Chapter 7
Quantum Two Body Problem,
Hydrogen Atom
7.B.2
The two-body problem
• The two-body problem: two spinless point objects in
3D interacting with each other (closed system)
• Interaction between the objects depends only on the
distance between them
• The operators describing their positions and
momenta satisfy these commutation relations:
[ Ri , Pj ]  i   ij
[ Ri , Rj ]  [ Pi , Pj ]  0
 ,   1,2


P  iˆPx  ˆjPy  kˆPz
R  iˆX   ˆjY  kˆZ
7.B.2
The two-body problem
• The Hamiltonian of the system:
2
2
 
P1
P2
H

 V R1  R2
2m1 2m2


• Let us introduce
operators:



m1 R1  m2 R2
RG 
m1  m2

 
PG  P1  P2
  
R  R1  R2


 m2 P1  m1 P2
P
m1  m2
• What commutation relations do these operators
satisfy?
7.B.2
The two-body problem
• Let us calculate this commutator:
[ X G , PGx ]
 m1 X 1  m2 X 2
 m1[ X 1 , P1x ]  m2 [ X 2 , P2 x ]

, P1x  P2 x  
m1  m2
 m1  m2

m1i  m2i
 i  [YG , PGy ]  [ Z G , PGz ]

m1  m2


m
P

m
P
2
1
x
1
2
x
• Similarly: [ X , Px ]   X 1  X 2 ,

m1  m2 

m2 [ X 1 , P1x ]  m1[ X 2 , P2 x ] m2i  m1i

 i

m1  m2
m1  m2
 [Y , Py ]  [ Z , Pz ]
7.B.2
The two-body problem
• On the other hand:
 m1 X 1  m2 X 2 m2 P1x  m1P2 x 
[ X G , Px ]  
,

m1  m2 
 m1  m2
m1m2 [ X 1 , P1x ]  m1m2 [ X 2 , P2 x ] m1m2i  m1m2i


0
2
2
(m1  m2 )
(m1  m2 )
 [YG , Py ]  [ Z G , Pz ]
• Similarly:
[ X , PGx ]  X 1  X 2 , P1x  P2 x 
 [ X 1 , P1x ]  [ X 2 , P2 x ]  i  i  0  [Y , Py ]  [ Z , Pz ]
7.B.2
The two-body problem
• Thereby, after transformation we have operators of
positions and momenta of two fictitious particles
• How do the operators for real particles depend on
the operators for fictitious particles?


 
PG  P1  P2



m1PG  m1P1  m1P2
+
 m2 P1  m1 P2
P
m1  m2



(m1  m2 ) P  m2 P1  m1 P2






m1PG  (m1  m2 ) P  m1P1  m1P2  m2 P1  m1P2

 
m1
PG  P  P1
m1  m2
7.B.2
The two-body problem
• Thereby, after transformation we have operators of
positions and momenta of two fictitious particles
• How do the operators for real particles depend on
the operators for fictitious particles?


 
PG  P1  P2
 m2 P1  m1 P2
P
m1  m2



(m1  m2 ) P  m2 P1  m1 P2



 m2 PG  m2 P1  m2 P2 +






 m2 PG  (m1  m2 ) P  m2 P1  m2 P2  m2 P1  m1P2
 
m2 
PG  P  P2
m1  m2
7.B.2
The two-body problem
• Let us now rewritethe Hamiltonian


 
P12
P22
H

 V R1  R2
2m1 2m2
2
2
   m2 

 m1 

PG  P  
PG  P 
m1  m2
m1  m2



 V


2m1
2m2

 
2
2 2
m1 PG
2m1 PG  P
P



2
2m1 (m1  m2 )
2m1 2m1 (m1  m2 )

 
2
2 2
m2 PG
2m2 PG  P
P



V
2
2m2 (m1  m2 )
2m2 2m2 (m1  m2 )
 

