# WKB Approximation, Bohr Sommerfeld quantization condition

So we are going to look at the last topic
in ah the Approximate Methods in Quantum Mechanics and namely the WKB Approximation which is
ah by the name Wenzel Kramers and Brillouin approximation .
And this approximation is valid for ah slowly varying potential ah slowly ah and linearly
varying potential ah I mean the the main requirement is that it has to be slowly varying. So, ah
suppose ah ah potential ah varies as ah something like this over a large distance. So, this
is your ah V of ah x versus x . So, this is x and this is V of x . So, it is that kind
of a potential . Now, ah let us assume that the particle has
energies which has a value which is like this represented by this ah E , ah now these are
called as ah the turning points ah. So, we will ah call these ah points as ah P and Q
ah P and Q are turning points. So, Vx is of course, ah ah a slowly varying potential all
right, and ah the particle has energy E. So, ah in the vicinity of ah this is a P the particle
actually slows down. And it ah so, the velocity of the particle is slow and the particle has
ah ah large velocities in this middle region and again it slows down and of course, ah
ah so here it ah it is slow and then of course, ah it you know it comes and then it goes back
. Ah. So, these regions are called as so, if
we draw two vertical lines and let them meet at this ah ah x axis then this is called as
the classically forbidden sorry this is called as a classically allowed region ,
and ah the regions beyond this so, these are ah Classically Forbidden region
and this is Classically Forbidden as well all right.
So, this is the situation ah here, so we are talking about a potential of this kind and
we certainly do not know the solution ah for this particular problem ah for the entire
problem, but even if we do not know the solution ah for this entire problem at least we can
get the asymptotic solutions, that is ah solutions for ah um ah regions which are ah far away
from this ah P and Q which are called as the turning points we will just ah define turning
points . So, these Turning Points are defined as x
1 and x 2 ah um and these turning points are defined by where the total energy becomes
same as the potential energy which means the kinetic energy goes to 0 . So, ah that is
the definition of turning points. So, the particle will come and ah turn ah from these
two points of course, the quantum mechanical particle has a tunneling probability around
these points as well. And so, we will get a ah dyeing solution or rather evanescent
wave in the classically forbidden region . ah And ah so, ah the other region where ah the
classically allowed region is where E is ah greater than ah E is greater than V of x and
that tells that ah . So, any x between ah so this is the classically allowed region
ok . And. So, near the turning points the kinetic
energy is small and, so, the particle ah spends a significant amount of time close to the
turning point and the motion gets slow and away from the turning point the kinetic energy
is large and the particle is said to be moving with large ah velocities . And which can be
ah understood quantum mechanically by calculating the probability density. So, the probability
density of the particle ah being close to the turning points is ah large which means
that the velocity is ah low. And ah at a faraway regions that is ah somewhere
the middle of this classically allowed region ah you will get ah less probability for the
particle to spend time there . And ah ah and as ah the particle moves close to the turning
point ah then it gets reflected from these ah points P and Q which are as I told you
that they are called as a turning points . ah And ah as it is also mentioned that ah this
method is best suited for a slowly varying potential and ah if the potential is not slowly
varying that is if its ah rapidly varying then of course, this method does not work
. So, let us ah get an analogy with ah optics
that ah this is ah the behavior of light in a ah um ah varying ah in a medium where the
refractive index is varying ah. So, if ah the refractive index varies too rapidly then
of course, the light gets reflected and so, this is ah so and if it is a very gradual
variation ah then of course, the light ah really does not get reflected.
So, ah in a slowly varying ah or rather we should write it as ah in a medium ,
which has slowly varying refractive index, ah the situation is ah pretty similar under
consideration for us in this ah WKB approximation all right .
So, this is the main idea of ah this ah problem that we are going to solve ah ah slowly varying
potential and ah ah um in terms of ah obtaining what are called as ah connection formula which
is applying the boundary conditions, and these are very special formula because usually we
have seen that the connection formula they actually ah connect the solutions at the boundaries.
