Where shall we? Shall we begin? So last term we discussed the dynamics of a harmonic oscillator,
which is a one dimensional system where the particle is bound by a potential V is a half CG squared.
And we found a number of important results. Most obviously there was quantisation of the energy, so discrete ness of the energy levels.
And very importantly there was the phenomenon of zero point energy that even when something was in its ground state,
it had the uncertainty principle obliged to have non-negligible kinetic energy and potential energy.
Indeed. So what we're going to do now is study a number of very simple one dimensional
potentials motion in a number of very simple one dimensional potentials,
which are themselves very artificial. These are step potentials.
So they so they change this continuously at some values of X from one value to another value.
So they're very artificial, but they the merit of them is that we can solve the governing well, the key equation,
the time independent Schrodinger equation and obtain the states a well-defined energy from which we can,
as we saw with the harmonic oscillator, recover the dynamics of the system.
We can solve this equation, this important equation, simply for these rather artificial potentials.
And we'll discuss at the end which of our results is artificial,
reflects the artificiality of the potentials and which are generic and ones that we can we can believe in.
So this is what we're going to start with is the square potential.
Well, which is this structure, the potential energy being plotted vertically, position being plotted horizontally.
Here is the origin. This is going to be a distance. A, this is a it's going to be symmetrical.
This is going to be a potential level v zero and this is going to be potential level nothing.
Right? So we set the zero as a potential energy to be at the bottom of the well.
And then there's then there's when you're more than distance away from the origin, you have some potential energy.
V0 is a constant. So this is highly artificial.
But let's see what quantum mechanics has to say about motion in in here and in particular.
So what we're going to look for is stationary states, that is to say, states of well-defined energy.
Which we write like. That's right. So these are states of well-defined energy.
We're interested in these because they enable us to to solve the time dependent Schrodinger equation trivially.
Once we know what all these things are, we can write down. And we know once we know what these things are and we know how to,
then we can express any arbitrary initial condition as a linear combination of these things,
and we can time evolve it in a simple way, as we saw last term.
So we're going to find these things and they're going to have wave functions which will call U of X.
So this is X. E is the wave function.
Now this potential is is symmetrical, is a is an even function of X, right?
So we have that that V of minus X is equal to V of x.
So it's an even function. The consequence of that is that we introduced the parity operation last term.
We had that P if you remember, X, PFC P is an operator, the parity operator.
Which makes out of a state of, say, the state that you would get or which has amplitude to be at minus x.
Well, this, that. What is this? This is the amplitude to be it X when you are in the state that P makes out of a.
And this was defined to be minus X upside.
In other words, it was defined to be the amplitude to be at minus X when you were in the state of PSI.
So the power is the operator makes a state which is the same as the state you first thought of.
Except if if every point is reflected through the origin.
We discussed this operator and because B is an even function of X, we have that PV.
X PV psi is going to be equal to minus x.
Minus X the upside, which because V is a is a is V is a function of position.
It's it's a function of the position. Operator This is going to be V of x x of psi.
In other words, it's going to be simply vivek's. Sorry, sorry, sorry, sorry.
Right. I need a minus X and a minus x so it can be view of minus x times of psi of minus x.
And that what's because the is an even function of x so v minus x is the same as v of x.
This is equal to this is this and this could be written as p.
So what does this mean? This means that because vs an even function of x, it computes with the operator.
With the operator p. In other words, p v equals nought.
Because this is rather a rigmarole.
I have to admit p commands with the because the is an even function which is a way of saying that the potential.
In symmetric. About the origin.
This is the bottom line. If you have a potential which is symmetric about the origins and even function of X, then it commutes with a parity operator.
The consequence of that is. So the Hamiltonian is of course is is as ever P squared over two m plus v of x p squared is an even function.
If you write this in the position representation, it's minus h bar squared d two by the x squared.
So this is an even function of x. So this commutes with the parity operator.
We've just figured that this competes with the parity operator. So we have the P comma H.
So this implies that peak homer H is zero. So the parity operator commutes with a Hamiltonian whose stationary states we would like to find,
whose igen states that just say the stationary states we would like to find.
So what does that imply when to operate his commute?
Remember, the fundamental rigmarole is that this implies there's a complete set of mutual aid and states.
Of P and H. That is to say we can if we wish is look for I can states of age which are states of well-defined parity and the
way functions of these states are well defined parity will either be even functions of X if the parity is even,
or there will be odd functions of x if the parity is odd. So we can.
So what this means is we can insist or we can look for.
