Nn ffpini n e 10 0 t0


where the derivative with respect to tn does not bring down the pn_i<j„_i term from the exponential because eq„-i is independent of time. The integral over pn_: can be replaced by reversing the steps which led up to (14.14):

and hence we obtain i

Qj-ip{qn,tn) -H (-iV„,q„) J dqn-i (qn,tn\qn-i,tn-i)H

x jdq0K{qn-x,tn-i\qo,to)i>{qoM) -H (-iVn,q„) j dqn-i (qn,tn\qn-i,tn-i)H

x Jdq0 {qn-i,tn-i\qo,to)Hip{qo,to) =H (-¿Vn,g„) J dq0K{q id ^n!

where the completeness relation was used in the third step. Hence the path integral (14.18) is fully equivalent to the Schrodinger equation, and we have an alternative way to describe quantum mechanics.

The free wave function which enters into Eq. (14.29) is the evolution of the state a under the free Hamiltonian Ho, given by

<Pa{Qn; tn) = J dq0 K(j{qn, tn; qoto)4>a{qo,t0) , (14.31)

where the free propagator is K0(q2, t2; q\,t\) = 0(<?2, ^ki, t\)0- These states can be regarded as plane wave states (although they could be atomic states, or some other basis of exact solutions of H0).

We will now prove Eq. (14.29) by demonstrating that it gives the same perturbation expansion for S as the previously derived expansion given in Eqs. (14.27) and (14.28). We will be satisfied to show this for the first three terms in the expansion. Using perturbation theory, the exact propagator (14.30) can be expanded in a power series in Hi,

where Kn is proportional to (Hi)n, and K0 is just the free propagator. Hence, we must prove

(%*)„ = N' J dqndqo<j>p{qn,tn)Kn(qn,tn-,qo,to)4>a(qo,to) ■ (14.33)

The first term in this series is

(Sj3a)0 =«'j dqndqo<t>p(qn,tn)Ko(qn,tn-,q0,t0)<t>a(qo,to)

= M' J dqn(t>*0{qn,tn)<f>a{qntn) =N'80a , (14.34)

where the free propagation of the wave function, as described in Eq. (14.31), was used in the last step. This will agree with the first term in Eq. (14.28) if we choose

0 0

Post a comment