Quantum Mechanics for Mathematicians

higherdstraightforward. We are calculating the first-order change inSdue to
an infinitesimal changeδγ= (δq(t),δq ̇(t))

δS[γ] =

∫t 2

t 1

δL(q(t),q ̇(t))dt

=

∫t 2

t 1

(

∂L

∂q

(q(t),q ̇(t))δq(t) +

∂L

∂q ̇

(q(t),q ̇(t))δq ̇(t)

)

dt

But

δq ̇(t) =

d

dt

δq(t)

and, using integration by parts

∂L

∂q ̇

δq ̇(t) =

d

dt

(

∂L

∂q ̇

δq

)

−

(

d

dt

∂L

∂q ̇

)

δq

so

δS[γ] =

∫t 2

t 1

((

∂L

∂q

−

d

dt

∂L

∂q ̇

)

δq−

d

dt

(

∂L

∂q ̇

δq

))

dt

=

∫t 2

t 1

(

∂L

∂q

−

d

dt

∂L

∂q ̇

)

δqdt−

(

∂L

∂q ̇

δq

)

(t 2 ) +

(

∂L

∂q ̇

δq

)

(t 1 ) (35.1)

If we keep the endpoints fixed soδq(t 1 ) =δq(t 2 ) = 0, then for solutions to

∂L

∂q

(q(t),q ̇(t))−

d

dt

(

∂L

∂q ̇

(q(t),q ̇(t))

)

= 0

the integral will be zero for arbitrary variationsδq.

As an example, a particle moving in a potentialV(q) will be described by a
Lagrangian

L(q,q ̇) =

1

2

m|q ̇|^2 −V(q)

for which the Euler-Lagrange equations will be

−

∂V

∂qj

=

d

dt

(mq ̇j) =mq ̈j

This is just Newton’s second law, which says that the force due to a potential
is equal to the mass times the acceleration of the particle.
Given a Lagrangian classical mechanical system, one would like to be able
to find a corresponding Hamiltonian system that will give the same equations of
motion. To do this, we proceed by defining (for each configuration coordinate
qj) a corresponding momentum coordinatepjby

pj=

∂L

∂q ̇j