3rd year

Sanju 9681634157
A single step barrier:- A single step barrier is shown in the figure and represented by ,
( )
The particle moves like a free particle in
the region
but as it approaches the
step potential towards the right
for
,it has to face a potential barrier of
height
Denoting the wave function in the region
( )
by
( )
respectively, we can write down
Schrodinger equation as follows
)
)
The energy of the particle may be greater or less than
separately .
Case 1
>
; we can consider the two case
Here we rewrite the equation (1) and (2) as,
( )
(
)
( )
(
The general solution of the above two equation
)
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Were;
Represents the incident wave;
Represents the wave reflected from barrier.
Represents the remitted wave into region
Represents the wave which would get reflection into barrier
(ii), which does not exist here.So
(
)
Determination of constants:- To evaluate the contents B and C in terms of A, we apply the
flowing conditions ,
(a) The wave function
must be continuous at the boundary
( )
( )
( )
(b) The 1st order derivative
must be continuous at
[
]
;
[
(
]
)
( )
Solving (8)and (9)
(
)
(
Solving
)
;
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Now incident wave
Probability current density for incident wave
6
| |
7
The wave reflect back is now
)
; corresponding current density,
| |
(
The wave transmitted into region (ii)
)
; corresponding current density
(
| |
)
The reflection coefficient in this case
| |
| |
| |
| |
| |
| |
(
(
)
)
(
)
The tram mission coefficient in this case
| |
| |
It can be noticed
| |
| |
| |
| |
(
)
(
[conservation of probability at boundary]
)
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Here we rewrite the equation (1) and (2) as
(
(
)
)
(
(
)
)
The general solution of the above equation,
Since
(
)
(
)
which shows
(
)
Where
Represents the incident wave;
Represents the ware reflected from barrier.
Represents non –oscillatory disturbance which penetrates the
potential barrier for indefinite distance.
Determination of constants:- To evaluate B,C in term of A , we use the following condition .
1.
must be continuous at boundary
( )
2.
( )
(
)
(
)
must be continuous at boundary
|
|
(
)
(
)
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(
)
Solving (23)and (24)
The solution is
Here incident wave function
Probability current density for incident wave
6
| |
7
Similarly we have for reflected
probability current density
| |
The reflection probability
| |
|
The transmitted wave
|
; probability current density ,
6
[
(
)
=0;
The trasmission coefficient in this case T=0;
Note,
7
(
)
]
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You can readily obtain the results in the 2
nd
case by replacing
.
Discussion:- case -1 i) Since
so the amplitude of the remitted ware is greater than
that of the incident wave. Also the de Broglie wave length is shorter in region1.
(ii) Since
T
hence the particle has some probability not being completely
transmitted , though its energy
. It contradict classical point of view
(iii) if
are exchanged R,T, remain unchanged i.e. we shall get the same result if
the particle incident on the step potential from the right .
Case-2
i) In this case we have
i.e. There is no absorption of the wave in the region (ii).
Actually the wave penetrating a small distance from the boundary into the region (ii) is
continually reflected till all incident energy is turned back in to the region I.
ii) Due to such reflection the amplitude of wave penetrating the region (ii) falls off
exponentially (this is similar to the total internal reflection of light).
3.ScatteringOverPotentialStep.cdf
3quantum-tunneling_en.jar
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POTENTIAL BARRIER OF FINITE HEIGHT AND WIDTH
Let us consider a particle of mass
and energy incident on a constant potential
barrier of height in the region
while the potential is zero everywhere else. The
nature of the potential and its mathematical from is shown below .
As the potential is independent of time we have to solve Schrodinger’s time
independent wave equation,
,
-
( )
to know the behaviour of the particle.
