130_notes.dvi

[p,x]ψ(x) =

̄h

i

(

ψ(x) +x

∂ψ(x)

∂x

−x

∂ψ(x)

∂x

)

=

̄h

i

ψ(x)

So, removing theψ(x) we used for computational purposes, we get the commutator.

[p,x] =

̄h

i

Later we will learn to derive the uncertainty relation for two variables from their commutator.
Physical variable with zero commutator have no uncertainty principle and we can know both of
them at the same time.

We will also use commutators to solve several important problems.

We can compute thesame commutator in momentum space.

[p,x]φ = [p,i ̄h

d

dp

]φ=i ̄h

(

p

d

dp

φ−

d

dp

pφ

)

=i ̄h(−φ) =

̄h

i

φ

[p,x] =

̄h

i

The commutator is the same in any representation.

• See Example 6.7.2:Compute the commutator [E,t].*


• See Example 6.7.3:Compute the commutator [E,x].*


• See Example 6.7.4:Compute the commutator [p,xn].*


• See Example 6.7.5:Compute the commutator of the angular momentum operators [Lx,Ly].*

Gasiorowicz Chapter 3

Griffiths Chapter 3

Cohen-Tannoudji et al. Chapter

6.6 Derivations and Computations

6.6.1 Verify Momentum Operator

p(op)

1

√

2 π ̄h

ei(p^0 x−E^0 t)/ ̄h=

̄h

i

∂

∂x

1

√

2 π ̄h

ei(p^0 x−E^0 t)/ ̄h

1

√

2 π ̄h

̄h

i

ip 0

̄h

ei(p^0 x−E^0 t)/ ̄h=p 0

1

√

2 π ̄h

ei(p^0 x−E^0 t)/ ̄h