Know. Okay.
So on Friday we began looking at operators, the connection between observables and operators.
So the observable is the primitive start is the starting point of our discussion.
And observable has a spectrum.
In other words, there are possible values you can get when you measure this observable to an observable is something you can measure.
So it has possible answers. And to each answer there is at least one state in which you are certain of getting that answer.
So a state where there is no ambiguity, there is no question there's nothing probabilistic about the result of that measurement out of those states.
In those numbers, we construct an operator, this animal here.
And one one good thing about this operator, one useful aspect of it is that if you squeeze it between the between the cat,
the state of your system and the associated bra, you get out the expectation value by.
Of the observable cue when we're in this state. So when there is uncertainty and the result of the measurement is probabilistic,
which normally will be the case for most states will be the case then this simple algebraic formula we showed last time,
I think that's where we finished that. That leads to the expectation value of that measurement.
So that's one way in which this operator Q is useful. You'll find as we go along that there are many other ways in which this operator.
Q which for the moment is going to have a hat to distinguish it from the observable Q which is a physical,
conceptual thing, and the operator, which is just some mathematical fiction which we're going to get used to gradually.
The distinction will blur. But I hope when you need to, you can distinguish between the physical thing.
So energy is the physical thing, and energy comes with an operator, which at the moment would be called hat.
Well, actually we did introduce that.
So the operator E hat is historical reasons called H and of course it is the operator, some over all possible energies of energy.
Energy. So these are the states of well-defined energy and these are the corresponding energies.
And this is the Hamiltonian.
In honour of the Irish mathematician who introduced this into classical physics, I called the corresponding operator into classical physics.
Okay. So any I guess you will have I hope you will recognise from endless lectures that if we have given a basis.
Any old basis, then every operator can be turned into a matrix.
Because given the basis, we can say, given any state find and this will be the sum a I, I can be written as this linear combination of basis vectors.
If we use any operator queue on on up side, we're going to get some other animal fae and we can expand Fae.
We can say that this is equal to the sum of i, i.
And then this becomes. Q operating on the sum of a j.
J. This being some David J. This being some Dave I.
Right. That's just substituting in here.
And then if I want to find out what b I is or actually what is, change this to k to make a slightly cleaner job.
This is just a dummy index. I can call it anything I like. Let's call it k.
If I want to find what b I is, I pick out to pick out of this sum over all the possible all the b case i,
i of course brought through with I so I brought through with I.
And that leads me to the conclusion that b I because this on this summit,
we're going to have an i k here, which is going to be nothing except when k is I.
So I get a b, I is equal to the sum of a j, the sum of I of I operate q j times a j because this is a complex number.
So, so when we break through by I, it doesn't get in the way because I is a linear function on the on the case.
So we can write this as the sum over j and i.
Q i. J a j where q.
I. J is by definition, the complex number that you get in this way by taking the JTH basis vector operating on it with the operator.
Q And then taking the DOT product, as it were, growing through with I.
So every operator can be represented by a matrix of complex numbers.
And of course, any one of these things is called any one of those numbers is called a matrix element.
And a lot of a lot of quantum mechanics, a lot of physics revolves around calculating matrix elements.
So it's a word that's often used. So it's a matrix made up of matrix elements.
These matrix elements are complex numbers. So if.
Now another point to make is if the basis ie is the basis of the eigenvectors of Q.
Now I forgot to last on Friday already. I think we saw I forgot to mention it just now.
I think on Friday we saw that these things well, we defined Q this way.
And with this definition it turned out that. Q I is an icon cat of Q and Q, I is an eigenvalue that was a consequence.
So these physically important states are as a consequence of this definition,
these physically important states become eigen cat's eigenvectors of the operation.
Q And these become the eigenvalues. So now we can say something different.
We can say Q is constructed out of its eigen kits and its eigenvalues in this manner was previously we had a physical statement that the
operator Q was constructed out of the states in which there's no ambiguity as to the measurement and the possible results of the measurement.
So if we use the eigen. Q I as our basis vectors, then this matrix becomes very simple.
Then Q J is going to be, of course, I.
Q Well, I'm going to put this in this. Q I. QJ But.
Q on. Q Jay is necessarily Q Jay Times.
Q Jay, this is so this becomes.
Q Jay Times. Q I times.
Q Jay, but this is Delta, right? Jay So this becomes.