R
 

R
7.B.2
The two-body problem
• Let us now rewrite the Hamiltonian
2
2

(m1  m2 ) PG (m1  m2 ) P
M  m1  m2



V
R
2
2(m1  m2 )
2m1m2
m1m2

2
2
m1  m2

PG
P
H

V R
2M 2

 
2
2 2
m1 PG
2m1 PG  P
P



2
2m1 (m1  m2 )
2m1 2m1 (m1  m2 )

 
2
2 2

m2 PG
2m2 PG  P
P



V R
2
2m2 (m1  m2 )
2m2 2m2 (m1  m2 )
 
 
 
7.B.2
The two-body problem
• Let us now rewrite the Hamiltonian
2
2

(m1  m2 ) PG (m1  m2 ) P


V R
2
2(m1  m2 )
2m1m2
2
2

PG
P
H

V R
2M 2
 
 
M  m1  m2
m1m2

m1  m2
• M: total mass, μ: reduced mass
• We converted our Hamiltonian into a sum of two
separate Hamiltonians for two fictitious particles:
H  HG  H r
2
PG
HG 
2M
2

P
Hr 
V R
2
 
7.B.2
The two-body problem
• The two parts of the Hamiltonian commute with
each other:
[HG , H r ]  0
• Therefore they both commute with the full
Hamiltonian
[HG , H ]  [H r , H ]  0
• Thus there should be basis common for all three
operators:
H G   EG 
H r   Er 
H   E
• In this case
E  EG  Er
   G  r
HG G  EG G
H r r  Er r
7.B.2
The two-body problem
• In the coordinate representations:


rG HG G  EG rG G


r H r r  Er r r


2

 G  G (rG )  EG  G (rG )
2M
 2
  

 
  V  r r (r )  Err (r )
 2

• The first equation is for a free particle and it is well
known how to deal with it
• The second equation is much more interesting from
a physical viewpoint
7.B.2
The two-body problem
• This equation describes the behavior of two
interacting particles in the center of mass frame and
also the behavior of a single fictitious particle in a
central potential
 2
  

 
  V  r r (r )  Err (r )
 2

• We can now drop the r subscript:
 2
  

 
  V  r  (r )  E (r )
 2

7.A.1
Particle in a central potential
• The Laplacian in spherical coordinates:
1 2
1  2
1 
1 2 


r  2  2 
 2
2
2 
r r
r  
tan   sin   
2
2
1 
1 L 



r


r r 2
r 2   2 
• Therefore:

 2 1 2

L2

 (r , ,  )  E (r , ,  )
r


V
(
r
)
2
2
 2 r r

2

r


 2
  

 
  V  r  (r )  E (r )
 2

7.A.2
Particle in a central potential
• Since the orbital angular momentum operator 
depends only on the angular coordinates: [ H , L ]  0
2
• And: [ H , L ]  0
• Since H, L2 and Lz commute there is a common
basis for all three of them

 2 1 2

L2

 (r , ,  )  E (r , ,  )
r


V
(
r
)
2
2
 2 r r

2

r


2
2
L  (r , ,  )  l (l  1)  (r , ,  )
Lz (r, ,  )  m (r, ,  )
7.A.2
Particle in a central potential
• Using the theory of the orbital angular momentum
operator:
m
 (r , ,  )  R(r )Yl ( ,  )
• And:
 2 1 d 2

l (l  1) 2
 
r
 V (r )  R(r )  ER(r )
2
2
2r
 2 r dr


 2 1 2

L2

 (r , ,  )  E (r , ,  )
r


V
(
r
)
2
2
 2 r r

2

r


2
2
L  (r , ,  )  l (l  1)  (r , ,  )
Lz (r, ,  )  m (r, ,  )
7.A.2
Particle in a central potential
• Using the theory of the orbital angular momentum
operator:
m
 (r , ,  )  R(r )Yl ( ,  )
• And:
 2 1 d 2

l (l  1) 2
 
r
 V (r )  R(r )  ER(r )
2
2
2r
 2 r dr

• This equation and its solutions depend on the
quantum number l as well as index k (that represents
different eigenvalues for the same l) and does not
depend on the quantum number m; thus:
 2 1 d 2

l (l  1) 2
 
r
 V (r )  Rk ,l (r )  Ek ,l Rk ,l (r )
2
2
2r
 2 r dr

7.A.2
Particle in a central potential
• The behavior of the R functions at the origin should
be sufficiently regular in order to represent a physical
solution
• The equation can be simplified via this substitution:
u k ,l (r )  rRk ,l (r )
Rk ,l (r ) 
u k ,l ( r )
r
  2 d 2 l (l  1) 2