So we can look for stationary states with wave functions.
You have x meaning, of course, x e that are either even functions.
You x equals equals you of minus x or odd functions.
U of x is minus u of minus x.
And this observation knowing what knowing that you're looking for an even function, say,
makes it much easier to find that function than if you don't know whether it's even resolved.
That's this is a this is a general observation.
And we'll just find a concrete example of it in a moment. So here is all.
Here is our potential. Again, a minus.
Say, what do we have?
We have here? What is the that the time independent Schrodinger equation.
That is to say, what is the. What is that? That is the equation which shows that h of psi is equal to FC in the position representation.
What is this equation important equation look like here. Well, the at this point here where the potential is zero,
h is p squared over two M so at this location here it becomes this becomes piece
going over to M which is minus H bar squared over to M D to u by the x squared.
So this is I suppose I should change this sorry in this context to you probably know.
Let's leave it with E that's what we were calling it the state e so this left hand part and that reduces
to just this in the position representation because we only have the kinetic energy at this location,
there is no potential energy. And on the right hand side, of course, we simply have e times u.
So we're trying to solve this, this equation, and we know all about the solutions of this equation.
This is just the simple harmonic motion equation of classical physics.
Essentially, it tells us so we know that you is cos k x or you is sine provide solutions of this equation.
Cos kicks is an even functions. So that must be the solution belonging to an even parity state.
And sine k x is an odd function of x, so it's an odd parity thing.
So this has solutions. U of x is equal to either cos k x or sine k x depending on parity.
And we have that k, right.
So when you double differentiate this, you're going to get minus K squared cos X.
So the minus sign deals with that.
We're going to have that minus K squared.
E is equal to this stuff here. In other words, we're going to have a K is equal to.
Yeah. No.
This is wrong. Sorry. Sorry, but it's the wrong way up. What am I doing?
Sorry. The derivative is here. Excuse me. So when we double differentiate, we'll get minus H plus squared k squared over to em.
We'll get a minus coming from the differentiation which will cancel this and we will have
that k squared is equal to 2me over each bar squared square root while k is is that.
So we have determined what the what the what the wave functions are of the stationary states in that interval from minus eight away.
And in terms of the energy, what we now need to do is.
Is think about the state of affairs here.
So, so. So this stuff is all true for monarchs.
Less than a what about the case when monarchs is greater than a.
So now the picture in this zone here.
What is the time independent Schrodinger equation look like? It still has kinetic energy minus h squared of a to m d to u by the x squared.
But now we have potential energy. How much? V zero.
So v zero times u. So this is this is the Hamiltonian operator operating on you and that's equal to e u.
And let's now say we're looking for bound states,
that's to say states where the energy is less than the potential energy out here so that the classically the particle will be confined inside here.
So we're going to look for bound states. You don't there are other states, too, but let's focus on the bound states.
That means that that E is less than V0.
So the particle classically is not allowed to get out of the the well.
What happens then? Well, then this equation becomes, uh.
D to you by the x squared is equal to.
If we put this onto this side, since v0 is by hypothesis bigger than you,
we have a negative right hand side and we can cancel the minus signs on the two sides.
So we if we write it as we have to v v0 minus the energy over h bar squared times u.
So again we have a double derivative is equal to some constant times the function.
But the difference is now that we do not have a minus sign here.
So instead of having sinusoidal solutions, we have exponential style solutions.
So the solution to this equation is that you is equal to a constant times e to
the plus or minus k times x where big k is the square root of this stuff here.
What are we going to. What about this sign ambiguity here? Right.
This there is a sign ambiguity here because it's the double derivative which has to be equal to a constant times u.
If we go for the minus sign in taking the double derivative two derivatives, we get down to minus signs.
We get a plus, obviously. So that's why this there's this ambiguity.
What do we do about that ambiguity?
Well, when X is greater than nought, we want the wave function to decrease as we head off to infinity, as X becomes larger and larger.
Well, in fact. Right, because we would like to be able to normalise the wave function.
We'd like to have the wave function mod squared integrated.
Overall space comes to one and that's not going to be possible if we have an exponential divergence.
So the consequence of that is that in this this zone here we take that U is proportional to E to the minus x because x is positive over there.
And that means the bigger x gets more we move over here, the smaller the wave function becomes.
If we're on this left side here where x is negative,
then we want to take you goes like either plus or minus E to the to the plus k x because x is negative in this zone here we and
the negativity of x gives the exponent of the of the of the exponential negativity so that the bigger that X becomes the modulus,
the more we move over over here, the, the smaller the wave function becomes.
