1549380323-Statistical Mechanics Theory and Molecular Simulation

112 Microcanonical ensemble

(

x(∆t)

p(∆t)

)

≈exp

(

∆t

2

F(x(0))

∂

∂p(0)

)

×exp

(

∆t

p(0)

m

∂

∂x(0)

)

×exp

(

∆t

2

F(x(0))

∂

∂p(0)

)(

x(0)

p(0)

)

. (3.10.24)

In order to make the notation less cumbersome in the proceeding analysis, we will
drop the “(0)” label and write eqn. (3.10.24) as

(

x(∆t)

p(∆t)

)

≈exp

(

∆t

2

F(x)

∂

∂p

)

exp

(

∆t

p

m

∂

∂x

)

exp

(

∆t

2

F(x)

∂

∂p

)(

x

p

)

. (3.10.25)

The “(0)” label will be replaced at the end.
The propagation is determined by acting with each of the three operators in suc-
cession onxandp. But how do we apply exponential operators? Let us start by asking
how the operator exp(c∂/∂x), wherecis independent ofx, acts on an arbitrary func-
tiong(x). The action of the operator can be worked out by expanding the exponential
in a Taylor series

exp

(

c

∂

∂x

)

g(x) =

∑∞

k=0

1

k!

(

c

∂

∂x

)k

g(x)

=

∑∞

k=0

1

k!

ckg(k)(x), (3.10.26)

whereg(k)(x) = dkg/dxk. The second line of eqn. (3.10.26) is just the Taylor expansion
ofg(x+c) aboutc= 0. Thus, we have the general result

exp

(

c

∂

∂x

)

g(x) =g(x+c), (3.10.27)

which we can use to evaluate the action of the first operator in eqn.(3.10.25):

exp

(

∆t

2

F(x)

∂

∂p

)(

x

p

)

=





x

p+∆ 2 tF(x)



. (3.10.28)

The second operator, which involves a derivative with respect to position, acts on the
xappearing in both components of the vector on the right side of eqn. (3.10.28):

exp

(

∆t

p

m

∂

∂x

)





x

p+∆ 2 tF(x)



=





x+ ∆tmp

p+∆ 2 tF

(

x+ ∆tmp

)



. (3.10.29)