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 ){ } :