Normal ordering normal modes

Abstract

In a soliton sector of a quantum field theory, it is often convenient to expand the quantum fields in terms of normal modes. Normal mode creation and annihilation operators can be normal ordered, and their normal ordered products have vanishing expectation values in the one-loop soliton ground state. The Hamiltonian of the theory, however, is usually normal ordered in the basis of operators which create plane waves. In this paper we find the Wick map between the two normal orderings. For concreteness, we restrict our attention to Schrodinger picture scalar fields in 1+1 dimensions, although we expect that our results readily generalize beyond this case. We find that plane wave ordered n-point functions of fields are sums of terms which factorize into j-point functions of zero modes, breather and continuum normal modes. We find a recursion formula in j and, for products of fields at the same point, we solve the recursion formula at all j.

1 Introduction

In perturbation theory about a translation-invariant vacuum, it is customary to decompose the quantum fields into operators $a_p^\dag$ and $a_p$ which create and annihilate plane wave excitations. The free vacuum is annihilated by $a_p$ and is the initial state in the perturbative expansion. This perturbation theory is simplest when the Hamiltonian is normal ordered, so that all $a^\dag _p$ appear to the left of all $a_p$.

At the same leading order,¹ the ground state of a quantum soliton is given by a coherent state formed by shifting the fields by the functions corresponding to their classical solutions [3,4,5]. The normal modes of the quantum soliton are, at linear order, described by quantum harmonic oscillators. The one-loop ground state of the soliton sector consists of the tensor product of the ground states of these oscillators [6]. If the fields are decomposed into the normal modes of the soliton, with operators $b^\dag _k$ and $b_k$ corresponding to the raising and lowering operators in the corresponding quantum harmonic oscillators, then the one-loop soliton ground state is, after the shift operator noted above, the state annihilated by all of the $b_k$.

Again a sensible perturbation theory exists which describes the spectrum of the one soliton sector. It is similar to that of the vacuum sector, except that treating zero modes requires special care [7,8,9]. In particular the calculation is simplest if the Hamiltonian is normal ordered by placing all $b^\dag _k$ to the left of $b_k$.

However, the Hamiltonian is usually given normal ordered in terms of plane waves, as is convenient for vacuum sector perturbation theory. Therefore the first step in soliton perturbation theory is to convert plane wave normal ordering to normal mode normal ordering. The goal of the present note is to describe how this can be done in the case of a scalar field theory in 1+1 dimensions. The fact that the theory is scalar and only in 1+1 dimensions does not appear to play a central role in our analysis, and so we expect that the approach in this paper can be trivially generalized to more complicated theories in more dimensions.

We find that the problem of converting plane wave normal ordering into normal mode normal ordering can be achieved in two steps. First, as will be described in Sect. 3, we show that plane wave normal ordered products of the form $:\phi ^n(x):_a$ can be decomposed into sums of products of factors of the form $:\phi _M^j(x):_a$ where

$$\begin{aligned} \phi (x)=\sum _M \phi _M(x) \end{aligned}$$

(1.1)

is a decomposition into different kinds of normal modes, such as even and odd breather modes. Next, in Sect. 4, we find that these factors can each be converted according to the Wick formula

$$\begin{aligned} :\phi ^j_M(x):_a= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\nonumber \\&\times \frac{j!}{2^m m!(j-2m)!}\mathcal {I}_M^m(x):\phi ^{j-2m}_M(x):_b \nonumber \\ \end{aligned}$$

(1.2)

where the contraction, except for the case of zero modes, is schematically

$$\begin{aligned} \mathcal {I}_M(x)= & {} \left\langle \frac{1}{2\omega _{k}}-\frac{1}{2\omega _p}\right\rangle \end{aligned}$$

(1.3)

with $\omega _k$ and $\omega _p$ the energy of a normal mode and plane wave respectively. Combining the decomposition (1.1) with the Wick formula (1.2) for each component, one arrives at our Wick’s theorem

$$\begin{aligned} :\phi ^j(x):_a= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}\mathcal {I}^m(x):\phi ^{j-2m}(x):_b \nonumber \\ \mathcal {I}(x)= & {} \sum _M \mathcal {I}_M(x) \end{aligned}$$

(1.4)

where $\mathcal {I}(x)$ can be found using the identity [9]

$$\begin{aligned} \partial _x \mathcal {I}(x)= & {} \int \frac{dk}{2\pi }\frac{\partial _x\left| g_{k}(x)\right| ^2}{2\omega _k} +\sum _i\frac{\partial _x\left| {g}_{i}(x)\right| ^2}{2\omega _i} \end{aligned}$$

(1.5)

in terms of continuum normal modes $g_k(x)$ and breathers $g_i(x)$ together with the condition that it vanish at as |x| tends to infinity.

We begin in Sect. 2 with a review of our formalism.

2 The setup

Table 1 Summary of notation

In this section we will review the one loop description of kinks developed in Refs. [10,11,12] using the formalism developed in Refs. [6, 13, 14], which has the advantage that it resolves the ambiguity noted in Ref. [15]. The key elements of our notation are summarized in Table 1.

For concreteness, we consider a theory of a real scalar field $\phi (x)$ and its canonical momentum $\pi (x)$ in 1+1 dimensions, described by a Hamiltonian

$$\begin{aligned} H= & {} \int dx {\mathcal {H}}(x) ,\quad {\mathcal {H}}(x)=\frac{1}{2}:\pi (x)\pi (x):_a\nonumber \\&+\frac{1}{2}:\partial _x\phi (x)\partial _x\phi (x):_a +\frac{M^2}{g^2}:V[g\phi (x)]:_a \end{aligned}$$

(2.1)

where M has dimensions of mass and g has dimensions of action${}^{-1/2}$. The perturbative expansion will be an expansion in $g^2\hbar$ and we will set $\hbar =1$. The plane-wave normal-ordering ${:}{:}_a$ will be defined momentarily.

We assume that the potential V has degenerate minima so that the classical equations of motion admit a time-independent kink solution

$$\begin{aligned} \phi (x,t)=f(x). \end{aligned}$$

(2.2)

In the Schrodinger picture of the quantum theory, the translation operator

$$\begin{aligned} \mathcal {D}_f={\mathrm{exp}}\left( -i\int dx f(x)\pi (x)\right) \end{aligned}$$

(2.3)

satisfies the identity [13]

$$\begin{aligned} :F\left[ \pi (x),\phi (x)\right] :_a\mathcal {D}_f\!=\!\mathcal {D}_f:F\left[ \pi (x),\phi (x)\!+\!f(x)\right] :_a \nonumber \\ \end{aligned}$$

(2.4)

for any functional F and maps the vacuum sector to the kink sector. For example, the kink ground state may be written

$$\begin{aligned} |K\rangle =\mathcal {D}_f{\mathcal {O}}|\Omega \rangle \end{aligned}$$

(2.5)

where $|\Omega \rangle$ is the free scalar vacuum state and ${\mathcal {O}}$ may be calculated in perturbation theory. As $|K\rangle$ is a Hamiltonian eigenstate, ${\mathcal {O}}|\Omega \rangle$ is an eigenstate of its similarity transform

$$\begin{aligned} H^\prime= & {} \mathcal {D}_f^{-1} H\mathcal {D}_f=Q_0+H_2+H_I\nonumber \\ H_2= & {} \frac{1}{2}\int dx\left[ :\pi ^2(x):_a+:\left( \partial _x\phi (x)\right) ^2:_a\nonumber \right. \\&\left. +M^2V^{\prime \prime }[gf(x)]:\phi ^2(x):_a\right] \end{aligned}$$

(2.6)

where $Q_0$ is the classical kink mass and $H_I$ consists of higher order terms in the g expansion. Note that gf(x) is dimensionless and so contains no powers of $\hbar$ and so no powers of g. We assume that the field has been shifted so that $\phi =0$ is a global minimum of the potential V.

Defining

$$\begin{aligned} {\tilde{V}}_p=\int dx M^2 V^{\prime \prime }[gf(x)] e^{-ipx} \end{aligned}$$

(2.7)

one easily evaluates

$$\begin{aligned} H_2= & {} -\frac{m^2}{4}\int \frac{dp}{2\pi } \frac{1}{\omega _p}\left( a^\dag _p a^\dag _{-p}+a_p a_{-p}\right) \nonumber \\&+\frac{1}{2}\int \frac{dp}{2\pi }\frac{m^2+2p^2}{\omega _p}a^\dag _p a_p\nonumber \\&+\frac{1}{2}\int \frac{d^2p}{(2\pi )^2}\frac{{\tilde{V}}_{p_1+p_2}}{\sqrt{\omega _{p_1}\omega _{p_2}}}\nonumber \\&\times \left( a^\dag _{p_1}a^\dag _{p_2}+2a^\dag _{p_1}a_{-p_2}+a_{-p_1}a_{-p_2}\right) . \end{aligned}$$

(2.8)

It is a mess. The perturbative ground state, which is annihilated by $a_p$, is not an eigenstate. As our shift function f(x) has explicit x dependence, this expression does not even have manifest translation invariance. This motivates us to search for a new basis of the operator algebra in terms of which $H_2$ is simple. More precisely, while (2.8) is complicated it is nonetheless quadratic in the operators a and $a^\dag$ and so it can be diagonalized by a Bogoliubov transform, which we now construct.

As $Q_0$ is $O(g^{-2})$ and $H_2$ is $O(g^0)$, these are the only terms which appear at one loop. In particular the one-loop kink ground state ${\mathcal {O}}_1|\Omega \rangle$ is an eigenstate of $H_2$. To find it, one expands the fields in terms of the fixed frequency $\omega$ solutions g(x) of the classical equations of motion for $H_2$

$$\begin{aligned} \phi (x,t)= & {} e^{-i\omega t}g(x),\quad M^2 V^{\prime \prime }[gf(x)]g(x)\nonumber \\= & {} \omega ^2g(x)+g^{\prime \prime }(x). \end{aligned}$$

(2.9)

This is a wave equation for a particle in a potential and its solutions are the normal modes of the field theory in the kink background. It generally has bound state and continuum solutions. We will refer to even and odd bound state solutions as $g_{BE,i}(x)$ and $g_{BO,i}(x)$ respectively, where the index i runs over distinct solutions if there is more than one. The corresponding frequencies, defined using (2.9), will be denoted $\omega _{BE,i}$ and $\omega _{BO,i}$. There will always be an even bound state solution corresponding to the translation symmetry, which we call

$$\begin{aligned} g_B(x)=\frac{1}{\sqrt{Q_0}} f^\prime (x). \end{aligned}$$

(2.10)

As it corresponds to a symmetry, it is a zero mode $\omega _B=0$. The other bound state solutions correspond to breather modes. Let $n_e$ and $n_o$ be the number of even and odd breather modes. We will name the continuum states $g_k(x)$ where k is defined by $\omega _k^2=k^2+m^2$ and the sign of k is fixed by demanding that asymptotically it becomes the corresponding plane wave. All of these solutions are clearly mutually orthogonal and we normalize them such that

$$\begin{aligned} \int dx g_{k_1} (x) g^*_{k_2}(x)= & {} 2\pi \delta (k_1-k_2),\quad \int dx |g_{B}(x)|^2\nonumber \\= & {} \int dx |g_{BE}(x)|^2=\int dx |g_{BO}(x)|^2=1.\nonumber \\ \end{aligned}$$

(2.11)

We also impose

$$\begin{aligned} g(-x)=g^*(x). \end{aligned}$$

(2.12)

Their inverse Fourier transforms

$$\begin{aligned} {\tilde{g}}(p)=\int dx g(x) e^{ipx} \end{aligned}$$

(2.13)

satisfy the completeness relations [16]

$$\begin{aligned}&\sum _I(-1)^I{\tilde{g}}_{I}(p){\tilde{g}}_{I}(q)\nonumber \\&\quad +\int \frac{dk}{2\pi }{\tilde{g}}_k(p){\tilde{g}}_{-k}(q)=2\pi \delta (p+q) \nonumber \\&\quad \times \sum _I g_{I}(x) g^*_{I}(y)+\int \frac{dk}{2\pi }g_k(x)g_{-k}(y)=\delta (x-y)\nonumber \\ \end{aligned}$$

(2.14)

where I runs over all $n_e+n_o+1$ bound state field labels $\{B,\{BE,i\},\{BO,i\}\}$ and the symbol $(-1)^I$ is equal to $-1$ for odd breathers and otherwise is 1.

