130_notes.dvi

12 Extending QM to Two Particles and Three Dimensions

12.1 Quantum Mechanics for Two Particles

We can know the state of two particles at the same time. The positions and momenta of particle 2
commutewith the positions and momenta of particle 1.

[x 1 ,x 2 ] = [p 1 ,p 2 ] = [x 1 ,p 2 ] = [x 2 ,p 1 ] = 0

The kinetic energy terms in the Hamiltonian are independent. There may be an interaction between
the two particles in the potential. TheHamiltonian for two particlescan be easily written.

H=

p^21

2 m 1

+

p^22

2 m 2

+V(x 1 ,x 2 )

Often, the potential will only depend on thedifference in the positions of the two particles.

V(x 1 ,x 2 ) =V(x 1 −x 2 )

This means that the overallHamiltonian has a translational symmetry. Lets examine an
infinitesimal translation inx. The original Schr ̈odinger equation

Hψ(x 1 ,x 2 ) =Eψ(x 1 ,x 2 )

transforms to

Hψ(x 1 +dx,x 2 +dx) =Eψ(x 1 +dx,x 2 +dx)

which can be Taylor expanded

H

(

ψ(x 1 ,x 2 ) +

∂ψ

∂x 1

dx+

∂ψ

∂x 2

dx

)

=E

(

ψ(x 1 ,x 2 ) +

∂ψ

∂x 1

dx+

∂ψ

∂x 2

dx

)

.

We can write the derivatives in terms of the total momentum operator.

p=p 1 +p 2 =

̄h

i

(

∂

∂x 1

+

∂

∂x 2

)

Hψ(x 1 ,x 2 ) +

i

̄h

Hp ψ(x 1 ,x 2 )dx=Eψ(x 1 ,x 2 ) +

i

̄h

Ep ψ(x 1 ,x 2 )dx

Subtract of the initial Schr ̈odinger equation and commuteEthroughp.

Hp ψ(x 1 ,x 2 ) =Ep ψ(x 1 ,x 2 ) =pHψ(x 1 ,x 2 )

We have proven that

[H,p] = 0

if the Hamiltonian has translational symmetry. Themomentum is a constant of the motion.
Momentum is conserved. We can havesimultaneous eigenfunctions of the total momentum
and of energy.