Nonlinear normal ordering

Suppose we have a (one-dimensional - although the principle is the same in any number of finite dimensions and can also be extended to field theory as in the paper) classical Hamiltonian of the form

(1)
\begin{align} H = \frac{p^2}{2} + V \left( x \right). \end{align}

We would like to quantize this using the Schroedinger ansatz $x \to \hat x$, $p \to \hat p = -i \hbar \partial_x$, where $\hat x$ denotes a multiplication operator on quantum states $\psi \left( x \right) \in L^2 \left( \mathbb{R} \right)$, so that we have

(2)
\begin{align} \hat x : \psi \left( x \right) & \to x \psi \left( x \right) \\ \hat p : \psi \left( x \right) & \to -i \hbar \partial_x \psi \left( x \right). \end{align}

(Note: sometimes people set $\hbar =1$ by convenient choice of units.)

Thus we want to obtain a quantum Hamiltonian $\hat H$ (a self-adjoint operator w.r.t. the $L^2$ inner product such that if we replace occurrences of $\hat x$ with $x$ and $\hat p$ with $p$, we recover the classical Hamiltonian $H$). We want to find the eigenfunctions of $\hat H$, which are the energy eigenstates:

(3)
\begin{align} \hat H \psi = E \psi. \end{align}

The eigenstate with lowest energy eigenvalue is the ground state; this is a quantum description of the physical system described by $H$ in its lowest energy configuration. (Physical interpretation of how a quantum state describes a physical system: the state $\psi \left( x \right)$ gives the probability density function for position $x$ via $PDF = \bar{\psi} \psi$.)

Notice however that there is an ambiguity in promoting $H$ to a quantum operator. We require $\hat H$ to be such that replacing all occurrences of $\hat x$ with $x$ and $\hat p$ with $p$ recovers $H$, but since $\hat x$ and $\hat p$ don't commute, there are actually infinitely many variations on the allowable $\hat H$'s we could define, all differing from one another by terms of order $\hbar$ (i.e. terms which are classically unimportant but which become important at quantum scales). This is the "factor ordering problem."

In problems where $V \left( x \right)$ is a quadratic form $\frac{1}{2} \langle x , Mx \rangle$ ($M$ a positive self-adjoint linear operator), a typical ordering for $\hat H$ is the "normal ordering," which factors $M$ into its unique positive self-adjoint square root $T$ and defines the "creation" and "annihilation" operators

(4)
\begin{align} a^* & = \frac{1}{\sqrt{2}} \left( T \hat x - i \hat p \right) \\ a &= \frac{1}{\sqrt{2}} \left( T \hat x + i \hat p \right) \end{align}

so that the quantized Hamiltonian

(5)
\begin{align} \hat H = a^* a + \frac{1}{2} \left( {\rm Tr \ } T \right) I \end{align}

admits the ground state

(6)
\begin{align} \psi \left( x \right) = \mathcal{N} \exp \left( - \frac{1}{2 \hbar} \langle x , Tx \rangle \right) \end{align}

with energy $\frac{1}{2} {\rm Tr \ } T$ (where$\mathcal{N}$ is a normalization constant).

But what if the potential $V \left( x \right)$ does not have the form $\frac{1}{2} \langle x , Mx \rangle$? We can still define a similar ordering (the idea was introduced by Moncrief and Ryan) which could be called the "nonlinear normal ordering." Notice that in order to do so, we just need to factorize $V \left( x \right)$ so that we can in turn factorize the Hamiltonian into something like the creation and annihilation operators we got in the linear case above ($V \left( x \right) = \frac{1}{2} \langle x , MX \rangle$).

We notice that if we could find a function $S \left( x \right)$ satisfying the differential equation

(7)
\begin{align} \frac{1}{2} \left( \frac{dS}{dx} \right)^2 = V \left( x \right), \end{align}

we could order $\hat H$ as

(8)
\begin{align} \hat H = \frac{1}{2} \left( \frac{dS}{dx} - i \hat p \right) \left( \frac{dS}{dx} + i \hat p \right), \end{align}

an expression whose two factors are analogous to the creation and annihilation operators $a^*$ and $a$ above. With this ordering, $\hat H$ admits the ground state

(9)
\begin{align} \mathcal{N} \exp \left[ -S \left( x \right) / \hbar \right]. \end{align}

Note that this is a zero-energy ground state (i.e. its energy eigenvalue is 0).

Now the question becomes the solution of 7. In one-dimensional cases such as we are sketching here, it is obviously simpler just to integrate 7 directly, but when we ascend to field theory, 7 will be replaced by a variational differential equation, and solving it directly won't be so simple. To construct a solution generally, we need a detour into Hamilton-Jacobi theory.