So, just ah ah epsilon left to the boundary to epsilon right to the boundary within the
limit epsilon going to 0 the solution should match and ah, but here of course, there is
no such ah things. In fact, ah very close to the turning points the solutions fail miserably
it is only asymptotically that is far away from the turning point the solutions give
good ah results . So, let us ah take a one dimensional problem
all right. So, ah how do we write Schrödinger equation I am ah simplifying it ah getting
once getting rid of one step ah you should write all steps . So, this is ah d square
side dx square plus a 2 m by h cross square ah E minus V of x psi x equal to zero . So,
that is the ah equation that we are all familiar with it is a one dimensional time independent
Schrödinger equation . This had to be solved ah so for ah V x less than E which means that
we are in the classically allowed region the total energy is larger than the potential
energy . So, this would be d 2 psi ah d x 2 plus a k square x psi x equal to 0 , where
ah the k of x which now depends of course, on x as opposed to earlier which ah it was
a constant ah um E minus V of x the x dependence of k comes from the x dependence of V and
ah of course, we know that if kx is a ah constant ah k of x is independent of x then the solutions
are simple and, but we still ah ah try to write down that solution. So, if k of x is
independent of x then the solution is that that psi of x goes as exponential ah plus
minus i k x, this is all known to you and these are ah ah well known solutions that
you have ah looked at these are the ah travelling waves free particle solutions and so on .
Now, of course, since K is varying slowly and I would highlight this word slowly ah
I just told you that the variation of ah k x is coming from V x so, if V x is a ah slow
ah function of x, k x is also a slow function of x .And ah now ah we may expect a solution
of the form which are ah um psi of x psi of x is equal to ah. So, this is like
exponential i u of x where u of x is nothing, but a plus minus ah k of x, dx .
Now, since k is not a constant it is ah variable or rather it varies on the space variable
x. So, it has to be integrated over and we have not specified ah the limits of ah integration,
but we could do that the lower limit is not important or rather even the limits are not
important we simply ah can just put it there, but let us just ah for now put a limit there
just is just any x arbitrary x. So, at the lowest level , one can expect the solution to have the form psi 0 just to say that it is a lowest level
approximation, it is ah exponential ah plus minus i ah k x dx and with an upper limit
put there. So, that is the lowest level approximation. So, if we ah name these equations as ah so,
this is equation 1 ah , this is equation 2 , ah this is equation 3 and call this as equation
4 . So, ah putting a 4 in ah um in say ah 2 ah or rather its put in actually 1 that
is the Schrödinger equation , one gets the Schrödinger equation now cast in the form
of ah psi 0 and plus this is a k square x ah minus plus i by 2 ah d k dx ah psi 0 of
x equal to 0 . Ah just let me tell you a priori that this
is mathematically very intensive and I will have to practice at least a few times in order
to get a hang of ah things ah, but these are ah not too difficult ah ah algebra these are
simple algebra, but ah you would still have to practice it there are a few change in variables
and there are ah knowledge about the special functions that are required for this .
So, ah this is the equation thats the Schrödinger equation for ah psi 0 ah if of course, d k
dx is ah neglected then we get the ah the equation to the lowest order which is d ah
d 2 psi naught dx 2 ah plus k square psi naught equal to 0 this is what we have been looking
at . So, ah lets ah write for a better solution
better means; better than the or rather than the zeroth order solution . So, psi of
x equal to some F of x and a psi naught of x where F of x either you call it a slow slowly
varying function or you can say that it is a monotonic function , ah um and a simple polynomial in x . So, call
this as equation 5 and let us call this as equation 6 .