(i)For region (i)
( )
: Let
be the wave function here,
(i) Equation ( ) becomes
√
Solving
(ii) For region (ii)
ware for becomes
( )
; ( )
:- Let
be the wave function. Taking
(
)
the
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√
(
)
Solution this equation = 1
( )
( )
In region (iii)
Let
:
be the wave function here ,
Schrodinger equation
( )
Since there is no particle comes from right in this region
( )
( )
The coefficients can be determined from the boundary conditions,
i)
ii)
is continuous at
is continuous at
Hence boundary conditions at
and
and
;
( )
( ) gives
( )
( ) gives
( )
(
)
( )
Boundary conditions at
Adding (7) and (8)
( )
( ) gives C
( )
( ) gives
( )
( )
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(
)
( )
Subtracting (8) form ( 7)
(
)
(
)
Adding 9 and 10
6
7
[cosh (
)
sinh(
)]
(
)
Subtracting (10) form ( 9)
[sin (
)
cosh (
)]
[cosh(
)
sinh (
)]
(
From equation (5)
From equation (6)
[sin (
0cos (
)
)
cos (
sin (
)]
)1
Solving (13) and (14)
[cos (
)
[
(
) sin (
] sin (
)
)]
)
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Refection probability
.
| |
/ sinh (
)
cos
(
)
4
5 sinh (
cos
(
)
4
5 sin
)
Transmission probability
| |
sinh (
)
4
4
5 sinh (
5 sinh (
(
)
)
)
Putting values of
| |
(
)
sinh (
)
Quantum mechanical tunnelling :- From the above discussion, we get
there is a
finite probability of transmission of the particle through the potential barrier of height
and width ‘ , even if
. This cannot be explained by classical theory. This
phenomenon of transmission of a particle through a potential barriers of finite width and
height when its energy is less than the barrier height, is called quantum mechanical
tunnelling.
 It can be noted that transmission probability depends on
barrier( ) T rapidly decreases with increase of E and ,
[very large ‘ ’ and
]
and also width of the
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sin (
)
4
5
(
| |
(
)
sin
(
)
)
Explanation of alpha decay:- Heavy elements are generally unstable .In attempt to gain
stability they may undergoes spontaneous disintegration with emission of particles
(
nucleus )
alpha-decay_en.jar
The particles are held inside a nucleus by strong attractive short range nuclear forces.
However when they are outside, there exists long range coulomb repulsive forces .The
variation of potential with distance from nucleus is shown below.
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alpha-decay_en.jar
Now maximum potential energy due to coulomb repulsion at
radius
(
)
effective nuclear
[
]
Now
particles emitted from most of the radioactive dements have energy from (510) Mev, much lower than .
Thus according to classical mechanics it is difficult to understand how the a particles of
lower energy can go over a potential barrier of higher energy. According to quantum
mechanics, we know that tunnelling through the potential barrier is possible due to wave
property of the particle, incident on a rectangular potential barrier even though the incident
energy might be too low for transmission according to classical theory .
e.g.The transmission coefficient is given by
| |
(
)
√
(
)
For
If we assume that -particle moves back and forth almost freely inside nucleus
| then number of collisions mode by the particle
of
and with a velocity
with the wall in one second ,
probability of escape per second
mean half line
sec
min
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Since T decreases with
hence a small change in value of
causes large change in the
value of T and hence explains wide variation in half lives of active radio elements.
This also explain why the observer - emission half- lives vary over such wide limits
,
-
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One dimensional rectangular potential well.:A 1D rectangular potential well is shown below .It consists of the three regions (i),(ii),(iii)
as,
[
We consider a particle of mass and energy to be initially in the potential well
( ) ]. The 1D time independents Schrodinger equation for these regions are,
(
)
( )
( )
(
)
( )
The energy E of the particle is considered less then so that
all are real quantities.
Since the wave function must vanish at
, the well –behaved solutions of the above
equation .
( )
( )
( )
The boundary condition at
At
,
(
)
(
require that
)
and
must be continuous at boundary.