Q Jay Times-delta. Jay So these matrix elements vanish and less Jay is equal to I.
When Jay is equal to I, we get the number. Q Jay, in other words, in this basis.
Q is represented by a diagonal matrix.
In other words, Q is going to look like The Matrix. Q.
Q AJ is going to be Q1, Q2, Q3, all these numbers down the diagonal and nothing everywhere else and so on.
Until we're bought, we'll run out, more to the point, run out of possible states in which Q has a well-defined value.
Okay. As a result of that, if we do this, if we if we take the complex conjugate.
No, no, no, but not do this. Yeah.
All right. Note if so. So the commissioner joined,
I think from I'm going to take it that you remember this from professor but as this lectures have you seen not joint of Q IJA of Q so the Matrix.
Q Now we've got three things now it's a bit confusing, isn't it?
We've got a physical quantity. Q Like the energy.
We've got an operator. Q hat, and we've got a matrix which is in one particular set of basis vectors is representing the operator.
So I'm a little bit short of notations. I've got a Q and a Q hat, but I will say I'm tempted to write to.
Right. Q i. J which sometimes means the particular complex number that you will find in the I throw in the Jth column of the Matrix.
Q But sometimes we use this notation. Q to imply the matrix that represents.
Q Do you see that there's a there's a slight overbooking of notation here, and it's it's it's universal in in in theoretical physics.
You can't well, nobody has a natty way of distinguishing distinguishing between the matrix and the, and the matrix elements.
So let me just write The Matrix. Q So the emission.
I under the Matrix. Q Is is Q Dagger and Q dagger is defined.
So the IGF element of it is equal to is the complex conjugate of the j element of the matrix.
Q Right. This means the complex conjugate.
So so the commission conjugate is you you take, you know, you swap rows and columns and you take the complex conjugate.
That's what happens with the individual elements. So let's see what happens here.
So we we can this property doesn't depend on what basis we look at it in.
So let's have a look at it there. So, so what is this?
Q. J So in the basis in the particular basis of the eigenvectors of.
Q what does this statement become? It becomes that.
Q dagger i j is equal to we figured out what that what q g is q g turned out to be.
Q Up there I Delta.
J or Delta. It doesn't matter. Right?
That's what we found. So that's. Q I.
J in this particular basis, no, i sorry. J I I've swapped I hope I've swapped it over.
And now I take the complex conjugate if.
If Q is real, then this becomes Q II times delta AJ is equal to Q.
AJ. So the permission now joint of Q will be Q itself, if it's possible, if all the elements in its spectrum are real.
And traditionally people have said it's obvious that an observable is a real number.
And I remember it was an undergraduate thinking, hang on a moment, that's ridiculous.
The impedance of a circuit, right, is something that I have to measure.
Yeah. Might be something you're doing.
One of the you might have done last year in some of the electronics practical measure, the importance of this circuit at this frequency.
It's clearly a complex number. So it's nonsense to say that observables have to be real cause they don't have to be real.
But if they are real, then the observable will be represented by an omission matrix.
So. So. If the spectrum. The spectrum is all real.
Then Q Hat is mission.
This is in the great majority of treatments. This is all back to front.
People say it's people say that every observable is going to be represented by or associated with a machine operator.
They then use some well-known theorem, which I'm sure you've met,
which says that every emission operator has real eigenvalues and orthogonal eigen caps.
And then therefore they say the eigen caps of these things are orthogonal.
That's not the way actually the flow of the logic of the the of the flow from the real physical world into the mathematical world works.
It's the other way. It's the real argument is that the eigen states in which the states sorry,
the states in which Q has a well-defined value, have to be mutually orthogonal.
Because why? Because. Q i.
Q j. This complex number is the amplitude to get QJ given.
Q. I. And if you know that the result of the measurement is going to be.
Q Why this this amplitude has to vanish for any QJ not equal to.
Q Why? So this whole functionality comes in is a physical requirement of the way we want to use the theory.
Then if the eigenvalues, if this are all real, it's a spectrum.
The possible results are all real. Then you end up with emission matrices, right?
But there's no need to be working with emission matrices. If if you want to work with the complex impedance as you're observable, that's not required.
But what you do need is this whole functionality result that is that is a consequence
of that's a logical necessity of the way we want to interpret the mathematics.
Okay. Now we can, of course, multiply operations together.
So something else we can do with operators is we've got two operators, O and Q,
we can define this animal by the rule that this multiplied object operating on any
state of SY is simply the result of using the operations in the sequence given.
