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Transcript
```Module 12 : 2-D and 3-D Quantum Problems
I.
Particle In A 2-D Box With Infinite Walls
Consider a particle trapped in a 2-D infinite box as shown below:
U (x) = 
y=b
U (x) = 0
U (x) = 0
U (x) = 
y=0
x=0
x=L
U (x) = 0
The Time Independent Schrodinger Equation for a multi-dimensional problem is
given by
 2    k 2  where k 2 
2 m E  U 
.
2
Using the Cartesian form of the Laplacian, we have inside the box
2  2 
2mE
 2   k 2  where k 2  2
2
x
y

and that  is zero outside the box.
Since each term of the potential energy depends on at most one spatial coordinate
(in this case no coordinates), we can use separation of variables to simplify our
work. Thus, we choose a product solution for the wave function where X and Y
are each functions of only one spatial coordinate.
 Xx  Yy 
Substituting into the Time Independent Schrodinger Equation, we have
2 X
2 Y
Y 2  X 2   k 2 XY
x
y
1 2 X 1 2 Y
2mE
2



k


 constant
X x 2 Y y 2
2
Each term depends on different spatial coordinates which are independent. Since
the equation must hold for any point inside the box, each term must be a constant!
Furthermore these constants must have the dimensions of inverse square meters
(i.e. wave number squared). Thus, it makes sense to write the individual constants
in a manner similar to what we did with the 1-D well as follows:
1 2 X
2
  kx
2
X x
1 2 Y
2
ky
2
Y y
We also note the additional requirement:
k x  k y  k 2  2mE
2
2
This is just the dot product of a vector (wave number) upon itself where

k  k x î  k y ĵ .

k
ky
kx
From our work on the 1-D infinite square well, we know the general solutions to
these two equations are given by
X  A Sin k x x   B Cos k x x  where 0  x  L
Y  D Sin k y y   E Cos k y y  where 0  y  b
Thus, the energy wave function is
A D Sin k x x Sin k y y  0  x  L and 0  y  b
0
otherwise

 
The boundary condition that  is zero for all values of y when x = 0 requires that
B = 0. The boundary condition that  is zero for all values of x when y = 0
requires that E = 0. The boundary condition that that  is zero for all values of y
when x = L requires that k x L  n π where n = 1, 2, 3, … The boundary condition
that that  is zero for all values of x when y = b requires that k y b  q π where q
= 1, 2, 3, … By plugging the results for the components of the wave number into
our energy equations, we obtain the energy eigenvalues:



2 k x  k y
h2 kx  ky
E n,q 

2m
8π2 m
2
2
2
2

h2 n2π2 q2π2 
E n,q  2  2  2 
8π m  L
b 
h c2
n 2 q2 
n  1,2,3,..
E n,q 

where



8 E rest  L2 b 2 
q  1,2,3...
To find the normalization constants, we integrate over all space in the box:
b
L



 2

 2

2
1 A Sin k x x dx D Sin 2 k y y dy  .



 0

 0



Although only the product of the constants A and D has physical meaning, it is
often common to normalize each individual function X and Y independently. In
this manner, you can think of the problem as simply being the product two 1-D
infinite well problems from Module 10. Doing this gives us, the following:
A
1
L
 Sin
2

2
L

2
b
k x x dx
0
D
1
b
 Sin
0
2
k y y  dy
Thus, the energy eigenfunctions for this problem are
 2
 nπx  qπy
Sin 

 Sin 
 0  x  L and 0  y  b
 n,q   b L  L   b 

0
otherwise

with associated energy eigenvalues of
h c2
n 2 q2 
n  1,2,3,..
E n,q 

where



8 E rest  L2 b 2 
q  1,2,3...
Since this is a 2-dimensional problem, there must be at least two quantum
numbers which determine an energy state’s wave function and energy eigenvalue
(one for each spatial coordinate). This is an important concept and has led us to
find additional variables which effect a systems energy but which are not spatial
in nature like spin!! It is also possible for two different energy wave functions
(energy levels) to have the same energy eigenvalues. This is called degeneracy.
One way in which this might happen is if the system possesses some symmetry.
