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.