This entire paper will be in the Schrodinger picture, and so the field $\phi (x)$ and $\pi (x)$ will be independent of time and of any choice of Hamiltonian, and of any decomposition of the Hamiltonian into a free and imaginary part. They satisfy canonical commutation relations exactly, in the full interacting theory.² We define the linear combinations

$$\begin{aligned} a^\dag _p= & {} \int dx\left( \sqrt{\frac{\omega _p}{2}}\phi (x)-\frac{i}{\sqrt{2\omega _p}}\pi (x)\right) e^{ipx},\ \ a_{-p}\nonumber \\= & {} \int dx\left( \sqrt{\frac{\omega _p}{2}}\phi (x)+\frac{i}{\sqrt{2\omega _p}}\pi (x)\right) e^{ipx}\nonumber \\ \phi _0= & {} \int dx \phi (x) g_B(x),\quad \pi _0=\int dx \pi (x) g_B(x)\nonumber \\ b^\dag _{BE,i}= & {} \int dx\left( \sqrt{\frac{\omega _{BE,i}}{2}}\phi (x)-\frac{i}{\sqrt{2\omega _{BE,i}}}\pi (x)\right) g^*_{BE,i}(x)\nonumber \\ b_{BE,i}= & {} \int dx\left( \sqrt{\frac{\omega _{BE,i}}{2}}\phi (x)+\frac{i}{\sqrt{2\omega _{BE,i}}}\pi (x)\right) g^*_{BE,i}(x)\nonumber \\ b^\dag _{BO,i}= & {} \int dx\left( \sqrt{\frac{\omega _{BO,i}}{2}}\phi (x)-\frac{i}{\sqrt{2\omega _{BO,i}}}\pi (x)\right) g^*_{BO,i}(x)\nonumber \\ b_{BO,i}= & {} \int dx\left( -\sqrt{\frac{\omega _{BO,i}}{2}}\phi (x)-\frac{i}{\sqrt{2\omega _{BO,i}}}\pi (x)\right) g^*_{BO,i}(x). \nonumber \\ \end{aligned}$$

(2.15)

From the canonical algebra satisfied by $\phi (x)$ and $\pi (x)$ one easily finds the algebra satisfied by their components

$$\begin{aligned} {}&[a_p,a_q^\dag ]=2\pi \delta (p-q),\quad [\phi _0,\pi _0]=i, \quad [b_{BE,i},b^\dag _{BE,j}]=\delta _{ij}\nonumber \\ {}&[b_{BO,i},b^\dag _{BO,j}] =\delta _{ij}, \quad [b_{k_1},b^\dag _{k_2}]=2\pi \delta (k_1-k_2) \end{aligned}$$

(2.16)

with all other commutators within each decomposition vanishing.

Using the completeness relations (2.14) one may invert (2.15) to arrive at two distinct expansions of the same fields. The first is in terms of plane waves

$$\begin{aligned}&\phi (x)=\int \frac{dp}{2\pi }\frac{1}{\sqrt{2\omega _p}}\left( a^\dag _p+a_{-p}\right) e^{-ipx},\quad \nonumber \\&\omega _p=\sqrt{m^2+p^2}\nonumber \\&\pi (x)=i\int \frac{dp}{2\pi }\sqrt{\frac{\omega _p}{2}} \left( a^\dag _p-a_{-p}\right) e^{-ipx}. \end{aligned}$$

(2.17)

Here the the mass m is defined by

$$\begin{aligned} m=M\sqrt{V^{\prime \prime }[0]}. \end{aligned}$$

(2.18)

The other decomposition is in terms of normal modes³

$$\begin{aligned} \phi (x)= & {} \sum _I \phi _{I}(x) +\phi _{C}(x),\quad \nonumber \\ \pi (x)= & {} \sum _I \pi _{I}(x)+\pi _{C}(x) \nonumber \\ \phi _B(x)= & {} \phi _0 g_B(x),\quad \nonumber \\ \phi _{BE,i}(x)= & {} \frac{1}{\sqrt{2\omega _{BE,i}}}\left( b_{BE,i}^\dag +b_{BE,i}\right) g_{BE,i}(x)\nonumber \\ \phi _{BO,i}(x)= & {} \frac{1}{\sqrt{2\omega _{BO,i}}}\left( b_{BO,i}^\dag -b_{BO,i}\right) g_{BO,i}(x),\quad \nonumber \\ \phi _C(x)= & {} \int \frac{dk}{2\pi }\frac{1}{\sqrt{2\omega _k}}\left( b_k^\dag +b_{-k}\right) g_k(x)\nonumber \\ \pi _B(x)= & {} \pi _0 g_B(x)\nonumber \\ \pi _{BE,i}(x)= & {} i\sqrt{\frac{\omega _{BE,i}}{2}}\left( b_{BE,i}^\dag - b_{BE,i}\right) g_{BE,i}(x) \nonumber \\ \pi _{BO,i}(x)= & {} i\sqrt{\frac{\omega _{BO,i}}{2}}\left( b_{BO,i}^\dag + b_{BO,i}\right) g_{BO,i}(x)\nonumber ,\quad \\ \pi _C(x)= & {} i\int \frac{dk}{2\pi }\sqrt{\frac{\omega _k}{2}}\left( b_k^\dag - b_{-k}\right) g_k(x). \end{aligned}$$

(2.19)

For each g here there is a corresponding $\omega$, defined by Eq. (2.9). In the case of the continuum modes in $\phi _C$, these are equal to

$$\begin{aligned} \omega _k=\sqrt{m^2+k^2}. \end{aligned}$$

(2.20)

It may appear surprising that m, which was defined in (2.18) using the other expansion, appears here. However (2.20) is merely our choice of parametrization of |k|, while its sign is fixed by (2.12). The only parametrization-independent content of (2.20) is that the minimum $\omega _k$ for a continuum mode is $\omega _0=m$ which arises because this delocalized mode has only measure zero support on the kink, and so its mass agrees with that of the lowest mass vacuum sector mode, m.

Normal ordering may be defined with respect to either decomposition. Plane wave normal ordering ${:}{:}_a$ places all $a^\dag$ to the left of each a. Normal mode normal ordering ${:}{:}_b$ places all $b^\dag$ and $\phi _0$ on the left of all b and $\pi _0$.

Finally one may simplify $H_2$

$$\begin{aligned} H_2= & {} Q_1+\frac{\pi _0^2}{2}+\sum _{i=1}^{n_e} \omega _{BE,i}b^\dag _{BE,i}b_{BE,i}\nonumber \\&+\sum _{i=1}^{n_o} \omega _{BO,i}b^\dag _{BO,i}b_{BO,i} +\int \frac{dk}{2\pi }\omega _k b^\dag _k b_k \end{aligned}$$

(2.21)

where $Q_1$ is the one-loop correction to the kink energy. One recognizes this system as the sum of a free quantum mechanical particle with position $\phi _0$ and momentum $\pi _0$ plus an infinite set of quantum harmonic oscillators. The one-loop vacuum therefore is annihilated by $\pi _0$ and also by all operators b

$$\begin{aligned} \pi _0{\mathcal {O}}_1|\Omega \rangle= & {} b_{BE,i}{\mathcal {O}}_1|\Omega \rangle \nonumber \\= & {} b_{BO,i}{\mathcal {O}}_1|\Omega \rangle =b_k{\mathcal {O}}_1|\Omega \rangle =0. \end{aligned}$$

(2.22)

We can now describe the entire Fock space of $H_2$. It consists of the vacuum, plus an arbitrary center of mass momentum proportional to $\pi _0$, where relativistic corrections appear only at higher orders in pertubation theory [10]. Also there is an infinite tower of oscillator excitations, created by $b_k^\dag$ with mass $\omega _k$.

Equation (2.22) implies that normal mode normal ordered operators $:A:_b$, with vanishing c-number component, have vanishing expectation values at one loop

$$\begin{aligned} \langle \Omega |{\mathcal {O}}_1^\dag :A:_b{\mathcal {O}}_1|\Omega \rangle =0. \end{aligned}$$

(2.23)

This is one motivation for considering normal mode normal ordering. Another is that it allows an efficient computation of states and energies beyond one loop [9].

3 Factorization

3.1 Factorization

The Hamiltonian H is plane wave normal ordered and a similarity transform by $\mathcal {D}_f$ preserves the normal ordering [13]. Therefore the Hamiltonian $H^\prime$ is also plane wave normal ordered. However for several applications, normal mode normal ordering is most efficient. In this paper we will study how to relate the two.