So, if you substitute ah these one ah putting 6 in 1 I get the same equation that is the
Schrödinger equation, now written in terms of the variable F or the function F which
is 1 over 2 k x ah dk x dx call this as equation number 7 ah now here of course, we have neglected
ah d 2 f dx 2 and the rationale being that we want a simple polynomial in the vicinity
of ah in the vicinity of the ah psi naught and moreover
it is a monotonic function, so it does not have a curvature. So, this ah are the simple
arguments that we can give ah in order to neglect d 2 f ah dx 2 .
So, the solution of ah 7 can be written as ah f of x equal to 1 by root over ah k x ok
. So, now with this this is only a trial solution
one can check that with this dF dx becomes equal to 1 divided by 2 k of x whole to the
power 3 by 2 and the dk dx . So, ah you can write a 1 over F of x df dx equal to a ah
k x to the power half ah 2 k x 3 by 2 ah and a ah d k dx and so, this half and this 3 by
2 will cancel and will give me a 1 over k x which is what is wanted. So, this is a solution
of ah so, thus ah 7 or rather 8 is a solution of 7 ok .
So, ah we have gotten a solution ah for the ah Schrödinger equation in some form, and
ah so this if we ah write 8 and put it back into 6 . So, just to your 6 is the equation
that ah you have ah made an unsearch for for psi of x which is a polynomial f of x into
the zeroth order ah solution . So, then this ah becomes equal to psi of x
which is just a slightly better approximation than the zeroth order approximation it is
equal to 1 by root over of kx exponential plus minus i k x dx ok . So, what is the difference
the difference is that that if you had ah a zeroth order approximation then the amplitude
would have been a constant here the amplitude depends on ah x and its of course, has also
ah the k in the denominator which of course, creates a problem that ah if that denominator
blows up as k goes to 0 which is what happens ah in the vicinity of the turning points ok.
So, this is ah because of this it is also called as the Phase integral method all right
. So, ah what is ah what are the solutions now we can ah write down this as equation
9 and hence the solution we can write down a full solution as C 1 divided by k x exponential
i k x d x with an upper limit here plus a C 2 divided by root over k x exponential minus
k x dx because the phase is integrated over that is why it is called as a Phase integral
method . So, that is the most general solution so far. So, that is the one ah order better
than or one level better than the ah solution that we had ah earlier proposed .
Now, ah this is for the classically allowed region this is what we had said that we ah
lets write down this ah for Vx less than E now we can ah go to the ah Vx greater than
E . So, this is number 2 this is the classically
forbidden if you wish let us write also classically
allowed , ah this is of course, classically forbidden for the simple reason that ah we
are talking about ah ah the kinetic energy being negative.
So, ah this one will give me ah just proceeding as earlier. So, this is D 1 root over ah k
prime x ah exponential x ah k prime x dx plus ah D 2 divided by a k prime x exponential
ah x k prime minus k prime x dx . So, that is the ah solution for this ah call it 10
and ah this is our ah ah 11 and a k prime is of course, given by ah um all right. So,
we could have written down the ah. So, there is a solution of is a solution of
ah d 2 psi dx 2 minus k prime ah square x psi of x with equal to 0. So, let us call
this as ah number 12 and ah where k prime x equal to 2 m by h cross square ah V of x
minus E root over of that . So, that is my equation number 13 .
So, my ah equations are 10 and ah 11 are, so Equations 10 and 11 are solutions ah um
to the first approximation I will put it in quote unquote which means; ah one level better
than the zeroth order approximation for a slowly varying potential all right.
So, ah, but these are fine this ah writing them down ah formally it is fine, but; however,
you would see that there are some problems in trying to connect the solutions on both
sides of the barrier and this is what could bring us to ah what are called as the connection
formula , let us ah try to analyze these things a little more details.
So, ah let us write ah graphically ah um so, this is the solution. So, this is what the
left ah curvature is, so this is the energy and this is ah of course, ah x equal to this
is x equal to a and ah. So, this is ah I am taking a thin layer which
is in the immediate vicinity of the turning point . So, this is my E and this is ah will
be called as region 1 and this will be called as region 2 and let us call this figure as
figure 1 and that same figure I would ah draw it on the other side where it looks like this
. So, then again the energy is like this . So, ah let me look at the other ah end of
the potential which is ah given by ah this ah ah the ah the curve of another slope the
other slope. So, again ah we have this turning point at x equal to a and ah we also take
ah two vertical lines in the vicinity of ah ah the turning point and ah .
