From Classical Mechanics to Quantum Field Theory

170 From Classical Mechanics to Quantum Field Theory. A Tutorial

v∈Ris the parameter of the group. The generator is[ 5 ]the unique selfadjoint
extension of

Kn(t)=

∑^3

j=1

nj(mXj|D−tPj|D), (2.103)

the constantm> 0 denoting the mass of the system andDbeing the G ̊ading or the
Nelson domain of the representation of (central extension of the) Galilean group
as we will discuss later.
(c)In QM, there are symmetries described by operators which are simulta-
neously selfadjoint and unitary, so they are also observables and can be mea-
sured. Theparityis one of them: (Pψ)(x):=ψ(−x)for a particle described
inL^2 (R^3 ,d^3 x). These are constants of motion (Ut−^1 PUt=P)ifandonlyif
they are dynamical symmetries (PUt=PUt). This phenomenon has no classical
correspondence.
(d)Thetime reversal symmetry, when described by an antiunitary operator
T is supposed to satisfy: THT−^1 =H. However, since it is antilinear, it gives
rise to the identity (exercise)Te−itHT−^1 =e−itT HT

− 1
, so thatTUt=U−tT as
physically expected. There is no conserved quantity associated with this operator
because it is not selfadjoint.

Exercise 2.3.69.
(1)Prove that if the Hamiltonian observable does not depend on time, it is a
constant of motion.

Solution. In this case, the time translation is described byUt=eitH and
trivially it commute withUs. Noether theorem implies the thesis.

(2)Prove that for the free particle inR^3 , the momentum alongx 1 is a constant of
motion as consequence of translational invariance along that axis. Assume that the
unitary group representing translations alongx 1 isUuwith(Uuψ)(x)=ψ(x−ue 1 )
ifψ∈L^2 (R^3 ,d^3 x).

Solution.The Hamiltonian isH= 21 m

∑ 3

j=1P
j^2. It commutes with the one-
parameter unitary group describing displacements alongx 1 , because as one can
prove, the said groups is generated byP 1 itself: Uu:=e−iuP^1. Theorem 2.3.67
yields the thesis.

(3)Prove that ifσ(H)is bounded below but not above, the time reversal symmetry
cannot be unitary.