( )
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(
)
(
)
[
]
( )
( )
Again from boundary condition at
(
)
(
(
)
)
(
(
)
[
)
]
(
)
From the above equations,
(
)
Both the conditions shows,
(
For
,
)
from equation (5)
cos
(
)
Showing the wave function inside the well is symmetric *
For
,
)
( )+
from equation (5)
sin
Showing the wave function inside the well is anti-symmetric *
( )+
BoundStatesInASquarePotentialWell.cdf
bound-states_en.jar
(
(
)
(
)
Sanju 9681634157
Eigenvalues:
From equations (9) and (12) we have;
The
(
)
(
(
)
)
(
)
(
)
signs represents the symmetric and anti-symmetric wave functions respectively.
For the symmetric case ;
(
)
(
(
)
)
sin (
tan (
)
cos (
(
)
)
)
(
)
For the symmetric case we can have,
(
From the values of
and ,
(
Denoting
)
)
(
)
we rewrite the equations
tan ( ) .......................... (20) (Symmetric functions)
cot ( ) ........................ (21) (Anti-symmetric functions)
And
(
)
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The solutions of the above two pair of equations gives the permissible energy eigen values
for the symmetric and the anti-symmetric cases.
The equations can only be solved graphically for
. The points of intersections of the
graph for any of the above equations((20) or (21)) with (22) gives the permissible energy
values.
The solutions are sketched graphically for two cases
√
√
[
√
]
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Hence there are two circles, intersection of these with the plots of
tan ( )
cot ( )
gives the permissible values.
From the above cases we have;
√
2 solutions(symmetric , antisymmetric ) for
3 solutions(symmetric , antisymmetric, symmetric ,) for
The number of possible energy levels increases with value of √
√
; more precisely
with product of √
From the figure it follows that there will be ,
One bound state(S)
if
Two bound states (S,A)
if
Three bound state (S,A,S)
N bound states (S A S A
where
√
√
if
)
if (
)
bound-states_en.jar
NOTE 1.
In the limiting case
the circle’s radius r also becomes infinite, and hence there will
be infinite number of solutions.
Function
asymptotes
will cross the curve
.
√
tan ( )
cot ( ) at the
Sanju 9681634157
(
)
It is the solutions for 1d infinite potential well.
NOTE 2.
In the region
, becomes imaginary. Solutions
hence becomes sinusoidal [in
the regions (i) (ii)] also. It means probability density is distributed all over the space and
state becomes unbound.
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The Linear Harmonic Oscillator (1D) :Wave equation for the oscillator :- The time independent Schr ̈ dinger wave equation for the
linear motion of a particle along the x- axis,
( )
(
Or,
)
............... (1)
Where E is the total energy, V is the potential energy and
particle.
is the wave function for the
For a linear S.H.O. along the x- axis with angular frequency
proportional to the displacement the potential energy,
under a restoring force
................... (2)
Substituting value of V in (1)
.
Or,
.
/ =0
/
= 0............................. (3)
This is the Schrodinger wave equation for the oscillator.
To simplify the above equation, we introduce a dimension less independent variable y,
; Where a = √
Now
=
And
Substituting the values of
in equation (3);
+.
Or,
/
.
/
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Or,
.
Or,
(
Where λ
/
)
=0
= 0......................... (4)
.................... (5)
Solution :- To solve the above equation let us try first an asymptotic solution i.e. when
Equations (4)reduces to,
.......... (6)
Which has approximate solution,
....................................... (7)
[since
-
&,
.
/
(
)
(
Or,
(
)
)
= 0................................. (8)
For large values of y equation (8) is reduced to equation (6). This suggests an accurate
solution of the equation (6) must be in the form,
( )
..................... (9)
Where ( ) is a finite polynomial in y.
To solve the equation (4) let us change the equation into a different equation for dependent
variable ( ).
(
.
.
.
)
/
/
.
/
/
(
)
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( )
.