And whether we want to take a plus sign, if we're looking for a state of positive, of even parity, then we want to take this plus sign.
So the wave function over here has the same same numerical value as the wave function, the corresponding point over there.
If we're looking for a state of odd parity, we take this minus sign.
So the wave function at negative x becomes minus the wave function at the corresponding point at positive x.
So, so that's, uh, that's where we are so far what we now need to do.
So we've now solved what have we done? We've solved the time independent Schrodinger equation everywhere except an x equals A in x equals -80.
But we haven't solved it at those points because at those points, if you go a bit to the, for example, at the point X equals A, if you go to the bit,
to the left of that point, then the wave function is supposed to be causal sine x and if you go a bit to the right,
it's going to be this exponential function. But at that point we must still have the time.
Independent Schrodinger Equation Satisfied. So what?
What does that require? Well, for the time independent Schrödinger equation to mean anything.
Even the second derivative of the wave function has to be well-defined at that point,
because the time independent writing equation equates the second derivative to some stuff.
So what we can say is that at x equals plus or minus a, we need that d to you by the x squared is defined or we can't solve the ties either.
Satisfy that. Satisfy the ties either.
Well, the rate of change. This is the rate of change of the gradient.
That is certainly not going to be defined if the gradient isn't continuous.
So that implies that do you buy the X is continuous?
And that's a non-trivial requirement because at the moment we've got the wave function in pieces and there's no obvious reason.
Thus we do do some engineering on our pieces. Why? The gradient is defined by one piece on one side of the barrier of the transition
should equal the gradient from a completely different function on the other side.
Okay. But so the gradient has to be continuous.
The gradient is certainly not going to be even it's not even going to be defined unless the wave function itself is continuous.
So we also require the same similar reasons.
But you is continuous. At these points.
So we have to insist. That you have a minus some tiny bit is equal to U of a plus some tiny bit.
And we have to insist that the you by the X evaluated today minus the tiny bit is equal to the U by the x.
Evaluated as a plus a tiny bit called Epsilon. That's what we've got to insist on.
What does that amount to that amounts to here?
If we're just to the left of a we have for the even parity solution we have that u is equal to cos k x an x is a minus epsilon,
but epsilon as small as we like.
So let's just make it equal to cos k that's got to equal the wave function on the right hand side which is some constant.
We'll call it but call that constant big a times E to the minus big k, times x plus a tiny bit.
Let's forget about the tiny bit because this is a continuous function. So we require that that's the continuity of U of x.
Similarly, the gradient just to the left of of x equals A is given by the derivative of cos.
So we're looking at minus K sine k and that's got to be equal to the gradient of a E to the minus X just to the evaluated at x equals A,
so that's minus big k a E to the minus K and that's the continuity of you by the X and the nice.
We also have to satisfy these continuity conditions that x equals minus AA.
But the nice thing about choosing is deciding that you're going to look for a wave function of well-defined parity over an even
functional odd function is that it's easy to persuade yourselves you probably want to sit down quietly and do this afterwards,
that if you satisfies these conditions on the right hand side at x equals plus a of the origin,
then you've also satisfied these conditions on the to the left of the origin.
These equations suffice to fix up the arrangements at both of the discontinuities and you don't have to deal with them separately.
That's the great advantage of of choosing way functions of well-defined parity.
So what do we what do we have here? We have a pair of equations and we have a number of unknowns.
As it stands, we do not know what little K is or Big K is, and we do not know what A is Big A is.
Right? Those are all unknowns. We've two equations and we need fundamentally to determine these unknowns.
Most important is to determine little K and Big K because they are related to the energy by formulae which are there's little K right at the top.
There is the square root of 2me average bar squared and big k halfway up is to m v0 minus c.
So little can big k are both related to the energy and once we found the energy
will know what both Big K and little K on is energy we're fundamentally after.
So that's what we want to focus on. Big is is of less interest.
So let's get rid of big A by dividing this equation through by this equation, then we will find so we divide equation two basically by equation one.
And that leads to the conclusion that minus let's do it down here minus k ten k from sign of a cosine is equal to minus big k nothing.
Everything else goes because we have an a e to the minus big K in both equations.
And let's ask what what this is. Let's try and relate this so big K and little K are both related to the energy
from which it follows that I could express Big K as a function of little K.
So let's do that. This is minus the square root of two and V0 minus E over bar squared,
which is equal to minus square root now to m e average bar squared is actually from we've maybe just lost it, unfortunately.
