Lecture 28: The Quantum Two-body Problem Phy851 Fall 2009

Lecture 28:
The Quantum Two-body Problem
Phy851 Fall 2009
Two interacting particles
• Consider a system of two particles with no
external fields
• By symmetry, the interaction energy can only
depend on the separation distance:
r r
P12
P22
H=
+
+ V R1 − R2
2m1 2m2
(
)
• From our experience with Classical Mechanics,
we might want to treat separately the
Center-of-mass and relative motion:
€
– Center-of-mass coordinate:
r
r
r
m1 R1 + m2 R2
RCM =
m1 + m2
– Relative coordinate:
r r r
R = R1 − R2
– This is recommended because the potential
depends only on the relative coordinate:
r r
V R1 − R2 = V (R)
(
€
)
Center-of-mass and relative momentum
• How do we go about finding the center-ofmass and relative-motion momentum
operators:
– Can we use:
r
r
r
m P + m2 P2
PCM = 1 1
m1 + m2
r r r
P = P1 − P2
?
• Answer: No, this is very wrong!
• Lets try instead to use what we know from
classical mechanics:
M = m1 + m2
r
r
r
r
d
m1V1 + m2V2
VCM = RCM =
dt
m1 + m2
r
r
PCM = M VCM
r
r
= m1V1 + m2V2
r
r r
PCM = P1 + P2
m1m2
µ=
m1 + m2
r d r r r
V = R = V1 − V2
dt
r
r
P = µV
m1m2 r r
=
V1 − V2
m1 + m2
(
r m2 P1 − m1 P2
P=
m1 + m2
)
€
Transformation to Center-of-mass
coordinates
•
We have defined new coordinates:
r
RCM
r
r
m R + m2 R2
= 1 1
m1 + m2
r r r
R = R1 − R2
• We have guessed that the corresponding
momentum operators are:
r
r r
PCM = P1 + P2
r m2 P1 − m1 P2
P=
m1 + m2
• To verify, we need to check the commutation
relations: m
m
[ X CM , PCM , x ] =
[ X , Px ] =
[X CM ,Px ] =
1
m1 + m2
[ X 1 , P1x ] +
2
m1 + m2
[ X 2 , P2 x ] = ih
m2
m1
[ X 1 , P1, x ] +
[ X 2 , P2, x ] = ih
m1 + m2
m1 + m2
m1m2
m2 m1
[X
,P
]
−
[X 2 ,P2,x ] = 0
1 1,x
2
2
( m1 + m2 )
( m1 + m2 )
[ X , PCM , x ] = [ X 1 , P1, x ] − [ X 2 , P2, x ] = 0
• So our choices for the momentum operators
were correct
r
RCM
r
PCM
Inverse Transformation
r
r
r r r
m1 R1 + m2 R2
R = R1 − R2
=
m1 + m2
r m2 P1 − m1 P2
r r
P=
= P1 + P2
m1 + m2
• The inverse transformations work out to:
r
r
µ r
R1 = RCM +
R
m1
r
r
µ r
R2 = RCM −
R
m2
r m1 r
r
P1 =
PCM + P
M
r m2 r
r
P2 =
PCM − P
M
Transforming the Kinetic Energy Operator
• Using the inverse transformations:
r
r
µ r
R1 = RCM +
R
m1
r
r
µ r
R2 = RCM −
R
m2
r m1 r
r
P1 =
PCM + P
M
r m2 r
r
P2 =
PCM − P
M
• We find:
r r m12 2
2m1 r r
P1 ⋅ P1 = 2 PCM +
P ⋅ PCM + P 2
M
M
r r
m22 2
2m2 r r
P2 ⋅ P2 = 2 PCM −
P ⋅ PCM + P 2
M
M
• So that:
P12
P22
m1 + m2 2
1 1
1  2
 P
+
=
PCM +  +
2
2m1 2m2
2M
2  m1 m2 
1 m1 + m2
1
1
=
=
+
µ
m1m2
m2 m1
2
PCM
P12
P22
P2
+
=
+
2m1 2m2 2 M 2 µ
The New Hamiltonian
r r
P12
P22
H=
+
+ V R1 − R2
2m1 2m2
(
)
• Becomes:
2
PCM
P2
H=
+
+ V (R )
2M 2µ
• Note that:
H = H CM + H rel
H CM
2
PCM
=
2M
H rel
P2
=
+ V (R )
2µ
• We call this ‘separability’
– System is ‘separable’ in COM and relative
coordinates
• When a system is separable, it means we can
solve each problem separately, and use the
tensor product to construct the full
eigenstates of the complete system
Eigenstates of the Separated
Systems
H CM
2
PCM
=
∈H
2M
(C )
• The eigenstates of this Hamiltonian are freeparticle eigenstates
r
pCM
r r
PCM pCM
(C )
H CM
r
pCM
3
r
pCM
+∞
I
€
(C )
(C )
= ∫ d pCM
r
r
= pCM pCM
(C )
(C )
2
r
pCM
=
pCM
2M
r
pCM
(C )
(C )
−∞
H rel
P2
=
+ V (R )∈ H
2µ
• Bound states:
H rel n, m
(R)
(R )
= En n, m
(R)
n is the ‘principle quantum number’
 labels energy levels
• Continuum states:
I
(R)
nmax d ( n )
H rel
= ∑ ∑ n, m n, m
n =1 m =1
(R)
r
k
(R)
r r
= E (k ) k
r r
+ ∫d k k k
+∞
3
−∞
(R)
(R)
Full Eigenstates
H CM
r
pCM
H rel n, m
r
H rel k
(C )
(R)
(R)
2
r
pCM
=
pCM
2M
= En n, m
r r
= E (k ) k
(C )
(R)
(R)
• We can form tensor product states:
H = H rel ⊗ H CM
r
r
pCM , n, m := pCM
(C )
r
r
r
pCM , k := pCM
(C )
I = I (C ) ⊗ I ( R )
I=I
⊗I
(R)
bound
+I
n, m
r
⊗ k
(R)
‘and’
(R)
(R)
I ( R ) = I bound
+ I continuum
(C )
⊗
(R)
(C )
‘or’
⊗I
(R)
continuum
Tensor Product States are
Eigenstates of the Full Hamiltonian
2
 pCM
 r
r
H pCM , n, m = 
+ En  pCM , n, m
 2M

