2.2 Plane wave solution

The next step in our program to set up a (spacetime covariant) functional quantum mechanics of closed strings, is to find the basic solutions of equation (7). As in the quantum mechanics of point particles, we seek first plane-wave solutions

 

$$\Psi (C ; 0) {\mathrm{const.}} \cdot \exp \left( \frac{i}{2} \oint _{C} x ^{\mu} d x^{\nu} P _{\mu \nu} (x) \right)$$ =

$${\mathrm{const.}} \cdot \exp \left( \frac{i}{2} \int _{0} ^{1} ds \, x ^{\mu} \, \frac{d x ^{\nu}}{d s} P _{\mu \nu} (s) \right) \quad .$$ = (12)

The overall constant in the above equation will be fixed by a suitable normalization condition. Apart from that, the wave functional (12) represents an eigenstate of the loop total momentum operator

$$\left( \oint _{C} dl (\bar{s}) \right) ^{-1} i \oint _{C} d... ...u} (\bar{s})} \Psi (C ; 0) = P _{\mu \nu} (C) \Psi (C ; 0)$$ (13)

with eigenvalues

$$P _{\mu \nu} (C) = \left( \oint _{C} dl (\bar{s}) \right) ^{-1} \oint_C dl(\bar{s}) P _{\mu \nu} (\bar{s}) \quad .$$ (14)

Note that, while $\sigma ^{\mu \nu} (C)$ is a functional of the loop C, i.e., it contains no reference to a special point of the loop, the area derivative $\delta / \delta \sigma ^{\mu \nu} (\bar{s})$ operates at the contact point $y ^{\mu} = x ^{\mu} (\bar{s})$. Thus, the area derivative of a functional is no longer a functional. Even if the functional under derivation is reparametrization invariant, its area derivative behaves as a scalar density under redefinition of the loop coordinate. In order to recover reparametrization invariance, we have to get rid of the arbitrariness in the choice of the attachment point. This can be achieved by summing over all its possible locations along the loop, and then compensating for the overcounting of the area variation by averaging the result over the proper length of the loop. This is what we have done in equations (13) and (14), thereby trading the string momentum density $P _{\mu \nu} (\bar{s})$, i.e. a function of the loop coordinate $\bar{s}$, with a functional loop momentum $P _{\mu \nu} (C)$. The same prescription enables us to define any other reparametrization invariant derivative operator. Hence, we introduce a simpler but more effective notation, and define the loop derivative as

$$\frac{\delta}{\delta C ^{\mu \nu}} \equiv \left( \oint _{C}... ...C} dl (s) \frac{\delta}{\delta \sigma ^{\mu\nu} (s)} \quad .$$ (15)

This represents a genuine loop operation withoutn reference to the way in which the loop is parametrized.

With the above remarks in mind, the functional Laplacian operator appearing in equation (7) is reparametrization and Lorentz invariant, and is constructed according to the same prescription: attach two petals at the loop point s, then compute the functional variation and, finally, average over all possible locations of the attachment point. In the notation (15), the loop laplacian reads

$$\frac{1}{2} \frac{\delta ^{2}}{\delta C ^{\mu\nu} \delta C _... ...a \sigma ^{\mu \nu} (s) \delta \sigma _{\mu \nu} (s)} \quad .$$ (16)

Then, if we factorize the explicit A-dependence of the wave functional

$$\Psi (C ; A) = \Phi(C) \exp \left( - i {\mathcal{E}} A \right)$$ (17)

the stationary functional wave equation takes the form

$$- \frac{1}{4 m ^{2}} \frac{\delta ^{2} \Psi [C ; A]} {\del... ...elta C _{\mu \nu}} = {\mathcal{E}} \, \Psi[C ; A] \quad .$$ (18)

Thus, we need to evaluate the second functional variation of $\Psi(C ; 0)$ corresponding to the addition of two petals to C, say at the point $y ^{\mu} = x ^{\mu} (\bar{s})$. The first variation of $\Psi(C ; 0)$ can be obtained from Eq. (12)

$$\delta \Psi (C ; 0) = \frac{1}{2} P _{\mu \nu} (\bar{s}) \delta \sigma ^{\mu \nu} (\bar{s}) \Psi (C ; 0) \quad .$$ (19)

Reopening the loop C at the same contact point and adding a second infinitesimal loop, we arrive at the second variation of $\Psi(C ; 0)$,

$$\delta ^{2} \Psi(C ; 0) = \frac{1}{4} P _{\mu \nu} (\bar{s... ...) \delta \sigma ^{\rho \tau} (\bar{s}) \Psi (C ; 0) \quad .$$ (20)

Then, Eq. (12) solves equation (7) if

$${\mathcal{E}} = \frac{1}{4 m ^{2}} \left( \oint _{C} dl (s... ... \oint _{C} dl (s) P _{\mu \nu} (s) P ^{\mu \nu} (s) \quad ,$$ (21)

which is the classical dispersion relation between string energy and momentum. Having determined a set of solutions to the stationary part of equation (7), the complete solutions describing the quantum evolution from an initial state $\Psi( C _{0} , 0)$ to a final state $\Psi (C , A)$ can be obtained by means of the amplitude (1), as follows

 

$$\Psi [C ; A] \sum _{C _{0}} K [C , C _{0} ; A] \Psi [C _{0} ; 0]$$ =

$$\left( \frac{m ^{2}}{2 i \pi A} \right) ^{3/2} \int [{\mathcal{D}... ...4 A} \left( \sigma (C) - \sigma (C _{0}) \right) ^{2} \right] \Psi [C _{0} ; 0]$$ =

$$\frac{1}{(2 \pi) ^{3/2}} \exp \frac{i}{2} \left[ \oint _{C} P_{\mu\nu} x ^{\mu} dx ^{\nu} \right. +$$ =

$$\qquad \qquad \qquad \qquad \qquad \left . - {A \over 2 m ^{2}} \... ...nt _{C} dl(s) \right) ^{-1} \oint _{C} dl (s) P _{\mu \nu} P ^{\mu \nu} \right]$$

$$\frac{1}{(2 \pi) ^{3/2}} \exp \frac{i}{2} \left[ \oint _{C} P _{\mu \nu} x ^{\mu} d x ^{\nu} - {\mathcal{E}} A \right] \quad .$$ = (22)

As one would expect, the solution (22) represents a ``monocromatic string wave train'' extending all over loop space.