Any Old Ordering

Consider a wide range of operator orderings:

(1)
\begin{align} \hat H = \frac{\hbar^2}{2} f(a) \partial_a g(a) \partial_a h(a) + V(a). \end{align}

The kinetic term can be expanded as

(2)
\begin{align} \frac{\hbar^2}{2} \Big[ a^{-1} \partial_a^2 + \Big( f \left( a \right) g' \left( a \right) h \left( a \right) + 2 f \left( a \right) g \left( a \right) h' \left( a \right) \Big) \partial_a + \Big( f \left( a \right) g' \left( a \right) h' \left( a \right) + f \left( a \right) g \left( a \right) h'' \left( a \right) \Big) \Big] \end{align}

We want to realize this as something of the form

(3)
\begin{align} \frac{\hbar^2}{2} \Big( F \left( a \right) \cdot G \left( a \right) \partial_a G \left( a \right) \partial_a \cdot \left[ F \left( a \right) \right]^{-1} \Big) + \Big( \textit{Quantum Potential } \Big). \end{align}

To do so, expand the effective kinetic term above, using the shorthand $F^{-1} \left( a \right)$ for $\left[ F \left( a \right) \right]^{-1}$ (not to be confused with the inverse function for $F \left( a \right)$):

(4)
\begin{align} & \frac{\hbar^2}{2} \Big( F \left( a \right) \cdot G \left( a \right) \partial_a G \left( a \right) \partial_a \cdot \left[ F \left( a \right) \right]^{-1} \Big) \\ = & \frac{\hbar^2}{2} \Big[ G^2 \left( a \right) \partial_a^2 + \Big( - G^2 \left( a \right) F^{-1} \left( a \right) F' \left( a \right) + G \left( a \right) G' \left( a \right) - G^2 \left( a \right) F^{-1} \left( a \right) F' \left( a \right) \Big) \partial_a - \Big( G \left( a \right) G' \left( a \right) F^{-1} \left( a \right) F' \left( a \right) - 2 G^2 \left( a \right) F^{-2} \left( a \right) \left[ F' \left( a \right) \right]^2 + G^2 \left( a \right) F^{-1} \left( a \right) F'' \left( a \right) \Big) \Big] \end{align}

So we have

(5)
\begin{align} \partial_a^2 & \implies G^2 \left( a \right) = a^{-1} \implies G \left( a \right) = a^{-1/2} \\ \partial_a & \implies G \left( a \right) G' \left( a \right) - 2 G^2 \left( a \right) F^{-1} \left( a \right) F' \left( a \right) = f \left( a \right) g' \left( a \right) h \left( a \right) + 2 f \left( a \right) g \left( a \right) h' \left( a \right) \\ & \implies - \frac{1}{2} a^{-2} - 2 a^{-1} \frac{F' \left( a \right)}{F \left( a \right)} = f \left( a \right) g' \left( a \right) h \left( a \right) +2 f \left( a \right) g \left( a \right) h' \left( a \right) \\ & \implies \frac{F' \left( a \right)}{F \left( a \right)} = - \frac{a^{-1}}{4} - \frac{a}{2} \Big( f \left( a \right) g' \left( a \right) h \left( a \right) + 2 f \left( a \right) g \left( a \right) h' \left( a \right) \Big) \end{align}

Interestingly, we are always brought to an effective kinetic term which is the Laplace-Beltrami operator for the metric $\left( a \right)$.

Note that $f \left( a \right) g \left( a \right) h \left( a \right) = a^{-1}$, so we have

(6)
\begin{align} \frac{F' \left( a \right)}{F \left( a \right)} = - \frac{a^{-1}}{4} - \frac{a}{2} \left( a^{-1} \frac{g' \left( a \right)}{g \left( a \right)} + 2 a^{-1} \frac{h' \left( a \right)}{h \left( a \right)} \right) \end{align}

Notice that the RHS only depends on logarithmic derivatives of $g$ and $h$ (denote these by $\rho \left( a \right) = \frac{g' \left( a \right)}{g \left( a \right)}$ and $\sigma \left( a \right) = \frac{h' \left( a \right)}{h \left( a \right)}$). Thus

(7)
\begin{align} &\frac{d}{da} \left[ \ln F \right] = - \frac{a^{-1}}{4} - \frac{1}{2} \frac{d}{da} \left[ \ln g + 2 \ln h \right] \\ & \implies F \left( a \right) = a^{-1/4} g^{-1/2} h^{-1} \end{align}

(note that because $F$ only appears in the differential operator (for the effective kinetic term) along with $F^{-1},$ we can drop the constant of integration). $F \left( a \right)$ now gives the transformation which puts our Hamiltonian in the form of an effective kinetic term on a curved background (with metric $(a)$) plus the usual potential modified by the addition of a quantum potential.

Also, we can read off the quantum potential to be

(8)
\begin{align} & \Big( f \left( a \right) g' \left( a \right) h' \left( a \right) + f \left( a \right) g \left( a \right) h'' \left( a \right) \Big) \\ & - \Big( - G \left( a \right) G' \left( a \right) F^{-1} \left( a \right) F' \left( a \right) + G^2 \left( a \right) F^{-2} \left( a \right) \left[ F' \left( a \right) \right]^2 - G^2 \left( a \right) \partial_a \left[ F' \left( a \right) F^{-1} \left( a \right) \right] \Big). \end{align}

Using $F' \left( a \right) F^{-1} \left( a \right) = \phi \left( a \right)$, as well as the definitions of $\rho \left( a \right)$ and $\sigma \left( a \right)$ we get

(9)
\begin{align} QP = a^{-1} & \Big( \rho \left( a \right) \sigma \left( a \right) + \sigma' \left( a \right) + \sigma^2 \left( a \right) \Big) + \left( - \frac{a^2}{2} \phi \left( a \right) - a^{-1} \phi^2 \left( a \right) + a^{-1} \phi ' \left( a \right) \right). \end{align}

Using the definition of $\phi \left( a \right)$ in terms of $\rho \left( a \right)$ and $\sigma \left( a \right)$ , we get

(10)
\begin{align} QP = \frac{5}{16} a^{-3} + a^{-1} \left( - \frac{ \rho^2 \left( a \right) }{4} - \frac{\rho ' \left( a \right)}{2} \right). \end{align}

Thus it is evident that the form of the quantum potential is completely determined by the logarithmic derivative $\rho \left( a \right) = \frac{g' \left( a \right)}{g \left( a \right)}$. We can now get quantum potentials other than just $a^{-3}$. It is probably reasonable to restrict the functions $f,g,h$ to have expansions of the form

(11)
$$c_n a^{-n} + c_{-n+1} a^{-n+1} + ... + c_{-1} a^{-1} + c_0 + c_1 a + c_2 a^2 + ...,$$

and to have no zeros or poles other than at $a = 0$.

page revision: 93, last edited: 26 Jul 2014 14:10