That is to say, you use you use. Q one up sy first which makes you some cat, which you then use R on, etc.
And when we, if we choose to look at this, if we ask, well, so what is the matrix of our.
Q So what's the matrix of this in sum basis in any basis now it's going to be i.
R What does this mean?
It means our q j and into here we can stick one of our identity operators the sum of M of M.
M. Right. We saw on Friday that this sum is the identity operator.
You can stick an identity operator anywhere into a product. And then this becomes I o hat and m Q Hat.
Oops. J And this now needs a sum of em.
And what is that? This is all I m this is Q MJ So this is just the usual for a matrix product.
So it's ah, I am Q. MJ.
And we will want to know. What the mission adjoint of this thing is.
We want to know what our CU dagger age is.
And so what is that going to be? It's going to be eye our hat.
Do I want to do this? I think I probably don't.
I think probably you've seen this done. I think what I what you've seen is done in the math physics lectures this year.
So I think we can just remind you that this is. Q Hat dagger or hat?
Dagger, right. The when you take the commissioner joint of a product of operators,
you reverse the order of the, of the things in the product and dagger the individual bits.
And I hope you've seen the demonstrate you'll find the demonstration in the book.
If you don't if you haven't seen it, you don't recall the demonstration from Professor Lewis lectures.
And this is all a bit dry and boring, isn't it? Okay.
One thing you may not have seen is functions of operators.
So in particular, for a given example, x,
the position x on the x axis is going to become an operator and we are going to want to evaluate functions
of X like the potential energy at the position x depends upon x and therefore is a function of x.
So in classical physics there is a potential function V of x that tells you the potential energy at the location x.
And since X is going to become an operator, V is going to become an operator, which is obtained by taking a function of an operator.
So we need to know what it means to take a function of an operator.
Another example is this going to be an operator associate with momentum.
The kinetic energy of a particle in classical physics is p squared over two M the momentum
squared of a over twice the mass because that's a half and V squared in classical physics.
So P squared is a function of P very simple one.
It's a function of P. So we need to know what it means to take a function of an operator.
When you do statistical mechanics, you will need to there is there is a there is a quantity, a density operator,
which for which you calculate the entropy of a system which involves a logarithm of of the density.
OPERATOR Right. So you need to take the logarithm of something. So we need to be able to take functions of operators.
So let's, let's decide what this means, right? So what we're going to be done, we're going to imagine we're given F of X, maybe.
So this is just a boring at the moment. This is just a boring number.
Suppose to be given a function. This is a boring number and that's a boring number.
Right? I'm just giving an ordinary function of a complex valued function of a complex valued number.
Say, all right. And let's imagine that we can tailor expand this so we can write this is f nought to value that F takes that nought plus f
one of x the first derivative right plus a half f to x squared over two factorial is the second derivative plus a third.
So write one of three factorial of sixth f3x cubed whoops over three factorial, etc.
Just so we get to imagine that we're going to imagine that our function can be Telesur is expanded in detail.
It might not be possible to expand it around the origin, but then we can expand it around some other place.
In some little neighbourhood, physicists always assume they can expand their functions, and sometimes that leads to major disasters right there.
Important bits of physics which happen only because you can't actually us there series expand everything in life but it's a good starting point.
Okay, so we given this function, now we want to know what F of Q is, right?
What is so what is F of Q?
My the answer to that is this it's the sum of f of q i.
Q i. Q y.
So this is the definition when we say a function of an operator, this is what we mean.
So what is it? This here is an operator which has. So it has the same iGen kits.
As it's argument. All right. So a function takes an argument. The argument is an operator.
This operator has Eigen Katz. So the function of an operator has the same like in cats by construction.
But the eigenvalues. Ah, the given function of the old eigenvalues.
And can you see that this is always, this is guaranteed to work because,
because we started with a function of let's even imagine this is a real valued function of a real value function on the real, on the real variable.
So then this is just going to be some real number for every this will be some real numbers.
So this is a perfectly well-defined thing. But actually it would all work perfectly fine with complex numbers.
Complex valued functions of a complex argument. So this is what we mean by a function of an operator.
I'm going to it's a it's a it's a problem. I mean, I'm leaving it as a problem.
You can now you can now show. So on some problem set, it's a problem to show that this definition is the same as.
