5.2 Integrating the path integral

As a consistency check on the above result, and in order to clarify some further properties of the path integral, it may be useful to offer an alternative derivation of equation (43) which is based entirely on the usual gaussian integration technique. As we have seen in the previous section, the Feynman amplitude can be written as follows

 

$$K [ x (s) , x _{0} (s) ; A ] = \int _{ x _{0} (s)} ^{x (s)} [D x ^{\mu} (\sigma)] [D p _{\mu \nu} (\sigma)] \times$$

$$\qquad \times \exp \left\{ \frac{i}{2 \hbar} \int _{X (\sigma)} p... ...n _{ab} \int _{\Sigma (\sigma)} d \xi ^{a} \wedge d \xi ^{b} H (p) \right\} \ .$$ (44)

In order to evaluate the functional integral (44), without discretization of the variables, we enlist the following equalities,

$$\int _{x _0 (s)} ^{x (s)} [D x ^{\mu} (\sigma)] \exp \left\{ \fra... ...bar} \int _{X (\sigma)} p _{\mu \nu} \, d x ^{\mu} \wedge d x ^{\nu} \right\} =$$

$$\quad = \int _{x _0 (s)} ^{x (s)} [D x ^{\mu} (\sigma)] \exp \lef... ...gma)} \! \! \! \! x ^{\mu} \, d p _{\mu \nu} \wedge d x ^{\nu} \right] \right\}$$

$$\quad = \delta \left[ d \left( p _{\mu \nu} d x ^{\nu} \right) \r... ...}{2 \hbar} \int _{\partial X (s)} p _{\mu \nu} x ^{\mu} d x ^{\nu} \right\} \ .$$ (45)

The functional delta function has support on the classical, extremal trajectories of the string. Therefore, the momentum integration is restricted to the classical area-momenta and the residual integration variables are the components of the area-momentum along the world-sheet boundary $p _{\mu \nu} (s)$. As a matter of fact, boundary conditions fix the initial and final string loops C ₀ and C but not the conjugate momenta. In analogy to the point particle case, the classical equations of motion on the final world-sheet boundary

$$d \left( p _{\mu \nu} d x ^{\nu} \right) \Big \vert _{x = x (s)} = 0$$ (46)

require that the three normal components of $p _{\mu \nu}$ be constant, i.e. $p _{\mu \nu} x ^{\prime \, \nu} (s)= \mathrm{const.}$ Hence, the functional integral over the boundary momentum reduces to a three dimensional, generalized, Gaussian integral

$$\int [D p _{\mu \nu} (\sigma)] \delta \left[ d \left ( p _... ... \right] (\dots) = \int [d p _{\mu \nu}] (\dots) \quad .$$ (47)

Moreover, the Hamiltonian is constant over a classical world-sheet and can be written in terms of the boundary $p _{\mu \nu}$. In such a way, the path integral is reduced to the Gaussian integral over the three components of $p _{\mu \nu}$ which are normal to the boundary

 

$$\! = \! \mathcal{N} \! \! \int \! [d p _{\mu \nu}] \! \exp \! \left\{ \! ... ...) d x ^{\nu} - \frac{A}{4 m ^{2}} p _{\mu \nu} p ^{\mu \nu} \right] \! \right\}$$ K [ x (s) , x ₀ (s) ; A ]

$$\! = \! \mathcal{N} \! \! \int \! [d p _{\mu \nu}] \! \exp \! \left\{ \! ... ... - C _{0}] - \frac{A}{4 m ^{2}} p _{\mu \nu} p ^{\mu \nu} \right] \! \right\} ,$$

and correctly reproduces the expression (43).