3.2 Diffusion Equation and Propagation Kernel

In the previous subsection, we have deduced the form of the propagation kernel (36), by evaluating the sum over histories as a Gaussian integral. With this result in hands, one can make contact with the more familiar formulation of quantum mechanics by showing that the propagation kernel $K$ satisfies a diffusion equation of the Schrödinger type. To see this, first we calculate

$${\partial\over\partial T}K( {\vec{\mbox{\boldmath {$x$}}}}... ...h {$q$}}}}\cdot {\vec{\mbox{\boldmath {$q$}}}}\right\} \quad ,$$ (38)

and

$$\Delta_x K( {\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\bo... ...h {$q$}}}}\cdot {\vec{\mbox{\boldmath {$q$}}}}\right\} \quad .$$ (39)

Comparing the two equations above, we conclude that

$${\hbar^2\over 2m}\Delta_x K( {\vec{\mbox{\boldmath {$x$}}}... ...ath {$x$}}}}- {\vec{\mbox{\boldmath {$x$}}}}_0 ; T ) \quad .$$ (40)

Finally, recalling the relation between the amplitude $K$ and the wave function, namely

$$\psi( {\vec{\mbox{\boldmath {$x$}}}} , t)= \int d^3 x_... ...t ) \psi({\vec{\mbox{\boldmath {$x$}}}}_0 , 0) \quad ,$$ (41)

we also conclude that $\psi( {\vec{\mbox{\boldmath {$x$}}}} , t )$ must satisfy the time-dependent Schrödinger equation.
At this point, it seems pedagogically instructive to reverse the procedure, and show that the propagation kernel can be determined by solving the diffusion equation (40) which we now assume as given. To this end, we make the following ansatz,

$$K( {\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\boldmath {$... ...box{\boldmath {$x$}}}}_0 ; T )/\hbar \right) \quad ,$$ (42)

in terms of two trial functions $F(T)$ and $M( {\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\boldmath {$x$}}}}_0 ; T )$. The overall normalization constant may be determined by the same boundary condition (35).
In order to determine the form of the trial functions, we demand that the tentative expression (42) satisfies equation (40). Thus, one finds

$${F(T)\over 2m} \Delta_x M ({\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\boldmath {$x$}}}}_0 ; T)=-{ d F(T)\over d T}$$ (43)

and

$${1\over 2m}\left(\vec\nabla M\right)\cdot \left(\vec\nabla M\right)=-{\partial M\over \partial T} \quad .$$ (44)

Note that equation (44) is just the classical Jacobi equation (23). Thus, without further calculations we can identify the phase of the kernel with the classical action:

$$M( {\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\boldmath {$... ...dmath {$x$}}}}-{\vec{\mbox{\boldmath {$x$}}}}_0\vert^2 \quad .$$ (45)

Next, in order to determine the dependence of the kernel pre-factor on $T$, we make use of equation (43). Therefore, we first apply the Laplacian operator to $S_{\mathrm{cl.}}( {\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\boldmath {$x$}}}}_0 ; T )$:

$${\partial\over\partial x^i}S_{\mathrm{cl.}}( {\vec{\mbox{\boldm... ... \left({\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\boldmath {$x$}}}}_0\right)_i$$ (46)

$${\partial^2 \over \partial x^k \partial x^i} S_{\mathrm{cl.}}( ... ...x$}}}}-{\vec{\mbox{\boldmath {$x$}}}}_0 ; T ) = {m\over T}\delta_{k i}$$ (47)

$$\delta^{k i}{\partial^2\over \partial x^k \partial x^i} S_{\math... ...{\boldmath {$x$}}}}_0 ; T )= {m\over T}\delta^k{}_k={3m\over T} \quad .$$ (48)

Then, we substitute the result into equation (43), which now takes the form

$${dF\over dT}=-{3\over 2T}F \quad\Rightarrow \quad {dF\over F}=-{3\over2}{dT\over T} \quad .$$ (49)

Finally, integrating the last equation, we obtain

$$F(T)={\rm const.}\times T {}^{-3/2} \quad .$$ (50)

Substituting all of the above in the original ansatz (42), leads to the following expression for the propagation kernel,

$$K( {\vec{\mbox{\boldmath {$x$}}}}-{\vec{\mbox{\boldmath {$... ...box{\boldmath {$x$}}}}_0 ; T )/\hbar \right) \quad .$$ (51)

Except for the normalization constant which is fixed by the initial condition, this is the same expression obtained from the path integral.