Or ,
Replacing
(
/
(
)
)
...................................(10)
,(n = 0,1,2 ,...............) Eq (10) becomes
2n
This is well known Hermites differential equation. The solution of the equation are called
Hermite’s Polynomials and are given by,
( )
(
)
............(11)
Eigen values :- For physical significance only those solutions of the Hermite eq (10) are
acceptable , for all values of for which ,
[ n = 0,1,2,........called quantum numbers]
Substituting ,
Or,
(
)
(
)
Or more generally,
=(n + )
...........................................(12)
From equation (12) we have the following conclusions,
1. The wave equations for the oscillator is satisfied only for discrete values of total
energies,
2. The lowest energy of the oscillator is obtained by putting n = 0 and it is,
This is called the ground state energy or the zero point energy of the harmonic oscillator.
(iii) The eigen values of the total energy depends only on one quantum number ‘n’ all
energy levels are hence non degenerate .
(iv) The successive energy levels are equally spaced; separation between two adjacent
energy level being
.
Wave functions :- For each value of ‘n’ there is a different wave function
general form ,
which have the
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( )
( )
( )
Or,
(
)
where a = √
is the normalization constant and is defined from the requirement,
Or,
( )
( )
∫
( )
∫
( )
,
-
( )
Or ∫
Using the relation of Harmait’s polynomial
( )
∫
( )
√
√
Or ,
=0
( )
√
0
1
√
( ).
1
First few Hermite polynomials corresponding to the eigen values and wave
functions are given below.
( )
n
0
1
( )
1
3
( )=2y
2
5
3
7
4
9
( )
( )
( )
( )
0
√
( )
( )
( )
( )
( )
[
[
√
√
1
] y
] (
√
√
[
0
[
( ).
1
] (
√
)
)
] (
)
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QuantizedSolutionsOfThe1DSchroedingerEquationForAHarmonicOsc.cdf
Correspondence with classical oscillator :- The classical probability of finding the particle
performing a linear S.H.M , within the distance between and
from its equilibrium
position is the ratio of the time ‘ ’ which particles takes to pass over distance dx in one
oscillation to the period of oscillation T;
i.e. Probability of finding it in dx,
( )
Now
(
=
=
)
(
(
√
√
√
)
)
(
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( )
i.e.
√
( )
√
............................... (1)
According to the quantum mechanics the probability – density,
( )
( )
( ) .................... (2)
For ground state (n =0),
( )
( )
. /
√
|
|
. /
√
This shows probability of finding particle is maximum at x=0 and decreases on either side
of x=0.
But equation (1) says the probability of finding the particle is maximum at
i.e. at
the end of the paths and minimum at
. Here there is contradiction in the two cases .
However for longer values of n the quantum mechanical probability density approaches to
classical probability density.
For n=0 and n =10 , n =50 the classical and quantum mechanical probability density are
shown below.
QuantumClassicalCorrespondenceForTheHarmonicOscillator.cdf
Sanju 9681634157
Algebraic method to solve S.H.O problem;
Raising and Lowering operators :-
Let us introduce two operators,
̂
̂
(
√
(
√
̂)
̂
̂)
̂
̂
̂
√
√
̂
̂
√
√
From the defination of ̂ ̂ we have the following conclusions ,
) ̂ ̂ are dimensionless operators.
