Derivation of Path Integral from Factor Ordering

Main

Steigl-Hinterleitner-ordered Hamiltonian:

(1)
\begin{array} {l c l} \hat H & = & \frac{1}{2} \hbar^{2} a^{-i} \partial_{a} a^{-j} \partial_{a} a^{-k} + V(a) \\ \ & = & \frac{1}{2} \hbar^{2} \left[ a^{-1} \partial_{a}^{2} - (j+2k)a^{-2} \partial_{a} +k(j+k+1)a^{-3} \right] + V(a) \\ \end{array}

where $V(a) = \frac{1}{2} \left( -a + \frac{ \Lambda}{3} a^3 \right)$.

Measure: $a^{i-k} da \implies \langle \psi_1 \vert \psi_2 \rangle = \int_{\mathbb R^+} a^{i-k} da \left[ \bar \psi_1 \psi_2 \right]$

Introduce modified momentum $\hat p = -i \hbar \left( \frac{\partial}{\partial a} + \frac{\Gamma}{2} \right), \Gamma = \left( i-k \right) a^{-1}$. W.R.T. inner product, $\hat p$ is self-adjoint (using integration by parts), and $\left[ \hat a , \hat p \right] = i \hbar$. Note that in the case $i-k = \frac{1}{2}, \Gamma = \frac{1}{2} a^{-1}$ and this matches the $\Gamma$ from Chaichian/Demichev (D'Olivo/Torres): $\frac{\partial}{\partial a} \left[ \ln a^{1/2} \right] = \frac{1}{a^{1/2}}\cdot \frac{1}{2a^{1/2}} = \frac{1}{2a}$.

Now let us write $\hat H = \frac{1}{2} \left( \hbar a^{-i} \partial_a a^{-j} \partial_a a^{-k} - a + \frac{\Lambda}{3} a^{3} \right)$ in terms of modified momentum $\hat p = -i \hbar \left( \frac{\partial}{\partial a} + \frac{\Gamma }{2}\right)$, dividing the kinetic term into a part ordered with the ordering $\left( \left( a^{-1} \hat p^{2} \right) \right) = \frac{1}{2}\left( a^{-1} \hat p^{2} + \hat p^{2} a^{-1} \right)$, and a quantum potential $\Delta V_{QP}$:

(2)
\begin{array} {l c l} -\frac{1}{2} \left( \hat p^2 a^{-1} + a^{-1} \hat p^{2} \right) &=& -\frac{1}{2} \left[ \left( -i \hbar \right)^2 \left( \partial_a + \frac{\Gamma}{2} \right)^2 a^{-1} + a^{-1} \left( -i \hbar \right)^2 \left( \partial_a + \frac{\Gamma}{2} \right)^2 \right] \\ \ &=& \hbar^2 \left[ a^{-1} \partial_a^2 + \left( -1 + i -k \right) a^{-2} \partial_a + \left( 1 - \left( i - k \right) + \frac{ \left( i-k \right)^2}{4} \right) a^{-3} \right]. \end{array}

Since

(3)
\begin{align} \hat H_{S/H} = \frac{\hbar^2}{2} \left[ a^{-1} \partial_a^2 + \left( -1 + i-k \right) a^{-2} \partial_a + k \left( -i +2 \right) a^{-3} \right] + V \left( a \right) \end{align}

we can define

(4)
\begin{align} H_{eff} =& -\frac{1}{2}a^{-1} p^{2} + V \left( a \right) + \frac{\hbar^2}{2} \left( k \left( -i +2 \right) - \left( 1 - \left( i -k \right) + \frac{\left( i - k \right)^{2}}{4}\right) \right) a^{-3} \\ =& -\frac{1}{2}a^{-1} p^{2} + V \left( a \right) + \hbar^2 \Delta V_{QP} \end{align}

so that

(5)
\begin{align} \left( \left( \hat H_{eff} \right) \right) = \hat H_{S/H} \end{align}

(using $\hat p = -i \hbar \left( \partial_a + \frac{\Gamma}{2} \right)$).

