Tip:
Highlight text to annotate it
X
In my last lecture, we had, at the end of the last lecture, we were considering the
solution of the Schrodinger equation, for the 3 dimensional isotropic oscillator. We
will, in this lecture continue our discussion on that and make a few comments about the
2 dimensional oscillator, which is also a problem of great interest in quantum mechanics
particularly in the theory of magnetism. And then, we will hopefully start on the Bra
Ket algebra formulation of quantum mechanics, so we continue our discussions where we left
off in the first in the previous lecture.
We were considering a spherically symmetric potential which is half mu omega square r
square, it is a spherically symmetric potential it is a spherically symmetric potential. In
my previous lecture, I wrote this as half mu omega square x square plus y square plus
z square; and solved the Schrodinger equation in Cartesian coordinate and obtain the Eigen
values as E is equal to n 1 plus n 2 plus n 3 h cross omega sorry I am sorry, I am sorry,
this we had obtained as E is equal to n 1 plus n 2 plus n 3 plus 3 by 2 plus 3 by 2
h cross omega. Where each of these integers n 1, n 2, n 3
can take values 0, 1, 2, 3 etcetera. Now, if I use this equation, then we can as we
had shown earlier, we can solve this equation 2 mu by h cross square E minus V of r psi
and we will we can use the method, we can use this spherical polar coordinates, r theta
phi to solve this equation. Since, the potential energy function depends
only on the r coordinate, we can write the solution as psi of r theta phi as R of r times
Y theta phi, where Y theta phi are the Eigen functions of the operator L square as we had
seen that that y theta phi are the spherical harmonics. Which are L square Y theta phi
was equal to l into l plus 1 h cross square Y l m, sorry this is also Y l m theta phi,
these are the spherical harmonies. So, if I substitute this, in this equation
then you will get the radial part of the Schrodinger equation, and if I define a function R of
r u of r such that, R of r is equal to u of r by r then u of r satisfies the following
equations.
This is the d 2 u by d r square plus 2 mu by h cross square E minus V of r minus V of
r minus l into l plus 1 h cross square by 2 mu r square u of r is equal to 0, this is
the radial part of the Schrodinger equation. This is one of the forms of the radial part
of the Schrodinger equation, and this simplification this simplification is always possible as
long as the potential energy function is spherically symmetric.
So, now we assume that V of r is half mu omega square r square, and then we write down as
we have done for the harmonic oscillator that xi is equal to gamma r gamma r, so this becomes
d u by d r becomes d u by d xi into d xi by d r which is gamma. If I differentiate this
again, so you get d 2 u by d xi square multiplied by gamma square, so I want to put this down
in the denominator, so plus please see this, 2 mu E by h let me write this down here, then
I will do it again. And this is half, my potential energy is half
mu omega square r square, so half into 2 cancel out, so you get mu square omega square by
h cross square r square, if I take 2 mu by h cross square out inside, so you get minus
l into l plus 1 r square u of xi equal to 0. So, I divide by gamma square, so I get
plus 2 mu E by h cross square gamma square, so I write this as capital lambda, where capital
lambda is defined to be equal to 2 mu E by h cross square gamma square and this becomes
please see, r square is xi square by gamma square.
So, this term becomes this term becomes mu square omega square by h cross square r square
is xi square by gamma square and there is one gamma square coming from here, so this
will become gamma to the power of 4. Now, I choose my gamma, so that this factor is
1, I have not yet this is the same method that we had used in solving the harmonic oscillator
problem, so I choose gamma, so that this is equal to under root of mu omega by h cross.
If I choose this value of gamma, then this coefficient becomes 1 and this becomes minus
xi square, and the last term l into l plus 1 gamma square r square, so that is xi square;
so minus l into l plus 1 gamma square r square, so that is xi square u of xi is equal to 0.
So, gamma square, so from here gamma square is mu omega by h cross, so this becomes 2
mu by E by h cross square mu omega by h cross, so mu mu cancels out, so you get 2 E by h
cross.
