NRQM10sol_1

NRQM10sol_1 - Physics 216 Solutions to Problem Set 1 Spring...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 216 Solutions to Problem Set 1 Spring 2010 1. Prove that ( x, t | T [ X ( t 1 ) X ( t 2 ) X ( t n )] | x , t ) = integraldisplay D [ x ( t )] x ( t 1 ) x ( t 2 ) x ( t n ) e iS [ x ( t )] / planckover2pi1 , where T is the time-ordered product symbol, S [ x ( t )] is the action [which depends on the path x ( t )], X ( t ) is the position operator in the Heisenberg picture, and x ( t ) is the eigenvalue of X ( t ) when acting on the position eigenstate | x, t ) . Assume that t t i and t t i for all i = 1 , 2 ,...,n . In general, the Heisenberg position operators X ( t ), evaluated at different times, do not commute. Therefore, a product of Heisenberg position operators, where each operator is evaluated at a different time, will depend on the order in which the operators appears. One particular order is one in which the largest time appears to the left and all the other operators appear in time order (i.e., time decreases as we move from left to right). This particular order corresponds to the time-ordered product, T [ X ( t 1 ) X ( t 2 ) X ( t n )] = X ( t i 1 ) X ( t i 2 ) X ( t i n ) , where t i 1 > t i 2 > > t i n . Consider the derivation of the path integral for ( x,t | T [ X ( t 1 ) X ( t 2 ) X ( t n )] | x ,t ) = ( x,t | X ( t i 1 ) X ( t i 2 ) X ( t i n ) | x ,t ) . The problem states that t t i 1 and t t i n . We can begin the derivation of the path integral in the same way as presented in the class lecture, by dividing up the time interval between t and t into slices: t t i n t i n 1 t i 2 t i 1 t. In contrast to the derivation presented in class, we do not assume that t i k = t + k (i.e., uniform time slices), since the t i k ( k = 1 , 2 ,...,n ) are fixed by the product of the Heisenberg position operators. Next, we insert a complete set of states, integraldisplay | x i k ,t i k )( x i k ,t i k | dx i k , immediately following each Heisenberg position operator (for k = 1 , 2 ,...,n ). Using X ( t i k ) | x i k ,t i k ) = x ( t i k ) | x i k ,t i k ) , where x ( t i k ) is the ( c-number) eigenvalue of X ( t i k ), it follows that: T [ X ( t 1 ) X ( t 2 ) X ( t n )] = integraldisplay dx i 1 integraldisplay dx i 2 integraldisplay dx i n x ( t i 1 ) x ( t i 2 ) x ( t i n ) ( x,t | x i 1 ,t i 1 )( x i 1 ,t i 1 | x i 2 ,t i 2 )( x i n ,t i n | x ,t ) . (1) 1 Since the x ( t i k ) are commuting c-numbers, we can rewrite eq. (1) as: T [ X ( t 1 ) X ( t 2 ) X ( t n )] = integraldisplay dx i 1 integraldisplay dx i 2 integraldisplay dx i n x ( t 1 ) x ( t 2 ) x ( t n ) ( x,t | x i 1 ,t i 1 )( x i 1 ,t i 1 | x i 2 ,t i 2 )( x i n ,t i n | x ,t ) ....
View Full Document

This document was uploaded on 10/31/2011.

Page1 / 20

NRQM10sol_1 - Physics 216 Solutions to Problem Set 1 Spring...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online