The Hamiltonian $H^\prime$, at nth order, for $n>2$ is

$$\begin{aligned} H_n=\frac{M^2g^{n-2}}{n!}V^{(n)}[gf(x)]:\phi ^n(x):_a \end{aligned}$$

(3.1)

where $V^{(n)}$ is the nth functional derivative of the potential V with respect to its argument. To calculate the soliton spectrum and energy corrections in perturbation theory, beginning with ${\mathcal {O}}_1|\Omega \rangle$, it is easiest to normal mode normal order the Hamiltonian. The plane wave normal ordering is defined in terms of $a^\dag$ and a and so, to evaluate these terms, we must use the plane wave expansion (2.17)

$$\begin{aligned} :\phi ^n(x):_a= & {} \int \frac{d^np}{(2\pi )^n}\frac{\mathrm{exp} \left( -ix\sum _{i=1}^n p_i\right) }{\sqrt{2^n\omega _{p_1}\ldots \omega _{p_n}}}:\prod _{i=1}^n\nonumber \\&\times \left( a^\dag _{p_i}+a_{-p_i}\right) :_a. \end{aligned}$$

(3.2)

To rewrite this in terms of normal mode operators, one need only insert (2.19) into the inverse of (2.17) to obtain the Bogoliubov transformations

$$\begin{aligned} a^\dag _p= & {} \sum _I a^\dag _{I,p}+a^\dag _{C,p},\quad \nonumber \\ a_{-p}= & {} \sum _I a_{I,-p}+a_{C,-p}\nonumber \\ a^\dag _{B,p}= & {} {\tilde{g}}_{B}(p)\left[ \sqrt{\frac{\omega _p}{2}}\phi _0-\frac{i}{\sqrt{2\omega _p}}\pi _0\right] ,\quad \nonumber \\ a_{B,-p}= & {} {\tilde{g}}_{B}(p)\left[ \sqrt{\frac{\omega _p}{2}}\phi _0+\frac{i}{\sqrt{2\omega _p}}\pi _0\right] .\nonumber \\ a^\dag _{BE,i,p}= & {} \frac{{\tilde{g}}_{BE,i}(p)}{2}\left( \frac{\omega _p+\omega _{BE,i}}{\sqrt{\omega _p\omega _{BE,i}}}b_{BE,i}^\dag +\frac{\omega _p-\omega _{BE,i}}{\sqrt{\omega _p\omega _{BE,i}}}b_{BE,i}\right) \nonumber \\ a_{BE,i,-p}= & {} \frac{{\tilde{g}}_{BE,i}(p)}{2}\left( \frac{\omega _p-\omega _{BE,i}}{\sqrt{\omega _p\omega _{BE,i}}}b_{BE,i}^\dag +\frac{\omega _p+\omega _{BE,i}}{\sqrt{\omega _p\omega _{BE,i}}}b_{BE,i}\right) \nonumber \\ a^\dag _{BO,i,p}= & {} \frac{{\tilde{g}}_{BO,i}(p)}{2}\left( \frac{\omega _p+\omega _{BO,i}}{\sqrt{\omega _p\omega _{BO,i}}}b_{BO,i}^\dag +\frac{-\omega _p+\omega _{BO,i}}{\sqrt{\omega _p\omega _{BO,i}}}b_{BO,i}\right) \nonumber \\ a_{BO,i,-p}= & {} \frac{{\tilde{g}}_{BO,i}(p)}{2}\left( \frac{\omega _p-\omega _{BO,i}}{\sqrt{\omega _p\omega _{BO,i}}}b_{BO,i}^\dag +\frac{-\omega _p-\omega _{BO,i}}{\sqrt{\omega _p\omega _{BO,i}}}b_{BO,i}\right) \nonumber \\ a^\dag _{C,p}= & {} \int \frac{dk}{2\pi }\frac{{\tilde{g}}_k(p)}{2}\left( \frac{\omega _p+\omega _k}{\sqrt{\omega _p\omega _k}}b_k^\dag +\frac{\omega _p-\omega _k}{\sqrt{\omega _p\omega _k}}b_{-k}\right) \nonumber \\ a_{C,-p}= & {} \int \frac{dk}{2\pi }\frac{{\tilde{g}}_k(p)}{2}\left( \frac{\omega _p-\omega _k}{\sqrt{\omega _p\omega _k}}b_k^\dag +\frac{\omega _p+\omega _k}{\sqrt{\omega _p\omega _k}}b_{-k}\right) . \end{aligned}$$

(3.3)

The key simplification comes from the fact that the modes from distinct oscillators commute with each other and they all commute with the zero modes. Thus, after inserting (3.3) into (3.2), one can separate the modes of each oscillator and the zero modes

$$\begin{aligned}&:\prod _{i=1}^n\left( a^\dag _{p_i}+a_{-p_i}\right) :_a\nonumber \\&\quad =\sum _{\{J^M|\cup _M J^M=[1,n]\}} \prod _M\left( :\prod _{i\in J^M}\left( a^\dag _{M,p_i}+a_{M,-p_i}\right) :_a\right) \nonumber \\ \end{aligned}$$

(3.4)

where M runs over $\{B,\{BE,i\},\{BO,i\},C\}$, the $J^M$ are disjoint and their union is [1, n].

For example, in the case of two point functions in the Sine-Gordon model, $n_e=n_0=0$ and $n=2$. Thus M runs over the labels B and C corresponding to the translation zero mode and the continuum. $J^M$ runs over the four subsets of $\{1,2\}$, leading to four summands

$$\begin{aligned}&:\prod _{i=1}^2\left( a^\dag _{p_i}+a_{-p_i}\right) :_a\nonumber \\&\quad = :\prod _{i=1}^2\left( a^\dag _{B,p_i}+a_{B,-p_i}\right) :_a+:\nonumber \\&\qquad \times \left( a^\dag _{B,p_1}+a_{B,-p_1}\right) :_a:\left( a^\dag _{C,p_2}+a_{C,-p_2}\right) :_a\nonumber \\&\qquad +:\left( a^\dag _{B,p_2}+a_{B,-p_2}\right) :_a:\left( a^\dag _{C,p_1}+a_{C,-p_1}\right) :_a\nonumber \\&\qquad +:\prod _{i=1}^2\left( a^\dag _{C,p_i}+a_{C,-p_i}\right) :_a. \end{aligned}$$

(3.5)

Note that in a local Hamiltonian, normal-ordered products appear in the combination (3.2) where this product is integrated over a kernel which is symmetric with respect to permutations of the $p_i$. Thus only the symmetric part of the product contributes to the Hamiltonian. This depends on the subsets $J^M$ only via their cardinalities $j_M=|J^M|$ which sum to n

$$\begin{aligned}&{\mathrm{S}}\left[ :\prod _{i=1}^n\left( a^\dag _{p_i}+a_{-p_i}\right) :_a\right] \nonumber \\&\quad =n! \sum _{\{j_M|\sum _M j_M=n\}} {\mathrm{S}}\left[ \prod _M\left( \frac{1}{j_M!}:\prod _{i=1+\sum _{N=1}^{M-1}j_N}^{\sum _{N=1}^{M}j_N}\left( a^\dag _{M,p_i}\!+\!a_{M,-p_i}\right) :_a\right) \right] \nonumber \\ \end{aligned}$$

(3.6)

where S symmetrizes all values of $p_i$. Where the letter M appears in the limits of the sum, it is understood that we have numbered the $n_o+n_e+2$ values of M from 1 to $n_o+n_e+2$. The ordering chosen does not matter.

For example, (3.5) becomes

$$\begin{aligned}&S\left[ :\prod _{i=1}^2\left( a^\dag _{p_i}+a_{-p_i}\right) :_a\right] \nonumber \\&\quad = S\left[ :\prod _{i=1}^2\left( a^\dag _{B,p_i}+a_{B,-p_i}\right) :_a\right] \nonumber \\&\qquad +2S\left[ :\left( a^\dag _{B,p_1}+a_{B,-p_1}\right) :_a: \left( a^\dag _{C,p_2}+a_{C,-p_2}\right) :_a\right] \nonumber \\&\qquad +S\left[ :\prod _{i=1}^2\left( a^\dag _{C,p_i}+a_{C,-p_i}\right) :_a\right] \end{aligned}$$

(3.7)

where the three terms correspond to $\{j_B=2,\ j_C=0\}$, $\{j_B=1,\ j_C=1\}$ and $\{j_B=0,\ j_C=2\}$. To avoid clutter, below the operator S will not be written explicitly, but we will write in the text when we symmetrize.

If we decompose $\phi (x)$ similarly to the plane wave operators

$$\begin{aligned} \phi (x)= & {} \sum _M \phi _M(x),\quad \nonumber \\ \phi _M(x)= & {} \int \frac{dp}{2\pi }\frac{1}{\sqrt{2\omega _p}}\left( a^\dag _{M,p}+a_{M,-p}\right) e^{-ipx} \end{aligned}$$

(3.8)

then we can use (3.6) to decompose

$$\begin{aligned}&:\phi ^n(x):_a=n! \sum _{\{j_M|\sum _M j_M=n\}} \prod _M\left( \frac{1}{j_M!}:\phi ^{j_M}_M(x):_a\right) .\nonumber \\ \end{aligned}$$

(3.9)

Note that the symmetrization is automatic here because of the symmetric kernel of the p integration in (3.2).

The normal ordering on the right hand side of (3.4) is defined to be whatever one obtains from (3.2) when all of the different oscillators are separated. This is well-defined. But is it a normal ordering?

3.2 The problem

To simplify this question, let us restrict our attention momentarily to the case $n_e=n_o=0$, as in the Sine-Gordon theory. The generalization to other values is trivial. Clearly, whatever ${:}{:}_a$ on the $a_{M}$ means, it is linear since the factorization above can be performed separately for each summand. So consider one summand in (3.5)

$$\begin{aligned} :a^\dag _{B,p_1}a^\dag _{B,p_2}:_a. \end{aligned}$$

(3.10)

The simplest guess would be that ${:}{:}_a$ places the $a_B^\dag$ on the left, and so the answer could be $a^\dag _{B,p_1}a^\dag _{B,p_2}$ or $a^\dag _{B,p_2}a^\dag _{B,p_1}$. The trouble is that these are not equal because

$$\begin{aligned}{}[a^\dag _{B,p_1},a^\dag _{B,p_2}]= & {} \left[ {\tilde{g}}_{B}(p_1)\left( \sqrt{\frac{\omega _{p_1}}{2}}\phi _0-\frac{i}{\sqrt{2\omega _{p_1}}}\pi _0\right) ,\nonumber \right. \\&\left. \times {\tilde{g}}_{B}(p_2)\left( \sqrt{\frac{\omega _{p_2}}{2}}\phi _0-\frac{i}{\sqrt{2\omega _{p_2}}}\pi _0\right) \right] \nonumber \\= & {} \frac{1}{2}\left( \sqrt{\frac{\omega _{p_1}}{\omega _{p_2}}}-\sqrt{\frac{\omega _{p_2}}{\omega _{p_1}}}\right) {\tilde{g}}_{B}(p_1){\tilde{g}}_{B}(p_2).\nonumber \\ \end{aligned}$$

(3.11)

Similarly, in the case of continuum modes

$$\begin{aligned}{}[a^\dag _{C,p_1},a^\dag _{C,p_2}]= & {} \int \frac{d^{2}k}{(2\pi )^{2}}\left[ \frac{{\tilde{g}}_{k_1}(p_1)}{2}\nonumber \right. \\&\times \left( \frac{\omega _{p_1}+\omega _{k_1}}{\sqrt{\omega _{p_1}\omega _{k_1}}}b_{k_1}^\dag +\frac{\omega _{p_1}-\omega _{k_1}}{\sqrt{\omega _{p_1}\omega _{k_1}}}b_{-k_1}\right) ,\nonumber \\&\times \frac{{\tilde{g}}_{k_2}(p_2)}{2}\left( \frac{\omega _{p_2}+\omega _{k_2}}{\sqrt{\omega _{p_2}\omega _{k_2}}}b_{k_2}^\dag \left. +\frac{\omega _{p_2}-\omega _{k_2}}{\sqrt{\omega _{p_2}\omega _{k_2}}}b_{-k_2}\right) \right] \nonumber \\= & {} \frac{1}{2}\left( \sqrt{\frac{\omega _{p_1}}{\omega _{p_2}}}-\sqrt{\frac{\omega _{p_2}}{\omega _{p_1}}}\right) \int \frac{d^{1}k}{(2\pi )^{1}}{\tilde{g}}_{k_1}(p_1){\tilde{g}}_{-k_1}(p_2).\nonumber \\ \end{aligned}$$