For instance if the box is a square with sides of length L then E1,3 and E3,1 state
would have the same energy as there is no difference to the system is you
exchange the x and y coordinates (rotate 90 degrees). This is called symmetrical
degeneracy. Many of the excited energy levels in the square system would be
degenerate. In other cases, degeneracy might occur just due to pure luck. This is
called accidental degeneracy. For instance, our original system is degenerate for
2 2
2 2
any group of (n,q) pairs where n b  q L  constant .
II.
Hydrogen Atom
A.
Coulomb Potential
To solve for any quantum problem, we first must find the potential energy
function for the problem. In the case of the hydrogen atom or any single electron
ion, the potential energy function is the Coulomb potential which can be written
in Cartesian coordinates as
U
 ke 2 Z
x 2  y2  z2
where ke 1.44 eV nm . Although it is technically correct to write the
Coulomb potential in Cartesian coordinates, it is a poor choice for solving the
Schrodinger equation since:
1)
We can’t use separation of variable when the potential has terms
containing more than one variable.
2)
The system and hence its boundary conditions has spherical symmetry.
2
Hence, we need to write the Coulomb potential using spherical coordinates as
U
 ke 2 Z
r
where Z = 1 for Hydrogen.
B.
Reduced Mass - 
In order to reduce the hydrogen atom from a two body problem to a one body
problem, we must replace the mass of the electron with the reduced mass (as
previously discussed in the module on the Bohr atom)
μ
mM
mM
where M is the mass of the proton and m is the mass of the electron.
C.
Laplacian in Spherical Coordinates
The Laplacian is much more complicated in coordinate systems other than
Cartesian, but it is necessary to use these coordinate systems for reasons
previously discussed. The Laplacian in various coordinate systems can be found
in math handbooks like CRC or Schaum’s. The Laplacian in spherical coordinate
systems is given as
1  2 
1
 
 
1
2
  2 r

 Sinθ   2 2
θ  r Sin θ 
r r  r  r 2Sinθ θ 
2
D.
Time Independent Schrodinger Equation
We can now put everything together to write the T.I.S.E. for the hydrogen atom.
1   2 Ψ 
1
 
Ψ 
1
 2Ψ
2μ 
ke 2 
Ψ
  2  E 
r

 Sinθ

θ  r 2Sin 2θ 
r 
r 2 r  r  r 2Sinθ θ 
 
We now write the wavefunction as a product of three single variable functions as
Ψ  R r Θθ Φ  .
Substituting into the T.I.S.E., we have
ΘΦ   2 R  RΦ  
Θ 
RΘ  2 Φ
2μ 
ke 2 
RΘΘ
  2  E 
r

 Sinθ

θ  r 2Sin 2 θ 
r 
r 2 r  r  r 2Sinθ θ 
 
1   2 R 
1
 
 
1
 2
2μ 
ke 2 

  2  E 
r

 Sinθ

θ  r 2Sin 2 θ 
r 
r 2 R r  r  r 2Sinθ θ 
 
We can isolate the phi dependence term by multiplying both sides by r Sin θ
and rearranging the terms.
2
2
Sin 2θ   2 R  Sin θ  
  1  2 
2μ r 2 Sin 2θ 
ke 2 


r

Sinθ



E






R r  r 
 θ 
θ   
r
2


Sin 2 θ   2 R  Sin θ  
  2μ r 2 Sin 2 θ 
ke 2 
1  2


r

Sin
θ

E








R r  r   θ 
θ 
r   
2

Because the left-hand side of the equation is independent of phi and must be equal
to the right-hand side for any value of phi, the equation must be equal to a
2
constant which we will denote by m l . Thus, we now have an ordinary
differential equation for phi and a partial differential equation involving theta and
r.
d 2Φ
2
  ml 
d
“Azimulthal Equation”
Sin 2 θ   2 R  Sin θ  
  2μ r 2 Sin 2 θ 
ke 2 
2
 E 
  m l
r

 Sin θ

2
R r  r   θ 
θ 
r 


We continue or process of isolating variables on the remaining partial differential
2
equation by dividing the equation by Sin θ to isolate theta.