 

 V (r ) uk ,l (r )  Ek ,l uk ,l (r )
2
2
2r
 2 dr

 2 1 d 2

l (l  1) 2
 
r
 V (r )  Rk ,l (r )  Ek ,l Rk ,l (r )
2
2
2r
 2 r dr

7.A.2
Particle in a central potential
• The behavior of the R functions at the origin should
be sufficiently regular in order to represent a physical
solution
• The equation can be simplified via this substitution:
u k ,l (r )  rRk ,l (r )
Rk ,l (r ) 
u k ,l ( r )
r
  2 d 2 l (l  1) 2

 

 V (r ) uk ,l (r )  Ek ,l uk ,l (r )
2
2
2r
 2 dr

 2 d 2

 
 Veff (r ) uk ,l (r )  Ek ,l uk ,l (r )
2
 2 dr

l (l  1) 2
Veff (r ) 
 V (r )
2
2r
7.A.3
Particle in a central potential
 1
m

(
r
)

u
(
r
)
Y
• Therefore:
k ,l , m
k ,l
l ( ,  )
r
• This function must be square-integrable:

2
k ,l , m
(r , ,  ) r sin drdd 

2
  Rk ,l (r ) r dr  Yl ( ,  ) sin dd  1
2
2
2
m
0
• Since the spherical harmonics are normalized

R
k ,l
0

(r ) r dr   uk ,l (r ) dr  1
2
2
2
0
• Quantum number l is called azimuthal, whereas
quantum number m is called magnetic
The hydrogen atom
7.C.1
7.C.3
• A system of a proton and an electron can form a
hydrogen atom
e2 1
• In this case the potential energy is: V ( r )  
40 r
• And the reduced mass of the system:
m p me
me


 me
m p  me 1 me / m p
• Therefore, the radial eigenproblem becomes:
2
  2 d 2 l (l  1) 2
e 1
 
uk ,l (r )  Ek ,l uk ,l (r )


2
2
2me r
40 r 
 2me dr
7.C.3
The hydrogen atom
• Let us make substitutions:
2
r
e me

 r
2
a0
40 
k ,l
2 Ek , l
40


2
e
me
• This yields a dimensionless equation:
 d 2 l (l  1) 2 
2
 2 

 uk ,l (  )  k ,l uk ,l (  )
2


 d
• Therefore, the radial eigenproblem becomes:
2
  2 d 2 l (l  1) 2
e 1
 
uk ,l (r )  Ek ,l uk ,l (r )


2
2
2me r
40 r 
 2me dr
7.C.3
The hydrogen atom
• Let us make substitutions:
2
r
e me

 r
2
a0
40 
k ,l
2 Ek , l
40


2
e
me
• This yields a dimensionless equation:
 d 2 l (l  1) 2 
2
 2 

 uk ,l (  )  k ,l uk ,l (  )
2


 d
• What are the asymptotes of the solutions?
 
d 2 u k ,l
d
2
  u k ,l
2
k ,l
uk ,l  Ae
 k ,l 
 Be
k ,l 
7.C.3
The hydrogen atom
• Let us make substitutions:
2
r
e me

 r
2
a0
40 
k ,l
2 Ek , l
40


2
e
me
• This yields a dimensionless equation:
 d 2 l (l  1) 2 
2
 2 

 uk ,l (  )  k ,l uk ,l (  )
2


 d
• What are the asymptotes of the solutions?
 0
d 2 u k ,l
d
2

l (l  1)