(3.12)

Thus the action of ${:}{:}_a$ on $a^\dag _{M,p}$ and $a_{M,-p}$ is more complicated than simply putting all $a_M^\dag$ on the left, since their order matters. This was not the case with the undecomposed plane wave oscillator modes because

$$\begin{aligned}{}[a^\dag _{p_1},a^\dag _{p_2}]= & {} [a^\dag _{B,p_1},a^\dag _{B,p_2}]+[a^\dag _{C,p_1},a^\dag _{C,p_2}]\nonumber \\= & {} \frac{1}{2}\left( \sqrt{\frac{\omega _{p_1}}{\omega _{p_2}}}-\sqrt{\frac{\omega _{p_2}}{\omega _{p_1}}}\right) \nonumber \\&\times \left( {\tilde{g}}_{B}(p_1){\tilde{g}}_{B}(p_2)+ \int \frac{d^{1}k}{(2\pi )^{1}}{\tilde{g}}_{k_1}(p_1){\tilde{g}}_{-k_1}(p_2)\right) \nonumber \\= & {} \frac{1}{2}\left( \sqrt{\frac{\omega _{p_1}}{\omega _{p_2}}}-\sqrt{\frac{\omega _{p_2}}{\omega _{p_1}}}\right) 2\pi \delta (p_1+p_2)=0 \end{aligned}$$

(3.13)

where we used the completeness relations (2.14) and the product of zero and a delta function vanishes at $p_1=p_2$ because this is the commutator of an operator with itself.

Conclusion: One may freely interchange the undecomposed plane wave mode operators $a^\dag$ and also a inside of ${:}{:}_a$, for example in Eq. (3.2). However, this shuffling fixes the order of the components $a^\dag _M$ and also $a_M$ in (3.4). In particular, the ordering of the components must be the same for all M, as this ordering is that chosen for the undecomposed operators.

Note that the symmetrized commutators vanish, and so this problem does not arise in the symmetrized products relevant to the computations of products of fields at the same point, as appear for example in the Hamiltonian.

3.3 A practical convention

In the previous section we learned that we need to make a choice. We need to choose the ordering of the $a^\dag _{p_i}$ and also of the $a_{-p_i}$ in (3.2). This choice does not affect our answer but it fixes the orderings of each component in (3.4). In this section we will choose an ordering which will facilitate the computations in the next section.

Let us define the shorthand

$$\begin{aligned} N_k(p_1\ldots p_k)=:\prod _{i=1}^k\left( a^\dag _{p_i}+a_{-p_i}\right) :_a. \end{aligned}$$

(3.14)

We choose the ordering defined by

$$\begin{aligned} N_0= & {} 1,\quad N_{k+1}(p_1\ldots p_{k+1})\nonumber \\= & {} a^\dag _{p_{k+1}}N_k(p_1\ldots p_k)+N_k(p_1\ldots p_k) a_{-p_{k+1}}. \end{aligned}$$

(3.15)

We remind the reader that the value of $N_k$ does not depend on this choice of ordering, as all $a^\dag$ commute with each other as do all a. However it does affect the definition of the normal ordering of the components.

Our strategy will be the following. First we will guess a formula for the normal ordering of the components

$$\begin{aligned}&N^M_k(p_1\ldots p_k)=:\prod _{i=1}^k\left( a^\dag _{M,p_i}+a_{M,-p_i}\right) :_a. \end{aligned}$$

(3.16)

Then we will show that, using the factorization formula (3.4) the guess yields the correct value of $N_k$. Recall that our definition of ${:}{:}_a$ on components is that it satisfies (3.4) and so once we have shown this, we will have verified that our guess indeed satisfies the definition and so corresponds to a valid convention.

Our guess is

$$\begin{aligned}&N^M_0=1,\quad \nonumber \\&N^M_{k+1}(p_1\ldots p_{k+1})=a^\dag _{M,p_{k+1}}N^M_k(p_1\ldots p_k)\nonumber \\&\quad +N^M_k(p_1\ldots p_k) a_{M,-p_{k+1}}. \nonumber \\ \end{aligned}$$

(3.17)

The factorization formula (3.4) in the case $n_e=n_o=0$ is

$$\begin{aligned}&N_k(p_1\ldots p_k)=\sum _{J\subset [1,k]} N_{|J|}^B(p_J)N_{k-|J|}^C(p_{[1,k]{\setminus } J}). \end{aligned}$$

(3.18)

Here we have adopted the shorthand $p_S$ for the ordered set of all $p_j$ with $j\in S$. The ordering is just the ascending order, since that appeared on the left hand side of the equation. We need to show that our guess (3.17) satisfies (3.18).

Our proof will be by induction. The base case, $k=0$ is trivial as the only term in the sum is $J=\varnothing$ and so (3.18) becomes $1=1$. Next assume that (3.18) is satisfied for some value of k and define

$$\begin{aligned} {\hat{N}}_{k+1}(p_1\ldots p_{k+1})=\sum _{J\subset [1,k+1]} N_{|J|}^B(p_J)N_{k+1-|J|}^C(p_{[1,k+1]{\setminus } J}) \end{aligned}$$

(3.19)

where the right hand side is defined using (3.17). We need to prove that ${\hat{N}}=N$ to complete the induction.

Each J either does or does not contain the element $\{k+1\}$ and so we may respectively divide the sum in two parts, redefining the dummy set J in the first sum by removing $\{k+1\}$

$$\begin{aligned} {\hat{N}}_{k+1}(p_1\ldots p_{k+1})= & {} \sum _{J\subset [1,k]}N_{|J|+1}^B(p_J,p_{k+1})N_{k-|J|}^C(p_{[1,k]{\setminus } J})\nonumber \\&+\sum _{J\subset [1,k]} N_{|J|}^B(p_J)N_{k+1-|J|}^C(p_{[1,k]{\setminus } J},p_{k+1})\nonumber \\= & {} \sum _{J\subset [1,k]} \left( a^\dag _{B,p_{k+1}}N_{|J|}^B(p_J,p_{k})\nonumber \right. \\&\left. +N_{|J|}^B(p_J,p_{k})a_{B,-p_{k+1}}\right) N_{k-|J|}^C(p_{[1,k]{\setminus } J}) \nonumber \\&+\sum _{J\subset [1,k]} N_{|J|}^B(p_J,p_{k})\left( a^\dag _{C,p_{k+1}}N_{k-|J|}^C(p_{[1,k]{\setminus } J})\nonumber \right. \\&\left. +N_{k-|J|}^C(p_{[1,k]{\setminus } J})a_{C,-p_{k+1}}\right) \nonumber \\= & {} a^\dag _{p_{k+1}}N_k(p_1\ldots p_k)\nonumber \\&+N_k(p_1\ldots p_k)a_{-p_{k+1}}=N_{k+1}(p_1\ldots p_{k+1})\nonumber \\ \end{aligned}$$

(3.20)

completing the induction.

In summary, we have shown that if we adopt the definition (3.17) for the plane wave normal ordering of component fields $a^\dag _M$ and $a_M$, then the factorization formula (3.18) is satisfied and so these components $N^M_k$ can be assembled to determine the plane wave normal ordered product $N_k$ of the undecomposed operators. Although our proof was for the case with no breather modes $n_e=n_o=0$, the equation (3.17) works in general and indeed the proof can be trivially generalized to show the compatibility of (3.17) and (3.4).

4 Recursion formulas

4.1 Zero modes

Define the coefficients $\alpha _{nm}$ by

$$\begin{aligned}&N^B_n(p_1\ldots p_n)=\left( \prod _{i=1}^n\sqrt{2\omega _{p_i}}{\tilde{g}}_B(p_i)\right) \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\phi _0^{n-2m}\nonumber \\ \end{aligned}$$

(4.1)

where $N^B_n$ was defined in (3.16). Then using (3.17) we can find the next product

$$\begin{aligned} N^B_{n+1}(p_1\ldots p_{n+1})= & {} \left( \prod _{i=1}^n\sqrt{2\omega _{p_i}}{\tilde{g}}_B(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left( a^\dag _{B,p_{n+1}}\phi _0^{n-2m}+\phi _0^{n-2m}a_{B,-p_{n+1}}\right) \nonumber \\ \end{aligned}$$

(4.2)

$$\begin{aligned}= & {} \frac{1}{2}\left( \prod _{i=1}^{n+1}\sqrt{2\omega _{p_i}}{\tilde{g}}_B(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left[ \left( \phi _0-\frac{i}{\omega _{p_{n+1}}}\pi _0\right) \phi _0^{n-2m}\right. \nonumber \\&\left. +\phi _0^{n-2m}\left( \phi _0+\frac{i}{\omega _{p_{n+1}}}\pi _0\right) \right] \nonumber \\= & {} \left( \prod _{i=1}^{n+1}\sqrt{2\omega _{p_i}}{\tilde{g}}_B(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left( \phi _0^{n-2m+1}-\frac{1}{2\omega _{p_{n_1}}}\phi _0^{n-2m-1}\right) .\nonumber \\ \end{aligned}$$

(4.3)

Dividing through by the product on the left one finds

$$\begin{aligned}&\sum _{m=0}^{\lfloor {\frac{n+1}{2}}\rfloor }\alpha _{n+1,m}\phi _0^{n-2m+1}\nonumber \\&\quad =\sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left( \phi _0^{n-2m+1}-\frac{n-2m}{2\omega _{p_{n+1}}}\phi _0^{n-2m-1}\right) . \end{aligned}$$

(4.4)

Finally matching terms with the same power of $\phi _0$ we arrive at the recursion relation

$$\begin{aligned} \alpha _{n+1,m}=\alpha _{nm}-\frac{n-2m+2}{2\omega _{p_{n+1}}}\alpha _{n,m-1} \end{aligned}$$

(4.5)

which, together with the initial condition $\alpha _{0m}=\delta _{m,0}$ fixes all of the coefficients $\alpha$.

The recursion relation (4.5) has a simple interpretation in terms of a Wick’s theorem. m is the number of contractions. The $(n+1)$st operator may either not contract, leading to the first term on the right hand side, or else it may contract. If it does contract, since there are m contractions in all, the first n operators have $m-1$ contractions. Therefore the $n+1$st operator may contract with any one of the $n-2(m-1)$ uncontracted operators, yielding the factor of $n-2m+2$ in the second term. Each contraction yields a factor of $-1/(2\omega _{p_{n+1}})$. Note that this contraction factor is not symmetric with respect to a permutation of the $p_i$, since it depends only on the $p_i$ with the highest value of i among the two contracted operators, which is $p_{n+1}$.