) ̂ ̂ are not Hermitian but ( ̂ )
=̂
) ̂ ̂ satisfy the communication relation
,̂ ̂ Proof : The product gives ,
̂ ̂ = (√
̂
̂
̂
̂
̂ ̂
̂
,̂ ̂ -
̂̂
,̂ ̂ -
̂
√
) (√
̂
̂
√
̂
(̂ ̂
̂ ̂)
̂
(̂ ̂
̂ ̂)
̂
̂
̂ ̂=
................ (5)
)Hamiltonian operator ̂ can be represented by ̂ ̂ as,
̂
( ̂ ̂ + )..................................(3)
)
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̂
(̂ ̂
)..................................(4)
The product gives ,
̂ ̂
(√
̂
̂
√
̂
̂
) (√
̂
̂
(̂ ̂
√
)
̂ ̂)
̂
̂ =(̂
Similarly, ̂
̂+ )
(̂ ̂
)
a) Commutation relation of ̂ ̂ with ̂ ,
[ ̂ ̂-
,̂ ̂
*, ̂
,̂
̂- ̂
, ̂ ̂ ̂-
̂ , ̂ ̂{ ,̂ ̂ -
̂- ̂
̂
Similarly [ ̂ ̂ - =
̂-
+
......................................... (6)
̂
is the eigen function of ̂ belonging to the
Lowering operator and Zero point energy :- If
eigen value E of teh oscillator , we can write ,
̂
Now [ ̂ ̂Or, ̂ ( ̂ )
̂( ̂ )
Or, ̂ ( ̂ )
̂
Or, ̂ ( ̂ )
Equation (2) says ̂
(
(̂ ̂
̂ ̂)
̂
̂
̂
)( ̂ ).................... (2)
is an Eigen function of ̂ with the eigen value
The operator ̂ is called the lowering operator.
–
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Repeated application of ̂ to the eigen functions will finally lead to the state of lowest
energy , the ground state say
i.e. ̂
̂
(̂ ̂
)
This ground state energy
............... (3)
is called zero point energy.
Raisng operator and energy eigen values:
Starting from the relation [ ̂ ̂ (̂̂
̂ ̂)
̂( ̂
)
̂( ̂
)
̂( ̂
̂
; we have;
̂
̂ (̂ )
)
̂
̂
̂
(
)̂
̂ is an eigen function of ̂ with the eigen value
operator.
For the ground state
̂( ̂
̂( ̂
̂(
̂
. Hence ̂ is called raising
, from equation (3);
)
(
)̂
=.
/
=
̂
) =
(̂
) =
(
̂
)
)
is called 1st excited state.
is the
corresponding energy eigen value.
Thus proceeding in the similar way, operating on
̂ *( ̂ )
+ =.
/
(̂ )
n times with ̂ , we obtain,
̂
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=
Here
.
/
is the energy eigen value in the nth state wave function.
Eigen function:
Let
be the ground state wave function. Hence the lowering operator,
̂
(
√
̂)
̂
̂
̂
√
√
operating on it will produce zero.; i.e.
̂
4√
Let
√
;
̂
,
̂
√
.
5
/
Solving,
is a constant that can be calculated from normalizing condition.
i.e.
∫ |
with
; where
√
|
;
∫
( )
Therefore normalize wave function,
√
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(
√
)
.
To calculate 1st excited state, we operate ̂ on
̂
as;
(
)
(
)(
√
can be determined from the normalizing condition.
Hence repeated operation by ̂ on
/
)
from left will produce ;
(
(
)
)
Normalizing it we can get the nth state for the linear harmonic oscillator.
Sanju 9681634157
Operators for Angular Momentum: - In classical mechanics the angular momentum ⃗ of a
particle is defined by, ⃗ = × ,
Where the position vector of the particle relative to some arbitrary rigin is
linear momentum.
is its
The components of ⃗ ,
;
;
The Expressions for the linear momentum operators,
̂ =̂ ̂
̂ ̂
[
]
̂ = ̂ ̂
̂ ̂
[
]
̂ =̂ ̂
̂ ̂
[
In a spherical polar coordinate,
̂
( sin
cot
̂
(
cos
̂
(
)
]
cos
)
cot sin
)
 The operator for the square of the total angular momentum can be obtained from the
above expressions as,
̂
̂
̂
̂
0
.
/
1
Three dimensional motion in a central field :- For a particle of mass m , the time
independent Schr ̈ dinger equation in three dimensions ,
0
1
( )
( )
In spherical polar coordinate ,
0
.
/
.
/
1
(
)
(
)
(
)
The above equation can be solved by the method of separation of variable when ‘V’ the
potential is central one, i.e only depends on the distance from the origin. In other words
(⃗⃗⃗)
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The Hydrogen atom :- In the hydrogen atom the charge of the proton +e and that of electron
–e ,
Electrostatic potential energy of the system , V =
;
The proton being 1836 time heavier than the electron, it is considered to be stationary with
the electron in motion around it .
