Transcript Of the states with n = 2?

Hydrogen Atom in
Wave Mechanics
The good news is that the Schroedinger equation for the hydrogen atom
has an EXACT ANALYTICAL solution! (this is one of the few problems in
Quantum Mechanics that does have such a solution – most problems in
QM cannot be solved exactly).
The bad news, however, is that the procedure of solving the equation
is extremely complicated. In a standard QM textbook it usually spans
over about forty pages loaded with math. In order to clearly understand
the solution procedure, one has to be familiar with elements of highly
advanced calculus, such as functions called the “spherical harmonics”
and things called the “Laguerre polynomials”. This is much beyond the
mathematical preparation level required for the Ph314 course. Therefore,
we will not go through the details of the solution procedure – we will only
review the general properties of the solutions, and you will be asked to
“accept without proof” a number of things.
Schroedinger Equation in three dimensions
 2   2  2  2
 2  2  2

2m  x
y
z
U ( x, y , z )  

  U ( x, y, z )  E

1
e2
40
x2  y2  z 2
The equation cannot be solved analytically in Cartesian coordinates.
However, if we use spherical polar coordinates, the equation becomes more complicated, but solvable (although it’s still not an easy
task!):
Now,    (r ,  ,  ), and
 2   2 2 
1
 
 
1
 2 

 2
 U (r , ,  )  E
 sin 
 2 2
 2 
2 
2m  r
r r r sin   
  r sin   
This equation has a separable solution t hat can be factored as :
 (r , ,  )  R(r )  ( )  ( )
The solutions of the Schroedinger Equation for the hydrogen atom
contains the variables r, , and , and fundamental constants, such
as the the electron mass and charge, the Planck Constant, speed of
light, and ε0 . In addition, three parameters emerge in a natural way
as “indices” or “labels” for the solutions. They are listed below:
n  the principal quantum number, which can take values of
n  1,2,3 . It is essentiall y the same as n in the Bohr model.
l  the angular momentum quantum number. Its allowed
val ues are l  0, 1, 2, 3,  , n  1.
ml  the magnetic quantum number. Its allowed values are
ml  0,  1,  2,  3,  ,  l.
Complete with these quantum numbers, the separated solutions of
the Schroedinger Equation can be wrtiien as:
 n,l ,m (r , ,  )  Rn,l (r )l ,m ( ) m ( )
l
l
l
The principal quantum number n specifies the energy level, just
as in the Bohr model. Te quantized energy levels obtained by
solving the Schroedinger Equation are given by exactly the same
formula as in the Bohr model:
me4
1
13.6 eV
En  
 2 
2 2 2
32  0  n
n2
Each quantum state of the electron in the Wave Mechanics Model
of the hydrogen atom can be described by a “triad” of integers:
(n, l, ml ). The energy of the state depends only on the first of these
numbers, but otherwise they are all different quantum states.
All states for n=1, 2, and 3 are shown below:
-1.5 eV
(3, 0, 0)
(3, 1, 1)
(3, 1, 0)
(3, 1, -1)
(2, 0, 0)
(2, 1, 1)
(2, 1, 0)
(2, 1, -1)
(3, 2, 2)
(3, 2, 1)
(3, 2, 0)
(3, 2, -1)
(3, 2, -2)
-3.4 eV
-13.6eV
(1, 0, 0)
Remember the notion of DEGENERACY we talked about
some time ago? What is the degeneracy of the n = 1 state?
Of the states with n = 2? Of the states with n = 3?
Radial probability densities
First, let’s shortly recapitulate what we told about probability density.
Recently, we talked about the probability of finding a particle in the
x region between x and x + dx in a 1-D quantum well. This probability
Is related to the particle-in-the-well wavefunction as:
P( x)dx   ( x) dx
2
In other words, the squared modulus of the wave function gives
the probability density.
In the case of a 2-D quantum well, we are interested in the probability of finding the particle in a “box”, or surface area element
defined by regions (x, x + dx) and (y, y + dy) . This probability is given
by:
2
P( x, y )dxdy   ( x, y ) dxdy
Again, the same rule is valid – the squared modulus of the wave
function is the probability density.
In order to find the probabilit y that the particle in a 1 - D well
can be found in a region of finite width, say, between certain
points x1 and x2 , one has to perform an integratio n :
x1
P( x1  x  x2 )    ( x) dx
2
x1
In the case of a 2 - D quantum well, in order to find the
probabilit y that the particle can be found in a finite  size
surface area region defined by points x1 and x2 , and
y1 and y2 , one hat to perform a double integratio n :
P( x1  x  x2 , y1  y  y2 ) 
y 2 x1
   ( x, y )
y1 x1
2
dxdy
In the case of the hydrogen atom the
“philosophy” is essentially the same.
The squared modulus of the wavefun2
ction:
 (r , ,  )
gives the probability density of finding
the electron at the location ( r ,  ,  ).
To compute the actual probability of
Finding the electron in a volume element
dV located at ( r ,  ,  ),
we multiply the probability density by
this volume element. In spherical polar
coordinates the volume element is:
dV  r 2 sin  drdd
A moment ago we said that the solution of the Schroedinger Equation
For the hydrogen atom can be separated into three functions, each
one depending only on a single variable – which can be written as:
 n,l ,m (r , ,  )  Rn,l (r )l ,m ( ) m ( )
l
l
l
Repeated from the preceding page :
 n,l ,m (r , ,  )  Rn,l (r )l ,m ( ) m ( ) and dV  r 2 sin  drdd
l
l
l
Therefore, the probability is:
2
2
2
2
 n ,l ,m (r , ,  ) dV  Rn,l (r ) l ,m ( )  m ( ) r 2 sin  drdd
l
l
l
Using this expression, one can calculate many “goodies”, or many
features of the hydrogen atom.
However, such calculations are not trivial, and we will use
them to determine only a single function of interest – namely, the
so-called radial probability density.
Imagine that the nucleus (proton) is
surrounded by a thin spherical “shell”
of radius r and thickness dr .
We want to determine the probability
of finding the electron within the volume
of this shell.
r
Nucleus
dr
Repeated from the last slide:
2
2
2
2
 n ,l ,m (r , ,  ) dV  Rn,l (r ) l ,m ( )  m ( ) r 2 sin  drdd
l
l
l
In order to find the probability in question, we have to integrate over
the other two variables,  and  :