1   2 R 
1
 
  2μ r 2
r

 Sin θ

R r  r   Sin θ θ 
θ   2
1   2 R  2μ r 2
r

R r  r 
2
2

ke 2  m l
 E 
 
2
r

 Sin θ
2

ke 2  m l
1
 
 
 E 
 

 Sin θ

2
r  Sin θ  Sin θ θ 
θ 

Since the left-hand side must equal the right-hand side of the equation regardless
of the values of r and theta, the equation must equal a constant which we will
denote by l (l  1) . This gives us our two final ordinary differential equations:
1 d  2 dR  2μ r 2
r

R dr  dr 
2

ke 2 
 E 
  l l 1
r


ml
1
d
d 

Sin
θ

  l l 1
dθ 
Sin 2θ  Sin θ dθ 
2
“Theta Equation”
E.
1.
Orbital Angular Momentum and Effective Potential
In order to obtain some physical understanding about the radial equation and to
put it into a form where we can find its solution in math handbooks, we are going
to do some algebraic rearrangement to the equation.
d2R
dR 2μ r 2
r
2r

dr
dr 2
2
2

ke 2  2 l l 1 
 E 
R 0

r
2μ r 2 

d 2 R 2 dR 2μ 
ke 2  2 l l 1 
R 0

 2  E 

r
dr 2 r dr
 
2μ r 2 
We also note that the potential of our electron for the Coulomb potential is
Ur   
k e2
r
Thus, we can rewrite the radial equation in a general form for any problem in
which the potential depends only on the radial variable and not just the Coulomb
potential as
d 2 R 2 dR 2μ 
 2 l l 1 

R 0.



 2 E U r 
dr 2 r dr
 
2μ r 2 
By dimensional analysis and our past knowledge of the Schrodinger equation in
one dimension, we know that the term in the bracket must represent a kinetic
energy due to motion in the radial direction.
K radial  K rotation  E  Ur 
Thus, the last term in the bracket must be kinetic energy due to rotation (i.e. due
to changes in theta and phi). Thus, we have the kinetic energy of rotation as
 2 l l  1
K rotation 
2μ r 2
We know from PHYS122 that the kinetic energy of rotation is given as
L2
K rotation 
2I
where L is the magnitude of the object’s orbital angular momentum and I is the
moment of inertia. For a point particle of mass , the moment of inertia is
I μ r2 .
Thus, our classical physics work gives us the rotational kinetic energy as
L2
.
K rotation
2μ r 2
Comparing our classical physics results with the result from the radial part of the
Schrodinger equation, we now observe relationship between the constant we used
to separate the theta and radial equations and the maginitude of the orbital angula
momentum.
L   l l  1
“Orbital Angular Momentum Eigenvalues”
If we again look at the radial equation at the top of the page, we see that the
rotational kinetic energy reduces the available energy for radial kinetic energy just
as if it was a potential energy. This is a common occurrence in a wide range of
physics problems involving radial forces including planetary motion. In these
cases, physicists often chose to talk about an “effective potential” which is the
sum of the actual potential due to forces and the kinetic energy due to rotation
which is called the centrifugal barrier potential or just the ”centrifugal barrier.”
L2  2 l l 1
U centifugal  
2I
2μ r 2
 2 l l  1
U effective  Ur  
2μ r 2
Thus, we can rewrite the radial equation in a more general form that is seen in
many different fields of physics as
d 2 R 2 dR 2μ

 2 E  U effective R  0 .
dr 2 r dr

As an example of how our work transcends various problems, you should note
that according to our formula the angular momentum of the moon can be thought
of as a centrifugal barrier (repulsive potential) which prevents the moon from
hitting the Earth. A similar argument explains why it is harder for a nucleus to
undergo alpha decay between two nuclear states of different angular momentum.
In alpha decay, the alpha particle must tunnel through the nuclear potential. In
order to conserve angular momentum in decay, the difference in the angular
momentum of two nuclear states must be offset by the angular momentum of the
emitted alpha particle. Because the effective potential is raised when the alpha
particle has angular momentum, the probability for tunneling is greatly reduced.
2.
Solution for the Coulomb Potential
We now wish to change our approach from looking at the general nature of the
radial equation to solving it for our specific case of the Coulomb potential.
d 2 R 2 dR 2μ 
k e 2  2 l l 1 
R 0

 2  E 

r
dr 2 r dr
 
2μ r 2 
a.
