130_notes.dvi

= i ̄hψ(x,t) +

(

i ̄ht

∂

∂t

−i ̄ht

∂

∂t

)

ψ(x,t)

= i ̄hψ(x,t)

Removing the wave function, we have the commutator.

[E,t] =i ̄h

6.7.3 Commutator ofEandx.

Again use the crutch of keeping a wave function on the right to avoidmistakes.

[E,x]ψ(x,t) =

(

i ̄h

∂

∂t

x−xi ̄h

∂

∂t

)

ψ(x,t)

=

(

i ̄hx

∂

∂t

−i ̄hx

∂

∂t

)

ψ(x,t) = 0

Since∂x∂t= 0.

6.7.4 Commutator ofpandxn

We can use the commutator [p,x] to help us. Remember thatpx=xp+ [p,x].

[p,xn] = pxn−xnp

= (px)xn−^1 −xnp

= xpxn−^1 + [p,x]xn−^1 −xnp

= x(px)xn−^2 + [p,x]xn−^1 −xnp

= x^2 pxn−^2 +x[p,x]xn−^2 + [p,x]xn−^1 −xnp

= x^2 pxn−^2 + 2[p,x]xn−^1 −xnp

= x^3 pxn−^3 + 3[p,x]xn−^1 −xnp

= xnp+n[p,x]xn−^1 −xnp

= n[p,x]xn−^1 =n

̄h

i

xn−^1

It is usually not wise to use the differential operators and a wave function crutch to compute
commutators like this one.Use the known basic commutators when you can.Nevertheless,
we can compute it that way.

[p,xn]ψ=

̄h

i

∂

∂x

xnψ−xn

̄h

i

∂

∂x

ψ=

̄h

i

nxn−^1 ψ

[p,xn] =

̄h

i

nxn−^1

It works pretty well for this particular case, but not if I havepto some power...