The time independent Schrodinger equation,
0
1
( )
Which is in spherical polar coordinate ,
0
Or,
.
.
/
.
/
.
1 ( )
/
/
]+
(
)
( )
To solve this equation, we use the method of separation of variables;
Putting,
(
)
( ) (
) and dividing by RY,
.
(
/
)
*
.
/
+ ............(3)
The left side of this equation is function of only; while the r.h.s. depends on θ ϕ, hence
both sides must be equal to a constant λ
Thus we get a radial equation.
.
(
/
)
..............(4)
And angular equation is given by,
.
/
λY
( )
Radial wave equation :- From equation (4) we have , after dividing by
.
Or,
+
Or,
+
(
/
0
)
(
)
0(
)
1
1
.....................................(6)
The above equation is the radical wave equation. It is equivalent to the motion in 1D with
effective potential,
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Angular wave equation :- The Angular equation (5) can be separated by putting ,
)
( ) ( ) equation ( )
Y(
Which gives ,
.
/
Dividing by QF,
.
Multiplying by sin
/
and re arranging ,
.
/
Taking both sides to be equal to a separation constant
...........................(6)
the
becomes ,
..........................................(7)
And the
- equation becomes,
.
Solution of
/
............................(8)
- equation :- The - equation.,
The solution of this equation,
( )
Since the wave function must be single valued , a change of
value,
Hence, (
)
( )
i.e.
which gives ,
(
)
‘
( )
by 2 must give the same
=
’ is known as magnetic quantum number
Physical significance:- The magnetic quantum number
is the measures of the zcomponenet of angular momentum.
The operator of the z- component of angular momentum,
̂
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̂
(
)
( ) ( ) ( )
=
(
)
=
(
)
=
Which shows
is the eigen value of ̂ for the wave function. Hence z – component of
angular momentum is quantized.
Solution of
The
Multiplying by
Putting
- equation(8)
.
/
.
/
,
.
/
( )
cos
Or ,
(
&
)
( ),
2(
Or, (
)
)
3
.
/
.
/
................................(10)
This is associated Legendre ‘s equation The series solution of this equation can be obtained
as a polynomial series by Frobenius method .
The condition for the series to be terminated,
(
) (k+
)
Where k,
are integers.
Let
is also the integer ;
Hence
(
),
is called azimuthal quantum number;
Since minimum values of k is 0 hence maximum value of values of , | |
The solution of equation (10) are associated Legendre Function l and order |
written as ,
| and
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(cos )
is normalizing constant.
ASSOCIATED LEGENDRE POLYNOMIALS
In mathematics, the associated Legendre polynomials are the canonical solutions of the general Legendre
equation
,
or equivalently
,
where the indices ℓ and m (which are integers) are referred to as the degree and order of the associated
Legendre polynomial respectively. This equation has nonzero solutions that are nonsingular on [−1, 1]
only if ℓ and m are integers with 0 ≤ m ≤ ℓ, or with trivially equivalent negative values. When in
addition m is even, the function is a polynomial. When m is zero and ℓ integer, these functions are
identical to the Legendre polynomials. In general, when ℓ and m are integers, the regular solutions are
sometimes called "associated Legendre polynomials", even though they are not polynomials when m is
odd.
The Legendre ordinary differential equation is frequently encountered in physics and other technical
fields. In particular, it occurs when solving Laplace's equation (and related partial differential equations)
in spherical coordinates. Associated Legendre polynomials play a vital role in the definition of spherical
harmonics.
Definition for non-negative integer parameters ℓ and m
These functions are denoted
, where the superscript indicates the order, and not a power of P.