So, the Schrodinger equation simplifies to simplifies to the radial part of the Schrodinger
equation, simplifies to d 2 u by d xi square plus lambda minus xi square minus l into l
plus 1 by xi square u of xi is equal to 0, where xi is equal to gamma r gamma r, gamma
is equal to under root of mu omega by h cross. Now, lambda is now the Eigen value, so lambda
is equal to 2 E by h cross omega, you must remember that xi now goes from 0 to infinity
and of course, u of 0 will be 0, because at the origin you have R of r is equal to u of
r by r. So, at r equal to 0, so u u of 0 will be 0,
these are the boundary conditions and u of infinity is 0, now we use the same trick,
same methodology as we used in the hydrogen atom problem, as xi tends to 0 this term dominates
and you will you can the solution of this term is d 2 u by d xi square minus l into
l plus 1 by xi square u xi. This will be u of xi will be xi to the power of l plus 1;
because if you differentiate this twice it will be l plus 1 into l xi to the power of
l minus 2 l minus 1, so that is what it, so as xi tends to 0 it behaves like this.
As xi tends to infinity it is this term which dominate, we had experienced this term in
the harmonic oscillator problem, so the the if you recollect that, so u of xi as xi tends
to infinity behaves as e to the power of minus half xi square. So, we this suggest that we
try out a solution just as we did in our hydrogen atom problem F of xi y of xi, where F of xi
is equal to xi to the power of l plus 1 e to the power of minus half xi square.
So, as you recall that we had u double prime is equal to F double prime y plus 2 F prime
y prime plus F y double prime, and we can calculate F prime from here, F double prime
from here and we have to make one more. These are rigorously correct solutions rigorously
correct solutions, we make another transformation that we define a new variable eta is equal
to xi square as we had done in the harmonic oscillator problem. If I substitute this for
F of xi and then make a transformation eta is equal to xi square.
Then the y satisfies this equation, eta d 2 y by d eta square plus c minus eta d y,
I leave that as an exercise to show this minus a y eta is equal to 0, where c will and we
shown to be equal to l plus 3 by 2 and a is equal to l by 2 plus 3 by 4 minus lambda by
4. So, once again what I have done is that that u of xi, I have written as xi to the
power of l plus 1 e to the power of minus half xi square into y of xi and then obtained
an equation for y of xi. And then we have since, there is a xi square
term here, we substituted a variable transform, we used a transformation eta is equal to xi
square then this function y of eta satisfies this equation. As we all know, this is the
familiar confluent hyper geometric equation, this is the confluent hyper geometric equation,
and the solutions are the solutions, the well behaved solutions are F 1 1 a, c eta and as
you know this is 1 plus a by c eta by factorial 1 plus a into a plus 1 and so on.
If this series is not terminated then for large values of eta it behaves as a, e to
the power of plus eta, so this infinite series once again is convergent. But, if it is not
terminated, if it is an infinite series, then for large values of eta it will behave as
e to the power of eta, that is e to the power of xi square.
So, e to the power of xi square times e to the power of minus xi square becomes e to
the power of plus half xi square and therefore, the function will. We cannot let that happen
and so therefore, a must be therefore, this must be a polynomial and since, in the numerator
you have a into a plus 1, then a into a plus 1 into a plus 2.
So, a must be integer, negative integer, so we must have for the infinite series to become
a polynomial polynomial, a must be equal to minus n r, where n r can be 0, 1, 2, 3. Because,
if let us suppose a is equal to minus 2 then the first term will be a into a plus 1, the
second term divided by something. Then it will be a into a plus 1 into a plus
2, so that will become 0 not only that term, but all subsequent terms will become 0 and
the series will become a polynomial. So, for the series to become a polynomial, a must
be a negative integer and therefore, this must be minus n r. So I obtain so I obtain
that that a must be a negative integer, so l by 2 plus 3 by 4 minus lambda by 4.