F of Q is equal to f nought times the identity plus f one times q plus f two over two Q times q plus.
Right. So if you if you've got the tensor is expansion, then you know what this stuff means, right?
Because we know what it is to multiply an operator on itself. We may not know what it is, take the logarithm of an operator,
but we do know what it is to multiply an operator on itself as many times as it usually will want, because we've defined multiplication of operators.
So this right hand side has a well defined meaning, and you should.
And it's an exercise to prove it's not not desperately difficult to prove that this animal on
the right that we're defining here has as eigenvectors these animals and its eigenvalues,
these animals, and therefore these two definitions coincide.
But this is the more general definition, because this doesn't assume that we can do any.
Taylor is expanding. This does. But when you can do a tiny series expansion or somehow express F in terms of algebra,
which has meaning for operators, which is just to allow which is which is to say only multiplication.
For example, you can't divide one operator by another operator. That doesn't necessarily mean anything, but you can multiply them together.
So when you when this definition works, then this one is the same as this one.
And that's an exercise that I would encourage you to be able to do.
But will not take time to do it now.
Because we're setting up this mathematical apparatus, and I'm sure you're all dying to do a bit of physics, and I am, too.
But we do have to cover a couple of little things here. Comitatus.
Oh, actually, perhaps. Perhaps we should.
We should this time. I moved over here. Okay.
So in some sense, the big news with operators is that a B hat is not necessarily equal to be had a hat.
You know, this already is in as much as, you know, the matrix multiplication doesn't compute generally.
So when you're multiplying matrices together, you don't expect the product this way in the product that way to agree.
And we've agreed that operators once once we take a particular basis, vector system of basis vectors can be represented by matrices.
So it's not surprising that there is this non-core mutability.
And the elementary techs claim this is the key thing about quantum mechanics.
They claim this is not the key thing about quantum mechanics, non commuting.
Things occur also in classical physics. And we'll see we'll we'll see that concretely as we go down the line.
However, it is a fact that these operators do not commute. And we we spend a great deal of time calculating this animal, which is which is AB minus B.
Okay. So the definition of A and B, A comma B in a square bracket is that it means just this.
Now we have some obvious results.
We have that A comma, B plus C, the comitato of A with B and C, the result of adding B to C is clearly the sum from this definition.
It follows that it is just this sum. The little one.
Oh, yeah. We have this obvious result that Abby is able to be a plus, a comma, B,
one of the reasons why we need to know the value as you will see why we need to know the value of a commentator is because we often need to swap.
We need to want to whatever. We often want to swap the order in which operators are around.
And the way to do it is to write that ap b.a plus this comitato, which is obviously true either way.
I think if it is, this adds in the thing that I should have had and takes away the thing that I've put in that I'm not entitled to have.
But it's obvious, right? And now finally, a less obvious result, which is that a B,
the product de B commuted with C is equal to A comma C with B standing by on the outside of the COMITATO plus excuse me,
plus A with C, comma B like this.
It's easy to prove this. I encourage you to prove it. I'm not going to take time to do it.
All you have to do is write down what this is from that definition and then insert two extra terms which cancel each other and you will.
You'll find you can arrange it like this. I would say it should be become a see you.
Absolutely right. Thank you very much for that. The other one I got, right?
Yeah. Okay. So what is this analogous to?
This is analogous to D by D C of a B.
If I have to do a differential of a product with respect to C, then that is equal to the A by de c.
B. Plus a. D.
B. D. C. Right.
This is the rule for differentiating a product. And can you see the mirror there?
The idea is that taking the comitato of something with C is analogous to taking the derivative of something with C and this is no accident.
This, for a mathematician in certain contexts is called a derivative.
And the and the and the rules that we are familiar with here is that you, you, you, first of all, if you have a product,
you can get the result by having this operation happen on the first thing while the second stands idly by.
And then you have two. You let the first one ads stand idly by and then you work on the second one.
So here we have you work on the first one second standing idly by, and then you work on the second one with the first one standing idly by.
The only material difference between these formulae is that this formula is left invariant.
If I move B over here, or if I move over there or whatever, I change the order here.
It won't make any difference because these are ordinary, boring multiplications of complex numbers.
But here it is. It does make a difference like this. A comma C is an operator.
It's the difference of two operators. So it's an operator. And,
and therefore it isn't clear that I can swap the order of this operator in
this operator in the order in which you write these things down is important.
