Going back through our derivation we have I integraldisplay D \u03a6 1 x D \u03a6 n x e

# Going back through our derivation we have i

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Going back through our derivation, we have I = integraldisplay D Φ 1 ( x ) D Φ n ( x ) ( 0 | e iH ( t n ) δt | Φ n ) ( Φ 2 | e iH ( t 2 ) δt | Φ 1 )( Φ 1 | e iH ( t 1 ) δt | 0 ) Φ j ( x j ) 148 Path Integrals
since the subscript on Φ is just it’s point in time, we know which Φ i ’s these correspond to. Let’s take the part with just Φ j integraldisplay D Φ j ( x ) braceleftBig e iH ( t n ) δt | Φ j ) Φ j ( x j ) ( Φ j | bracerightBig = φ ˆ ( x j ) integraldisplay D Φ j ( x ) | Φ j )( Φ j | (13.27) So we get to replace Φ( x j ) by the operator φ ˆ ( x j ) stuck in at the time t j . Then we can collapse up all the integrals to give integraldisplay D Φ( x,t ) e iS [Φ] Φ( x j ,t j ) = (Big 0 vextendsingle vextendsingle vextendsingle φ ˆ ( x j ,t j ) vextendsingle vextendsingle vextendsingle 0 )Big (13.28) If you find the collapsing-up-the-integrals confusing, just think about the derivation backwards. An inser- tion of φ ˆ ( x j ,t j ) will end up by | Φ j )( Φ j | , and then pull out a factor of Φ( x j ,t j ) . Now say we insert two fields integraldisplay D Φ( x,t ) e iS [Φ] Φ( x 1 ,t 1 )Φ( x 2 ,t 2 ) (13.29) The fields will get inserted in the appropriate matrix element. But if you check, you will see that the ear- lier field will always come out on the right of the later field. So we get integraldisplay D Φ( x ) e iS [Φ] Φ( x 1 )Φ( x 2 ) = (Big 0 vextendsingle vextendsingle vextendsingle T braceleftBig φ ˆ ( x 1 ) φ ˆ ( x 2 ) bracerightBigvextendsingle vextendsingle vextendsingle 0 )Big (13.30) In general, integraldisplay D Φ( x ) e iS [Φ] Φ( x 1 ) Φ( x n ) = (Big 0 vextendsingle vextendsingle vextendsingle T braceleftBig φ ˆ ( x 1 ) φ ˆ ( x n ) bracerightBigvextendsingle vextendsingle vextendsingle 0 )Big (13.31) Thus we get time ordering for free in the path integral! Why does this work? There are a few cross checks you can do. As an easy one, suppose the answer were integraldisplay D Φ( x ) e iS [Φ] Φ( x 1 )Φ( x 2 ) = (Big 0 vextendsingle vextendsingle vextendsingle φ ˆ ( x 1 ) φ ˆ ( x 2 ) vextendsingle vextendsingle vextendsingle 0 )Big (13.32) Well, the left hand side doesn’t care whether I write Φ( x 1 )Φ( x 2 ) or Φ( x 2 )Φ( x 1 ) , since these are classical fields. So what would determine what order I write the fields on the right? We see it must be something that makes the fields effectively commute, like the time-ordering operator. We’ll do another check once we get a little more experience playing with the path integral. From now on, we’ll just use φ ( x ) instead of Φ( x ) , for the classical fields. 13.6.1 Current shorthand There’s a great way to calculate path integrals using currents. Let’s add a source term to our action. So define Z [ J ] = integraldisplay D φ exp braceleftbigg iS [ φ ] + i integraldisplay d 4 xJ ( x ) φ ( x ) bracerightbigg (13.33) Then, Z [0] = integraldisplay D φe i integraltext d 4 x L [ φ ] = ( 0 | 0 ) (13.34) Next, observe that d dJ ( x 1 ) integraldisplay d 4 xJ ( x ) φ ( x ) = φ ( x 1 ) (13.35) So, i dZ dJ ( x 1 ) = integraldisplay D φ exp braceleftbigg iS [ φ ] + i integraldisplay d 4 xJ ( x ) φ ( x ) bracerightbigg φ ( x 1 ) (13.36) And thus i dZ dJ ( x 1 ) | J =0 = integraldisplay D φ exp { iS [ φ ] } φ ( x 1 ) = ( 0 | φ ( x 1 ) | 0 ) (13.37) 13.6 Time-ordered products 149
Similarly, ( i ) n d n Z dJ ( x 1 ) dJ ( x n ) | J =0 = ( 0 | T { φ ( x 1 ) φ ( x n ) }| 0 ) (13.38) So this is a nice way of calculating time-ordered products.

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