So, ah we have already told that this point ah x equal to a is called as the turning point,
ah look at the solutions that we have written down ah they had a 1 by ah root over k x in
the denominator root over ah k prime x in the denominator both of them vanish as ah
v becomes equal to e. So, in which case these solutions are not ah ah valid ah in the vicinity
of these turning points. So, they are ah the solutions that we have obtained are valid
solutions ah away from the turning points away means far away from the turning point.
So, let us just write far away from the turning points .
So, let us also demarcate these regions 1 and 2 here and regions 1 and 2 here so ah
so, we call the left region as ah region 1 and the right region as region 2 all right.
So, ah in the neighborhood of the turning points, the potential energy variation is
approximately linear . So, near x equal to a which are the
turning points, we can write ah 2 m by h cross square E minus V of x which is nothing, but
the ah k square is ah minus alpha x minus a.
So, if you look at ah the left figure then we have ah alpha is less than 0 and in the
right figure alpha is ah greater than 0. So, these are the respective slopes of the potential
energy profile ah as we have drawn . And ah so, if we substitute ah these ah form
into the Schrödinger equation, that is this form ah let us call this as ah equation 15
ah in continuation with our earlier notations ah. So, we have ah putting 15 into 1 we have
a d 2 psi d x 2 minus alpha x minus a psi of x equal to 0. So, ah this is ah just like
a linear potential and the solutions are ah called as the AIRY functions which are written
as AI and BI ah. And that is so, these are exact solutions are available , but let us
just ah get them a little more ah in a ah familiar form and ah .
So, let us have a variable transform ah in which we write z equal to alpha to the
power 1 by 3 x minus a and ah so, ah d 2 psi dz 2 its equal to a del del z of ah del psi
del z ah or we can simply write it as ah del 2 psi del z 2 , ah this and this is equal
to del del z of del psi del x ah and del x del z ok .
And ah this is nothing, but equal to del 2 psi del z del x and del x del z plus a del
psi del x ah del 2 x del z 2 ah from the given condition del x del z is simply equal to alpha
to the power minus 1 third and del 2 ah. So, this is minus one third not half pardon me
for this . So, del 2 x del z 2 equal to 0. So, the second term is equal to 0 and the
first term only contributes, and if you do this ah this ah simplification or
this variable transform then we get this equation as d 2 psi z i I am skipping one step which
you can fill it up and, this is equal to minus z psi z equal to 0 and this is the ah linear
potential. So, this is a Schrödinger equation for a linear potential call this equation
number 16 where of course, you should remember this z equal to alpha to the power 3 x minus
a as written at the top of this slide ok. So, ah we have to now ah write down the solutions
in terms of the airy functions; however, the airy functions are less lesser known functions
than the bessel function. So, we will ah establish also a relationship between the ah bessel
functions and the airy functions . So, at this moment we are a priori ah introducing
on an ad hoc basis this Airy functions, but later on they will be shown ah to be the solutions
of this equation 16 and Bessel functions you must have ah already seen in the context of
either electrodynamics or ah quantum mechanics ah say particle in a spherical box .