So these these rules should be kind of should be you should make sure you understand where they come from.
You should memorise them.
And broadly speaking, once you once you've got these three rules on board, you never need to look inside a commentator and use this relationship here.
It's bad practice, by and large, when you're doing computations to expand commentators to see what's inside them.
In the same way, I would say as this rule here of course, comes from looking at A, B evaluated at C plus delta C minus A,
B evaluated at C, all over delta C limit, all this stuff, you know, using this stuff, you can prove this.
But we once you've got the rules of calculus, you don't do this expanding stuff anymore.
You just, you know, that's what lies underneath it. That's the justification.
But you don't go back to that every time you have to do a calculation, every time you have to differentiate the x of x cubed,
you do not write that this is x plus delta x cubed minus minus x cubed all over
delta x cubed and come to the conclusion that it's about three x squared to you.
So please don't resist the temptation to to expand out to commentators, write the contents of a commentator out.
There are times when ultimately you have to do that, but most of the time you don't and try and avoid doing it by using these rules here.
Okay. Okay.
I'm going to need one result which combines these statements and those statements we're going to need very shortly to calculate what F, B, comma A is.
So I've got it. So I will want the commentator concretely.
This is going to be V of X and I'm going to want to take the commentator with the momentum operator and these things.
These all need hats, I suppose. Yeah. And those things up there need hats, but you're managing them all.
So this is I'm going to want to, I'm going to want to calculate something like this.
So let's see what this comes to. In order to see what it comes to,
I'm going to imagine that I can expand F in this manner so I can write this as a f nought
times the identity plus f one times B plus f two over two times B squared plus blah blah.
Right. The title sir is expansion of F around the origin commuted with a so now I can use
that second rule there that second rule to do the commentator of this product.
The comment this is a boring number, right? This is a number and this is the identity operator.
So this isn't a number. That's a number, but this is the identity operator.
And the identity operator obviously commutes with everybody because I times a is going to
be this is going to be a same as a times I is going to be a so the commentator so so I
use the second rule to say that the commentator of this sum with a is the sum of the
commentators of this thing with a vanishes and this thing with a so that's going to be F one,
B hat, comma, a hat. This comes outside the comitatus.
Or maybe I should have added that to the rule list there because it's a boring number.
But I think it's it's kind of an obvious principle plus f two over two factorial a B squared comma a plus
F three over three factorial B cubed comma a plus plus plus plus plus plus plus right until your board.
So that's that's the middle rule used. Now we use the last rule to say that this is f one.
Well this is, this is just to repeat, but this B squared is B times B so I can expand this into F two it into B hat, comma, a hat, b hat plus.
Right? So it was b b commuting with a so I worked on the first B while the second B stood idly by and now
I have to put down the first B standing idly by and have the second B worked on by a plus dot,
dot, dot, plus F three, etc. Right.
Which is going to involve three terms because it'll be b b b committee with a, so it'll be three things to consider.
And this is as far as I can go in general. But in an important case, if the hat A hat commutes with B.
So if this. Commentator So be hat a hat commentator is an operator.
This is the difference in two operators.
So if this operator commutes with B hat, then this B camera and this B camera and this one could all be taken outside.
And I have that. So under this condition, that F of B hat commuting with a hat is equal to B hat, comma, a hat times F one plus, f two plus.
And you can, you see it will be F three over two because the f three would have been over three factorial, but we would have had three terms.
Oh, sorry. This is going to be times B this silly me, this is going to be times b hat.
This is going to be times B squared plus. So this is what this will all reduce to, which can be more conveniently written as as d f by DP.
So this is an operator. Oops. Sorry. Yeah.
It doesn't matter which order I put it in. This is an operator. And that Taylor series is the Taylor series for Def by the x.
So I can write this stuff. Here is the F by D, B, and then here is my B camera.
And I was momentarily panicked about having written this in front of this.
But we've agreed that this operator computes with B, that was the condition under which we were making this further development.
And if this thing commutes with B, it commutes with every function of B in particular,
it commutes with the F IDB, which is a function of B, so it doesn't matter which order I put this in.
So this is a function. Which means it has the same I can cats.
So that's a result we're going to want. And there's one other thing that now needs to be discussed.
Which is the physical implications of a commuting would be so if they had a B hat.
Equals nought. We say commuting observables.
Then the mathematicians assure us we have a theorem.
And the theorem is that in this case.
There is a complete set.