4.2 Solving the recursion formula

Recall that to compute the Hamiltonian we only need the symmetrized $N_n$. In this case the choice of $\omega _{p_i}$ is irrelevant, it is only important that no N have two $\omega _{p_i}$ with the same i. Said differently, adding an antisymmetric piece to N will not change the symmetrized N and so will not change H. We can thus shift N to be of the form

$$\begin{aligned}&N^B_n(p_1\ldots p_n)\nonumber \\&\quad =\left( \prod _{i=1}^n\sqrt{2\omega _{p_i}}{\tilde{g}}_B(p_i)\right) \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }a_{nm}\phi _0^{n-2m}\prod _{i=1}^m\left( \frac{-1}{2\omega _{p_i}}\right) \nonumber \\ \end{aligned}$$

(4.6)

where a is a pure number which simply counts the number of ways to make m contractions. a satisfies the recursion relation

$$\begin{aligned} a_{n+1,m}=a_{nm}+(n-2m+2)a_{n,m-1}. \end{aligned}$$

(4.7)

As the contractions are interchangeable, $a_{nm}$ contains a factor of 1/m!. This is multiplied by the number of choices for the jth contraction, which is $(n-2j+2)(n-2j+1)/2$, for each j from 1 to m. In all one finds

$$\begin{aligned} a_{nm}= & {} \frac{1}{m!}\prod _{j=1}^m\frac{(n-2j+2)(n-2j+1)}{2}\nonumber \\= & {} \frac{1}{2^m}\frac{n!}{m!(n-2m)!}. \end{aligned}$$

(4.8)

In the case with no breathers, the decomposition of the fields (3.9) becomes

$$\begin{aligned} :\phi ^n(x):_a=\sum _{j=0}^n \frac{n!}{j!(n-j)!} :\phi ^{j}_B(x):_a :\phi ^{n-j}_C(x):_a. \end{aligned}$$

(4.9)

Assembling the results above, we have evaluated the first factor in (4.9)

$$\begin{aligned} :\phi ^{j}_B(x):_a= & {} \int \frac{d^jp}{(2\pi )^j}\frac{e^{-ix\sum _ip_i}}{\sqrt{2^j\omega _{p_1}\ldots \omega _{p_j}}}N^B_j(p_1\ldots p_j)\nonumber \\= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{1}{2^m}\frac{j!}{m!(j-2m)!}\phi _0^{j-2m}\nonumber \\&\times \int \frac{d^jp}{(2\pi )^j}e^{-ix\sum _ip_i}\left( \prod _{i=1}^j{\tilde{g}}_B(p_i)\right) \prod _{i=1}^m\left( \frac{-1}{2\omega _{p_i}}\right) \nonumber \\= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}g_B^{j-2m}(x)\mathcal {I}_B^m(x)\phi _0^{j-2m}\nonumber \\= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}\mathcal {I}_B^m(x)\phi _B^{j-2m}(x) \end{aligned}$$

(4.10)

where we have introduced the contraction factor

$$\begin{aligned}&\mathcal {I}_B(x)=g_B(x){\hat{g}}_B(x),\quad \nonumber \\&{\hat{g}}_B(x)=-\int \frac{dp}{2\pi }e^{-ipx}\frac{{\tilde{g}}_B(p)}{2\omega _p}. \end{aligned}$$

(4.11)

4.3 Odd breathers

Similarly to the plane wave ordered products $N_n(p)$ we will define the normal mode ordered products

$$\begin{aligned} B^{BO}_n=:\left( b^\dag _{BO}-b_{BO}\right) ^n:_b. \end{aligned}$$

(4.12)

Our goal in this section is to learn how to expand $N_n(p)$ in terms of $B^{BO}_n(k)$.

Using the identity

$$\begin{aligned} B^{BO}_n=\sum _{k=0}^n (-1)^k\frac{n!}{k!(n-k)!} b_{BO}^{\dag n-k} b_{BO}^k \end{aligned}$$

(4.13)

one readily derives the anticommutator

$$\begin{aligned} \{b^\dag _{BO}-b_{BO},B^{BO}_n\}=2B^{BO}_{n+1}-2nB^{BO}_{n-1} \end{aligned}$$

(4.14)

and the commutator

$$\begin{aligned} \left[ b^\dag _{BO}+b_{BO},B^{BO}_n\right] =2nB^{BO}_{n-1} \end{aligned}$$

(4.15)

which will be useful momentarily.

Proceeding as for the zero mode, we define coefficients $\alpha _{nm}$ by

$$\begin{aligned}&N^{BO}_n(p_1\ldots p_n)=\left( \prod _{i=1}^n\sqrt{\frac{\omega _{p_i}}{\omega _{BO}}}{\tilde{g}}_{BO}(p_i)\right) \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}B^{BO}_{n-2m}.\nonumber \\ \end{aligned}$$

(4.16)

Then using (3.17)

$$\begin{aligned}&N^{BO}_{n+1}(p_1\ldots p_{n+1})=\left( \prod _{i=1}^n\sqrt{\frac{\omega _{p_i}}{\omega _{BO}}}{\tilde{g}}_{BO}(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left( a^\dag _{BO,p_{n+1}}B^{BO}_{n-2m}+B^{BO}_{n-2m}a_{BO,-p_{n+1}}\right) \nonumber \\&=\frac{1}{2}\left( \prod _{i=1}^{n+1}\sqrt{\frac{\omega _{p_i}}{\omega _{BO}}}{\tilde{g}}_{BO}(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left( \{b^\dag _{BO}-b_{BO},B^{BO}_{n-2m}\}+\frac{\omega _{BO}}{\omega _{p_{n+1}}}\nonumber \right. \\&\times \left. [b^\dag _{BO}+b_{BO},B^{BO}_{n-2m}]\right) \nonumber \\&=\left( \prod _{i=1}^{n+1}\sqrt{\frac{\omega _{p_i}}{\omega _{BO}}}{\tilde{g}}_{BO}(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left( B^{BO}_{n-2m+1}+(n-2m)\nonumber \right. \\&\times \left. \left( -1+\frac{\omega _{BO}}{\omega _{p_{n+1}}}\right) B^{BO}_{n-2m-1}\right) \end{aligned}$$

(4.17)

and so

$$\begin{aligned}&\sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}\left( B^{BO}_{n-2m+1}\!+\!(n-2m)\left( -1\!+\!\frac{\omega _{BO}}{\omega _{p_{n+1}}}\right) B^{BO}_{n-2m-1}\right) \nonumber \\&\quad =\sum _{m=0}^{\lfloor {\frac{n+1}{2}}\rfloor }\alpha _{n+1,m}B^{BO}_{n-2m+1}. \end{aligned}$$

(4.18)

Matching coefficients we obtain the recursion relation

$$\begin{aligned} \alpha _{n+1,m}=\alpha _{nm}+(n-2m+2)\left( -1+\frac{\omega _{BO}}{\omega _{p_{n+1}}}\right) \alpha _{n,m-1}. \nonumber \\ \end{aligned}$$

(4.19)

So far we have not used symmetrization, and so our recursion relation may be applied to computing any n-point function. Again, for calculating n-point functions at the same point, as in our interaction terms, we may shift $N^{BO}$ by an operator which vanishes when symmetrized

$$\begin{aligned} N^{BO}_n(p_1\ldots p_n)= & {} \left( \prod _{i=1}^n\sqrt{\frac{\omega _{p_i}}{\omega _{BO}}}{\tilde{g}}_{BO}(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }a_{nm}B^{BO}_{n-2m}\prod _{i=1}^m\left( -1+\frac{\omega _{BO}}{\omega _{p_i}}\right) .\nonumber \\ \end{aligned}$$

(4.20)

Note that the product on the right can be rewritten

$$\begin{aligned} \prod _{i=1}^m\left( -1+\frac{\omega _{BO}}{\omega _{p_i}}\right) =\left( 2\omega _{BO}\right) ^m\prod _{i=1}^m\left( -\frac{1}{2\omega _{BO}}+\frac{1}{2\omega _{p_i}}\right) \nonumber \\ \end{aligned}$$

(4.21)

so that it resembles the contraction terms in (4.6). Proceeding as above, the recursion formula satisfied by the $a_{nm}$ is again (4.7) and so the $a_{nm}$ are given by (4.8).

$$\begin{aligned} :\phi ^{j}_{BO}(x):_a= & {} \int \frac{d^jp}{(2\pi )^j}\frac{e^{-ix\sum _ip_i}}{\sqrt{2^j\omega _{p_1}\ldots \omega _{p_j}}}N^{BO}_j(p_1\ldots p_j) \nonumber \\= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{1}{2^m}\frac{j!}{m!(j-2m)!}B^{BO}_{j-2m}\nonumber \\&\times \int \frac{d^jp}{(2\pi )^j}e^{-ix\sum _ip_i}(2\omega _{BO})^{(2m-j)/2}\nonumber \\&\times \left( \prod _{i=1}^j{\tilde{g}}_{BO}(p_i)\right) \prod _{i=1}^m\left( -\frac{1}{2\omega _{BO}}+\frac{1}{2\omega _{p_i}}\right) \nonumber \\= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}\mathcal {I}_{BO}^m(x)\frac{g_{BO}^{j-2m}(x)B^{BO}_{j-2m}}{(2\omega _{BO})^{(j-2m)/2}}\nonumber \\= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}:\phi _{BO}^{j-2m}(x):_b\mathcal {I}_{BO}^m(x) \end{aligned}$$

(4.22)

where we have defined the contraction factor

$$\begin{aligned} \mathcal {I}_{BO}(x)= & {} g_{BO}(x){\hat{g}}_{BO}(x),\quad {\hat{g}}_{BO}(x)\!=\!\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{BO}(p)\nonumber \\&\times \left( -\frac{1}{2\omega _{BO}}+\frac{1}{2\omega _p}\right) . \end{aligned}$$

(4.23)

The contraction factor is similar to $\mathcal {I}_B(x)$ except that the contraction contains two terms $1/(2\omega _{BO})$ and $1/(2\omega _p)$ with a relative sign. These are respectively the contraction arising from the normal mode normal ordering and the plane wave normal ordering. In the case of $\mathcal {I}_B(x)$ the normal mode normal ordering was fundamentally different, as it was a rule for the placement of the canonical variables $\phi _0$ and $\pi _0$ and not for the oscillator modes.

The occurrence of a difference of contractions in $\mathcal {I}_{BO}$ is reminiscent of the general contraction defined in Ref. [18].

