What is the action of 
on (x,p)? Let us begin with its
action on x. In order to perform this operation, we need two useful
identities (of which the first is actually a consequence of the second):
In order to see this, simply expand the left side in a Taylor series:
since all derivatives higher than the first order derivative go to 0 when acting on x. The second identity concerns the action of the same operator on a function g(x): Here, we need all orders of the Taylor series:
where 
indicates the kth derivative of g. However, the last line
is just the Taylor series expansion of g(x+c). Thus,
The same identities hold for the action of 
on p and on h(p).
Note that the action of 
on p or h(p) has no effect, i.e.
it acts like an identity operator and the same for on x or g(x).
Using these identities, we can now evaluate the action of on x:
The first operator that acts on x has no effect, since it only involves momentum derivatives. Thus,
The second operator changes x to 
:
The last operator has no effect on x but acts on the p that was introduced
in the second step. It changes p to 
. Thus,
We note that the x and p that appear in the above expression are
the initial conditions (x(0),p(0)) and that 
produces
the approximate evolution 
, which takes the form of the
second order Taylor expansion:
Similarly, the action of on p is
The first operator changes p to :
The next operator changes F(x) to 
and has no effect on p:
Finally, the last operator changes p to in both places
where it appears in the above expression,
Note that the argument of the second force in the above expression is just, .
Also, the x and p that appear in the above are, again, the initial conditions,
(x(0),p(0)), so that we can write
Combining the expressions for and 
, we see that the
produces the velocity Verlet algorithm that we introduced earlier in the lectures.
However, it has been derived in a powerful way, which manifestly shows that the
algorithm is symplectic and time-reversible.
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Mark Tuckerman