2
P(r ) dr  r Rn,l (r ) dr  l ,ml ( ) sin  d   ml ( ) d
2
2
2
0
2
0
This formula looks scary, right? Fortunately, however, it is not scary
at all, because in the separated wavefunction each function is individually normalized, so that both integrals are equal to unity! Hence, the
radial probability density takes a simple form:
P(r )  r Rn,l (r )
2
2
and the probabilit y of finding the electron within the shell volune is :
2
P(r )dr  r Rn ,l (r ) dr
2
First few radial wave functions Rnl (r )
of the hydrogen atom :
The radial probabilit y density functions
for the three lowest states of hydrogen :
P(r )  Rnl (r )
2
Practical example:
(from the final exam in the Fall 2006 Ph314 course).
Since the total wave function of hydrogen is a three-dimensional
function, plotting the overall probability density function is a much
more complicated task than plotting the radial probability. But the
pictures shown here can give you a pretty good idea of how these
Functions look like for various quantum states of the electron.
Imagine each plot
to be rotated by
360° about the
vertical axis (z).
Another possible
way of illustrating
how the 3-D probability density functions look like is to
plot the surfaces of
constant |ψ(r, , )|2
We already know what role the quantum number n plays:
it describes the energy of a given state, and the mathematical form of this energy is identical as in the Bohr
model.
But what do the other two quantum numbers, l and ml ,
tell us? What is their physical meaning?
The first of these numbers is related to the electron’s
angular momentum. Namely, the length of the electron’s
angular momentum vector is given by:

L  l (l  1) 
(sorry, this is another mathematical
equation I have to ask you to
“accept without proof”).
For instance, in a state with l  1, the electron' s angular momentum


vector length is L  2 ; for l  3, L  3(3  1)   12 
As was said, for states with the quantum number n, l can take
values of l = 0, 1, 2, … n-1. As follows from the above, there are
states of the the same energy, but with different angular momenta.
This is not a surprising situation in physics. There is a classical
analog in the planetary motion:
There are many orbits of the same energy, but with different
ellipticity. The circular orbit corresponds to the largest angular
momentum. With increasing ellipticity,
the angular momentum
decreases.
Lower ang. momentun
than for circular orbit
Circular orbit:
Maximum L
Lowest ang. momentum
of the three orbits shown
A comet:
Very elongated orbit
At this part of the
orbit, the comet’s
motion is very
slow.
Comets, which are celestial bodies with extremely elongated orbits,
spend most time both close to the Sun, or far away from the Sun,
in the peripheral region of the Solar System.
Corresponding
situation in hydrogen
atom
The physical meaning of the ml quantum number: it’s the length of
the projection of the angular momentum on the z axis.
The projection can only take
values that are integer multiples
of ℏ.

Example : L  l (l  1) 
What' s the maximum possible
value of ml ?

ml  L (projectio n cannot be
longer tha n the vector itself! )
ml2  (l 2  l ), so ml  l is OK.
But could ml be equal to l  1?
Let' s check :
(l  1) 2  l 2  2l  1  l 2  l
The projection would be longer
than the vector, so such a situa tion is not possible.