Let's bring it back into focus right at the top there. It's to me average bar squared is in fact K squared.
So to me average plus squared is K squared.
So what we want to do is write this as two and V nought over squared minus k squared.
So here we have an equation now that this equals this which has only one unknown, namely K.
So the left side is a function of K. The right side is a function of K.
Any values of K for which you these two sides are equal are are provide solutions to our time independent
Schrodinger equation and provide wave functions for stationary states for states of well-defined energy.
To solve this to solve this equation, the way to go is to divide through by this.
Well, obviously cancel the minus signs divide through by k both sides and this then becomes
ten k is equal to the square root of 2mv nought over bar squared k squared minus one.
And it's good now to multiply the top and the bottom of this by a squared and write this as
the square root of w over k squared minus one where w is two and v0a squared over squared.
Why do I want to do that? K is obviously dimensionless because this is a wave number.
The wave function was sine qua. The argument of sine must always be dimensionless.
So this. This is dimensionless. That's obviously dimensionless.
Therefore this must be dimensionless.
You can explicitly check that this is dimensionless for the reason I've defined w is that this thing is dimensionless.
It's a dimensionless measure of the depth and width of the potential.
Well, right. It depends on the depth of the potential. Well, it depends on the width of the potential well.
And it's the dimensionless. The mathematics is telling us that this is how you quantify how you know what kind of
potential well you've got where you've got a very deep one or a very shallow one.
So how are we going to solve this equation here?
Well, the way to go is is to plot both sides of the equation graphically.
All right. So the tangent is is to plot both sides of the equation graphically.
And see what you get, see at what points they meet.
So if this is a plot, this is a being plotted this way.
Then if I plot ten K, it starts off at zero and rises like this.
And when K becomes equal to pi over two, it zooms off to infinity.
Right. That's what tangents do. And down here, it goes in symmetrically.
What does this thing do when K is nothing?
That thing is obviously infinity because it becomes W over nothing square rooted.
So this. This this is the left hand side. I better write that down.
So this here is the left hand side. It's ten K.
The right hand side is coming down from infinity and it's going to go to zero when K is equal to or k squared is equal to W.
So this right hand side is cruising down from infinity and it's going to go to zero when K is root w.
And from this we see it's obvious now that no matter what the value of W is.
So as you change the width and depth of the potential, you move this point to the right of the left.
But no matter where you put it from nowhere to infinity, it will cross.
There will be this intersection here. These two curves always cross.
And where they cross gives you a value of K and therefore a value of K,
which is a solution to the to the time in the independent Schrödinger equation.
So what we've just learned is that this well, always.
As a bound state. It doesn't matter how shallow the water is or how narrow the well is,
it always has a bound state because these two curves always cross this tangent on the left.
Right has has all the branches.
There's also a branch. So tangent goes off, comes along here, goes off to pulse plus infinity.
And then somewhere down here it's when, when, when K is equal to pie upon to plus a bit it becomes minus.
It comes in from minus infinity and repeats itself.
So this is the LHC second branch. And this second branch may or may not cut this curve of the right hand side again.
Right. So if we made a smaller than here, this is where.
This is what? This is where a. This is this is the place pie.
If if we would make this less than that, we wouldn't get a second solution.
But if we have a bigger than this, we do get a second solution. So it may have.
Other even parity.
The stationary states. So depending on the depth of the potential, we have one, two, three, four,
etc. in stationary states with of even parity, we've only dealt with even parity case.
If we want to look at the old parity states, let's just begin to do this.
But this is basically an exercise for the problems. Then our conditions are a wave function for the odd parrot stage states.
We've lost it somewhere but it's go away parity away function to the old parity states is is the sign
kegs right up there right so in the middle it's sign kegs at the edge it's still a E to the minus X.
So our continuity conditions become that sign K is equal to a E to the minus K, and that's the continuity.
Of the wave function itself at x equals a the derivative of this is going to give me k cos k is equal to minus big k a e to the minus k.
We divide this equation by this equation analogously to what we did before.
We are going to get a k cotangent of k, a lot of a is equal to minus.
Is equal to minus b k. So now we have only one minus sign.
Previously we had a pair of minus signs which cancel. Now we have a one minus sign because we differentiate a sign.
We didn't differentiate a cosine the and correspondingly we have a cotangent instead of a tangent.
So this equation can be graphically solved. Uh, that's the exercise.
And we find that we may get nought one two solutions for K and for every solution we have an odd parity.
