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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 onlydepend on the separation distance:
• From our experience with Classical Mechanics,we might want to treat separately theCenter-of-mass and relative motion:– Center-of-mass coordinate:
– Relative coordinate:
– This is recommended because the potentialdepends only on the relative coordinate:
€
H =P12
2m1
+P22
2m2
+ Vr R 1 −
r R 2( )
21
2211
mm
RmRmRCM +
+=
rrr
21 RRRrrr
−=
€
Vr R 1 −
r R 2( ) = V (R)
Center-of-mass and relative momentum
• How do we go about finding the center-of-mass and relative-motion momentumoperators:– Can we use:
• Answer: No, this is very wrong!
• Lets try instead to use what we know fromclassical mechanics:
21 PPPrrr
−=
2211 VmVmrr
+=
21
2211
mm
PmPmPCM +
+=
rrr
CMCM VMPrr
=
21 mmM +=21
21
mm
mm
+=µ
21
2211
mm
VmVmR
dt
dV CMCM +
+==
rrrr
21 PPPCMrrr
+=
21 VVRdt
dV
rrrr−==
?
VPrr
µ=
( )2121
21 VVmm
mm rr−
+=
21
2112
mm
PmPmP
+
−=
r
Transformation to Center-of-masscoordinates
• We have defined new coordinates:
• We have guessed that the correspondingmomentum operators are:
• To verify, we need to check the commutationrelations:
• So our choices for the momentum operatorswere correct
21
2211
mm
RmRmRCM +
+=
rrr
21 RRRrrr
−=
21 PPPCMrrr
+=21
2112
mm
PmPmP
+
−=
r
hiPXmm
mPX
mm
mPX xxxCMCM =
++
+= ],[],[],[ 22
21
211
21
1,
hiPXmm
mPX
mm
mPX xxx =
++
+= ],[],[],[ ,22
21
1,11
21
2
€
[XCM ,Px ] =m1m2
m1 + m2( )2[X1,P1,x ]−
m2m1m1 + m2( )2
[X2,P2,x ] = 0
0],[],[],[ ,22,11, =−= xxxCM PXPXPX
Inverse Transformation
• The inverse transformations work out to:
21
2211
mm
RmRmRCM +
+=
rrr
21 RRRrrr
−=
21 PPPCMrrr
+=21
2112
mm
PmPmP
+
−=
r
Rm
RR CM
rrr
11
µ+=
Rm
RR CM
rrr
22
µ−=
PPM
mP CM
rrr+= 1
1
PPM
mP CM
rrr−= 2
2
Transforming the Kinetic Energy Operator
• Using the inverse transformations:
• We find:
• So that:
Rm
RR CM
rrr
11
µ+=
Rm
RR CM
rrr
22
µ−=
PPM
mP CM
rrr+= 1
1
PPM
mP CM
rrr−= 2
2
2122
21
11
2PPP
M
mP
M
mPP CMCM +⋅+=⋅
rrrr
2222
22
22
2PPP
M
mP
M
mPP CMCM +⋅−=⋅
rrrr
2
21
2221
2
22
1
21 11
2
1
222P
mmP
M
mm
m
P
m
PCM
++
+=+
µ2222
22
2
22
1
21 P
M
P
m
P
m
P CM +=+
1221
21 111
mmmm
mm+=
+=
µ
The New Hamiltonian
• Becomes:
• Note that:
• We call this ‘separability’– System is ‘separable’ in COM and relative
coordinates
• When a system is separable, it means we cansolve each problem separately, and use thetensor product to construct the fulleigenstates of the complete system
( )RVP
M
PH CM ++=
µ22
22
( )212
22
1
21
22RRV
m
P
m
PH
rr−++=
relCM HHH +=
M
PH CMCM 2
2
= ( )RVPHrel +=
µ2
2
Eigenstates of the SeparatedSystems
• The eigenstates of this Hamiltonian are free-particle eigenstates
• Bound states:
• Continuum states:
€
r p CM
(C )
( )CCMCM M
PH H∈=
2
2
)()( C
CMCM
C
CMCM pppPrrrr
=
)(2
)(
2C
CMCMC
CMCM pM
ppH
rr=
( ) ( )Rrel RV
PH H∈+=
µ2
2
)()(,,
R
n
R
rel mnEmnH =
)()()(
RR
rel kkEkHrrr
=
)(3)( C
CMCMCMC pppdI
rr∫+∞
∞−
=
n is the ‘principle quantum number’ labels energy levels
)(3
1
)(
1
)()(max
,,Rn
n
nd
m
RR kkkdmnmnIrr
∫∑∑+∞
∞−= =
+=
Full Eigenstates
• We can form tensor product states:
)(2
)(
2C
CMCMC
CMCM pM
ppH
rr=
)()(,,
R
n
R
rel mnEmnH =
)()()(
RR
rel kkEkHrrr
=
CMrel HHH ⊗=
)()(,:,,
RC
CMCM mnpmnp ⊗=rr
)()(:,
RC
CMCM kpkprrrr
⊗=
)()( RC III ⊗=)()()( R
continuumRbound
R III +=
‘and’
‘or’
)()()()( Rcontinuum
CRbound
C IIIII ⊗⊗ +=
Tensor Product States areEigenstates of the Full Hamiltonian
)()(2
,2
CR
CMnCM mnpEM
p⊗
+=
r
mnpEM
pmnpH CMn
CMCM ,,
2,,
2 rr
+=
)()()()()()( ,,RC
CMRrel
RC
CMCCM mnpHmnpH ⊗⊗ +=
rr
€
= HCM(C ) r
p CM(C )( ) ⊗ n,m (R )
+r p CM
(C )⊗ Hrel
(R ) n,m (R )( ))()()()(
2
,,2
R
n
C
CM
RC
CMCM mnEpmnpM
p⊗⊗ +=
rr
mnpH CM ,,r
• Proof:
€
= HCM(C ) + Hrel
(R )( ) r p CM
(C )⊗ n,m (R )
mnpEM
pmnpH CMn
CMCM ,,
2,,
2 rr
+=
Example: Hydrogen Atom
• For the hydrogen system (e + p) we have:
• Switch to relative and COM coordinatesgives:
• The eigenstates of HCM in H(C) are free-particle eigenstates:
• The non-trivial task is to find the eigenstatesof Hrel in H(R):
pep
p
e
e
RR
e
m
P
m
PH rr
−−+=
0
222
422 πε
R
eP
M
PH CM
0
222
422 πεµ−+=
R
ePHrel
0
22
42 πεµ−=
relCM HH +=
)(2
)(
2C
CMCMC
CMCM pM
ppH
rr=
€
r p CM
(C ){ } :