In constructing the Euclidean signature path integral, we wish to obtain the propagator for the Wick+conformal rotated Schrodinger equation (obtained through the substitution $s=-it$)

(6)
\begin{array} {l c l} \hbar \frac{\partial}{\partial s} \psi (a) &=& \hat H_{S/H} \psi (a) \\ \ &=& \left[ \frac{\hbar^2}{2}a^{-i}\partial_a a^{-j} \partial_a a^{-k} + V(a) \right] \psi (a) \\ \ &=& \left( \left( \hat H_{eff} \right) \right) \psi (a) \end{array}

The evolution operator for this equation is $\mathcal{U} \left( s, s_0 \right) = \exp\left[\frac{\left( s-s_0 \right)}{\hbar} \left( \left( \hat H_{eff} \right) \right) \right]$, so we are looking for the propagator

(7)
\begin{align} K \left( a'',t'' | a',t' \right)=\lim_{N \to \infty} \int_{0}^{\infty} \prod_{n=1}^{N-1} a_{n}^{i-k} da_{n} \prod_{m=1}^{N} \left< a_{m} | \mathcal{U} \left( s_{m}, s_{m-1} \right) | a_{m-1} \right> \end{align}

where

(8)
\begin{array} {l c l} \left< a_m | \mathcal{U} \left( s_{m} , s_{m-1} \right) | a_{m-1} \right> &=& \left< a_m | \exp \left[ \frac{\epsilon}{\hbar} \left( \left( \hat H_{eff} \right) \right) \right] | a_{m-1} \right> \\ &=& \left< a_m | 1 + \frac{\epsilon}{\hbar} \left( \left( \hat H_{eff} \right) \right) + \mathcal{O} \left( \epsilon \right) | a_{m-1} \right> \\ &\doteq& \left< a_m | a_{m-1} \right> + \frac{\epsilon}{\hbar} \left< a_m | \left( \left( \hat H_{eff} \right) \right) | a_{m-1} \right> \end{array}

which expands as

(9)
\begin{align} &\left< a_m | a_{m-1} \right> + \frac{\epsilon}{\hbar} \left< a_m | \left[ -\frac{1}{2} \frac{a^{-1} \hat p^2 + \hat p^2 a^{-1}}{2} + \frac{1}{2} \left( -a + \frac{\Lambda}{3} a^3 \right) + \hbar^2 \Delta V_{QP} \right] | a_{m-1} \right> \\ & \ \ \ \ \ =\left( a_m \right)^{(k-i)/2} \delta \left( a_m - a_{m-1} \right) \left( a \right)^{(k-i)/2} + \frac{\epsilon}{\hbar} \left\{ - \frac{1}{4} \left< a_m | \left( a^{-1} \hat p^2 + \hat p^2 a^{-1} \right) | a_{m-1} \right> \right. \\ & \ \ \ \ \ \ \ \ \ \ \ \left. + \frac{1}{2} \left< a_m | \left( -a + \frac{\Lambda}{3}a^3 \right) | a_{m-1} \right> + \hbar^2 \left< a_m | \Delta V_{QP} | a_{m-1} \right> \right\} \\ & \ \ \ \ \ = \left( a_m a_{m-1} \right)^{(k-i)/2} \int_{-\infty}^{\infty} \frac{dp}{\left( 2 \pi \hbar \right)} \exp \left( \frac{i}{\hbar} p \left( a_m - a_{m-1} \right) \right) \times \\ & \ \ \ \ \ \ \ \ \ \ \ \ \left\{ 1- \frac{\epsilon}{\hbar} \left[ \frac{1}{2} \left( \frac{a_{m} ^{-1} + a_{m-1}^{-1}}{2} \right) p^2 - \left( \frac{V \left( a_m \right) + V \left( a_{m-1} \right)}{2} \right) \right. \right. \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. \left. - \hbar^{2} \left( \frac{\Delta V_{QP} \left( a_{m} \right) + \Delta V_{QP} \left( a_{m-1} \right)}{2} \right) \right] \right\} \\ & \ \ \ \ \ \doteq \left( a_m a_{m-1} \right)^{(k-i)/2} \int_{-\infty}^{\infty} \frac{dp}{\left( 2 \pi \hbar \right)} \exp \left( \frac{i}{\hbar}p \Delta a_m + \frac{\epsilon}{\hbar} \bar H \left( a_m \right) \right) \end{align}

where $\Delta a_m \equiv \left( a_m - a_{m-1} \right)$ and $\bar f \left( a_m \right) \equiv \frac{f \left( a_m \right) + f \left( a_{m-1} \right) }{2}.$

The part of the integral involving $p$ is

(10)
\begin{align} \int_{-\infty}^{\infty} \frac{dp}{\left(2 \pi \hbar \right)} \exp \left[ \frac{i}{\hbar}p \Delta a_m - \frac{\epsilon}{\hbar} \left( \frac{a_{m}^{-1} + a_{m-1}^{-1}}{4}p^2 \right) \right] \end{align}

which can be evaluated with the help of

(11)
\begin{align} \int_{-\infty}^{\infty} dp \exp \left( -Ap^2 + i B p \right) = \int_{-\infty}^{\infty} dp \quad e^{-Ap^2} \cos Bp. \end{align}

Using

(12)
\begin{align} &\int_{-\infty}^{\infty} \exp \left[ - \left( ax^2 +2bx +c \right) \right] \cos \left( px^2 +2qx + r \right) dx \\ &= \frac{\sqrt{\pi}}{\sqrt[4]{a^2 +p^2}} \exp \frac{a \left( b^2-ac \right) - \left( aq^2 - 2bpq +cp^2 \right) }{a^2 + p^2} \cos \left\{ \frac{1}{2} \arctan \frac{p}{a} - \frac{p \left( q^2 - pr \right) - \left( b^2 p - 2abq + a^2 r \right)}{a^2 + p^2} \right\} \end{align}

we obtain for the momentum integral above

(13)
\begin{align} \sqrt{\frac{\pi}{A}} \exp \left( - \frac{B^2}{4A} \right) = \sqrt{ \frac{ 2 \pi \hbar }{ \epsilon } } \left( \frac{a_{m}^{-1} + a_{m-1}^{-1}}{2} \right) ^{-1/2} \exp \left( - \frac{ \left( \Delta a_m \right)^2}{\epsilon \hbar \left( a_{m}^{-1} + a_{m-1}^{-1} \right)} \right) \end{align}

so after integration over $p$ the short-time propagator becomes

(14)
\begin{align} \left( a_m a_{m-1} \right)^{(k-i)/2} \frac{1}{ \sqrt{ 2 \pi \hbar \epsilon }} \left( \frac{a_{m}^{-1} + a_{m-1}^{-1} }{2} \right) ^{-1/2} & \exp \left\{ - \frac{ \epsilon}{\hbar} \left( a_{m}^{-1} + a_{m-1}^{-1} \right) ^{-1} \left( \frac{ \Delta a_m}{ \epsilon} \right)^2 \right\} \times \\ & \exp \left[ \frac{\epsilon}{\hbar} \left( \bar V \left( a_m \right) + \hbar \bar{ \Delta V}_{QP} \left( a_m \right) \right) \right] \end{align}

Note that using the binomial theorem,

(15)
\begin{align} \left( ab \right)^{1/4} &= \frac{1}{\sqrt{2}} \left( a + b \right)^{1/2} \left[ 1 - \left( \frac{a-b}{a+b} \right)^2 \right]^{1/4} \\ &=\frac{1}{\sqrt{2}} \left( a + b \right)^{1/2} \left[ 1 - \frac{1}{4} \left( \frac{a-b}{a+b} \right)^2 + \frac{1}{2} \left( - \frac{1}{4} \cdot \frac{3}{4} \right) \left( \frac{a-b}{a+b} \right)^4 + \frac{1}{6} \left( - \frac{1}{4} \cdot \frac{3}{4} \cdot \frac{7}{4} \right) \left( \frac{a-b}{a+b} \right)^6 + \cdots \right] \end{align}

so that since $\left( a^{-1} + b^{-1} \right)^{1/2} = \left( \frac{a+b}{ab} \right)^{1/2},$ we can write

(16)
\begin{align} \left( a_{m} a_{m-1} \right)^{\left( k-i \right) /2} &= \left( a_{m} a_{m-1} \right)^{\left( k-i \right) /2 + 1/4} \left( a_{m} a_{m-1} \right)^{-1/4} \\ &= \left( a_{m} a_{m-1} \right)^{\left( k-i \right) /2 + 1/4} \left\{ \frac{1}{\sqrt{2}} \left( a_{m}^{-1} + a_{m-1}^{-1} \right)^{1/2} \times \right. \\ & \ \ \ \ \ \ \left. \left[ 1 - \frac{1}{4} \left( \frac{a_m - a_{m-1}}{a_m + a_{m-1}} \right)^2 + \frac{1}{2} \left( - \frac{1}{4} \cdot \frac{3}{4} \right) \left( \frac{a_m - a_{m-1}}{a_m + a_{m-1}} \right)^4 + \cdots \right] \right\} \end{align}

and the short-time propagator becomes

(17)
\begin{align} &\left( a_{m} a_{m-1} \right)^{ \left( k-i \right)/2 } \frac{1}{ \sqrt{2 \pi \hbar \epsilon } } \left( \frac{a_{m}^{-1} + a_{m-1}^{-1}}{2} \right)^{-1/2} \exp \left[ - \frac{ \epsilon }{\hbar} \left( \frac{1}{2} \cdot \frac{a_m + a_{m-1}}{2} \left( \frac{ \Delta a_m}{\epsilon} \right)^2 - \frac{\left( \Delta a_m \right)^4}{4 \left( 2 \bar{a}_m \right) \epsilon^2} \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \exp \left[ \frac{\epsilon}{\hbar} \left( \bar{V} \left( a_m \right) + \hbar^2 \bar{\Delta V_{QP} } \left( a_m \right) \right) \right] \\ &= \left( a_m a_{m-1} \right)^{ \left( k-i \right)/2 + 1/4 } \left[ 1 - \frac{1}{4} \left( \frac{\Delta a_m}{2 \bar{a}_m } \right)^2 + \cdots \right] \cdot \frac{1}{\sqrt{2 \pi \hbar \epsilon }} \exp \left[ - \frac{ \epsilon }{\hbar} \left( \frac{\bar{a}_m}{2} \left( \frac{ \Delta a_m}{\epsilon} \right)^2 - \frac{\left( \Delta a_m \right)^4}{4 \left( 2 \bar{a}_m \right) \epsilon^2} \right) \right] \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times \exp \left[ \frac{\epsilon}{\hbar} \left( \bar{V} \left( a_m \right) + \hbar^2 \bar{\Delta V}_{QP} \left( a_m \right) \right) \right]. \end{align}

The full propagator is thus

(18)
\begin{align} K \left( a'', t'' | a', t' \right) &= \lim_{N \to \infty } \int_{0}^{ \infty} \prod_{n=1}^{N-1} a_{n}^{i-k} da_{n} \prod_{m=1}^{N} \left( a_m a_{m-1} \right)^{ \left( k-i \right) /2 +1/4 } \left[ 1 - \frac{1}{4} \left( \frac{\Delta a_m}{2 \bar{a}_m} \right)^{2} + \cdots \right] \times \\ & \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\sqrt{2 \pi \hbar \epsilon}} \exp \left[ - \frac{ \epsilon }{\hbar} \left( \frac{\bar{a}_m}{2} \left( \frac{ \Delta a_m}{\epsilon} \right)^2 - \frac{\left( \Delta a_m \right)^4}{4 \left( 2 \bar{a}_m \right) \epsilon^2} - \bar{V} \left( a_m \right) - \hbar^2 \bar{\Delta V}_{QP} \left( a_m \right) \right) \right] \\ &=\left( a' a'' \right)^{ \left( k-i \right) /2 +1/4 } \lim_{M \to \infty} \left( \frac{1}{2 \pi \hbar \epsilon} \right)^{N/2} \int_{0}^{\infty} \prod_{n=1}^{N-1} a_{n}^{1/2} da_n \times \\ & \ \ \ \ \ \ \ \ \ \ \ \prod_{m=1}^{N} \left[ 1 - \frac{1}{4} \left( \frac{\Delta a_m}{2 \bar{a}_m} \right)^2 - \frac{3}{32} \left( \frac{\Delta a_m}{2 \bar{a}m} \right) ^4 + \mathcal{O} \left( \left( \frac{\Delta a_m}{2 \bar{a}_m} \right)^6 \right) \right] \times \\ & \ \ \ \ \ \ \ \ \ \ \ \exp \left[ - \frac{\epsilon}{\hbar} \left( \frac{\bar{a}_m}{2} \left( \frac{\Delta a_m}{\epsilon} \right)^2 - \frac{\left( \Delta a_m \right)^4}{4 \left( 2 \bar{a}_m \right) \epsilon^2} - \bar{V} \left( a_m \right) - \hbar^2 \bar{\Delta V }_{QP} \left( a_m \right) \right) \right] \end{align}

We next show that the series preceding the exponential can be replaced with an additional quantum potential in the action. The main result needed to do so is

Lemma 1. For n > 0,

(19)
\begin{align} \frac{1}{\epsilon^{1/2}} \int_{0}^{\infty} & a^{1/2} da \left( b-a \right)^{n} e^{-b \left( b-a \right) ^{2} / 2 \hbar \epsilon} = \\ &2^{\frac{n-1}{2}} \hbar^{ \frac{n+1}{2}} \left( \frac{\epsilon}{b} \right) ^{\frac{n}{2}} \left( \left( -1 \right) ^n +1 \right) \Gamma \left( \frac{n+1}{2} \right) + C_1 \left( b \right) + C_2 \left( b \right), \end{align}

where for all m > 0,

(20)
\begin{align} &C_1 \left( b \right) = \mathcal{O} \left( \epsilon ^{\frac{n}{2}+ \frac{1}{4}} \right) \\ &\lim_{\epsilon \to 0} \frac{C_2 \left( b \right)}{\epsilon^m} = 0 \end{align}

Proof:

(21)
\begin{align} &\frac{1}{\epsilon^{1/2}} \int_{0}^{\infty} a^{1/2} da \left( b-a \right)^{n} e^{-b \left( b-a \right) ^{2} / 2 \hbar \epsilon } = \\ &\frac{b^{1/2}}{\epsilon^{1/2}} \int_{0}^{\infty} da \left( b-a \right)^n e^{-b \left( b-a \right)^2 / 2 \hbar \epsilon } - \frac{1}{\epsilon^{1/2}} \int_{0}^{\infty} da \left(b^{1/2} - a^{1/2} \right) \left( b-a \right)^n e^{-b \left( b-a \right)^2 / 2 \hbar \epsilon } \end{align}

Using an obvious change of variables, the integral forms of gamma functions

(22)
\begin{align} &\int_{0}^{\infty} x^{\nu - 1} e^{-x} dx = \Gamma \left( \nu \right) \\ &\int_{0}^{u} x^{\nu - 1} e^{-x} dx = \gamma \left( \nu, u \right), \end{align}

their functional relation $\gamma \left( \alpha , x \right) = \Gamma \left( \alpha \right) - \Gamma \left( \alpha , x \right),$ and the asymptotic representation

(23)
\begin{align} \Gamma \left( \alpha , x \right) = x^{ \alpha - 1} e^{-x} \left( \sum_{m=0}^{M-1} \frac{(-1)^m \Gamma \left( 1 - \alpha + m \right)}{x^m \Gamma \left( 1 - \alpha \right)} + \mathcal{O} \left( | x | ^{-M} \right) \right), \end{align}

the first integral can be performed to yield

(24)
\begin{align} \frac{b^{1/2}}{\epsilon^{1/2}} & \int_{0}^{\infty} da \left( b-a \right)^n e^{-b \left( b-a \right)^2 / 2 \hbar \epsilon } = 2^{\frac{n-1}{2}} \hbar^{\frac{n+1}{2}} \left( \frac{\epsilon}{b} \right)^{\frac{n}{2}} \left( \left( -1 \right)^n + 1 \right) \Gamma \left( \frac{n+1}{2} \right) \\ & - \hbar \epsilon^{1/2} b^{\frac{2n-3}{2}} e^{-\left( \frac{b^3}{2 \hbar \epsilon} \right)} \left( \sum_{m=0}^{M-1} \frac{\left( -1 \right)^m \Gamma \left( \frac{1-n}{2} + m \right) \left( 2 \hbar \epsilon \right)^m}{b^3 \Gamma \left( \frac{1-n}{2}\right)} + \mathcal{O} \left( \left( \frac{2 \hbar \epsilon}{b^3} \right)^M \right) \right). \end{align}

Since the second term decays faster than $e^{-\left( \frac{b^3}{2 \hbar \epsilon} \right)}$ as $\epsilon \to 0,$ we can use the fact that $e^{-x^{-1}} < \left( \frac{n}{2} \right)^n x^{n} \quad \forall n > 0$ to identify the second term as the desired $C_2 \left( b \right).$

The second integral can be bounded using the fact that $\left| b^{1/2} - a^{1/2} \right| \le \left| b - a \right|^{1/2}$:

(25)
\begin{align} \left| - \frac{1}{\epsilon^{1/2}} \int_{0}^{\infty} \left( b^{1/2} - a^{1/2} \right) \left( b - a \right)^n e^{- \frac{b \left(b-a \right)^2}{2 \hbar \epsilon}} da \right| & \le \frac{1}{\epsilon^{1/2}} \int_{0}^{\infty} \left| b - a \right|^{n+\frac{1}{2}} e^{- \frac{b \left(b-a \right)^2}{2 \hbar \epsilon}} da \\ & \le \frac{2}{\epsilon^{1/2}} \int_{0}^{\infty} u^{n+\frac{1}{2}} e^{- \frac{b u^2}{2 \hbar \epsilon}} du \\ &= \epsilon^{\frac{n}{2} + \frac{1}{4}} \left( \frac{2 \hbar}{b} \right)^{\frac{2n + 3}{4} } \Gamma \left( \frac{2n+3}{4} \right), \end{align}

so that it satisfies the requirements of $C_1 \left( b \right),$ and the proof is complete. $\Box$

The next lemma will allow us to identify the form the quantum potential due to the series in $\left( \frac{\Delta a_{n+1}}{2 \bar{a}_{n+1}} \right)^2$ will take.

Lemma 2.

(26)
\begin{align} I = \frac{1}{\sqrt{ 2 \pi \hbar \epsilon}} \int_{0}^{\infty} a_{n}^{1/2} da_n \left( - \frac{1}{4} \left( \frac{\Delta a_{n+1}}{2 \bar{a}_{n+1}} \right)^2 \exp \left( - \frac{\epsilon}{\hbar} \cdot \frac{a_{n+1}}{2} \left( \frac{\Delta a_{n+1}}{\epsilon} \right)^2 \right) \right) \end{align}
(27)
\begin{align} \lim_{\epsilon \to 0 } \frac{I}{\epsilon} = - \frac{\hbar}{16} a_{n+1}^{-3}. \end{align}

Proof:

page revision: 310, last edited: 11 Mar 2015 17:38