QuantumPhysics.dvi

9.2 Noether’s Theorem

(1) Ifδqi(t) is a symmetry transformation, there exists aconserved chargeC, given by

C=

∑

i

piδqi−X pi=

∂L

∂q ̇i

(9.4)

The chargeCis conserved, namely we haveC ̇ =dC/dt= 0, for any pathqi(t) which obeys

the Euler-Lagrange equations. The result is proven by straightforward calculation, and the

use of the Euler-Lagrange equations.

(2) The conserved charge generates the infinitesimal transformation as follows,

δqi={qi,C} (9.5)

This result is proven by starting from the δL= X ̇ (which holds without the use of the

Euler-Lagrange equations), using it to compute∂X/∂piand∂X/∂qi, and then using those

derivatives to evaluate{qi,C}. Note that conservation meansC ̇= 0,

C ̇ =dC

dt

=

∂C

∂t

+{C,H}= 0 (9.6)

even thoughCmay haveexplicit time dependence.

A first simple example is provided by translations of the dynamical variablesqi. The

finite transformation is ̃qi(t,α) =qi(t) +αviwhereviare constants. As a result,δqi(t) =vi,

andδq ̇i= 0. Variation of the Lagrangian gives,

δL=

∑

i

∂L

∂qi

vi (9.7)

which expresses the derivative ofLwith respect to the combination

∑

iqi(t)vi. For this to be

a symmetry, we needδLto be a total time derivative of a local quantityX, without using

the Euler-Lagrange equations. But this can happen only ifδL=X= 0. in turn expressing

the fact that the Lagrangian does not actually depend on the combination

∑

iqi(t)vi. As a

result, the associated conserved charge is given by

Cv=

∑

i

piδqi=

∑

i

pivi (9.8)

expressing the fact that the associated momentum in that direction is conserved.

A second example is provided by time translations, whose finite transformation is ̃qi(t,α) =

qi(t+α). The associated infinitesimal transformation isδqi(t) = ̇qi(t), and we also have

δq ̇i(t) = ̈qi(t). Now compute the variation of the Lagrangian,

δL=

∑

i

(

∂L

∂qi

q ̇i+

∂L

∂q ̇i

q ̈i

)

=

dL

dt

−

∂L

∂t

(9.9)