• Proof:
r
H pCM , n, m
= (H
(C ) r
= H CM
pCM
(C ) r
€
= HCM pCM
(
(C )
CM
(C )
(C )
2
r
pCM
=
pCM
2M
)
+H
⊗
(R )
rel
n, m
⊗ n,m
(C )
)
r
pCM
(C )
⊗ n,m
(R)
(R) r
+ H rel
pCM
(R )
r
+ pCM
⊗ n, m
(R)
2
 pCM
 r
= 
+ En  pCM
 2M

r
+ pCM
(R)
(C )
(R )
(C )
⊗
n, m
(R)
(
(R )
⊗ H rel
n,m
(C )
⊗ En n, m
⊗ n, m
(C )
2
 pCM
 r
r

H pCM , n, m = 
+ En  pCM , n, m
 2M

(R)
(R )
)
Example: Hydrogen Atom
• For the hydrogen system (e + p) we have:
Pp2
2
e
P
e2
H=
+
−
r r
2me 2m p 4πε 0 Re − R p
• Switch to relative and COM coordinates
gives:
2
PCM
P2
e2
H=
+
−
= H CM + H rel
2 M 2 µ 4πε 0 R
• The eigenstates of HCM in
particle eigenstates:
{
€
r
pCM
(C )
}:
H CM
r
pCM
H(C)
(C )
are free-
2
r
pCM
=
pCM
2M
(C )
• The non-trivial task is to find the eigenstates
of Hrel in H(R):
H rel
P2
e2
=
−
2 µ 4πε 0 R