4.4 Even breathers

The normal ordering of even breathers is identical to that of odd breathers except for a few sign differences. Defining

$$\begin{aligned} B^{BE}_n=:\left( b^\dag _{BE}+b_{BE}\right) ^n:_b \end{aligned}$$

(4.24)

and using the identity

$$\begin{aligned} B^{BE}_n=\sum _{k=0}^n\frac{n!}{k!(n-k)!} b_{BE}^{\dag n-k} b_{BE}^k \end{aligned}$$

(4.25)

one finds

$$\begin{aligned} \{b^\dag _{BE}+b_{BE},B^{BE}_n\}= & {} 2B^{BE}_{n+1}+2nB^{BE}_{n-1},\quad \nonumber \\ \left[ b^\dag _{BE}-b_{BE},B^{BE}_n\right]= & {} -2nB^{BE}_{n-1}. \end{aligned}$$

(4.26)

Then defining

$$\begin{aligned} N^{BE}_n(p_1\ldots p_n)=\left( \prod _{i=1}^n\sqrt{\frac{\omega _{p_i}}{\omega _{BE}}}{\tilde{g}}_{BO}(p_i)\right) \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\alpha _{nm}B^{BE}_{n-2m}.\nonumber \\ \end{aligned}$$

(4.27)

The same computation as in the odd case yields the recursion relation

$$\begin{aligned} \alpha _{n+1,m}=\alpha _{nm}+(n-2m+2) \left( 1-\frac{\omega _{BO}}{\omega _{p_{n+1}}}\right) \alpha _{n,m-1}. \nonumber \\ \end{aligned}$$

(4.28)

Comparing (4.19) and (4.28) one sees that the contractions of even and odd breathers differ by an overall sign.

In the symmetric case one may shift $N^{BE}$ to

$$\begin{aligned} N^{BE}_n(p_1\ldots p_n)= & {} \left( \prod _{i=1}^n\sqrt{\frac{\omega _{p_i}}{\omega _{BE}}}{\tilde{g}}_{BE}(p_i)\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }a_{nm}B^{BE}_{n-2m}\prod _{i=1}^m\left( 1-\frac{\omega _{BE}}{\omega _{p_i}}\right) .\qquad \end{aligned}$$

(4.29)

where $a_{bm}$ again satisfies (4.7) and so we conclude that

$$\begin{aligned} :\phi ^{j}_{BE}(x):_a=\sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}:\phi _{BE}^{j-2m}(x):_b\mathcal {I}_{BE}^m(x) \nonumber \\ \end{aligned}$$

(4.30)

where we have defined the contraction factor

$$\begin{aligned}&\mathcal {I}_{BE}(x)=g_{BE}(x){\hat{g}}_{BE}(x),\quad \nonumber \\&{\hat{g}}_{BE}(x)=\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{BE}(p) \left( \frac{1}{2\omega _{BE}}-\frac{1}{2\omega _p}\right) . \end{aligned}$$

(4.31)

The relative sign in the recursion relation has indeed translated into a relative sign in the contraction factor with respect to $\mathcal {I}_{BO}$. As $g_{BO}(x)$ is imaginary and $g_{BE}(x)$ is real due to our convention (2.12), the relative sign may be absorbed by taking the complex conjugate of g(x) in the definition of $\mathcal {I}(x)$. We will now see in the continuum case that this definition arises quite naturally.

4.5 Continuum modes

Define

$$\begin{aligned} B^{C}_n(k_1\ldots k_n)=:\prod _{i=1}^n\left( \frac{b^\dag _{k_i}+b_{-k_i}}{\sqrt{2\omega _{k_i}}}\right) :_b. \end{aligned}$$

(4.32)

Using the identity

$$\begin{aligned} B^{C}_n(k_1\ldots k_n)\!=\!\sum _{J\subset [1,n]}\!\!\left( \prod _{j\in J}\frac{ b^{\dag }_{k_j}}{\sqrt{2\omega _{k_j}}}\right) \!\left( \prod _{j\in [1,n]{\setminus } J} \frac{b_{-k_j}}{\sqrt{2\omega _{k_j}}}\right) \nonumber \\ \end{aligned}$$

(4.33)

one finds the commutator

$$\begin{aligned}&\left[ \frac{b^\dag _{k^\prime }-b_{-k^\prime }}{\sqrt{2\omega _{k^\prime }}},B^{C}_n(k_1\ldots k_n)\right] \nonumber \\&\quad = -\frac{1}{2\omega _{k^\prime }}\sum _{J\subset [1,n]}\left[ \sum _{j^\prime \in J}2\pi \delta (k_{j^\prime }+k^\prime )\prod _{j\in J{\setminus } j^\prime }\right. \nonumber \\&\qquad \times \frac{b^{\dag }_{k_j}}{\sqrt{2\omega _{k_j}}}\prod _{j\in [1,n]{\setminus } J} \frac{b_{-k_j}}{\sqrt{2\omega _{k_j}}}\nonumber \\&\qquad +\sum _{j^\prime \in [1,n]{\setminus } J}2\pi \delta (k_{j^\prime }+k^\prime )\prod _{j\in J} \nonumber \\&\qquad \times \frac{b^{\dag }_{k_j}}{\sqrt{2\omega _{k_j}}}\prod _{j\in [1,n]{\setminus } J{\setminus } j^\prime } \left. \frac{b_{-k_j}}{\sqrt{2\omega _{k_j}}}\right] \nonumber \\&\quad =-\frac{2}{2\omega _{k^\prime }}\sum _{j^\prime \in [1,n]}\nonumber \\&\qquad \times 2\pi \delta (k_{j^\prime }+k^\prime )B^{C}_{n-1}(k_1\ldots {\hat{k}}_{j^\prime }\ldots k_n) \end{aligned}$$

(4.34)

and similarly the anticommutator

$$\begin{aligned}&\left\{ \frac{b^\dag _{k^\prime }+b_{-k^\prime }}{\sqrt{2\omega _{k^\prime }}},B^{C}_n(k_1\ldots k_n)\right\} \nonumber \\&\quad = 2B^{C}_{n+1}(k_1\ldots k_n, k^\prime ) +\frac{2}{2\omega _{k^\prime }}\sum _{j^\prime \in [1,n]}\nonumber \\&\qquad \times 2\pi \delta (k_{j^\prime }+k^\prime )B^{C}_{n-1}(k_1\ldots {\hat{k}}_{j^\prime }\ldots k_n). \end{aligned}$$

(4.35)

We will need the integrals of these identities, where the integral over $k^\prime$ is performed using the Dirac delta function

$$\begin{aligned}&\left[ \frac{1}{2}\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{k^\prime }(p_{n+1})\frac{\omega _{k^\prime }}{\omega _{p_{n+1}}}\nonumber \right. \frac{\left( b^\dag _{k^\prime }-b_{-k^\prime }\right) }{\sqrt{2\omega _{k^\prime }}},\nonumber \\&\qquad \left. \int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\alpha _{nm}^{k_1\ldots k_{n-2m}} B^{C}_{n-2m}(k_1\ldots k_{n-2m})\right] \nonumber \\&\quad =-\frac{1}{2\omega _{p_{n+1}}} \sum _{j^\prime =1}^{n-2m}\int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}{\tilde{g}}_{-k_{j^\prime }}(p_{n+1})\alpha _{nm}^{k_1\ldots k_{n-2m}}\nonumber \\&\qquad \times B^{C}_{n-2m-1}(k_1\ldots {\hat{k}}_{j^\prime }\ldots k_{n-2m}) \end{aligned}$$

(4.36)

and

$$\begin{aligned}&\left\{ \frac{1}{2}\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{k^\prime }(p_{n+1})\frac{\left( b^\dag _{k^\prime }+b_{-k^\prime }\right) }{\sqrt{2\omega _{k_{n+1}}}},\nonumber \right. \\&\qquad \times \left. \int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\alpha _{nm}^{k_1\ldots k_{n-2m}} B^{C}_{n-2m}(k_1\ldots k_{n-2m})\right\} \nonumber \\&\quad =\int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{k^\prime }(p_{n+1})\alpha _{nm}^{k_1\ldots k_{n-2m}} \nonumber \\&\qquad \times B^{C}_{n-2m+1}(k_1\ldots k_{n-2m},k^\prime ) \nonumber \\&\qquad +\sum _{j^\prime =1}^{n-2m}\int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}{\tilde{g}}_{-k_{j^\prime }}(p_{n+1})\alpha _{nm}^{k_1\ldots k_{n-2m}}\nonumber \\&\qquad \times \frac{1}{2\omega _{k_{j^\prime }}}B^{C}_{n-2m-1}(k_1\ldots {\hat{k}}_{j^\prime }\ldots k_{n-2m}) \end{aligned}$$

(4.37)

for arbitrary matrices $\alpha _{nm}$.

We will define the matrices $\alpha _{nm}$ by

$$\begin{aligned} N^{C}_n(p_1\ldots p_n)= & {} \left( \prod _{i=1}^n\sqrt{2\omega _{p_i}}\right) \nonumber \\&\times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\alpha _{nm}^{k_1\ldots k_{n-2m}}B^{C}_{n-2m}(k_1\ldots k_{n-2m}).\nonumber \\ \end{aligned}$$

(4.38)

Then (3.17) implies

$$\begin{aligned}&N^{C}_{n+1}(p_1\ldots p_{n+1})=\frac{1}{2}\left( \prod _{i=1}^{n+1}\sqrt{2\omega _{p_i}}\right) \nonumber \\&\qquad \times \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor } \left( \left\{ \int \frac{dk^\prime }{2\pi }{\tilde{g}}_{k^\prime }(p_{n+1})\nonumber \right. \right. \frac{\left( b^\dag _{k^\prime }+b_{-k^\prime }\right) }{\sqrt{2\omega _{k_{n+1}}}},\nonumber \\&\qquad \left. \left. \times \int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\alpha _{nm}^{k_1\ldots k_{n-2m}} B^{C}_{n-2m}(k_1\ldots k_{n-2m})\right\} \right. \nonumber \\&\qquad +\left[ \int \frac{dk^\prime }{2\pi }{\tilde{g}}_{k^\prime }(p_{n+1})\frac{\omega _{k^\prime }}{\omega _{p_{n+1}}}\nonumber \right. \frac{\left( b^\dag _{k^\prime }-b_{-k^\prime }\right) }{\sqrt{2\omega _{k^\prime }}},\nonumber \\&\qquad \times \left. \left. \int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\alpha _{nm}^{k_1\ldots k_{n-2m}} B^{C}_{n-2m}(k_1\ldots k_{n-2m})\right] \right) \nonumber \\&\quad =\left( \prod _{i=1}^{n+1}\sqrt{2\omega _{p_i}}\right) \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\nonumber \\&\qquad \times \left[ \int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{k^\prime }(p_{n+1})\alpha _{nm}^{k_1\ldots k_{n-2m}}\right. \nonumber \\&\qquad \times B^{C}_{n-2m+1}(k_1\ldots k_{n-2m},k^\prime )\nonumber \\&\qquad +\sum _{j^\prime =1}^{n-2m}\int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\left( \frac{1}{2\omega _{k_{j^\prime }}}-\frac{1}{2\omega _{p_{n+1}}}\right) {\tilde{g}}_{-k_{j^\prime }}(p_{n+1})\alpha _{nm}^{k_1\ldots k_{n-2m}}\nonumber \\&\qquad \left. \times B^{C}_{n-2m-1}(k_1\ldots {\hat{k}}_{j^\prime }\ldots k_{n-2m}) \right] . \end{aligned}$$

