂ of Z is equal to E × Z.0730
We just expressed the Schrӧdinger equation as an Eigen value problem.0739
If you given some wave function and if you operate on it with this thing, this we have not given a name to it yet, it is actually called a Hamiltonian operator.0743
If we operate on this wave function, we actually end up getting the wave function back multiplied by some constant.0751
The constant happens to be the total energy of the system.0759
This is profoundly beautiful.0761
We expressed the Schrӧdinger equation as an Eigen value problem.0772
That is exactly what it is, as an Eigen value problem.0782
We have presented with a particular Schrӧdinger equation for a particular system.0787
Our task is to find the function Z, the wave function and the associated Eigen values that happened to coincide with these operators.0792
The operator is the same, it is the Hamiltonian operator.0802
We are going to take this operator and see if we can find a function and the Eigen values that are associated with it.0805
For each function, there is some energy for the system.0811
When the system is in this particular energy state, the particle is behaving this way according to the wave function, that is all we are doing.0816
The rest of it is just math, it really is.0824
When we have solved the equation, we will not only have found a wave function and Z of X but also the total energy of the particle.0829
The total energy of the particle in the state is represented by the wave function.0877
In the state that is represented by the particular wave function the Z of X.0890
Once again, the operator is called the Hamiltonian.0906
HR is called the Hamiltonian operator.0910
Let us go ahead.0918
Our Hamiltonian operator happens to equal - H ̅²/ 2 × the mass, the second derivative +0928
this potential energy function that is the Hamiltonian operator right there.0941
Notice, if we have - H ̅²/ 2M D² DX² + V of X × some C is equal to the energy × the Z.0946
Notice, this is potential energy, this part right here.0970
This is the total energy is equal to the potential energy + the kinetic energy.0981
Therefore, this makes the kinetic energy operator.0991
I will use E kinetic.0998
Operators are quantum mechanical.1008
The operators, this and this, or this thing together, are quantum mechanical.1022
The energy is classical mechanical, that is the relationship.1030
There is an association.1036
Classical quantities, things like energy, momentum, angular momentum, position, things like that, are represented in quantum mechanics,1039
But operators that is the whole idea.1052
Operators are quantum mechanical.1057
Scalars like the energy are classical mechanical.1060
When there is some quantity in classical mechanics like momentum, position, energy, angular momentum, whatever it is,1076
In quantum mechanics those are represented by the kinetic energy operator.1083
The position operator, the momentum operator, they are represented by operators.1087
The reason they are is because all the information is contained about a particular system of quantum mechanical system is contained in the wave function.1092
In order to extract information from that function, we have to operate on it.1103
If I want to find the angular momentum of a particle, I'm going to take the angular momentum operator of the wave function.1108
If I want to find the position, I'm going to take the position operator of that function1114
and it is going to give me some certain information that I can do something with.1118
Operators are quantum mechanical.1126
Scalars are classical mechanical.1128
Such associations are the very heart of quantum mechanics.1131
Now given what is above,1136
Let me do this on the next page, that is fine I can do it here.1140
Given what is above, in another words this thing that we just did.1144
Let us define the kinetic energy operator that is equal to -HR²/ 2N DDX².1151
This is the kinetic energy operator.1171
If I have a wave function and I want to know what the kinetic energy is at a given moment,1178
I will go ahead and apply the kinetic energy operator to that wave function and it tells me something.1182
When I say apply it to that wave function, I will be more specific about that when we actually talk about1188
how we are going to extract information that we can actually measure and see from the wave function.1194
But that is what we are doing.1200
We defined V which is just V of X which means multiply the function by this function V of X.1203
This is the potential energy operator.1219
In other words, if I want to know what the potential energy of a particular particle is,1222
of an electron in this particular system whatever happens to be, I take the wave function and multiply it by the potential energy function.1228
This is the potential energy operator.1239
In this particular case, the operation is just multiply by.1241
It is fine, let me go ahead and in the next page.1250
The Hamiltonian operator that is equal to the sum of the kinetic and potential energy operators, that is equal to.1252
I think I can go ahead and put on the next page.1260
Let us do this.1270
H ̅ that is equal to –H ̅²/ 2M D² DX² + VX this is the total energy operator.1273
Or just the energy operator.1295
These are operators.1304
Let us go ahead and fiddle with this a little bit.1307
From the classical point of view, classically the kinetic energy of a particle you know was equal to ½ MV² that is also equal to this.1311
If you take ½ MV² it is going to be equal to the momentum squared divided by twice the mass.1323
We will go ahead and just start playing with this formally, mathematically.1333
Formally just means we are working with this symbolically.1338