So, if I multiply this by minus 1, so you get minus l by 2 minus 3 by 4 plus lambda
by 4 is equal to n r, I take this to that side and I multiply by 4, so I get lambda
is equal to 4 n r plus 2 l 2 2 l plus 3. If you recollect that lambda of we had put equal
to lambda was equal to 2 E by h cross omega, so this is equal to 2 E by h cross omega 2
E by h cross omega and therefore, we will have E is equal to if I divide by 2. So, 2
n r plus l plus 3 by 2 h cross omega, so these are the rigorously correct energy Eigen values,
where n r and l both takes the values 0, 1, 2, 3 l also take the value 0, 1, 2, 3 etcetera.
So, as in the previous case where we had used Cartesian system of we of course, must be
at the same Eigen values, so you will have you will have the ground state the ground
state will be n r equal to 0 and l equal to 0 and so therefore, E will be equal to 3 by
2 h cross omega. And what will be the wave function, the wave functions we had denoted
that since l is equal to 0, so m will be equal to 0, so my angular part will be Y 0 0 theta
phi, so this will be just a 1 over under root of 4 pi.
So, this will be the angular part and because my wave function is R of r times Y theta phi
y l m theta phi, l is 0 then m will be also 0 and and the, and R of r R of r will be u
of r by r, if I divide by, so this is the same as so. If I multiply by gamma, so u of
xi by xi and this will be with normalization constant, this will be xi to the power of
l e to the power of minus half xi square F 1 1 a, c, xi square.
So, I write this down once again once again R of r will be xi to the power of l e to the
power of minus half xi square F 1 1 a, c, xi square, if a is 0 this is a normalization
constant, a is 0, so this is 1 and l is 0, so this is, so this is just n into e to the
power of minus half r square gamma square r square, this is my wave function. Let me
do one more case and then we will end there, the second Eigen function will be n r equal
to 0 first excited states. And l is equal to 1, if l is equal to 1 the
Eigen values will be 5 by 2 h cross omega, so your radial part of the wave function,
the radial part of the wave function will be n xi to the power of l, that is r to the
power of l, so that is r gamma r to the power of 1 a is 0. And n r is 0, so a is 0, so this
is one so e to the power of minus half xi square, but since Y l is 1, so m can take
values 1, 0 minus 1. So, we will have this multiplied by Y 1 1
Y 1 0 and Y 1 and minus 1, so this will be as we had obtained earlier 3 fold degenerate
state, so what we have tried to show you is that for the 3 dimensional oscillator problem,
isotropic oscillator problem. One can solve the radial part of the Schrodinger equation,
in terms of the confluent hyper geometric function.
And one you give me the value of a and c, that is you give me the values of n r and
l, and I will be able to immediately give you a is equal to minus n r and c is equal
to l by, so a is equal to minus n r and c is equal to l plus 3 by 2.
So, if you give me the values of a and c, I can immediately write down the polynomial
and I can write down the corresponding wave function, so it is really very straight forward
you give me any value of a, you give me any value of N r any value of N r and any value
of l, I can immediately write down the wave function, within a normalization within a
normalization factor. And that can be easily calculated by using
the normalization condition, that 0 to infinity R of r mod square r square d r, from r equal
to 0 to infinity this must be equal to 1. So, finally, we consider the last problem
in the series and that is the 2 dimensional harmonic oscillators.
Now, if I have the 2 dimensional, we conclude this lecture by considering the 2 dimensional
oscillators, and in which we will have the V as half of mu omega square x square plus
y square. So, we have to solve the 2 dimensional Schrodinger equation in terms of x and y or
in terms of rho and phi, were rho phi are circular are the are the coordinates like
this rho x is equal to rho cos phi and y is equal to rho sin phi.