We have an old parity stationary state. Let's have a look at the uncertainty principle in this example.
Just to remind you, last term, quite early on, we showed we considered the case.
So this is last as a summary, last term. We consider the case where obsessive X is proportional to E to the minus X squared over four sigma squared.
So we considered a wave function whose spatial form was a Gaussian such that when you mod squared this in order to get the probability distribution,
you found that it was a Gaussian with dispersion sigma. So.
So this thing here is the expectation of X squared.
Right. That's what the sigma squared is from that formula. And what did we find?
We found fundamentally by doing a for a transform, that if that's what the if that's what the wave function looks like,
then the probability well, the amplitude to find this is.
So this is x ABC. Right.
The wave function looks like that. Then the amplitude to get a certain momentum,
if you would make a momentum measurement was looking like E to the minus p squared over four sigma p squared.
Right where this thing becomes the expectation value of p squared.
So the variance, the expectation values momentum in this state is zero.
The expectation of the square of the momentum is going to be this sigma p squared.
And what we found was that Sigma Times Sigma P using wedge bar on to this was a concrete
example of the uncertainty principle where the smaller the uncertainty in X is,
the bigger the uncertainty in the momentum.
And correspondingly the smaller momentum in P is, the bigger the uncertainty X has to be because the product is always the same.
In this particular Gaussian example. So we have some idea that the, the, the uncertainty in X Times, the uncertainty in P, we,
we say to ourselves is inherently never smaller than this number upon to it may be bigger than h bar upon two easily.
It often is bigger than H, but usually is bigger than H bar to. But this is kind of as small as it gets.
So that's that was what we did last term. So let's have a look at it in this particular case.
Right. So let's look at the ground state.
Remember this? Oh, yeah. Another point to remind you is that the ground state wave function of the harmonic oscillator.
So the the ground state of a harmonic oscillator actually has a wave function, which is this Gaussian.
The time that we did this calculation. This Gaussian was just picked out of the air.
But for the ground state of the harmonic oscillator actually does have this this thing here.
This fits this example. So let's have a look at the wave function of the ground state of this of this that we have here.
What you might argue to yourself, you might say, okay, so what we can say is that P is less than or on the order of.
Well, peace. Quiet over to him. The energy is less than zero, right?
Because it's a bound. It's bound. So it's less than zero.
So what we can say is that P. P squared is less than two and V zero.
Yeah. Sorry. That's meant to be a less than seems reasonable, doesn't it?
The particle can't have any more kinetic energy than the energy it requires to escape because we know it's bound.
The next thing you might say is, well, so what's the what's what's what's x squared?
What's the uncertainty in X? Well, you might say to yourself, Well, look, this particle is trapped in this potential.
Well, that goes from plus A to minus A.
So what's the uncertainty in in X, in x squared?
Well, this must be on the order of a squared.
Seems reasonable, doesn't it. Less than on the order of.
Well I'll put it in a factor two or four or just to be sure that where the lesson is is, is holding because the thing is trapped by potential.
Well it extends only from plus A to minus A, so it has a range of two ways they can run in.
So we say X squared is definitely less than two squared. It's almost certainly it's clearly less than that, significantly less than that.
So what does that give me? And and this this square is the momentum is so.
So we have that that X squared.
P squared. I mean, I can put expectation values around that, too, I think is less than on the order of.
Whatever it is. 8 a.m. v not a square.
But that is w is somewhere. Yeah. Look, w.
This is basically w because that was to Venus.
So this Venus squared is uh, something like 4mv not a squared is is h bar squared.
If I done this right, because I'm getting an answer, which I'm. Thomas W Yes, exactly.
That's right. The dimensionless number. W Right. The point is right that W could be made as small as you like.
We've agreed. And you slope have bound state.
So this naive argument suggests that we're going to violate the uncertainty principle because we can make this as small as we like.
So what's the problem? This argument's bogus. What's bogus about it?
A hole in the wall. It's not all in the well as you make w smaller and smaller.
Let's draw what the wave function looks like.
So if we have let's, let's have a nice big value of w a big value w means a nice spicy well with right high walls.
And then we will have a ground state wave function, which is a cosine and some little bits like this.
And indeed the probability to find the particle outside the well outside, plus or minus, they will be not very much.
But as you make this smaller and smaller and smaller, you bring this in making a smaller and and or you lower these making this curvature smaller.
So that curvature of that wave function is a reflection of the kinetic energy, which is a reflection of of nought.
We we come to the we have a nice we have a narrow, a techy witchy thing like this.
Then you have barely you have an almost a straight line across here.
I can't draw it. Well, almost a straight line across here. And then enormous long exponential decays.
And the particles almost certain not to be in the potential. Well, that's a very remarkable conclusion to come to, right.
That you can trap a particle with a potential well, which is almost it has very little probability of actually being in.
But that's what the theory says.
Something else we can we can show is we can we can say we can argue as follows supposing we given some other potential.
Well, here is some other potential. Well, I'm sorry.
It's meant to be it's meant to be an even function of X, right?
So it's meant to be the same on both sides. Symmetrical and somebody also.
So does this potential well have a bound state then?
You reason as follows let us inscribe a nice square potential.
Well, there are many square potential worlds you can inscribe into it. Right. But there's one.
Then you can argue that this potential well is narrower and shallower than the one you were given.
So less likely to trap a particle. But this potential world has a state has a bound state.
So this. Well. Shallower.
Narrower. But it has a bound state.
At least one bound state because it's a square. Well, and we understand about square wells.
So you reason that this wider and deeper well also has a bound state.
So we we're making an inference from this very special, rather artificial squared.
Well, potential about all one dimensional square wells. What happens now?
Let's let's conclude with a special case.
An important special case. Which is the infinitely deep.
Well, woops. So now we let V0 become arbitrarily large and we have a, we have a,
well it looks like this, it just goes up and up and up and up and up and up and up and up.
What happens then? Well, let's go to our graphical solution.
What have we done? We've made we've left a constant.
I mean, we've left at some values and we've allowed V zero to become arbitrary large, which means we've made W arbitrarily large.
What's the implication of that for our solution of this equation?
So we're where we're solving this equation for w arbitrarily large.
So so we're saying that the roots are going to become where ten K the even parity roots are going to be where ten K equals infinity.
And we know what that means. We know that K must be pi upon two, three, pi upon two, etc.
So we've done, we've made W goes to infinity, which implies that the governing equation, the equation that determines K becomes ten K equals infinity,
which implies that K is equal to either pi upon two or three pi upon to, etc., etc., etc. and all of these values are going to be okay.
Graphically, what's happening is that this curve, right?
This was the curve of the right hand side for finite.
For finite W is going to just become a line along here at infinity and it's going to cross these various branches of the tangent at infinity,
which means that these pi upon two, three pi upon two points, and it's going to cross every single one of them all the way, all the way out.
Right? So we can have an infinite number of state of even parity stationary states and they're going to have those values of K.
And if when you when you study the corresponding problem for the AUD, for the AUD parity states,
which we just vaguely discussed here, it's going to be the same deal.
We're going to be solving an equation which says that Cot K is going to be infinite and that means that K is going to be is going to be pi to pi.
Three pi. So on. So these are going to be the even parity states.
And then we will find okay is equal to pi, two pi, three pi, etc. for the quad parity states.
So the infinitely deep well will have an infinite number of solutions.
And what do the wave functions look like. Well the wave functions are going to be cos K X where, where K is equal to some number of odd pi by twos.
So the wave function of the even parity states is going to vanish.
This is going to be cos k x and the wave function is going to vanish here and here by virtue of these conditions there.
And the same thing will happen for the odd parity states.
For example, the first odd parity state is going to have a wave function which which looks like this.
It's going to it's going to be sine.
K x. The best piece of chalk.
It's going to be sign kegs where K is equal to a number of pies.
And therefore the the sign vanishes. So it's going to vanish.
So this is a concrete example,
but it leads to the general it inspires the general principle that a wave function vanishes adjacent to an infinite region of infinite potential.
But. Physically.
What's happened is that is that this well, this big cake has grown bigger and bigger and bigger and bigger,
and therefore this exponential has grown steeper and the curvature of it has grown larger and larger and larger and larger.
So we've gone through, as we raise the walls,
the transition at the edge of the well goes from being sort of nice and nice and easy with a small value of K.
Through. Big a value about a bigger value of K to ultimately an infinite value of K,
which allows it to continuously go from a finite slope right round to two no slope.
And that's why it's a general property of the solutions of the time independent Schrodinger
equation that as you approach the edge of an infinite region of infinite potential,
the wave function vanishes and in anticipation of that infinity.
So another strange aspect, you know, another physically strange thing that the theory is telling us is that the wave function,
that the particle has negligible probability of being found in the neighbourhood of this region where it's strictly forbidden.
It anticipates the fact that it's going to be forbidden.
And you won't find it even near this dangerous place. Okay, it's time to stop.