We are going to define the kinetic energy operator is equal to this momentum operator squared/ 2M.1340
Let us go ahead and multiply here, the momentum operator squared is going to equal 2M ×1356
the kinetic energy operator and it is going to equal 2M × the kinetic energy operator which happens to be - H ̅²/ 2M D² DX².1365
The 2M cancels and we get ourselves a square of the linear momentum operator which is going to equal - H²² DX².1380
We are able to derive another operator.1396
These exponent here on the operator, it is nothing more than sequential operation.1403
Here is operator² is equivalent to doing P again.1408
If you saw the linear momentum operator cubed, it will just be P again.1414
The squared is just a symbol for sequential operation.1424
This does not mean this is symbolic for the operation.1428
It does not mean take the linear momentum operator, whatever you get square it.1431
That is not what it means.1436
When you see an operator squared, it means operate sequentially.1440
Let us go ahead and break this down.1444
Squared is equal to -H² D² DX².1448
Let us go ahead and separate this out.1458
- I H DDX in other words we are going to factor this out and - I H DDX.1468
This linear momentum operator squared actually factors out.1475
If I multiply these two, if I operate and operate, I end up getting this thing.1479
Therefore, we can go ahead and define the straight linear momentum operator in the X direction which is one of these – I H ̅ DDX.1484
This is the linear momentum operator,1497
if I have some quantum mechanical system and if I have another wave function for it.1505
If I want to know something about a linear momentum, whatever is that I want to know, I operate on the wave function with this operator and it gave me some information.1510
And again, what I mean by operating on it is not just going to be a direct operation.1521
We will see what it is, we are still going to play with this mathematically but is essentially what I’m doing.1526
I will leave that for the time being.1534
Let us see, now we have our linear momentum operator.1538
We have our Hamiltonian which is the total energy operator.1542
We have our kinetic energy operator.1545
We have a potential energy operator.1548
We have a majority of things that we need to actually get started.1549
Let us go ahead and finish off with an example here.1553
We will close this lesson out.1557
One more, here we go.1561
Let P = -I H ̅ DDX and in this particular case our functions are going to be sin X.1564
We want to show that the operator squared of F does not equal the operation on F².1571
In the last lesson we show that operators do not commute.1581
We want to also make sure that you understand the difference between squaring operator and taking something and getting a function and squaring that.1585
They are not the same thing.1592
Let us go ahead and do this.1594
The operator squared of F is equal to,1597
Again, we set it as a sequential operation on F.1601
It is going to equal this one, that is the far left one and it is going to be – I H ̅ DDX of sin X.1606
We end up with this being - I H ̅, the derivative of sin X is cos of X.1617
I think I’m going to it up here.1629
We have the - I H ̅ of the –I H ̅ of cos of X and we end up with - -, we end up with I² H ̅².1635
I’m sorry I forgot my derivative operator.1658
Let me go back and erase this, it is – I H DDX of - i H ̅ cos X.1662
When I take the derivative of cos X, I get - sin X - -is + I and I is I²,1674
H ̅ and H ̅ is H ̅².1685
I² is -1.1691
Let us skip a few steps.1693
- and - is going to be positive, I and I is going to be I².1695
I end up I² is -1 and – and - is going to cancel and become a positive.1702
We are just left with an I² H ̅² and sin X.1705
Let us go ahead and do P of F and we will square that.1716
The P of F is going to be – I H ̅ DDX sin X².1721
It is going to be – I H ̅ cos X² is going to equal -1 × -1 is positive.1736
I × I is going to be I².1752
H ̅² cos² X is equal to – H ̅² cos² X.1756
Clearly, this and this do not equal each other.1767
In general, some operator raised to some exponent and then applied to F absolutely1772
does not equal that operator applied to F and then raised to the exponent.1783
This is a symbolic representation of how many times you are going to operate in sequence.1789
This squared right here is an actual mathematical operation.1793
These are not the same thing.1796
We have to be very careful about our symbolism, about our mathematics.1797
Profoundly important.1805
Just take your time and work slowly.1806
Now that we have introduce this notion of an operator, there are a lot of symbols floating around.1809
In any case, we will go ahead and leave it that.1818
Thank you so much for joining us here at www.educator.com.1821
We will see you next time for a continue discussion of operators in Quantum Mechanics, bye.1823
1 answer
Mon Feb 23, 2015 6:43 PM
Post by Carly Sisk on February 23, 2015
Hey Professor Raffi!
Just to clarify what Anhtuan was saying; the final answer should be positive (hbar^2 * sinx), correct?
1 answer
Wed Jan 28, 2015 12:32 PM
Post by Anhtuan Tran on January 28, 2015
Hi Professor Hovasapian,
On your last example, when you were doing P^2f, your second last step was i^2 * hbar^2 (-sinx). So the answer should be hbar^2 * sinx, because you already took into account the minus sign from sinx and minus sign from i^2. Therefore, there should be no i^2 left at the end.
Could you please double check it? Thank you.
2 answers
Sat Sep 27, 2014 6:27 AM
Post by Raj Singh on September 26, 2014
Where did the i in (-iHbar d/dx) came from when deriving linear momentum operator? additionally if you multiply 2 -i wouldnt you get positive i or positive 1?