So, you have here x, y 2 dimensional and this is my rho and this is the angle phi, these
are known as the circular coordinates. So, in this if I take rho and phi as my coordinate
system, then I can write this down as my coordinate system, then I can write this down as half
mu omega square rho square. So, the Schrodinger equation is in 2 dimension
either I can write this down as delta 2 psi by delta x square plus delta 2 psi by delta
y square plus 2 mu by h crosses square E minus half mu omega square multiplied by x square
plus y square psi of x, y is equal to 0, and I leave this an exercise, you again use the
method of separation of variables. You write this down as psi x, y as X of x
Y of y you will obtain Hermite-Gauss function and the energy Eigen values will be n 1 plus
half plus n 2 plus half, so n 1 plus n 2 plus 1 h cross omega; these will be the Eigen values
n 1 is equal to 0, 1, 2, 3, n 2 will also be equal to 0, 1, 2, 3 etcetera.
And the corresponding wave functions would be N of n 1 H n 1 of xi multiplied by N of
n 2 H of n 2 eta e to the power of minus half xi square plus eta square. So, these are the
Hermite-Gauss function, but since I can use also the cylindrical coordinates.
And we will obtain 1 over rho 1 over rho delta by delta rho rho delta psi by delta rho plus
1 over rho square delta 2 psi by delta phi square plus 2 mu by h crosses square E minus
V of rho, V depends only on the rho coordinates psi of rho phi. I can again use the method
of separation of variables and phi goes from 0 to 2 pi, so the phi variables, so I can
write this down as psi of rho, phi is equal to the radial part times the phi part.
And one will obtain a term like this, so if I I will obtain a term like this 1 over phi
d 2 phi by d phi square as we had obtained in the angular momentum case problem, we will
set this equal to minus m square and we will get phi of phi is equal to the normalized
E root 2 pi e to the power of I m phi, where m is equal to 0 plus minus 1 plus minus 2
etcetera. So, we will have a equation which will involve
only the rho term and and I leave it as an exercise for you, if you define any function
which is such that, R of rho is equal to u of rho by square root of rho. Then u of rho
will be, will satisfy the following equation, d 2 u by d rho square plus lambda minus rho
square minus m square minus 1 by 4 by rho square u of rho is equal to 0.
So, the phi part will separate out and you make a substitution like this, it is fairly
straight forward and one will obtain in equation like this, once again we look at the limiting
form of rho tending to 0. So, this term will dominate and you will have d 2 u by d rho
square minus m square minus 1 by 4 by rho square u of rho and the solution of this equation
will be will be rho to the power of m plus 1. So, it will come out to be rho to the power
of m plus half, so if I differentiate it once, so it will become m plus half times m minus
half, so that will be m square minus 1 by 4.
So, therefore, the the the small rho behavior will be u rho, the same methodology will be
rho to the power of m plus half actually mod m mod m, because here m square appears, times
the large wave behavior will be again e to the power of minus half rho square and multiplied
by y of rho. Then because of this term you introduce another
variable which is eta is equal to rho square and this we take G of rho and we can write
down just as we did for u double prime of rho will be equal to G times y double prime
plus 2 G prime y prime plus G double prime y of rho, but the primes do not differentiation.
And if you use this transformation, that eta is equal to rho square, and then you will
find that the solution is again can be written down in terms of confluent hyper geometric
equation. And for the confluent hyper geometric equation to get terminated, the solution will
be F 1 1 a, c, eta and a will be equal to minus lambda by 4 plus m plus 1 by 2 is equal
to minus n. For this to terminate a must be equal to a
negative integer, so if you multiplied by negative sign, so you get lambda by 4 minus
m plus 1 by 2 is equal to n, so lambda will be equal to, this is equal to 2 E by h cross
omega, so this will be 4 n plus 2 modulus m plus 1. So, E will be, if I divide by 2,
so this will be 2 n 2 n plus m plus 1 h cross omega, so if n is 0 m is 0 then you have the
ground state the ground state energy Eigen value as we had obtained earlier.
But, these solutions are of extreme importance in considering the effect of the magnetic
field and so, what we have done is that we have to summarize, that we have to consider
a 2 dimensional oscillator. We can solve this by using the method of separation of variables
and then or we can use the circular coordinates, where x is equal to rho cos phi and y is equal
to rho sine phi. If we do that the corresponding del square operator in 2 dimensional becomes
this. We can use the method of separation of variables,
so this becomes d 2 phi by so, this term will separate out, if you multiply this equation
by rho square, and then divide this equation by R times phi it you will have a term which
depends only on phi and the remaining terms will depend on rho.
