QuantumPhysics.dvi

Figure 16: The evolution operator given as a summation over paths,qa=qandqb=q′

The problem will be, however, that the HamiltonianHis not generally diagonal in the position basis
(as was clear from the free particle Hamiltonian already). Of course we could use the momentum
basis, in which the free particle Hamiltonian is diagonal, but a general Hamiltonian will not be
diagonal in this basis either. The trick is to use a mixed basis. Consider for example a general
potential type Hamiltonian, of the form,

H(Q,P) =

1

2 m

P^2 +V(Q) (14.24)

It is manifest that we have〈q|H(Q,P)|p〉=〈q|p〉H(q,p), whereH(q,p) is just the classical Hamil-
tonian. Now, given a general Hamiltonian, H(Q,P), it can always be rearranged so that allP
operators are on the right of allQoperators. It is in this form that we can use it to define the
classical HamiltonianH(q,p) by,

〈q|H(Q,P)|p〉=〈q|p〉H(q,p) (14.25)

We are now in a position to complete our calculations. We shall define the classical Hamiltonian
by

〈q|U(ε)|p〉=〈q|p〉

(

1 −i

ε

̄h

H(q,p) +O(ε^2 )

)

=〈q|p〉exp

{

−i

ε

̄h

H(q,p)

}

(14.26)

and use this expression to compute the matrix element,

〈qn|U(ε)|qn− 1 〉 =

∫

dpn

2 π ̄h

〈qn|U(ε)|pn〉〈pn|qn− 1 〉