So, ah this is your equation that you want to solve dz square minus z psi z ah equal
to 0. So, let us ah first look at z greater than 0 which ah also means that ah x is ah
greater than a. So, that is ah the lets do another transformation, psi of z equal to
ah root over z phi z . So, then ah this if you substitute this ah equation 17 into 16
then, we get a differential equation which is equal to in terms of phi ah its nothing,
but just rewriting the Schrödinger equation this and minus z cube plus 1 by 4 ah phi of
z equal to 0 ah. So, again we make another transformation introduce a variable called as xi which is
equal to 2 third z to the power 3 by 2 and that, ah lets us or arrive at this differential
equation ah plus as ah xi ah d phi d xi minus ah xi square plus 1 over 9 phi of xi equal
to 0. So, this is called as the modified Bessel
function and the solutions are ah i to the powe[r]- i ah for 1 third and
this is i of minus 1 third xi. And so, the solution is obtained as psi z equal to root
over z i plus minus 1 third xi which is also equal to z to the power half i plus minus
1 third ah 2 third xi to the power 3 by two. So, that is the solution for z greater than
0 . Now, what happens for z less than 0 once again
we shall do this ah ah transformation that we will call a eta equal to minus 1 over z.
So, ah the d to psi d eta 2 plus eta psi of eta equal to 0 is the solution which we can
write it as ah we can write down these ah equation numbers of course, ah ah so, these
are ah ah. So, this is 18 and this is 19 this is 20 and this is 21 . So, this is equal to
ah 22, 22 and ah now making ah another transformation that psi of eta. So, all these transformations
are making sure that ah we are ah trying to simplify the situation as ah ah much as possible
. And introduce another variable called as ah
zeta which is equal to 2 third eta to the power 3 by 2 now we are doing it for z less
than 0, and that gives a differential equation with this ah for phi in terms of these zeta,
which is this ah this is not ah this is that ah zeta and plus ah zeta d phi d zeta plus
ah zeta square minus 1 by 9 ah phi of this this is not xi, but it is zeta this is equal
to 0 . So, this is called as the Bessel function and the solutions are J plus minus 1 third
. So, the solutions are written as ah psi of
z which is equal to ah z to the power half J plus minus 1 third and 2 third, now it is
a mod z to the power 3 by 2 and this is equation number 26. So, ah we can write down the solutions
so, the summary of these solutions are psi of z equal to root over of z ah C 1 I ah minus
1 third ah zeta plus C 2 I 1 third zeta, that is ah equation number 27. So, this is for
z greater than 0 and ah there is the other solution is psi of z is root over z plus ah
C ah 3 J minus 1 third ah zeta plus a C 4 J 1 third ah zeta . So, this is ah zeta and
this is equation 28 and this is for z less than 0.
So, this is the situation so far , that we have been trying to solve ah ah this equation
and ah this ah in presence of a linear potential or a slowly varying potential now it has become
a linear potential because as you ah come closer to the turning point ah whatever may
be the variation if you come very close it looks like ah ah linear potential, and this
is what ah we have obtained ah. So, far in terms of the modified Bessel function and
the Bessel function . Now, there are airy functions which are ah
simply ah called as so, these are Airy functions, and these airy functions are written in terms
of the Bessel functions and the modified Bessel functions, which are in this form a ah minus
ah I minus 1 third minus 1 third ah xi minus I 1 third ah xi ah z greater than 0 and there
is a B i ah z or this is equal to so, B i will just ah. So, that is ah for z greater
than 0 and it is again the A i ah z that is equal to 1 third z to the power ah so this
is not 1 third this is really half ah. So, this is half and J minus 1 third ah xi
minus J 1 third ah for z less than 0 and similarly a Bi z which is equal to 1 by ah root 3 ah
z to the power half ah and there is a I minus 1 third ah xi minus I 1 third ah xi with a
plus sign here which is for z greater than 0 and this is equal to B i z ah this is 1
by root 3 z to the power half and there is a ah J ah J minus 1 third ah xi minus J 1
third ah xi ah this is for z less than 0 . So, ah just a bit graphically we are so, this
is equal to so, this is . So, this is my A i so, this is x greater ah i mean x ah this
way and so these are. So, this is your Ai function and the Bi function looks like ah
so this is the Bi function and so on . So, ah we can ah start from the ah um to get
the asymptotic forms of these ah Ai and Bi we could start from the Bessel functions and
look at their asymptotic behavior rather we would write down straight away the asymptotic
behaviors of this. And so, that tells us that ah the a a i z ah that goes as 1 by 2 root
pi z to the power minus 4 exponential minus xi for ah z going to plus infinity ah.