2
u k ,l
uk ,l  C l 1  D l
7.C.3
The hydrogen atom
• Taking into account the asymptotes the solution
could besought in the following form:
l 1  k ,l 
u k ,l (  )   e
y k ,l (  )
 d 2 l (l  1) 2  l 1 k ,l 
2
l 1 k ,l 
 2 
   e
yk ,l (  )  k ,l  e
y k ,l (  )
2


 d
 d 2 l (l  1) 2 
2
 2 

 uk ,l (  )  k ,l uk ,l (  )
2


 d
7.C.3
The hydrogen atom
• Taking into account the asymptotes the solution
could besought in the following form:
l 1  k ,l 
u k ,l (  )   e
y k ,l (  )
 d 2 l (l  1) 2  l 1 k ,l 
2
l 1 k ,l 
 2 
   e
yk ,l (  )  k ,l  e
y k ,l (  )
2


 d
 d 2  l 1 k ,l 
 2
2 l (l  1)  l 1 k ,l 
 2   e
  e
yk ,l   k ,l  
y k ,l
2

 

 d 
dyk ,l

d
k ,l  l 
l 1 k ,l 
( e
y k ,l )  e
  
 k ,l yk ,l  yk ,l (l  1) 
d
 d

7.C.3
The hydrogen atom

d  k ,l  l  dyk ,l
e







y

y
(
l

1
)
k
,
l
k
,
l
k
,
l



d 
d



2


d
yk ,l 2dyk ,l
 k ,l  l

e
  

((
l

1
)



)
k
,
l
2

d

d



e

d 2 y k ,l
d
2
 k ,l 
  (l  1)l

2
  yk ,l 
 k ,l   2k ,l (l  1)  

  
l
 2
l (l  1)  l k ,l 
  e
  k ,l   2 
y k ,l
 

dyk ,l
 2((l  1)  k ,l  )
 2(1  k ,l (l  1)) yk ,l  0
d
7.C.3
The hydrogen atom
• Let us look for the solution in the following form of a
jmax
polynomial:
j
jmax
dy k ,l
d

dcj j
j 0
d
y k ,l (  )   c j 
j 0
j
d
  c j j j j1 1
 cj
max
d
j 1
j 0
j
  c j 1 ( j  1) 
jmax
jmax
j 0
2
d y k ,l
d

2

d 2 y k ,l
d
2
jmax 1
jmax  2
j 1
j 0
j 1
c
(
j

1
)
j


 j 1
 2((l  1)  k ,l  )
dyk ,l
d
j
c
(
j

1
)(
j

2
)

 j 2
 2(1  k ,l (l  1)) yk ,l  0
7.C.3
The hydrogen
atom
j 1
j 1
j 1
c
(
j

1
)(
j

2
)


2

c
(
j

1
)

 j2
k ,l  j 1
j max  2
max
j 0
j 0
 2(l  1)
jmax 1
jmax
j 0
j 0
j
j
c
(
j

1
)


2
(
1


(
l

1
))
c

 j 1
 j 0
k ,l
• Equating the coefficients of like powers of ρ yields:
c j 1 ( j  1)( j  2(l  1))  c j 2(1  k ,l (l  1  j ))  0
c j 1  c j
2(k ,l (l  1  j )  1)
( j  1)( j  2(l  1))
• The polynomial terminates at:
c jmax 1  c jmax
2(k ,l (l  1  jmax )  1)
( jmax  1)( jmax  2(l  1))
0
7.C.3
The hydrogen atom
• Therefore:
k ,l (l  jmax  1)  1
• Let us recall that:
k ,l
E k ,l  
(l  jmax
• Defining:
c jmax 1  c jmax
l  jmax  1
2 Ek , l
40


2
e
me
• Combining the two equations:
1
k ,l 
1
2
 e  me


2 
 1)  40  2
2
2(k ,l (l  1  jmax )  1)
( jmax  1)( jmax  2(l  1))
0
7.C.3
The hydrogen atom
• Therefore:
k ,l (l  jmax  1)  1
• Let us recall that:
k ,l
E k ,l  
• Defining:
(l  jmax
l  jmax  1
2 Ek , l
40