(4.39)

Summarizing, we find

$$\begin{aligned}&\sum _{m=0}^{\lfloor {\frac{n+1}{2}}\rfloor } \int \frac{d^{n-2m+1}k}{(2\pi )^{n-2m+1}}\alpha _{n+1,m}^{k_1\ldots k_{n-2m+1}}B^{C}_{n-2m+1}(k_1\ldots k_{n-2m+1})\nonumber \\&\quad =\sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor } \left[ \int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{k^\prime }(p_{n+1})\nonumber \right. \\&\qquad \times \alpha _{nm}^{k_1\ldots k_{n-2m}} B^{C}_{n-2m+1}(k_1\ldots k_{n-2m},k^\prime )\nonumber \\&\qquad +\sum _{j^\prime =1}^{n-2m}\int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\left( \frac{1}{2\omega _{k_{j^\prime }}}-\frac{1}{2\omega _{p_{n+1}}}\right) \nonumber \\&\left. \qquad \times {\tilde{g}}_{-k_{j^\prime }}(p_{n+1})\alpha _{nm}^{k_1\ldots k_{n-2m}}B^{C}_{n-2m-1}(k_1\ldots {\hat{k}}_{j^\prime }\ldots k_{n-2m}) \right] \nonumber \\&\quad =\sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\int \frac{d^{n-2m+1}k}{(2\pi )^{n-2m+1}}{\tilde{g}}_{k_{n-2m+1}}(p_{n+1})\nonumber \\&\qquad \times \alpha _{nm}^{k_1\ldots k_{n-2m}} B^{C}_{n-2m+1}(k_1\ldots k_{n-2m+1})\nonumber \\&\qquad +\sum _{m=1}^{\lfloor {\frac{n}{2}}\rfloor +1}\sum _{j^\prime =1}^{n-2m+2}\int \frac{d^{n-2m+2}k}{(2\pi )^{n-2m+2}} \left( \frac{1}{2\omega _{k_{j^\prime }}}-\frac{1}{2\omega _{p_{n+1}}}\right) \nonumber \\&\qquad \times {\tilde{g}}_{-k_{j^\prime }}(p_{n+1})\alpha _{n,m-1}^{k_1\ldots k_{n-2m+2}}\nonumber \\&\qquad \times B^{C}_{n-2m+1}(k_1\ldots {\hat{k}}_{j^\prime }\ldots k_{n-2m+2}) \end{aligned}$$

(4.40)

where ${\hat{k}}_{j^\prime }$ indicates that $k_{j^\prime }$ is omitted. Matching yields the recursion relation

$$\begin{aligned}&\alpha _{n+1,m}^{k_1\ldots k_{n-2m+1}}\nonumber \\&\quad ={\tilde{g}}_{k_{n-2m+1}}(p_{n+1})\alpha _{nm}^{k_1\ldots k_{n-2m}}\nonumber \\&\qquad +\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{-k^\prime }(p_{n+1})\left( \frac{1}{2\omega _{k^\prime }}-\frac{1}{2\omega _{p_{n+1}}}\right) \nonumber \\&\qquad \times \sum _{j^\prime =1}^{n-2m+2}\alpha _{n,m-1}^{k_1\ldots k_{j^\prime -1}k^\prime k_{j^\prime }\ldots k_{n-2m+1}}. \end{aligned}$$

(4.41)

Symmetrizing we may write

$$\begin{aligned} N^{C}_n(p_1\ldots p_n)= & {} \left( \prod _{i=1}^n\sqrt{2\omega _{p_i}}\right) \sum _{m=0}^{\lfloor {\frac{n}{2}}\rfloor }\int \frac{d^{n-2m}k}{(2\pi )^{n-2m}}\nonumber \\&\times \left( \prod _{i=1}^{n-2m}{\tilde{g}}_{k_i}(p_i)\right) a_{nm}B^{C}_{n-2m}(k_1\ldots k_{n-2m})\nonumber \\&\times \int \frac{d^m k^\prime }{(2\pi )^m}\prod _{i=1}^{m}\left( {\tilde{g}}_{-k^\prime _i}(p_{n-2m+2i-1}){\tilde{g}}_{k^\prime _i}(p_{n-2m+2i})\nonumber \right. \\&\left. \times \left( \frac{1}{2\omega _{k^\prime _i}}-\frac{1}{2\omega _{p_{n-2m+2i}}}\right) \right) \end{aligned}$$

(4.42)

where again $a_{nm}$ satisfies (4.7) and so is given by (4.8). We therefore conclude

$$\begin{aligned} :\phi ^j_C(x):_a= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\mathcal {I}_C^m(x)\int \frac{d^{j-2m}k}{(2\pi )^{j-2m}}\nonumber \\&\times \left( \prod _{i=1}^{j-2m}g_{k_i}(x)\right) \nonumber \\&\times \frac{j!}{2^m m!(j-2m)!}B^{C}_{j-2m}(k_1\ldots k_{j-2m})\nonumber \\= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}\mathcal {I}_C^m(x):\phi ^{j-2m}_C(x):_b\nonumber \\ \mathcal {I}_C(x)= & {} \int \frac{dk}{2\pi }g_{-k}(x){\hat{g}}_k(x),\quad \nonumber \\ {\hat{g}}_k(x)= & {} \int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_k(p)\nonumber \\&\times \left( \frac{1}{2\omega _{k}}-\frac{1}{2\omega _p}\right) . \end{aligned}$$

(4.43)

While the algebra leading up to our result seemed more complicated than in the case of the bound states, our final result is essentially the same. The only difference is that $\mathcal {I}_C$ is integrated over normal modes k. However, even in the case of breathers, there will be a sum over breather modes i, and so this distinction is superficial.

4.6 Wick’s Theorem

Finally we may use our decomposition (3.9) to reassemble $:\phi ^j(x):_a$ by inserting (4.10), (4.22), (4.30) and (4.43)

$$\begin{aligned} :\phi ^j(x):_a= & {} \sum _{m=0}^{\lfloor {\frac{j}{2}}\rfloor }\frac{j!}{2^m m!(j-2m)!}\mathcal {I}^m(x):\phi ^{j-2m}(x):_b ,\quad \nonumber \\ \mathcal {I}(x)= & {} \sum _M \mathcal {I}_M(x). \end{aligned}$$

(4.44)

The contraction factor $\mathcal {I}(x)$ can be efficiently evaluated using the completeness relations (2.14)

$$\begin{aligned} \mathcal {I}(x)= & {} g_B(x){\hat{g}}_B(x)+\sum _i g_i(x){\hat{g}}_i(x)+\int \frac{dk}{2\pi }g_{k}(x){\hat{g}}_{-k}(x)\nonumber \\ \end{aligned}$$

(4.45)

$$\begin{aligned}= & {} -\int \frac{dq}{2\pi }e^{-iqx} \int \frac{dp}{2\pi }\frac{e^{-ipx}}{2\omega _p} \nonumber \\&\quad \times \left[ {\tilde{g}}_B(q){\tilde{g}}_B(p)+\sum _{i=1}^{n_e}{\tilde{g}}_{BE,i}(q){\tilde{g}}_{BE,i}(p)\right. \nonumber \\&-\sum _{i=1}^{n_0}{\tilde{g}}_{BO,i}(q){\tilde{g}}_{BO,i}(p)\nonumber \\&\left. +\int \frac{dk}{2\pi }{\tilde{g}}_k(q){\tilde{g}}_{-k}(p)\right] \nonumber \\&+\int \frac{dk}{2\pi }\frac{1}{2\omega _k}g_{k}(x)\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{-k}(p)\nonumber \\&+\sum _{i=1}^{n_e}\frac{1}{2\omega _i}g_{BE,i}(x)\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{BE,i}(p)\nonumber \\&-\sum _{i=1}^{n_o}\frac{1}{2\omega _i}g_{BO,i}(x)\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{BO,i}(p)\nonumber \\= & {} -\int \frac{dq}{2\pi }e^{-iqx} \int \frac{dp}{2\pi }\frac{e^{-ipx}}{2\omega _p} 2\pi \delta (p+q)+\int \frac{dk}{2\pi }\nonumber \\&\times \frac{1}{2\omega _k}g_{k}(x)\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{-k}(p)\nonumber \\&+\sum _{i=1}^{n_e}\frac{1}{2\omega _i}g_{BE,i}(x)\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{BE,i}(p)\nonumber \\&-\sum _{i=1}^{n_o}\frac{1}{2\omega _i}g_{BO,i}(x)\int \frac{dp}{2\pi }e^{-ipx}{\tilde{g}}_{BO,i}(p)\nonumber \\= & {} \int \frac{dp}{2\pi }\left[ -\frac{1}{2\omega _p}+e^{-ipx}\left( \sum _{i=1}^{n_e}\frac{1}{2\omega _i}g_{BE,i}(x){\tilde{g}}_{BE,i}(p)\nonumber \right. \right. \\&-\sum _{i=1}^{n_o}\frac{1}{2\omega _i}g_{BO,i}(x){\tilde{g}}_{BO,i}(p)\nonumber \\&\left. \left. + \int \frac{dk}{2\pi }\frac{1}{2\omega _k}g_{k}(x){\tilde{g}}_{-k}(p)\right) \right] . \end{aligned}$$

(4.46)

It is finite. The first term is x-independent, and integrates to infinity. Now we take the derivative with respect to x, and so the first term vanishes

$$\begin{aligned} \partial _x \mathcal {I}(x)=\sum _{i=1}^{n_e+n_0}\frac{\partial _x\left| g_{i}(x)\right| ^2}{2\omega _I} +\int \frac{dk}{2\pi }\frac{\partial _x\left| g_{k}(x)\right| ^2}{2\omega _k}. \end{aligned}$$

(4.47)

Notice that the relative sign between the even and odd breathers has disappeared, because $g_i(x)^2$ is positive in the even case and negative in the odd case. This, together with the boundary condition that $\mathcal {I}(x)$ vanishes as x goes to infinity, completely determines $\mathcal {I}(x)$.