Asymptotic Solution for Large Radial Distances
For large r, the differential equation is approximately,
d 2 R 2μ E

R 0
dr 2
2
d 2R
2μ E
 2 R
2
dr

2μ E
d2R
2

C
R
where
C

since E  0

dr 2
This is an equation which we have solved several times previously. Since the
wave function must stay finite at large values of r, the only acceptable solution is
the exponential decay solution.
R r  e
b.

2μ E
r

for large r
Ground State Wavefunction
Let us see if our asymptotic wavefunction is a possible solution to the hydrogen
atom for all radial locations in some cases. We do this by using a trial
wavefunction based upon our asymptotic results and plugging it into the full
radial equation. Let us define the radial wave function in terms of a constant delta
as
R Ae

r
δ
and find the necessary radial derivatives as
dR
1
 R
dr

d2R 1
 R.
dr 2  2
Plugging our results into the radial Schrodinger equation, we have
r2
2r
2μ
  2
2



 2
 2 l l 1 
2
 E r  k e r 
 0
2μ 

2
2
 2 μ E 1  2  2μ k e
  r   l l  1  0
 2  2  r  
2

 
 
 
In order for the polynomial to be true for all values of r, the coefficients for each
term must be zero!! This gives us three conditions that must be met for the trial
function to be a solution for the hydrogen atom. The 2nd order coefficient gives us
 2μ E 1 
 2  2 0
 
 
2
E
2μ 2
The linear coefficient gives us
 2μ k e 2 2 

   0
2



μ k e2 
 .


Solving for delta, we see that the constant is actually the first Bohr radius.
2

 a o  0.529 x10 10 m
2
μke
By combining results for the linear and 2nd order coefficients, we can determine
the energy eigenvalue associated with our trial wavefunction.
1
E
2μ
2
 μ k e2

 

μ 2 k 2 e4
  
  13.6 eV
2
2


Thus, we see that our trial wavefunction will be a solution of the hydrogen
atom only if the energy is the ground state energy of the hydrogen atom.
We know look at the constant term.
l l  1  0
This term will only be zero if l  0 or l   1 . We will see later that the theta
equation restricts l to non-negative values so we have found that l  0 . This
implies that the electron in the ground state has zero angular momentum
since  00  1  0 in contradiction to the Bohr atom. The ground state
electron actually has some intrinsic angular momentum. The Schrodinger
equation doesn’t correctly predict this. Instead, we would need to use the
Dirac equation from relativistic quantum mechanics.
R r A e

r
ao
with l  0 and E   13.6 eV
“Radial Wavefunction for Ground State of Hydrogen”
c.
Asymptotic Solution for Small Radial Distances
Let us start by rewriting our radial differential equation as
d 2R
dR 2μ  2
 2 l l 1 
2
R 0 .
r
2r
 2  E r  ke r 
dr
2μ 
dr 2
 
2
For small values of r, we can simplify the differential equation as
d2R
dR
r
2r
 l l 1R  0 .
2
dr
dr
2
One of the important things that we see when we write the equation in this form is
that for small radius values the radial function is primarily effected by angular
quantum number l and not the wavefunctions energy eigenvalue E!!
Because the electron must have a zero probability of being found at r = 0, it is
reasonable to choose a power solution for R as our test wavefunction of the form
R r  A r c where c is a constant.
Thus, we can now compute the required derivatives as
dR
R
 c A r c-1  c
dr
r
d 2R
R
c- 2





c
c

1
A
r

c
c
1
dr 2
r2
Plugging our results into the Schrodinger equation, we find that
r 2 c c  1
R
R
 2 r c  l l 1R  0
2
r
r
 c c  1 2 c  l l 1R  0
c
2

 c  2 c  l l  1 R  0
c
2

 c  l l 1 R  0
 cc  1  l l  1 R  0
In order for this to be true at all values of small r, the coefficient must be zero.
Thus, we have shown that c  l . This tells us that the radial wavefunction near the
nucleus depends on the electron’s angular momentum!!
R r  A r l
It is important that you realize how much information can be obtained
without actually solving the full radial differential equation. This approach
to problem solving is an important tool in your future work!!
d.
