# Parenrightbigg 2 1 2 integraldisplay d 4 x

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parenrightbigg 2 1 2 integraldisplay d 4 x integraldisplay d 4 y [9 D xx D xy D yy + 6 D xy 3 ] + O ( λ 3 ) (82) These diagrams are the bubbles x y and x y Expanding Eq. (79) including terms up to O ( λ 2 ) in the numerator and denominator, we find (Big 0 vextendsingle vextendsingle vextendsingle T braceleftbig φ 0 ( x 1 ) φ 0 ( x 2 ) e i integraltext L in t bracerightbig vextendsingle vextendsingle vextendsingle 0 )Big (Big 0 vextendsingle vextendsingle vextendsingle T braceleftbig e i integraltext L int bracerightbig vextendsingle vextendsingle vextendsingle 0 )Big = D 12 g 2 integraltext bracketleftbig 1 8 D 12 D xx D xy D yy + 1 12 D 12 D xy 3 + bracketrightbig 1 g 2 integraltext bracketleftbig 1 8 D xx D xy D yy + 1 12 D xy 3 bracketrightbig (83) Since 1 1 + g 2 x = 1 g 2 x + O ( g 4 ) , we can invert the denominator in perturbation theory to see that the bubbles exactly cancel. Hamiltonian derivation 11
More generally, the bubbles will always factor out. Since the integrals in the expansion of the numerator corresponding to the bubbles never involve any external point, they just factor out. The sum over all graphs, in the numerator, is then the sum over all graphs with no-bubbles multiplying the sum over the bubbles. In pictures, + + x 1 x 2 x y + + (84) = parenleftbigg + + parenrightbigg × 1 + + + (85) The sum over bubbles is exactly (Big 0 vextendsingle vextendsingle vextendsingle T braceleftbig e i integraltext L in t bracerightbig vextendsingle vextendsingle vextendsingle 0 )Big . So, ( Ω | T { φ ( x 1 ) φ ( x 2 ) }| Ω ) = (Big 0 vextendsingle vextendsingle vextendsingle T braceleftbig φ 0 ( x 1 ) φ 0 ( x 2 ) e i integraltext L int bracerightbig vextendsingle vextendsingle vextendsingle 0 )Big no bubbles (86) where “no-bubbles” means that every connected subgraph involves an external point. 3.6 Position space Feynman rules We have shown that the same sets of diagrams appear in the Hamiltonian approach as in the Lagrangian approach: each point x i in the original n -point function ( Ω | T { φ ( x 1 ) φ ( x n ) }| Ω ) gets an external point and each interaction gives a new vertex whose position is integrated over and coefficient is given by the coefficient in the Lagrangian. As long as the vertices are normalized with appropriate permutation factors, as in Eq.(27), the combinatoric factors will work out the same, as we saw in the example. In the Lagrangian approach, we saw that the coefficient of the diagram will be given by the coefficient of the inter- action multiplied by the geometrical symmetry factor of the diagram. To see that this is also true for the Hamiltonian, we have to count the various combinatoric factors: There is a factor of 1 m ! from the expansion of exp ( i L int ) = 1 m ! ( i L int ) m . If we expand to order m there will be m identical vertices in the same diagram. We can also swap these vertices around, leaving the diagram looking the same. If we only include the diagram once in our final sum, the m ! from permuting the diagrams will cancel the 1 m !