So, this phi term we set equal to this constant and the phi solution will become as we have
obtained in the angular momentum case, 1 over root 2 pi e to the power I m phi; so with
that, we obtain we make a transformation something like this.
We look at the small rho behavior small rho behavior will be rho to the power m plus half,
large rho behavior will be dominated by this term, so this will be e to the power of half
rho square and therefore, we will try out the solution like this. And then we will substitute
eta is equal to rho square; we will I leave it as an exercise, it is a very simple exercise,
and you will get the confluent hyper geometric equation. And that confluent hyper geometric
equation the equation the function must be terminated otherwise, it would behave if it
is, if this series is not terminated at large values of eta, it will behave as e to the
power of eta. So, therefore, this would behave as e to the
power of rho square, because eta is equal to rho square rho square, if you multiply
this here it will become e to the power of half rho square, so the wave function will
diverge at rho equal to infinity, we cannot let that happen. And therefore, a must be
a negative integer and therefore, which is given by minus lambda by 4 plus m plus 1 by
2 must be a negative integer. And so therefore, if I these are the Eigen
values of the problem and the Eigen functions can be written down in terms of the confluent
hyper geometric function a, c rho square times e to the power of minus half rho square, times
rho to the power of n plus half, let me write it down carefully.
So, the wave functions, so the final energy Eigen values will be E is equal to 2 n plus
m plus 1 h cross omega where n is equal to 0, 1, 2, 3 m is equal to mod m 0, 1, 2, 3
etcetera. And the wave functions will be psi of r phi or rho phi will be rho to the power
of m plus half, so the if I divide by square root of r then half goes away E to the power
of minus half rho square F 1 1 a, c, rho square where a is equal to minus of n. And so and
c is equal to mod m plus 1, so this are the and the phi dependence of course, there is
a normalization constant here, and phi dependence will be 1 over root 2 phi to the power of
i m phi, this is the phi dependence, this is the rho dependence. These are the rigorously
correct solutions for the harmonic oscillator, for the 2 dimensional oscillator problem and
as I told you that the ground state corresponds to n equal to 0 and m equal to 0, so this
is equal to 1. So, this completes this completes the solutions
of the 1 dimensional Schrodinger equation and sometimes the 2 dimensional Schrodinger
equation and of course, the 3 dimensional Schrodinger equation. We have solved the hydrogen
atom problem the 3 dimensional oscillator problem, the diatomic molecule problem, the
linear harmonic oscillator problem. And also we have considered tunneling through
a barrier and things like that, so this is one aspect of the very important aspect of
quantum mechanics, that we have considered, we also considered the free particle problem
and propagation of wave packets. In the following few lectures, we will consider yet another
formulation of quantum mechanics it is it is by Dirac in and it is for understanding
this, one must be very familiar with the rules of matrix algebra.
So, we will carry along, so we will try to tell you all these steps for example, let
us consider the harmonic oscillator problem, V of x linear harmonic oscillator problem
and we have the potential energy distributions like this. So, we have a parabolic index potential
and we had obtained that these are the Eigen, these are the discreet Eigen.
So, this corresponds to n equal to 0 n equal to 1 and this is equal to n equal to 2 and
similarly, n equal to 3 this was denoted by the wave function this psi 0 of x, psi 1 of
x, psi 2 of x and so on. Now, we consider this linear harmonic oscillator, these are
the different state of the oscillator. Now, what Dirac said was each state of a dynamical
system is represented by A ket vector and we denote this by A ket like this. So this
vector this state I denote by ket 0, this state by ket 1, I have hydrogen atom state
also, we have solved the hydrogen atom problem each has a specific quantum number, a set
of quantum numbers. Now, each state of the hydrogen atom I represent
by A ket vector, so let us suppose this state I represent by ket vector A, and another state
by ket vector B. So, each state of a dynamical system is represented by A ket vector such
that, any linear combination of these two vectors is also another vector in the same
space. I will detail this little later, corresponding to each ket vector there is a bra vector A,
such that we can form a scalar product which is bra A B and this quantity is a complex
number.