Of mutual aid cats. We'll call these mutual Asian cats.
Just I. That is to say, for each and every one of these,
it is true that a hat on I is equal to a i i and simultaneously be hat on I is equal to some number by when I when to operate is commute.
There's a theorem that states this. What does that mean for the real physical world?
What that says in the real for the real physical world is there is a complete set of states,
these states in which the result of making a measurement of a is definitely known
and simultaneously the result of making it a measurement of B is certainly known.
So there is a complete set of states in which there is no ambiguity.
There is nothing probabilistic about the result of measuring either of these quantities.
It's very important to bear in mind that in complete, we're not merely saying that there is a state or ten states with this property.
There are enough states with this property that any state can be written as a linear combination.
Whatever GI of these objects right there complete.
That's what completeness means that any state. So there is a complete set of states in which there's absolute certainty.
It does not mean that.
The fact that I definitely there's no uncertainty in the value that B takes implies that there's no uncertainty in the value that takes.
That is not that does not follow from the commuting of A and B as we will see it.
It may well be the case is that there are estates in which B definitely has a value, for which A, the outcome of measurement of A is uncertain.
So it's so the the result of two observables commuting their operators commuting is slightly technical because it involves this complete statement.
It is that there is a complete set of states in which the outcomes of the measurements of both observables are certain.
Okay. Now if a comma be not equal to zero, what does this mean?
All it means is that there is at least one cat.
Such that a comma be.
There may be an infinite number of cats such that a bee operates on them and produces nothing.
But there is one. There is at least one. If you say that these operators don't compute,
you're saying or asserting that there is at least one cat where the commentator operating on it doesn't produce nothing.
So it. If so, what does this imply?
It implies that there is no complete.
So it's a very weak and not emotionally striking result that there just isn't a complete set of states in which they're both.
They both have definite values. There may be a very large number of states in which they do have definite values simultaneously.
So it is not a statement that you can't know the value of this simultaneously.
With the value of that, we'll come across a counterexample next term, I guess a very important counterexample.
So don't run away. It's a very it's a very, very widely held misconception that if two operators don't commute,
you can't know the value of the one and the value of the other.
That's just not true. The statement is that there isn't a complete set of states with that nice property.
Okay. We've just got time to start on the next really important section, which is about time evolution.
Maybe. Maybe it's time to move over here. Okay.
So physics is about prophecy. It's prophecy that works. It's about predicting the future.
That's what it's about. And therefore the core of it is equations of motion, Newtonian mechanics.
We think of usually as to do with F equals Emma.
It's, it's making it seem to what the acceleration is when you can calculate the acceleration and you know the,
and you know the initial position and velocity,
you can predict where your by your missile is going to be at some future time or your planet is going to be at some future time and so on.
Right? That's what it's all about. So at the core of quantum mechanics sits its time evolution equation.
And I'm not going to immediately justify this, I'm just going to write it down.
It's the time. What's the time?
Dependent. Shredding a.
This is the core of the subject. This is where the physics sits and it's high bar what I call the sci fi d t is equal to h upside.
This is why it's because it appears in this central, crucial, vital equation.
The Hamiltonian sits here. That's why the Hamiltonian matters.
Right? Its status in life is unique because it uniquely tells you about the future.
And that's what physics is about. Okay.
And this is the state. Any system.
So it's completely non-negotiable for a state which purports to describe a real physical object.
It has to satisfy this equation. It tells you how the state evolves in time.
It's, of course, a very abstract object at the moment. It won't be telling you much.
And at the moment, I can't connect it.
We will be connecting it very shortly. But just at the moment, I can't connect this for most of you to classical mechanics.
Those of you who've done did the seven short option will recognise this perhaps just a little bit as having something to do with Hamilton's equations.
But we will. So the justification, the physical justification is that this is the dominant equation.
Will, will, will come by and by. But ultimately, there's no way this can be derived from anything you already know.
This cannot be derived out of classical physics.
Classical physics can be derived out of this because classical physics provides an approximation to this.
Right. And the assertion is that nature evolves things according to this equation.
And whether that's true or not can only be determined by experiment. It's got nothing to do with mathematics, and it's got nothing.
And it cannot be justified on the basis of classical physics, ultimately.
But if this is a valid statement, it should it should produce the right Newtonian equations of motion.
I will show you that it does produce the right Newtonian equations of motion, because Newtonian mechanics is an approximation to quantum mechanics.
