156 CHAPTER10. SYMMETRYANDDEGENERACY
Then
TH ̃[
∂
∂x
,x]ψ(x) = H ̃[
∂
∂x′
,x′]ψ(x′)
= H ̃[
∂
∂x
,x]ψ(x′)
= H ̃[
∂
∂x
,x]Tψ(x) (10.16)
Fromthisequation,weseethattheoperatorTcommuteswiththetheHamilitonian
operator
[T,H ̃]= 0 (10.17)
Nownoticethat
TF(x) = F(x+a)
= F(x)+
∂F
∂x
a+
1
2
∂^2 F
∂x^2
a^2 +...
=
∑∞
n=0
an
n!
∂n
∂xn
F(x)
= exp
[
a
∂
∂x
]
F(x)
= exp[iap/ ̃ ̄h]F(x) (10.18)
whichmeansthat T istheexponentialofthe momentumoperator. SinceT com-
muteswithH ̃foranydisplacementa,itfollowsthatthemomentumoperatorpalso
commuteswithH
[p, ̃H ̃]= 0 (10.19)
Thisis easytocheck forthefreeparticle Hamiltonian,since H ̃ = p ̃^2 / 2 m, andp ̃
commuteswithp ̃^2.
InLecture8,wefoundanequationofmotionforexpectationvalues
d
dt
=
i
̄h
<[Q,H]> (10.20)
ItfollowsthatifanHermitianoperatorcommuteswiththeHamiltonian,thecorre-
spondingobservableisconserved:
d
dt
= 0 (10.21)
Inthespecialcaseofthefreeparticle,wethereforehaveconservationofmomentum
d
dt
<p>= 0 (10.22)