So that, we say that each state of a dynamical system, I represent by A ket vector there
can be many ket vector associated with the state, one can take a linear combination of
these two ket vectors. Corresponding to each states ket vector there is a bra vector A
such that, we can form a scalar product A B, which is a complex number which is a complex
number and a such that the complex conjugate of this number is denoted by B A it may appear
a little abstract, but I will give you an example in a moment.
Now, let us suppose I put B is equal to A, so then this this tells me, this is an axiom
then I have bra A ket A which is a number, so this is A times A. So, which tells us that
the scalar product of A ket vector with its own bra is a real number, I further say that
not only this is a real number, but this is always positive definite, that is this is
greater than or equal to 0, unless A is a null ket.
Now, what is a null ket, a ket vector A is a null ket, when if pre multiplied by bra
B, if this is 0 for any bra B, then ket A is said to be null ket sorry similarly, bra
B is A null bra, if and only if bra B ket A is 0 for any ket A.
Let me make you, try to make you understand the concept that I have developed using matrices,
now I consider a 2 dimensional space, so I have a column vector representing a vector
a ket vector A, I represent this by say a b. Another vector B I represent this by c
d I forgot to mention here, that if A B is 0, then A vector a ket A and ket B are said to
be orthogonal to each other, this scalar product is said to be 0. If bra A ket A is 1, then
we say that the ket vector is normalized. So, let me let me go back to this 2 dimensional
space, where a ket vector is represented by a, b 2 numbers and bra B also, ket B also
belongs to the same same space 2 dimensional space.
So, you can write down A plus B will be again in this another vector in the 2 dimensional
space, so this is 1, 2, 3, 4, so this will be 4 plus 4 here and 5 here. Now, corresponding
to this bra A corresponding to this the bra A will be the complex conjugate of this and
this this is known as the dual vector space. Corresponding to a column vector that is always
a rho vector, and if I multiply these two bra A ket A it is a number, so you will have
a star, b star, a, b, so this will be mod a square plus mod b square. So, this is always
greater than 0, it can be 0 if and only if, a is 0 and b is 0, so then that is a null
ket. Because, if you have a null ket you can multiply
this pre multiply by any bra and this will be 0, now here bra B is equal to c star d
star and the same analysis I can do, now if I have bra A ket A. If this is 1, that is
if mod a square plus mod b square is equal to 1, we say that the ket is normalized.
So, let me give you an example, so let me take a ket A which is given by 2 dimensional
space, 1 by root 2 1 by 1, 1 1, so these are both real numbers, so bra A is equal to 1
over root 2 1 1, so bra A ket A will be 1 over 2 1 1, 1 1, so that is 2 and that is
1. So, we say this is a normalized ket, now we again we take another ket B and we write
down 1 over root 2 1 minus 1 as you can see, that this ket is also normalized.
However, if I now do B A then 1 over root 2 multiplied by 1 over root 2 is 1 over 2
and bra B is 1 minus 1 and ket A is 1 1, so this is 0, so we say that ket A and ket B
are orthogonal to each other.
Let me give you another example, so let me write down there is another vector in the
same space c, which is equal to 1 over root 2 1 minus i and I leave this is as an exercise,
where I is equal to square root of minus 1, that this is a normalized ket. And if I take
another ket D is equal to 1 over root 2 1 plus I then this and this arte orthogonal
to each other, but if I take A ket A which is 1 over root 2 1 1 then that is not orthogonal
to either of this vectors. So, we will continue our discussion on bras
and kets in our next lecture, but before that we will I will request all of you to brush
up your matrix algebra, because the operator algebra that we will develop, it will be very
easy to understand that; if we if we have a knowledge of the algebra involved with matrices,
thank you.