2
e
me
• Combining the two equations:
1
k ,l 
1
2
 e  me


2 
 1)  40  2
2
2
 e  me

n  l  jmax  1 E0  
 13.6 eV
40  2

• We obtain:
E0
E k ,l   2
n
2
7.C.3
Spectrum
• Since
l  0,1,2,... jmax  0,1,2,...
• Conventionally and conveniently n is used to label
the energy spectrum
• n is called a principal quantum number
• A given value of n characterizes an electron shell
• Defining:
2
 e  me

n  l  jmax  1 E0  
 13.6 eV
40  2

• We obtain:
E0
n  1,2,3,...
E k ,l   2
n
2
7.C.3
Spectrum
• Since
l  0,1,2,... jmax  0,1,2,...
• Conventionally and conveniently n is used to label
the energy spectrum
• n is called a principal quantum number
• A given value of n characterizes an electron shell
• Defining:
n  l  jmax
• We obtain:
2
 e  me

 1 E0  
 13.6 eV
 40  2
E0
n  1,2,3,...
En   2
n
2
7.C.3
Spectrum
• Since
l  0,1,2,... jmax  0,1,2,...
• There is a finite number of values of l associated
with the same value of n:
l  n  1  jmax
l  0,1,2,..., n  1
• Each shell contains n sub-shells each
corresponding to a given value of l
• Defining:
n  l  jmax
• We obtain:
2
 e  me

 1 E0  
 13.6 eV
 40  2
E0
n  1,2,3,...
En   2
n
2
7.C.3
Spectrum
• Since
l  0,1,2,... jmax  0,1,2,...
• There is a finite number of values of l associated
with the same value of n:
l  n  1  jmax
l  0,1,2,..., n  1
• Each shell contains n sub-shells each
corresponding to a given value of l
• Since
m  l ,l  1,..., l  1, l
• Each sub-shell contains (2l + 1) distinct states
associated with the different possible values of m for
a fixed value of l
7.C.3
Spectrum
• The total degeneracy of the energy level with a value
of En is:
n 1
(n  1)n
g n   (2l  1)  2 l  1  2
 n  n2
2
l 0
l 0
l 0
n 1
n 1
• Conventionally, different values of l are
(spectroscopically) labelled as follows:
• Subshell notations:
n  1, l  0  1s
n  2, l  0  2 s
n  2, l  1  2 p
n  3, l  0  3s
...
l0s
l 1 p
l2d
l 3 f
l4 g
7.C.3
Spectrum
l0s
n  1, l  0  1s
n  2, l  0  2 s
n  2, l  1  2 p
n  3, l  0  3s
...
l 1 p
l2d
l 3 f
l4 g
7.C.3
Spectrum
7.C.3
Eigenfunctions
• Let us synopsize all the transformations and
assumptions for the eigenfunctions
 (r , ,  )  R(r )Yl ( ,  )
m
   G  r
2
Rk ,l (r ) 
e me
r
 r

2
40 
a0
u k ,l ( r )
r
l 1  k ,l 
u k ,l (  )   e
y k ,l (  )
jmax
y k ,l (  )   c j 
j
j 0
c j 1  c j
2(k ,l (l  1  j )  1)
( j  1)( j  2(l  1))
k ,l 
1
l  jmax
1

1 n
7.C.3
Eigenfunctions
• Let us also recall the normalization conditions:
2

(
r
,

,

)
r
sin drdd 
 n ,l , m
2

  Rn,l (r ) r dr  Yl ( ,  ) sin dd  1
2
2
m
2
0

R
n ,l
0

(r ) r dr   un,l (r ) dr  1
2
2
2
0
• Now, using all this information, let us calculate the
eignefucntions for the problem
n  l  jmax  1 Eigenfunctions l  jmax  0
7.C.3
• We will start with the ground level, a nondegenerate
1s subshell
 n,l ,m (r , ,  )  1,0,0 (r , ,  )  R1,0 (r )Y00 ( ,  )


u1,0 (r )
r

Y ( ,  ) 

