To get insight on the functional form of the
eigenfunctions and the precise energies let
us now again start with the Schrodinger equation,
it is psi = E psi, but now pre multiply the
equation on both sides by b instead of b dagger.
So the Schrodinger equation is h bar omega
b dagger b + half psi = E psi and we pre multiply
both sides of the equation by the operator
b. This gives h bar omega b b dagger b + b
/ 2 psi = E times b of psi, we want to make
the left hand side of this equation look like
the Hamiltonian.
We now use the commutation relation and write
this bb dagger as 1 + b dagger b. So, this
becomes h bar omega b dagger b + 3 / 2 and
we write the b psi outside is equal to E times
b psi. And if we make the operator on the
left hand side look like the Hamiltonian operator,
then we get h bar omega b dagger b + half
b of psi = E - h bar omega b of psi. So, we
see that if psi is an eigenfunction of the
Hamiltonian, then b of psi is also an eigenfunction
of the Hamiltonian.
Because here we have h of b of psi = E - h
bar omega b of psi, but, the b of psi has
an eigenvalue which is less than the eigenvalue
of psi / h bar omega. This operator b acts
on an eigenfunction of the Hamiltonian and
gives a new eigenfunction with eigenvalue
lower by h bar omega. And since this lowers
the energy, it is called the ladder down operator
we now make an argument that for a harmonic
oscillator with potential energy v of x = half
k x square.
Where k is positive the total energy is thus
positive and this implies that the lowest
eigenvalue of the Hamiltonian must be greater
than zero. So, the eigenvalues are have a
lower bound and all values are greater than
that lower bound. Now suppose that psi 0 is
the lowest energy eigenfunction. So, H of
psi 0 is let us say E 0 times psi 0 and we
have seen that b of psi 0 is also an eigenfunction
of the Hamiltonian with an eigenvalue E 0
- h bar, omega. So, that implies that there
is another eigenfunction which has energy
lower than E 0.
So, there is a contradiction here on the one
hand we are saying that psi 0 is the lowest
energy eigenfunction and on the other hand,
there seems to be another eigenfunction b
of psi 0, which is having an energy lower
than E 0 this contradiction can be resolved
and this equation can be satisfied if b psi
0 = 0 and b psi 0 = 0 implies that one over
square root of 2, I am just writing the definition
of the operator b here this is equal to 0.
So, psi 0 satisfies the following equation
here. And this implies that d / dq of psi
0 = - q times psi 0, this is a differential
equation for psi 0 and to solve it, we can
take d psi 0 / psi 0 so we do a separation
of variables is equal to - qdq integrating
this equation gives ln of psi 0 = - q square
/ 2 plus a constant of integration, which
implies that psi 0 is equal to some Constant
e to the power of - q square / 2.
So, this gives us a functional form for the
lowest eigenfunction of the harmonic oscillator
Hamiltonian. Furthermore, the eigenvalue of
psi 0 the lowest eigenfunction is h of psi
0 = h bar omega the Hamiltonian b dagger b
+ half times psi 0 and we have seen that b
operating on psi 0 = 0. So, that is h bar
omega 0 + half psi 0. So, the eigenvalue is
simply h bar omega / 2 times psi 0 and the
eigenvalue is simply h bar omega / 2.
So far we have seen that if psi is an eigenfunction
of the harmonic oscillator Hamiltonian, then
b dagger psi is also an eigenfunction with
eigenvalue e + h bar omega. And we have seen
that b of psi is also an eigenfunction with
a lower eigenvalue e - h bar omega. Moreover,
we have seen that the lowest eigenfunction
has the functional form c e to the power of
- q square / 2 and this has an eigenvalue
8 psi 0 = h bar omega / 2 psi 0.
The question now is what is the functional
form of the other eigenfunction besides the
lowest eigenfunction of the harmonic oscillator
and for this we can operate with the ladder
up operator on the lowest eigenfunction of
the harmonic oscillator and get all the other
eigenfunctions. So, let us do that now. So,
we operate with b dagger on the lowest eigenfunction
psi 0.
And we write this explicitly, c is the constant
associated with the psi 0 and the square root
of 2 is part of the b dagger operator d / dq
/ q e to the power of - q square / 2. And
when we take the derivative and write this,
we get c over square root of 2 - e to the
power of - q square / 2 - 2q / 2 + q times
e to the power of - q square / 2 and on simplifying
this becomes c over square root of 2q e to
the power of - q square / 2 + q e to the power
of - q square / 2.
Which is c over square root of 2 2q multiplied
/ e to the power of - q square by to the functional
form for the first excited eigenfunction,
b dagger psi 0, we can write this as psi 1
is thus c over square root of 2 2q e to the
power of - q square / 2. Let us find the next
higher eigenfunction and for that we operate
with the b dagger operator on psi 1 that is
in other words b dagger on b dagger of psi
0. And that is some c prime / square root
of 2 - d / dq + q, operating on q times e
to the power of - q square / 2.
That is equal to c prime / square root of
2 - q e to the power of - q square / 2 - 2q
/ 2 - e square e to the power of - q square
/ 2 + q square e to the power of - q square
/ 2 we take e to the power of - q square / 2
common because this is there in all the terms
and then this becomes c prime / square root
of 2 q square - 1 + q square e to the power
of - q square / 2, and that gives c prime
/ square root of 2 2q square - 1 e to the
power of - q square / 2.
And this is the functional form of psi 2,
or the second excited state of the harmonic
oscillator. We notice that the eigenfunctions
of the harmonic oscillator have certain pattern
in their functional form. So if you look at
psi 0, first, you see that this is simply
the Gaussian function e to the power of - q
square / 2 the psi 1 here is a Gaussian function
e to the power - q square / 2 multiplied by
a polynomial 2q.