Hamilton-Jacobi Theory

In classical mechanics, we try to find the trajectory of a particle over time, $x \left( t \right)$, which minimizes the action $\int L \left( x, \dot x, t \right) dt$. For the Hamiltonian formalism, we define the canonical momentum $p \equiv \frac{\partial L}{\partial \dot x}$ and Hamiltonian $\dot x p - L$, and solve the system

(10)
\begin{align} \dot x = \frac{\partial H}{\partial p} \\ \dot p = - \frac{\partial H}{\partial x} \end{align}

As an aid to integrating this system, suppose we have a solution $S \left( x , t \right)$ to the differential equation (the Hamilton-Jacobi equation)

(11)
\begin{align} \frac{\partial S}{\partial t} = - H \left( x , \frac{\partial S}{\partial x}, t \right), \end{align}

where on the right-hand side $H \left( x, p, t \right)$ is the Hamiltonian (allowing for the possibility of a Hamiltonian with explicit time-dependence, even though this rarely occurs, and never in our cases of interest. We will specialize to the case of Hamiltonians with no explicit time-dependence in a moment.)

Then take

(12)
\begin{align} p = \frac{\partial S}{\partial x} \end{align}

and solve for the integral curve of

(13)
\begin{align} \dot x = \left. \frac{\partial H}{\partial p} \right|_{p = \frac{\partial S}{\partial x}} \end{align}

using the initial data for the problem at hand. By construction the integral curve satisfies the first of Hamilton's equations 10. To show that it also satisfies the second, differentiate both sides of 11 with respect to $x$:

(14)
\begin{align} \frac{\partial}{\partial x} \left[ \frac{\partial S}{\partial t} \right] &= \frac{\partial}{\partial x} \left[ - H \left( x, \frac{\partial S}{\partial x}, t \right) \right] = \left. \frac{\partial}{\partial x} \left[ - H \left( x, p, t \right) \right] \right|_{p = \frac{\partial S}{\partial x}} \\ \frac{\partial}{\partial t} \left[ \frac{\partial S}{\partial x} \right] & = - \left. \frac{\partial H}{\partial x} \right|_{p = \frac{\partial S}{\partial x}} \\ \dot p &= - \left. \frac{\partial H}{\partial x} \right|_{p = \frac{\partial S}{\partial x}} \end{align}

Also note that we can construct a solution to the Hamilton-Jacobi equation 11 as follows. Let $x$ denote the integral curve to 13 for initial data $x_0$ at time $t_0$ and final data $x_1$ at time $t_1$. Thus evaluating on this history $x$, $S \left( x , t \right)$ now depends on $t$ through $x$ as well as directly, so we have

(15)
\begin{align} \frac{d}{dt} \left[ S \left( x , t \right) \right] &= \frac{\partial S}{\partial t} + \frac{\partial S}{\partial x} \cdot \frac{\partial x}{\partial t},\\ &= - H \left( x , \frac{\partial S}{\partial x}, t \right) + \frac{\partial S}{\partial x} \cdot \frac{\partial x}{\partial t} \end{align}

or specializing to the case where the Hamiltonian has no explicit time-dependence,

(16)
\begin{align} \frac{d}{dt} \left[ S \left( x , t \right) \right] = \frac{\partial S}{\partial x} \cdot \frac{\partial x}{\partial t} - H \left( x , \frac{\partial S}{\partial x} \right). \end{align}

Now using the fact that $x$ is the integral curve of

(17)
\begin{align} \dot x = \left. \frac{\partial H}{\partial p} \right|_{p = \frac{\partial S}{\partial x}}, \end{align}

16 becomes

(18)
\begin{align} \frac{dS}{dt} &= p \cdot \dot x - H \left( x , p \right) \\ &= L \left( x, \dot x \right) \end{align}

so that

(19)
\begin{align} S \left( x_1 \right) - S \left( x_0 \right) = \int_{t_0}^{t_1} L \left( x , \dot x \right) dt \end{align}

So if we take $S \left( x_0 \right)$ to be the absolute minimizer of the action

(20)
\begin{align} \int_0^{\infty} L \left( x , \dot x \right) dt \end{align}

for initial data $x \left( 0 \right) = x_0$, we have a solution to 11. Also note that there is no explicit time-dependence in this definition of $S$, so it is actually a solution to the zero-energy Hamilton-Jacobi equation

(21)
\begin{align} H \left( x , \frac{\partial S}{\partial x} \right) = 0. \end{align}

Note that for a Hamiltonian of the form $H = \frac{p^2}{2} + V \left( x \right)$, the zero-energy Hamilton-Jacobi equation is

(22)
\begin{align} \frac{1}{2} \left( \frac{dS}{dx} \right)^2 = - V \left( x \right) \end{align}

which only differs from 7 by the sign of $V \left( x \right)$. So in order to find a ground state 9 for the ordering 8, we can solve the Hamilton-Jacobi equation not for the Hamiltonian $H = \frac{p^2}{2} + V \left( x \right)$ that we are quantizing, but that for the associated "upside down potential" Hamiltonian $H = \frac{p^2}{2} + V \left( x \right)$ (Note: the other equivalent way to phrase what we are doing is to say that we solve the imaginary-time HJE.)

This is why in order to find the ground state for the nonlinear-normal-ordered Hamiltonian for physics in a Lorentzian signature, we need to solve the Dirichlet problem for the corresponding problem with Euclidean signature - because we are using the construction ending in 20 to obtain the solution $S \left( x \right)$ for the upside-down-potential Hamilton-Jacobi equation 7.

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