0
r
a0
r
 e c Y ( ,  ) 
e
r
a0 r

 c0
Rn,l (r ) r dr   
e
 a0
0
2

a0 c0
1
4
2

n
0
0 0
• Normalizing:

e
0
0
2

2
r

a0
 1, 0 
r
y1, 0 (  )Y00 ( ,  )
c0
c Y ( ,  )  e
a0
0
0 0
 2
c0
 r dr 
2

a0

2 
e
0

2r
a0

r
a0
1
4
r dr 
2
c0 
a0
2
a0 c0
4
2
7.C.3
Eigenfunctions
• We obtained the eigenfunction of the ground state!
1,0,0 (r , ,  ) 
1
a03

e
r
a0
• It is completely spherically symmetric
7.C.3
Eigenfunctions
• What is the probability density of finding an electron
in an elementary volume?
2
d Pn,l ,m (r , ,  )  n,l ,m (r , ,  ) r 2 sin drdd
3
2
2
 Rn ,l (r ) r dr Yl ( ,  ) sin dd
2
m
• The probability of finding an electron between r and
2 2
r + dr is proportional to
Rn,l (r ) r dr
• For the ground state this probability is thus
2r

proportional to
a
f (r )  r 2 e
f ' (r )  2re

2r
a0
2
2r

e
a0

2r
a0
0
 2re

2r
a0

r 
1  
 a0 
7.C.3
Eigenfunctions
• The maximum of this probability occurs at
r  a0
• Parameter a0 is known as the Bohr radius
a0  0.052 nm
Niels Henrik
David Bohr
(1885 – 1962)
f ' (r )  2re

2r
a0
2
2r

e
a0

2r
a0
 2re

2r
a0

r 
1  
 a0 
7.C.3
Eigenfunctions
• The ground state function can be used to generate
the rest of the eigenfunctions
• E.g.,
l 1 j  n
c j 1  2c j
n( j  1)( j  2(l  1))

r 
1 
e
 2, 0, 0 (r ,  ,  ) 
8a03  2a0 
1
1
r
2,1, 1 (r , ,  ) 
e
3 a
8 a0 0
1

r
2,1,0 (r , ,  ) 
e
4 2a03 a0
r
2 a0


r
2 a0
sin e i
r
2 a0
cos 
7.C.3
Eigenfunctions
• The most general expression:
3
 2  (n  l  1)!

 n ,m ,l (r ,  ,  )  

3
 na0  2n(( n  1)!)
e

r
na0
l
 2r  2l 1  2r  m

 Ln l 1 
Yl ( ,  )
 na0 
 na0 
Edmond Nicolas
Laguerre
(1834 – 1886)
• Where
q


p
p d   x d 
x q 
Lq  p ( x)  (1)   e   (e x )

 dx    dx 

p
• Is the associated Laguerre polynomial
7.C.3
Eigenfunctions
L00  1
L20  2
L10   x  1
L12  6 x  18
L  x  4x  2
L  12 x  96 x  144
L10  1
L30  6
L11  2 x  4
L13  24 x  96
2
0
2
2
2
2
L12  3 x 2  18 x  18
Edmond Nicolas
Laguerre
(1834 – 1886)
L32  60 x 2  600 x  1200
p
q


p
p d   x d 
x q 
Lq  p ( x)  (1)   e   (e x )

 dx    dx 

• Is the associated Laguerre polynomial
7.C.3
Eigenfunctions
• The radial parts of the eigenfunctions:
R1,0 (r , ,  ) 
1

2
e
3
0
r

2 a0
a
r
R2,1 
e
3 a
24a0 0
r
a0
R2,0

r 
1 
e

2a03  2a0 
1

r
2 a0
7.C.3
Eigenfunctions
• Probability density plots for the wave functions: