In the previous segment we saw that the 
Schrodinger equation for an infinite 
potential well without even writing down 
the Schrodinger equation itself. 
So of course it doesn't always happen 
this way, so for more complicated 
potentials we actually have to do some 
calculations, and then this segment I'm 
going to present an example of a general 
method actually that is used to solve the 
Schrodinger equation, and also consider 
the so-called finite potential well. 
So the, potential that we're actually 
going to study is, is illustrated here so 
here is, formal sort of mathematical 
expression for this potential and this is 
the, it's a illustration so, for, x. 
So this is one dimensional potential. 
For X is greater than A over 2 or smaller 
than minus A over 2. 
We have a finite wall, if you want, with 
a height of U naught and in between minus 
A over 2 and A over 2 we have a potential 
well. 
And so the question that we are going to 
be interested in is whether or not we 
have levels, energy levels, allowed 
states within the potential well, which 
implies that the energy of these states 
must be smaller than this maximum U 
naught but larger than 0. 
So these are positive energies here in 
this formulation. 
Now, clearly we have three regions that 
appear in this potential. 
So one region will be to the left of the 
point minus A over 2. 
The second region, which is the potential 
well itself, will be between minus A over 
2 and A over 2. 
And the third region is X greater than A 
over 2 Two. 
And for each of the originals, we can 
write down the corresponding Schrodinger 
equation. 
Let's say, for the regions one and three, 
for absolute value of x greater than a 
over two we have the kinetic energy plus 
the potential energy u naught is equal to 
the energy that we're actually looking 
for e times psi of X. 
And for the region of the potential well 
we formally have just the free 
Schrodinger equation since the potential 
here is equal to 0. 
Now to solve this Schrodinger equation 
the main challenge in some sense of 
solving the Schrodinger equation is 
actually would be to match the solutions 
in different regions. 
And by match I mean the full link. 
We're going to demand that both the wave 
function itself and it's first derivative 
are continuous everywhere. 
In particular in these matching points 
minus a over 2 and a over 2. 
And so here I have formally represented 
the as so, so plus zero basically means 
to the right of a certain point. 
And minus zero means to the left of a 
certain point. 
And the reason one, our weight function 
or let's say its derivative to be 
continuous is because if we did not have 
this continuity. 
Let's say in this point if we let's say 
solve our Schrodinger equation three and 
origin two and department we don't bother 
to mention them, then, in general so our 
way function in this case is going to be 
something like this let's say. 
And there will be a jump, find a jump. 
In, well, in the wave function is 
derivative, and if we calculate the 
second digit which is here, it uses the 
jump, we're going to have an infinite 
secondary digit which will be, which 
wouldn't be able to compensate in our 
Schrodinger equation. 
So there will be essentially a delta 
function coming up from this, contiu-, 
from this discontinuity, and, well, we 
don't have it in our problem. 
In our problem we don't have a delta 
problem. 
So, therefore, we must demand that the 
wave function is derivative so in our 
continuous everywhere. 
And it turns out that this along with 
these, with this condition would be would 
be enough to find the constraints which 
will determine the actual energy level. 
So now the latter constraint namely that 
the wave function should decay at 
infinity is essentially, physically a 
meaningful constraint. 
So if we were looking for a bound state 
of our particle, so we want the 
probability of finding this particle at 
infinity to be zero. 
And we want the probability to be sort of 
localized in the vicinity of the 
potential. 
And so this is where the second 
constraint is coming from. 
Now to simplify this equation Ev equals U 
naught is that, well we can rewrite it. 
It, let's say the first on we can 
introduce a new parameter gamma, which 
is, a 2m over x squared, U-not, minus, 
epsilon minus energy E, so let me 
actually called gamma squared, 
And, well, we can, we an just put the 
energy, in, in the left-hand side and 
multiply everything by minus. 
2 m over h squared, and this would give 
rise to the full equation. 
Psy 2 prime minus gamma psi, gamma 
squared psi is equal to zero. 
And for the second equation we can 
introduce the parameter is the key, and, 
we're going to have this with psi 2k plus 
k squared cy equals to zero. 
So let me just rewrite in this slightly 
nicer way. 
So here are basically the two equations I 
just wrote and here are the parameters. 
Both these equations are actually very 
simple equations so, this one has a 
solution either power plus minus gamma x. 
You can just plug it in and see the well, 
the the inevitable of the exponential and 
the exponential is the exponential itself 
and well, clearly is going to satisfy 
this equation. 
And the second equation well which is the 
free Schrodinger equation, unsurprisingly 
gives us just plain waves as the solution 
either problem, plu-, plus minus i K 
times x. 
So and again, the challenge is going to 
be to match those guys in each of these 
points, minus a over 2 and a over 2. 
Now, it's not really necessary for the 
solution of the problem but, I would like 
to use the symmetry of the problem here 
and this would give me an excuse in some 
sense to introduce the symmetry concepts 
and quantum mechanics in a much Broader 
context. 
So namely I would like to point out the 
following fact. 
if, the Hamiltonian commutes with an 
operator a, this can be an operator of 
any physical quantity, now, then the 
solutions to the Schrodinger equation, 
that is an equation which is an 
Eigenvalue problem for, for the 
Hamiltonian. 
So then, the solutions can also be chosen 
to have definite value of 8, which means 
that the same solutions are going to be 
the Eigen states of the Eigenvalue 
problem for this operator E whatever it 
is. 
So in general if the 2 operators were not 
to commute, so I wouldn't necessarily be 
able, have been able to find As a set of 
way of function which would both A and E, 
the energy definite. 
So, but in the presence of a symmetry, 
you want this possibility exist. 
Now, what does that have to do with our 
problem? 
So in our problem, we have a sort of 
rather obvious Symmetry which is 
essential inversion symmetry of the 
potentials of a flip x if it x goes to 
minus x. 
So the potential is unchanged and so the 
kinetic energy also doesn't change so its 
a second derivative with respect to x, so 
therefore the Hamiltonian one can 
determine that Hamiltonian is indeed 
symmetric or invariant under this. 
Under this operation, and so for 
[INAUDIBLE] that forces this inversion is 
this operator i, which basically thinks 
adds to minuses. 
And this operature has two item values 
the barrier. 
Either plus or minus 1. 
So the functions feature high end 
function of this inversion operature 
either even or odd function of the 
coordinate. 
And so what, what we are saying in this, 
sort of, very complicated language is 
that we can find the Eigenstates of our 
problem the solutions to the Schrodinger 
equation which would be either, odd or 
even functions. 
That's all we are saying and, well, 
another statement that I would like to 
make is that if there is a symmetry in 
the problem, you'd better use it. 
So it always simplifies things. 
Never makes things more complicated, 
always simplifies things. 
And so that's what I'm going to do here 
as well. 
So in general if I were to ignore, this 
symmetry. 
So the solution, to the[UNKNOWN] 
equations we just discussed in this, 
region of the will would be arbitrary 
linear combination of two plane wave. 
Now I'm saying so see there is this 
symmetry in the problem. 
So I would like instead of just having 
this arbitrary efficiency c 1 and c 2, I 
would like to consider separately 
solutions with definite parity. 
Either even functions which is basically 
cosign with the c 1 equals to c 2 or a 
sign which is a node function. 
In which case c1 is equal to a complex 
conjugate. 
and these are purely measuring constants. 
So, in the following I am going to 
specifically consider to focus on this 
even solution just for simplicity but you 
can find the general analysis This in a 
number of books so this is the problem we 
are solving by the way. 
Is a absolutely classical problem, but 
classical not in the sense of classical 
physics, but in the sense that it appears 
in just about every course or every book 
on quantum physics. 
Now having determined the solution we're 
interested in, in region two and now have 
to 
To determine the solutions in regions one 
and three that we're going to be focusing 
on. 
And by the way, you, the symmetry of the 
problem, we can actually pick just one of 
these regions. 
So we don't really have to use, in this 
case, both matching points, we can do 
well just using one point. 
And as we discussed, a general solution 
to the Schrodinger equation, to this 
equation in the region three is given by 
a linear combination of these two 
exponentials, which are no longer 
oscillating functions, but are functions, 
that, are either, decaying, a rapidly 
decaying function, or a rapidly, growing 
function. 
And, so to, simplify to find the proper 
solution in the region three, let's say, 
for x is greater than a over 2 We have 
recalled, the constraint and infinity 
namely that the wave function must remain 
finite. 
Well actually I should correct. 
It should actually go to zero as we go to 
infinity. 
So otherwise if we don't impost this 
constraint, if we allow, in this case 
this, churum in the wave function, the 
probability of a particle to leak. 
in some sense to infinity would explode 
exponentially, which doesn't make any 
sense. 
So therefore we simply draw the B term in 
this in this wave function, and only 
we'll focus on the on this wave function 
in the region 3. 
So now we're in, in a position to 
actually match the solutions. 
We know the solution in each region. 
So here I do present a solution in all of 
the regions, including region one. 
So this guy's for region one. 
This guy's for region two, and this guy's 
for region three. 
But as I said, due to symmetry we can 
focus on only one matching point. 
And this is what I'm going to do. 
So now we can ride the wave function to 
left at this point, which is going to be 
c times cosign key a over two and, it 
must be equal to the way function to the 
right, which is this guy, which is a 
ethopro minus, gamma, a over two. 
So this is the first matching condition 
and the second matching condition is 
going to be the derivative of this 
function. 
So the derivative of cosine is minus 
sine, so were going to have minus C times 
k sine of k a over 2, to the left. 
Must be equal to minus A gamma e to the 
power minus gamma a over 2, to the right. 
And now what I'm going to do, I'm 
going to divide so let me call this 
equation, equation number two. 
And this is going to be equation, 
equation number one. 
And so if we divide equation number two 
by equation number one. 
Which we're allowed to do, what we're 
going to get in the in the left hand 
side, we're going to have k times sine or 
a cosine is a tangent of ka over 2. 
And in the right hand side, so the 
minuses will go away and the exponentials 
are going to go away, we're going to get 
Just gamma. 
So here I have the same equation written 
a exclusively in a nice way. 
And also I recall the definitions of 
gamma and k, that I made in a previous 
slide. 
So k itself is the square root of 2mE 
over h squared. 
And the energy is really what we are 
looking for. 
And energy appears both in the left hand 
side and in the right hand side, in a 
rather non-linear way. 
And as a matter of fact there is no way, 
we cannot solve this equation 
analytically. 
So this is, no linear algebraic equation 
and it doesn't have a formal sort of 
closed solution, that is four. 
So which by the way is quite amazing that 
the problem is simple as this finite 
potential. 
Well in one dimensional quantum mechanics 
which is just about the simplest problem 
you can think of It cannot really be 
solved, so this gives you something, it 
tells you something about the complexity 
now, of the technical complexity of the 
Schrodinger equation. 
Now to precede further towards a bit of 
numerical analysis or asymptotic 
analysis. 
It is always a good idea, in such cases, 
to get rid of the physical, dimensional 
quantities such as energy or this 
potential height which have physical 
dimension and move to dimensionless 
parameters. 
And so here the nature of the 
dimensionless parameters is going to be 
Well, the argument of this tangent which 
is EE over 2. 
And the parameter introduced here. 
Which is basically to be able to write 
our equation in a nice and dimensionless 
way. 
The reason why dealing with 
dimension-less quantities is much more 
convenient than with. 
quantities in case it has non-trivial 
physical dimensions is because, for the 
former, we can talk about them being 
large or small. 
So we can see that something is small or 
large as comparing it with one. 
While for a dimensional quantity it 
doesn't really make sense to talk about 
large or small unless we Specify what we 
actually mean by that for instance is one 
meter large or small. 
Well does we don't know unless we compare 
with the lets say size of an atom and 
which case it's clearly large or with the 
size with the size of the universe, in 
which case one meter is really small. 
So here one, once we get these parameter 
si it allows us to define basically a 
deep. 
And shallow potential so we will call the 
potential deep if this guide, the, of 
this squad squared is much larger than 
one and the potential is going to be 
called shallow if it's much smaller than 
one. 
And by the way, in the, full length two 
video segments we're going to be focusing 
on, shallow potentials which are well 
mottled by delta function. 
So, naively you would say, you would 
think, that delta function is something 
which is actually very deep because it 
goes to infinity if you, approach, x 
equals zero, let's say if we're talking 
about potential u of x, which is let's 
say minus u, delta of x. 
But, well, delta of x and potential delta 
of x is really, it means, to represent, a 
Potential with the very with the very 
small radius and the radius in order for 
the integral the delta function be equal 
to 1 so the radius of the potential, the 
size of the potential should scale with 
the depth of the potential as so. 
So as a goes to zero, well you not is 
sort of proportional to y and 1 over a. 
But here we see that the definition of 
the shallow potential is u naught a 
squared times m over a squared must be 
much more than one which means that the 
Delta potential is actually closer to 
shallow potential which is an important 
observation and will take advantage of 
later on. 
But in its stage let me just say that we 
have these two special cases and even 
though we cannot solve exactly this 
equation, for arbitrary side, we can do 
two separate analysis of this equation in 
such two special cases. 
And here, I present this analysis. 
So this is again the equation, the same 
equation, so the red tangent of x is 
plauded here, so basically applauded the 
functions on the left of this equation 
and on the right of this equation. 
And you see by the way that when x 
becomes larger than psi. 
So this right hand side becomes purely an 
imaginary constant which implies that 
there is definitely no solution to this 
equation as the tangent here does not 
become measured. 
So so it makes sense only to plot the 
well the this function for a axis smaller 
than psi. 
So here I have an example of deep 
potential where I say a set psi to 220 
and so I have all these red, red curves 
corresponding to the tangent and this 
blue curve corresponds to the right hand 
side and where, when, wherever I see it 
crossing between these two guys, so let's 
see, here, here or here or here, etc. 
So, I have levels. 
In this case, I can count the number of 
levels that are allowed in this quanta. 
Well, one, two, three, four, five, six, 
seven levels are allowed. 
Well for the symmetric wave functions if 
size equal to 20. 
Now lets consider relatively shallow 
potential that this on the other hand, so 
lets say with si equals 1. 
In this case I only have to plot this 
functions for x as more than 1 and there 
is just 1 crossing. 
Between the blue curve and the red curve, 
which means that there is just one single 
level in the shallow potential well. 
Which by the way is a very important 
result that we're going to confirm on the 
case of a one dimensional delta 
potential, and we're going to see how 
this result is modified in higher 
dimension. 
Now, the last comment I'm going to make 
is that in principal I could have 
continued this analysis of this 
self-consistency equation and there's a 
lot to do to determine the actual 
numerical value of the energy levels to 
see what happens with the anti-symmetric 
levels with the old wave functions. 
And I'm not going to do this right now in 
the lecture because admittedly, it's 
probably not the most exciting thing in 
quantum mechanics so I will encourage you 
to take a look at various text books and 
other lectures online, for example these 
set of lectures by professor Michael 
Fowler at University of Virginia. 
and also you will see a few problems 
going up on this solution in your 
homework. 
So you will have a chance to look into 
this problem in more detail and 
understand the solution at a deeper 
level.