Solving the Radial Schrodinger Equation in General
Eisberg and Resnick shows in Appendix N how to solve the general radial
equation. I will leave the details for your reading, but I do want to point out some
important facts that emerge from the process.
i)
Each solution has a polynomial whose form depends on both
the orbital angular quantum number and a second constant n
which is called the “principle quantum number.”
ii)
The constant n must be a non-zero positive integer and must be
larger than the orbital angular quantum number in order for
the radial equation to have a finite solution.
iii)
The energy eigenvalue associated with each radial
wavefunction depends only on principle quantum number and
has the value E n  
13.6
eV .
n2
The general form of the radial wavefunctions found by actually solving the radial
Schrodinger equation can written in terms of the product of our asymptotic
solutions and an unknown function. The general form of the radial wavefunctions
in terms of the ratio of the radial distance to the Bohr radius is given by
l
 r 
 r   n a 
R r   H (r/a o )   e  o 
ao 
l
n
where H(r/ao) is a polynomial whose form depends on two constants n and l .
A few of the radial wavefunctions are provided below. For more functions look at
Table 8.4 on page 280 of Serway/Moses/Moyer 3rd Edition as
n
R nl r 
l
3
2
1 0
r

1
ao
  2e
 ao 
3
2
r
 1   r   2ao
2 0 
  2 - e
 2ao   a o 
2 1
 1 


 2ao 
3
2
r

r
2ao
e
3ao
F.
Azimuthal (Phi) Equation
d 2Φ
2
  ml 
d
We have seen this equation previously. In order for this equation to stay finite as
we vary phi from minus infinity to positive infinity, it is necessary that m l  0
2
(i.e. m l must be real). Thus, the solution is of the form of a complex exponential
  A e i ml  .
Since we also require that the wave function be single valued at any point in
space, the argument for the complex exponent must repeat anytime we increase or
decrease phi by an integer multiple of two pi. Evaluating the function at phi equal
zero gives us
   0  A e i 0  A     2 
   2 π   A ei ml 2  A
e i ml 2 1
m l 2 π  (integer) 2π
m l  ......  2,  1, 0,1, 2,.....
We can determine the normalization constant by solving the equation
2π
Φ
2
d 1
0
2π
A
0
2
e -ml e ml d  1
2π
A
2
 d 1
0
A 2  1
2
A

1
2π
1 i ml 
e
2π
“Normalized Phi Wavefunction”
G.
Theta Equation
We know return to the theta equation which can be written as
2
1 d
d  
ml 
 Sin θ
  l l 1
  0 .
Sin θ dθ 
dθ  
Sin 2 θ 
The solutions to this equation will depend on two quantum numbers, l and m l .
To find the solution to this equation, we define a new variable, x, as x  Cos θ .
We also find that
dx
  Sin θ . Using these results and the chain rule of
dθ
Calculus, we can rewrite the theta equation as
2
1 dx d 
dx d  
ml 
 Sin θ
  l l 1
  0
Sin θ dθ dx 
dθ dx  
Sin 2 θ 
2

- Sin θ d 
ml 
2 d 
 - Sin θ
  l l 1
  0 .
Sin θ dx 
dx  
Sin 2θ 
2
d 
d  
ml
2
  - Sin θ
  l l  1 
dx 
dx  
Sin 2θ

  0

We now use the fact that Sin θ 1 Cos θ 1 x to change the equation to
2
2
2
2
d  2
d  
ml
  x 1   l l 1 
dx 
dx  
1-x2
2

d 
ml
2 d 
 1 - x    l l  1 
dx 
dx  
1- x2

  0

2
d 2
d 
ml
1- x  dx 2  2 x dx  l l 1 1 - x 2

2

  0


  0

This equation is known as Legendre’s Associated Differential Equation (see
Eq. 28.49 page 165 of the Schaum’s Math handbook ). The solutions of this
equation are well know special functions of engineering and physics (See
Mathematical Methods for Physicists by Arfkin or similar textbooks). Eisberg and
Resnick also go through the general solution technique in Appendix N.
The general results for the solution of Legendre’s Associated Differential
Equation are the following:
i)
ii)
The equation only has finite solutions if l is a non-negative integer
such that l  m l .
The wavefunctions are normalized Associated Legendre
ml
Polynomials, Pl x  , which are defined by the quantum number
l and m l .
2l 1 l  m l ! ml
P x 
2 l  m l ! l
θ lml 
H.
Spherical Harmonics - Yl θ, 
m
We can now combine our results for theta and phi to produce another famous set
of special functions called spherical harmonics.
Ylml θ,    lml  ml 
2l  1 l  m l ! ml
Pl Cosθ  eiml
4π l  m l !
Spherical harmonics are tabulated in Table 8.3 on page 269 of
Serway/Moses/Moyer 3rd Edition. You may also obtain them using relationship
28.50 for the Associated Legendre Polynomial on page 165 in the Schaum’s Math
Handbook:
1  x 

2
ml
l
P
ml
2
2l l !
l
d ml l 2


x

1
dx ml l
0
Example: Find Y1
Solution:
0
First find P1
1  x 

2 0
0
1
P
d 2
x  1
2 (1!) dx
P10 
1 d 2
2x

x  1
2 dx
2
P10  x  Cosθ
Now plug your result into the relationship for the spherical harmonics.
I.
Y10 θ,   
2(1)  1 1  0! 0 i (0)
P e
4π 1  0! 1
Y10 θ,  
3
1 3
Cosθ 
Cosθ
4π
2 π
Z-Component of the Angular Momentum
We know from classical physics that orbital angular momentum is defined by the
equation
  
L  r p .
From the definition, the x-component of the angular momentum is found to be
Lz  x p y - y px .
If we now use the operators for position and linear momentum that we discovered
when developing the Schrodinger equation several modules ago, we have
L̂ z   i  x



 

i  y i   x  y  .
y
x
y
x 

However, we can use the chain rule of Calculus and the relationships between
Cartesian coordinates and spherical coordinates (see Eisberg and Resnick
Appendix M) to rewrite the z-component of the angular momentum’s operator as
L̂ z   i 


Applying the operator to our spherical harmonics, we find that
L̂ z Ylml θ,  
2l 1 l  m l ! ml
 iml 

Pl Cosθ   i 
e 
4π l  m l !



L̂ z Ylml θ,     i  i ml 
2l  1 l  ml ! ml
P Cosθ  eiml
4π l  ml ! l
L̂ z Ylml θ,    m l  Ylml θ,   .
Thus, we see that the spherical harmonics are eigenfunctions for the z-component
of the angular momentum operator with eigenvalues of m l  . Another interesting
fact that comes out of this study is that there is an uncertainty relationship
between the various components of the orbital angular momentum. You can know
with perfect precision the magnitude of the angular momentum and the projection
of the angular momentum upon one axis, but not its other two components!!
We summarize these results as follows:
i)
The eigenfunctions for both angular momentum ands the zcomponent of angular momentum are the spherical harmonics.
ii)
The z-component of the orbital angular momentum is determined
by the magnetic quantum number, m l , by the relationship
Lz  ml 
iii)
The location of the angular momentum vector is quantized in space
due to the relationships that we have developed. It traces out a
cone in the x-y plane at an angle with respect to the z-axis given by
Cosθ 
Lz
ml

L
l l  1
z
ml

 l l  1
y
x
J.
Energy Degeneracy
We found in our solution of the hydrogen atom that the energy depended
only upon the principle quantum number, n, and not upon the orbital
angular momentum quantum number, l , or the magnetic quantum
number, m l . Thus, the system has two types of degeneracy!!
The first degeneracy which is related to the energy being independent of the
magnetic quantum number is due to the azimuthal symmetry of the systems. Since
we could have chosen any direction in space for the z-axis, our energy results are
independent of any phi dependence. This degeneracy is removed if the symmetry
is broken by a non-spherical potential (such as a magnetic field) being applied to
the system. This removal of degeneracy can be seen by additional lines in the
emission spectra which originate when an external magnetic field is applied
(Zeeman Effect). In fact even when no external field is applied, very fine line
splitting is observed in precise experiments because the hydrogen atom does have
a small internal magnetic field being applied to the electron due to coupling
between the electron’s intrinsic angular momentum (spin) and its orbital angular
momentum (fine structure) as well as magnetic effect due to the spin of the proton
(hyper-fine structure). We will take up the effects of magnetism and the evidence
for intrinsic angular momentum in the next module.
The second type of degeneracy has to due with the orbital angular momentum
quantum number. This type of degeneracy is due to the 1/r nature of the Coulomb
potential. For other quantum problems including multi-electron atoms where
electrons make the potential non-central or for systems involving potentials
which are radial but not 1/r as in the case of Van der Waals forces, this energy
does depend on the angular quantum number.
III.
Calculating Probabilities in 3-D
The basic method for calculating the probability that a particle is in some 3-D
region of space is the same as in Module 3 except that the integral is now over a
P

2
 dv .
space
To be able to evaluate the integral in 3-D, we must first write an infinitesimal
volume of space in spherical coordinates. To determine the formula for a
differential volume element, we incrementally adjusting each coordinate
independently to obtain a differential length, dS. These three differential lengths
form the sides of cube whose differential volume is obtained by
dv  dS1 dS 2 dS3
A.
Cartesian Coordinates
For Cartesian coordinates, we find that the three differential lengths are
dS1  dx , dS 2  dy , and dS3  dz
with the differential volume given by
dv  dS1 dS 2 dS3  dx dy dz .
Thus, the probability for finding the particle in a region of space where
x 1  x  x 2 , y 1  y  y 2 , and z 1  z  z 2 is given by
x 2 y2 z2
P
     dz dy dx .
*
x1 y1 z1
If the wavefunction has been written as a product of single variable wavefunctions
as in the case of the particle in a box (i.e. Ψ  XYZ ), the probability equation
can be written in the form
y2
x2

z2


P  X dx Y dy Z dz
2
x1
B.
2
y1
2
z1
Spherical Coordinates
In spherical coordinates, we find that the three differential lengths are
dS1  dr , dS 2  r dθ , and dS3  r Sinθ d .
with the differential volume given by
dv  dS1 dS2 dS3  r 2 Sinθ dr dθ d .
You should draw a picture of a spherical coordinate system and work
through this Calculus until you can quickly derive this result whenever you
need it.
Thus, the probability for finding the particle in a region of space where
r1  r  r2 , θ 1  θ  θ 2 , and  1     2 is given by
θ2  2
2
r2

P r R
2
l
n
dr

2
Ylml Sinθ dθ d
θ1  1
r1
Another common thing to do with this result is to use the fact that
dCosθ    Sinθ
and rewrite the equation as

P r R
r1
Cosθ1   2
2
r2
2
l
n
dr
  
Ylml
2
dCosθ  d .
Cos θ 2  1
In many problems, we are interested in the probability that the particle is any
where in some radial region of space. In this case our integrations across theta and
phi are all possible values (i.e. 0    2  and 0  θ   ). In this case, our
previous work gives us

1   2
2
r2
P r R
2
l
n
dr

Ylml
2
dCosθ d .
-1  0
r1
Since the spherical harmonics are normalized wavefunctions, we have
1  2

Ylml
2
dCosθ  d  1 .
-1  0
Therefore, the radial probability for finding a particle at a distance between r1 and
r2 is
2
r2

P  r 2 R ln dr .
r1
You should recall that in Module 3 we defined the probability density function by
the equation
x2

P  Ρ dx .
x1
By comparing our radial probability equation to our work on Module 3, we define
the radial probability density function as
2
P  r 2 R ln .
It is important to realize that the radial probability density can go to zero
when r goes to zero even though the radial wavefunction doesn’t (consider
our ground state work earlier). The probability of finding the electron
depends not only on its probability of occupying a specific location
(determined by the square of the radial wavefunction) but also upon the
number of available locations (determined by the size of the sphere through
r2).
To find the values for r where the probability density is a maximum or minimum
(i.e. most probable and least probable radial positions for finding the particle),
you can either plot the probability density function using Excel, Mathcad, Maple,
etc. or you take advantage of your knowledge of Calculus I and set the derivative
of the radial probability density with respect to r equal to zero and solve for the
roots.
d
P 0
dr
“Condition for Maximum or Minimum Radial Probability”
You know that the probability density is zero r = 0 and r = . For other possible
results, you can plug your root back in to the probability density equation to see if
it is zero or a maximum.
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