And again, if you look at to it is again the
Gaussian function e to the power of - q square
/ 2 multiplied by another by another polynomial
and the order of the polynomial is equal to
the quantum number of the function. So, the
psi 2 has a polynomial of order 2 and the
psi 1 has a polynomial of order 1 and the
psi 0 has a 0 order polynomial or just a constant
multiplying the Gaussian function.
The polynomials which are part of the functional
form of the harmonic oscillator Eigen functions
are called Hermite polynomials and some of
the lower ones have the following functional
forms. So, H 0 of q is just 1 H 1 of q = 2q
H 2 of q = 4q square - 2, H 3 of q = 8q cube
- 12q. And these forms of the Hemite polynomials
can be very easily looked up in any textbook
on spectroscopy or quantum mechanics.
The last thing to remember is that the variable
q, which is the dimension less coordinate
was introduced by us to simplify the derivation
and q is actually related to x in the following
manner h bar x. So, all of these functional
forms are actually functions of x psi of x.
And for example, the lowest eigenfunction
psi 0 of x would become c e to the power of
- m omega / h bar x square / 2, when we write
it in terms of x by replacing q with x, this
can be further simplified as c e to the power
of - alpha x square / 2, where alpha = m omega
/ h bar in terms of k the alpha is equal to
square root of km / h bar.
The eigenfunctions of the harmonic oscillator
Hamiltonian have the following functional
form. So, if you take nth eigenfunction, this
is some constant multiplied by a polynomial
of degree n multiplied by the Gaussian function
e to the power of - alpha x square / 2, where
alpha = m omega / h bar, we have seen that
this polynomial of degree n are the Hermite
polynomials and we have looked at the functional
forms of some of these.
Now, the question is what is this constant
c Now, the vjust comes from the normalisation
of the eigenfunction. So, for example, let
us do this for the lowest eigenfunction. So,
the lowest eigenfunction is psi 0 = c e to
the power of – alpha x square / 2 and to
get C we impose the normalization condition
that is psi 0 star psi 0 dx limits from - infinity
to infinity = 1. This implies that c square
- infinity to infinity e to the power of - alpha
x square / 2 multiplied / e to the power of
- alpha x square / 2 dx = 1.
And that means c square - infinity to infinity
e to the power of - alpha x square dx = 1.
The function here is just the Gaussian function
and the integral of this from - infinity to
infinity is a standard integral with value
square root of pi over alpha. So, therefore,
c square times square root of pi divided by
alpha = 1 and this implies that c = alpha
divided by pi to the power of 1 / 4.
The normalisation condition for all other
eigenfunctions can be similarly found and
they in general depend on the quantum number
of the eigenfunction, but the condition to
normalise them is always the same, which is
minus infinity to infinity psi n star psi
n dx = 1, this is the normalisation condition.
The normalisation constant depends on n and
can be found by applying this condition.
We will now look at the shapes of some of
the lowest eigenfunctions of the harmonic
oscillator. So, let us start with the lowest
eigenfunction psi 0, which is a normalisation
and N 0 e to the power of - alpha x square
/ 2, N 0 is a constant, which we can obtain
by using the normalisation condition that
we discussed. The functional form of this
function is just the Gaussian function. So,
if this is the x axis, then the function has
the typical Gaussian shape like this. So,
this is how psi 0 looks.
Let us look at the first excited vibration
wave function. That is psi 1 is equal to the
normalisation constant multiplied / 2 root
alpha x e to the power of - alpha x square
/ 2. We note here that this is a function
which is a product of 2 functions. This is
a linear function. And this is a Gaussian
function. So if we were to plot these, here
is the x axis, the linear function is a straight
line this 2 root of alpha x is just a straight
line like this. And the Gaussian function,
which is the second part, is a function like
this.
And if you take the product of these 2 functions,
then we get a function which looks like this
which has 0 value at x = 0 and it has a negative
value first and a positive value, and it looks
like this. So this is the shape of the psi
1 eigenfunction. Let us look at the next eigenfunction
psi 2, which is N 2 times 4 for a square root
of alpha x square - 2 e to the power of - alpha
x square / 2. So we notice that here we have
a quadratic function multiplied by a Gaussian
function.
And the first function is a parabola. But
the value of the parabola when x = 0 is - 2.
So if you were to plot that parabola, it looks
something like this, where the value here
is this match is - 2 and the Gaussian function
is again like this, this is the x axis. So,
the product of this function will be like
this where it has value initially positive
then negative then positive again and then
goes like this.
So, this is the shape of the psi 2 eigenfunction
psi 3 has the functional form in 3 multiplied
by now a cubic polynomial so, square root
of alpha x cube – 12 alpha x multiplied
by the Gaussian function. So, here is cubic
multiplied by Gaussian, the cubic function
looks something like this. And the Gaussian
function looks like this. So the product of
the 2 gives a function which has 0 values
at as the x = 0.
And it has a negative value initially it becomes
positive goes through the 0. And it has a
function for shape. And it has a shape which
looks like this. So this is psi 3. If we write
these functions and their energies, along
with the potential energy function, then the
entire picture looks something like this.
Here is the potential energy the lowest function
looks like that this is the energies are all
equally spaced.
So, I can draw the energies like this is psi
0 psi 1 psi 2 psi 3 and if I draw the shapes,
they look like this and like that and like
the energies of these different states are
h bar omega / 2 1 + half h bar omega / 2 and
for the psi 2 it is 2 + half h bar omega / 2
and here it is 3 + half h bar omega / 2 This
gives a complete picture of the eigenvalues
and eigenfunctions of the harmonic oscillator.