Right. Okay.
Now, suppose let's. This is.
This is kind of a scary equation, right? So let's.
Let's try and find some circumstance in which we can solve this.
Right. So suppose our system has well-defined energy.
In other words, the state of sci at time t.
Well, the state of upside is equal to d where h e is equal e.
Write a state of well-defined energy has to be an eigen function of the energy operator.
H with eigenvalue e. That's that's what it is.
So let's suppose that we in our system happens to have well-defined energy.
Then it will then it will have to solve this equation and we'll have each by the e by d t is equal to g h e is equal to e e.
So the rate of change of e is simply proportional to E, and we know how to solve that equation.
We spot it just from ordinary, old fashioned calculus.
We spot that this implies that e at time t is equal to e to the minus i e t over bar e of zero.
So I feel I feel entitled to write this down on the basis of just boring classical classical mathematics,
which says that, that if we know that the X by d t, no, I shouldn't do it there.
If I know that the x by d t where x is some variable is equal to x, that implies that x of t is equal to x of nought e to the.
Right. So this result for many a result inspires me to write down that I can now trivially check
by differentiating this right hand side that it satisfies this differential equation.
Right. Because when I because when I differentiate this right hand side, this thing is not a function of time.
It's the it's the value that the state of well-defined energy takes a time t equals nought.
So it has no time derivative. So the time derivative comes merely from this, which is a totally boring exponential of a bunch of real numbers.
Well, with part from the AI. All right, so we know how to differentiate this.
So it's easy to evaluate the time derivative of this, and it's trivial to check that.
Then it's that. Then e satisfies this equation. So what does this tell us?
This is a very important result. It tells us that the time evolution of states of well-defined energy is really dead trivial.
They basically don't change.
All that happens is the phase goes around in increments at a constant rate, each over each bar with a frequency of reach bar,
which is, of course, incredibly for typical systems like this is incredibly large because bar is so small, it's on the bottom there.
So this frequency is stupendous for an object like that.
So this thing has some energy and its wave function is zooming around at some hysterical rate.
That's all that's happening.
The beautiful thing is that this enables us to solve the general problem, because if I have if I have a CI, I want to solve this.
So I've got now some system that's not in a state of well-defined energy. And we'll see that real systems never are in states of well-defined energy.
But then I can surely write this as a linear combination with coefficients that depend on time of states of well-defined energy.
Right? These are a complete set of states because they're. Yeah, we've been through this.
This is just boring, right?
So I can put I simply put this and that's this expression, this expansion into both sites of my time dependent Schrodinger equation.
And we discover that, that we discover that h bar the up side by d t is equal to h bar brackets.
We have to differentiate this stuff so it's a and dot e and t plus a and times the time derivative of
this times key and by d t was that equal to that's equal to on this side h enter the sum and e n.
I've missed some of her. And indeed I have. I missed out a son over. Thank you.
Just about here. I'm kind of conscious of that horrible clock.
And. But, uh. Well, okay, why don't we just write this?
Why don't we just write carry this on and write this as a sum over n of a and h e n.
But. This term.
This term here cancels this term here i hpr and so far d n by d t is h e n so these terms all cancel those terms leading to the conclusion.
So when I when I look at this stuff is equal to this stuff.
I've cancelled this to the right side now. So there's nothing. And the left side has this stuff has a dot.
So I've got the conclusion that the sum of N of a and dot e and of t equals nought bra through with an e i of t.
And that leads to the conclusion that I. Dot equals nought.
So the AI's. The AI a constant.
So we have a solution. This enables us to.
To write down the solution to the general problem.
We have that upside of t is equal to the sum of some constants and which you can determine from the initial conditions times e.
N of t. But I can explicitly write that out because I know how this thing evolves in time.
This is the sum and of nought e to the minus i e and of t e n.
T of ball times. N of nought.
So once. So this is the really, really this is a fabulously important equation.
So this part of it is needs to be purchased the back of the retina.
And it it's the key to everything because it tell and what it tells us is once we know what these states,
the well-defined energy are and the approved energies, we can trivially evolve in time the dynamical state of our system and predict the future.
We have everything. That's it. So a large part, a huge part of this subject revolves around finding what these states of well-defined energy are,
because they have this enormous predictive power.
They are the miracle. They are sort of one to drug.
They solve the problem, they do it. So we'll talk some more about them tomorrow.