Their most straightforward definition is in terms of derivatives of ordinaryLegendre polynomials (m ≥ 0)
The (−1)m factor in this formula is known as the Condon–Shortley phase. Some authors omit it. The
functions described by this equation satisfy the general Legendre differential equation with the
indicated values of the parameters ℓ and m follows by differentiating m times the Legendre equation
for Pℓ:[1]
Moreover, since by Rodrigues' formula,
the
can be expressed in the form
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similarly
This equations allows extension of the range of m to:
.
Note: the differentiable term has the highest order term
; hence it is at the best
times differentiable i.e.
Hence
Physical significance :- The square of the total angular momentum operator ,
̂
,
.
/
Operating on total wave function,
̂
=
=
The solution for.
Or ̂
0
.
0
/
.
0
1
/
.
1 using
/
1
From equation (9);
.
/
(
)
λ QRF
)
= (
This shows the wave function is an eigen function of ̂ with the eigen value (
where being orbital angular momentum number.
This measures the experimental values of square of orbital angular momentum.
The magnitude of angular momentum vector is ,
(
)
|⃗ |
)
√(
The radial equation :- The radial equation given by equation (6)
)
;
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(
+
Or ,
+
0
( ))
=0
.
/–
(
)
1
Solution of radial wave equation for ground state:The radial part of time – independent Schrodinger wave equation for the hydrogen atom
+
0
.
/–
(
)
1
.....................(1)
Where is the orbital angular momentum quantum number and other symbols have their
usual meanings.
For ground state
( )
+
.
/
...................(2)
We consider the simple solution of (2) consistent with all physical requirements,
( )
.
( )
+
0
1
Since R (r)
Or, .
/
.
L.H.S. of the above Equation is independent of
both sides of the above equation must be zero.
( )
/
and must be true for all values of
Hence
....(2)
(1st Bohr radius )
Combining (1) and (2) E =
The radial wave equation is the only equation that contains of a term involving the total
energy E. The Energy E can be positive or negative
Where E is positive the electron is unbound to the proton. In this case the solutions of the
wave equation give finite wave functions only for the total energies given by,
=
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Where n =1,2,3,... is called total quantum number or principal quantum number . This is
associated with total energy of the electron.
A summary of three quantum numbers
with their permissible values are as follows:
1) Total quantum number :2) Orbital angular momentum quantum number ,
HydrogenAtomRadialFunctions.cdf
3) Orbital magnetic quantum no.
Hydrogen atom wavefunctions.mp4
Compute the expectation value of r for the ground state of hydrogen
atom. The normalized ground wave function is
radius of the Bohr orbit.
Ans:-
( )
Normalising condition gives ,
( ) ( )
∫
∫
∫ . /
π
. /
( ) ∫
( )
Or,
√
( )
( )
, being the
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u (r) =
√
( )
( ) ( )
Now <r> = ∫
( ) ( )
∫
=
∫
( ) ( ) ∫
⌈( )
=
=
Most probable position of the eletron : The probability dp of locating the electron between
and
is,
,
Now ( )
√
. /
=
=
( )
Where
( )
Now for a maximum ,
( )
[
[
]
]
This equation , gives
i.e. Probability of finding the electron of the H- atom at the ground state is maximum at
r= ,
the 1st Bohr radius.
For normalized wave function of hydrogen atom for 1s state is,
(
)
.
/
is the Bohr radius.
Find the expectation value of potential energy of the electron 1s state.
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Now
(
∭
.
)
(
/
)
∭
∫
∫
(
(
∫
)
)
Note
 Find the expectation value of coulomb force on an electron in the above case
We know F =
<F>
∭
(
)(
∭ *√
+ (
)
+
∫
(
(
)
)
sin
∫
sin
∫
)
The unnormalized hydrogenic 2p state wave function for m = 0 is
Determine the normalization constant . also calculate <r>.
Let the wave function be,
(
)
is
cos
∭
Or, ∭
Or,
∫
( )
sin
∫
sin
∫
cos
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Or,
⌈( )
(
)
Or,
.
Or,
=1
=1
Or, A =
√
∭
∫
∫
⌈
⌈
sin
∫
∫
.
sin
∫
∫