This is ah 1 by root over pi mod z to the power minus 4 a sin ah xi plus ah pi by 4
for z going to minus infinity. So, that is the behavior of the airy functions ah for
ah the arguments to be in that region which is shown, there and a B i z has this ah 1
by root pi z to the power minus 4 ah z to the power minus 1 by 4 e to the power xi for
z going to plus infinity and it goes to 1 by root pi ah mod z going to minus 1 by 4
cosine of xi plus pi by 4 for ah z going to minus infinity .
Ah. So, ah this means that the airy functions the Ai the first kind these are called airy
functions of first kind these are called airy functions of second kind and they have these
asymptotic behaviors which ah the ah for z going to infinity ah the airy functions of
the first kind goes as exponential minus xi the other has a four x xi z going to minus
infinity it has a sin ah or it sin vary sinusoidally whereas, ah for the Bi function for z going
to infinity it goes as e to the power xi and ah for the other one it goes as cosine .
So, ah if we ah write down finally, the ah equation ah which we had obtained in terms
of the z. So, this is our ah equation whose solutions we have we could have written it
down simply, but we wanted to still ah um introduce this ah Bessel functions ah which
are. So, finally, these are these airy functions, so, ah with a and b as ah the coefficients
it is z . So, a and b are the ah the so, this is the final solutions of this slowly varying
potential a and b are amplitudes which are ah needed to be obtained from the boundary
condition. Now, this boundary condition is a little tricky
for the simple reason that we will just ah explain ah. So, now, this psi of z that goes
as a Ai z which goes as a by 2 root pi z to the power minus just look at this ah ah the
last slide for z going to plus infinity and it goes as a by root pi mod z whole to the
power minus 1 by 4 sin of xi plus pi by 4 for z going to minus infinity.
So, the solution is that that 1 by 2 z to the power minus half exponential minus xi
it goes over to ah z to the power minus 1 by 4 sin of xi plus pi by
4 . So, this is coming from the ah airy function of the first kind where we have put b equal
to 0 b equal to 0 is that because of the square integrability of the wave function because
bi diverges ah for ah ah um. So, this term is ah is increasing. So, we have just ah put
the coefficient of that equal to 0 ah because otherwise you will land up with this ah problem
of the square integrability of the wave function mod psi square dx ah integrated over all space
will not be equal to 1 and the. So, this basically what I we are trying to
say is that this is the solution in region 2 and this is the solution in region 1 and
this solution in region 2 should go to solution in region 1 and the other way around that
is solution of region 1 is connected by to the solution of region 2 by the same formula
is not allowed for the simple reason is that, ah as you change as you trying to go from
region 1 to region 2 ah this could cause a change in the phase of the sin and you know
the sin ah will ah if it changes its phase by a pi by 2 psi becomes cosine, and cosine
is not connected to an exponentially dying solution which are we are going to see cosine
is actually related to the exponentially growing solution . So, this is the reason that it
is a ah it is a connection formula it is called as a connection formula and it is purely ah
unidirectional all right . So, ah let us ah look at the other case ah
. Now, try to connect solutions from region 1 to region 2 . So, this was actually
from region 2 to region 1 that is what ah you see. So, we can we can write it ah since
we know so, region 2 to region 1 and now we are trying to connect it from region 1 to
region 2 ah we will put a equal to 0 and we will right psi z equal to b ah Bi z that is
the airy function of the second kind and just to remind you that, this B i z it goes as
1 by root over pi z to the power minus 1 over 4 exponential of xi for z going to plus infinity
and this is equal to 1 by root over of pi ah z going to minus this cosine of xi plus
pi by 4 and ah z going to minus infinity . So, now I will connect this as the other ah
connection formula and cosine of xi plus pi by 4 , this is in region 1 should connect
to z to the power minus 1 by 4 exponential xi. So, this is in region 2 this is called
as the Connection formula ok . So, this is the second connection formula
so, let us call it as a first connection formula .
And this is the ah second connection formula . So, just to go over it again that ah the
recipe for connecting solutions from region 2 . So, ah if you ah look at the region 2
so, region 2 is actually the classically ah forbidden ah region . So, region 2 is actually
the ah classically forbidden region where e is less than V of x and region, ah 1 is
the classically allowed region where ah e is ah ah greater than ah V of x ah of course,
ah that is the situation changes when you are talking about ah ah the two sides of the
barrier ah . So, ah in any case ah these are the connection
formulas that are or these are the ah the the boundary conditions that are ah used , now
this boundary conditions are distinctly different than the boundary conditions that we have
seen for ah ah all ah these other problems . The last thing that we would do in this
regard is that ah we will derive . The Bohr Sommerfeld quantization condition using the connection formula ok . So, ah we
draw this picture once again and ah we call it ah. So, we have ah let us say we have ah
this varying like this where ah we have ah the energy is like this. So, this is the energy
and this is my x 1 this is my x 2 this is my region 1 and this is my region 2 , ah that
is the energy ah so, this ah ah. So, this is a xi 2 a this is a xi 1 a. So, this is
region 1 and this is again ah xi 2 B and ah this is xi 1 B and so on . So, these are points
A and B and ah these are x 1 and x 2. So, let us define some new parameters xi 1 a equal
to x 1 to x k of x dx. So, this is between x 2 greater than x greater
than x 1 xi 1 ah xi 2 a rather , xi 2 a equal to a k prime x these are the phases that ah
we have seen x less than x 1 x 2 some x 1. So, this is x 1 x 2 x 1 and this is x 1 to
x ah which is x is some ah arbitrary variable. So, xi 2 B this is equal to x 2 x 2 and dx
k of x and a xi 2 B equal to x 2 to x k prime of ah x dx for x greater than x 2 and this
is of course, x 1 less than x less than x 2 .
So, these are my new variables and what we are trying to do is the following . So, ah
at A ah we have these ah going to the connection formula will have a 2 k prime x exponential
minus xi 2 a it is connected with 1 by k of x cosine of xi 1 a minus pi by 2 or pi by
4 sorry this is pi by 4 and at any point r . So, that is inside the region . So, the
wave function is like C 1divided by root over K of at r cosine of x 1 to xr ah and ah kx
dx thats the phase ah multip[lied]- minus pi by 4 called this one as alpha. So, this
is equal to C 1 root over K of r cosine alpha .
Similarly, at B ah the connection says that it is a k prime x exponential minus xi 2 B
ah xi 2 B it goes as 1 by root over k of x cosine of xi 1 ah ah B minus pi by 4 . So,
at any arbitrary point r psi of r is equal to some C 2 divided by K of r cosine of x
r to x 2 ah K of x dx minus pi by 4 and, this one we are going to cast it in the form of
alpha . So, let us call this as ah beta . And ah Cos of beta equal to cosine of ah xr.
So, ah just ah rewriting the Cos beta term as a pi by 4 plus ah x 2 divided by x r and
a K x dx ah. So, see that ah we have just used the fact that the Cos of minus theta
equal to Cos theta because Cos is an even function .
Ah. So, this can be a slightly ah modified in the form of a pi by 4 plus ah x 1 to xr
k of x dx plus x 2 to x 1 ah k of x dx this is just simply writing down the x 2 to x r
by splitting it into 2 terms , and ah this is nothing, but equal to cosine of ah alpha
plus pi by 2 minus x 1 to x 2 ah kx dx and this is nothing, but equal to cosine of alpha
minus eta. So, cosine beta is cosine alpha ah minus eta where eta equal to x 1 to x 2
k dx ah minus a pi by 2. So, this is nothing, but equal to cosine alpha
cosine eta plus a sin alpha sin eta . So, at a general ah point the wave function is
ah C 1 equal to ah so, it is ah so, psi of r . So, psi of r equal to C 2 divided by k
of r cosine alpha, cosine eta plus a sin alpha sin eta and ah. So, this wave function has
to be ah equal to this wave function that we have ah written down.
So, C 1 so, equating the coefficient C 1 equal to C 2 cosine eta and 0 equal to C 2 sin eta
ah since C 2 is not equal to 0 . Then a sin eta has to be equal to 0 which is equal to
sin n pi for n equal to ah 0, 1, 2 and so on, sin n pi .
So, that tells that eta equal to n pi equal to x 1 to x 2 k dx minus pi by 2 and I can
multiply by h cross and then it becomes x 1 to x 2 ah h cross k dx which is equal to
n plus half ah h cross pi , where h cross is equal to h over 2 pi. Now I can write this
down equal to p dx ah x one to x 2 equal to n plus half h cross pi which a little bit
of ah algebra it shows that it is 2 p dx x 1 to x 2 this is equal to n plus half h simply
h. So, this ah can be written as ah x 1 to x 2 p of x dx plus x 2 to x 1 p of x dx which
is equal to 2 x 1 to x 2 p of x dx which is nothing, but equal to a closed integral of
p p of x ah dx . So, that tells that the left hand side of
this equation ah the condition that we have gotten from here is equal to ah closed integral
of px dx. So, that is the Bohr Sommerfeld quantization condition that a p x dx is equal
to n plus half h and for ah n to be large this is simply equal to P x dx equal to n
h and this is the ah we will call it BS QC which is called as the Bohr Sommerfeld quantization
condition . So, just give one ah simple example
the example; is that ah the ah in a certain ah system the theta is like a sawtooth wave
as a function of t. So, this is theta of t that is the angular variable and the r simply
is just a constant at a value . So, d theta dt equal to omega and dr dt equal to 0.
So, that tells that ah my P ah now its x dx we just simply write it as ah pq dq ah, where
ah Q is the canonical ah coordinate and P is the ah canonical momentum which are nothing,
but equal to ld theta which is equal to l l is a constant here ah d theta from the full
range which is 0 to 2 pi which is equal to 2 pi l. So, ah P q dq is nothing, but equal
to nh which is equal to 2 pi l. So, l becomes equal to nh by 2 pi, which is equal to nh
cross and this is called as the Bohrs quantization .
So, this is one of the Bohrs postulate where he said that the angular momentum is ah quantized
in terms of ah h cross and those are the allowed orbits in which the electrons are allowed
to move around the nucleus where they do not ah emit electromagnetic radiation and they
are called as the stationary orbits . So, ah just to go back rerun the whole thing
again there is a lot of mathematics that we have done, but what we have finally, said
is that ah for a slowly varying potential ah the solutions that you write down fails
miserably close to the turning points because the amplitudes are proportional to 1 over
ah k x or 1 over root over kx ah where k is the ah the wave vector, which is ah ah obtained
from this ah energy and the potential ah ah energy relation the total energy and the potential
profile relation and ah . But these are good solutions away from the
turning points, if they are good solutions away from the turning point there has to be
a way to connect the solutions into the from the classically forbidden region to the classically
allowed region, that gives us one ah very important factor that you cannot do this ah
arbitrarily that is connect the two solutions and write it as equality, ah it has to be
done with care that sometimes you can go from region 2 to region 1 via 1 relation.
But to come back from region 1 to region 2 you will write another relation, and these
are called as the connection formula this connection formula have important applications
where one can actually ah compute what is called as a Bohr Sommerfeld Quantization condition
we have ah showed a simple case where ah we recover the Bohrs postulate starting from
the ah quantization ah condition, which are ah which are of course, the byproduct or artifact
of the ah the connection formula .