5 Applications to the Sine-Gordon soliton

5.1 Normal ordering the Hamiltonian

In the Sine-Gordon theory the interaction Hamiltonian density in $H^\prime$ is [19]

$$\begin{aligned} {\mathcal {H}}_I= & {} \frac{m^2}{\sqrt{\lambda }}\text {sin}(\sqrt{\lambda }f(x))\sum _{n=1}^{\infty }\frac{(-\lambda )^n}{(2n+1)!} :\phi ^{2n+1}(x):_a\nonumber \\&-\frac{m^2}{\lambda }\text {cos}(\sqrt{\lambda }f(x))\sum _{n=2}^{\infty }\frac{(-\lambda )^n}{2n!} :\phi ^{2n}(x):_a. \end{aligned}$$

(5.1)

The contribution arising from bound states is

$$\begin{aligned} {\mathcal {H}}_B= & {} -\frac{m^2}{\lambda }\text {cos}(\sqrt{\lambda }f(x))h_e+\frac{m^2}{\sqrt{\lambda }}\text {sin}(\sqrt{\lambda }f(x))h_o\nonumber \\ h_e= & {} \sum _{n=2}^{\infty }\frac{(-\lambda )^n}{2n!} :\phi _B^{2n}(x):_a\nonumber \\ ,\quad h_o= & {} \sum _{n=1}^{\infty }\frac{(-\lambda )^n}{(2n+1)!} :\phi _B^{2n+1}(x):_a. \end{aligned}$$

(5.2)

Using (4.10) the plane wave normal ordering may be evaluated explicitly

$$\begin{aligned} h_e= & {} \sum _{n=2}^{\infty }(-\lambda )^n \sum _{m=0}^{n}\nonumber \\&\times \frac{1}{2^m m!(2n-2m)!}\mathcal {I}_B^m(x)\phi _B^{2n-2m}(x). \end{aligned}$$

(5.3)

To simplify this sum, we will include the terms at $n=0$ and $n=1$, which are present in the Hamiltonian although they are not the only terms at their orders. These terms only affect the noninteracting part of the Hamiltonian, which is known to be the Poschl-Teller Hamiltonian. So we define

$$\begin{aligned} {\tilde{h}}_e= & {} \sum _{n=0}^{\infty }(-\lambda )^n \sum _{m=0}^{n}\frac{1}{2^m m!(2n-2m)!}\mathcal {I}_B^m(x)\phi _B^{2n-2m}(x)\nonumber \\= & {} \sum _{p=0}^{\infty } \sum _{m=0}^{\infty } \frac{(-\lambda )^{p+m}}{2^m m!(2p)!}\mathcal {I}_B^m(x)\phi _B^{2p}(x)\nonumber \\= & {} \text {cos}\left( \sqrt{\lambda }\phi _B(x)\right) \mathrm{exp}\left( -\lambda \mathcal {I}_B(x)/2\right) . \end{aligned}$$

(5.4)

The $n=0$ and $n=1$ terms that we have added are

$$\begin{aligned} {\tilde{h}}_e-h_e=1-\lambda \left( \frac{\mathcal {I}_B+\phi _B^2(x)}{2}\right) . \end{aligned}$$

(5.5)

Similarly, including the $n=0$ term,

$$\begin{aligned} {\tilde{h}}_o= & {} \sum _{n=0}^{\infty }(-\lambda )^n\sum _{m=0}^{n}\nonumber \\&\times \frac{1}{2^m m!(2n-2m+1)!}\mathcal {I}_B^m(x)\phi _B^{2n-2m+1}(x)\nonumber \\= & {} \sum _{p=0}^{\infty }\sum _{m=0}^{\infty }\frac{(-\lambda )^{p+m}}{2^m m!(2p+1)!}\mathcal {I}_B^m(x)\phi _B^{2p+1}(x)\nonumber \\= & {} \frac{1}{\sqrt{\lambda }}\text {sin}\left( \sqrt{\lambda }\phi _B(x)\right) \mathrm{exp}\left( -\lambda \mathcal {I}_B(x)/2\right) \end{aligned}$$

(5.6)

where

$$\begin{aligned} {\tilde{h}}_o-h_o=-\phi _B(x). \end{aligned}$$

(5.7)

Substituting this back into (5.2) and letting the tilde remind the reader that the noninteracting (quadratic in $\phi$ and below) terms are incorrect, we find

$$\begin{aligned} {\tilde{{\mathcal {H}}}}_B= & {} -\frac{m^2}{\lambda }\text {cos}\left( \sqrt{\lambda }\left( \phi _B(x)+f(x)\right) \right) \mathrm{exp}\left( -\lambda \mathcal {I}_B(x)/2\right) . \nonumber \\ \end{aligned}$$

(5.8)

where

$$\begin{aligned} {\tilde{{\mathcal {H}}}}_B-{\mathcal {H}}_B= & {} m^2\text {cos}(\sqrt{\lambda }f(x))\left( -\frac{1}{\lambda }+\frac{\mathcal {I}_B+\phi _B^2(x)}{2} \right) \nonumber \\&-m^2\text {sin}(\sqrt{\lambda }f(x))\frac{\phi _B(x)}{\sqrt{\lambda }} . \end{aligned}$$

(5.9)

As the terms in (5.9) are at most $O(\lambda ^0)$, and ${\mathcal {H}}_I$ anyway does not include all terms at this order, we will not be interested in these terms. The terms at these leading orders are given in this example in Ref. [19], and more generally are of the form (2.21). In particular they include no interactions and the tadpole vanishes in the full expression.

Equation (5.8) has a straightforward interpretation. The combination $\phi _B(x)+f(x)$ is just the $\mathcal {D}_f$ translated field, brutally truncated to the zero mode part. The prefactor and the cosine term are thus just the original Sine-Gordon action, translated and truncated. However we see that the plane wave normal ordering is now gone, indeed it was our goal to eliminate it, and instead there is an exponential of a contraction term. Thus plane wave normal ordering is equivalent to multiplication by the exponent of the bound state contraction. Of course only the bound state contraction appeared because we have truncated our Hamiltonian by only considering the bound component of the field. Our result is trivially normal mode normal ordered as it only involves the operator $\phi _0$.

More generally we may expect the exponential to include the sum of the contractions of the various normal modes

$$\begin{aligned} {\tilde{{\mathcal {H}}}}_I= & {} -\frac{m^2}{\lambda }:\text {cos}\left( \sqrt{\lambda }\left( \phi (x)+f(x)\right) \right) :_b\nonumber \\&\times \mathrm{exp}\left( -\lambda \mathcal {I}(x)/2\right) \nonumber \\ \mathcal {I}(x)= & {} \sum _M \mathcal {I}_M(x)=\mathcal {I}_B(x)+\mathcal {I}_C(x). \end{aligned}$$

(5.10)

The appearance of contractions in an exponential in (5.10) is also similar to the generalized Wick’s theorem postulated in Ref. [18]. It would be useful to understand this connection more precisely, as the generalized Wick’s theorem may provide a simple extension of our results to more complicated and interesting models.

5.2 Application: time independent perturbation theory

We have already noted that Eq. (5.10) is only correct beyond $O(\lambda ^0)$, as we did not include all terms at lower orders. Recalling that f is of order $\lambda ^{-1/2}$, the cos term may be expanded as in (5.1)

$$\begin{aligned} {\tilde{{\mathcal {H}}}}_I= & {} \left( \frac{m^2}{\sqrt{\lambda }}\text {sin}(\sqrt{\lambda }f(x)) \sum _{n=0}^{\infty }\frac{(-\lambda )^n}{(2n+1)!} :\phi ^{2n+1}(x):_b\nonumber \right. \\&\left. -\frac{m^2}{\lambda }\text {cos}(\sqrt{\lambda }f(x))\sum _{n=0}^{\infty }\frac{(-\lambda )^n}{2n!} :\phi ^{2n}(x):_b\right) \nonumber \\&\times \sum _{j=0}^{\infty }\frac{\left( -\lambda \mathcal {I}(x)\right) ^j}{2^j j!}. \end{aligned}$$

(5.11)

In particular the $O\left( \lambda ^{1/2}\right)$ terms in ${\mathcal {H}}_I$ are

$$\begin{aligned} {\mathcal {H}}_3\!=\!-\sqrt{\lambda }m^2\text {sin}(\sqrt{\lambda }f(x))\left( \frac{:\phi ^3(x):_b}{6}\!+\!\frac{:\phi (x):_b\mathcal {I}(x)}{2}\right) . \nonumber \\ \end{aligned}$$

(5.12)

This was reported already in Ref. [9], but now using Wick’s theorem a calculation that originally required several days has been reduced to less than an hour.

Using (5.11) we can easily go to higher orders. For example at $O(\lambda )$ we find that ${\mathcal {H}}_I$ contains

$$\begin{aligned} {\mathcal {H}}_4=-\lambda m^2 \text {cos}(\sqrt{\lambda }f(x))\left( \frac{:\phi ^4(x):_b}{24}+\frac{\mathcal {I}(x):\phi ^2(x):_b}{4}+\frac{\mathcal {I}^2(x)}{8} \right) .\nonumber \\ \end{aligned}$$

(5.13)

To apply this formula, first expand the soliton ground state

$$\begin{aligned} {\mathcal {O}}|\Omega \rangle =\sum _{n=0}^\infty |0\rangle _n \end{aligned}$$

(5.14)

where $|0\rangle _0={\mathcal {O}}_1|\Omega \rangle$ is the one-loop state described above and each successive term comes with another power of $\lambda ^{1/2}$. Then, similarly expanding the Hamiltonian the Schrodinger equation can be expanded order by order, beginning with

$$\begin{aligned} H_2|0\rangle _0= & {} Q_1|0\rangle _0,\quad \nonumber \\ H_3|0\rangle _0= & {} -(H_2-Q_1)|0\rangle _1,\quad \nonumber \\ (H_4-Q_2)|0\rangle _0= & {} -H_3|0\rangle _1-(H_2-Q_1)|0\rangle _2. \end{aligned}$$

(5.15)

The first equation was solved in Ref. [16] and the second in [9]. However $Q_2$ and $|0\rangle _2$ are still unknown. We can see from the third equation that $Q_2$ receives three contributions, one from each term in the decomposition of $|0\rangle _0$. From (5.13) we see that the $|0\rangle _0$ contribution to the two-loop correction to the soliton mass $Q_2$ is

$$\begin{aligned} Q_2^{(0)}=-\frac{\lambda m^2}{8} \int dx \text {cos}(\sqrt{\lambda }f(x))\mathcal {I}^2(x). \end{aligned}$$

(5.16)

The $|0\rangle _1$ contribution $Q_2^{(1)}$ can be found by acting (5.12) on the expression for $|0\rangle _1$ in [9]. To compute the $|0\rangle _2$ contribution $Q_2^{(2)}$ one needs to use the translation invariance of the ground state as described in [9]. The full calculation appears in Ref. [20].

5.3 Another application: correlation functions

The unsymmetrized Wick’s theorem can be used to calculate Schrodinger picture correlation functions of plane wave normal ordered operators, should one be interested in such objects

$$\begin{aligned} F(x_1\ldots x_n)={}_0\langle 0|:\phi (x_1)\ldots \phi (x_n):_a|0\rangle _0. \end{aligned}$$

(5.17)

We can decompose $\phi (x)=\phi _B(x)+\phi _C(x)$. As each $\phi _B(x)$ and $\phi _C(x)$ commute, this product can be factorized into a sum over all subsets S of [1, n]

$$\begin{aligned}&F(x_1\ldots x_n)\nonumber \\&\quad =\sum _{S\subset [1,n]}{}_0\langle 0|:\prod _{i\in S}\phi _C(x_{i}):_a:\prod _{i\in [1,n]{\setminus } S}\phi _C(x_{i}):_a|0\rangle _0.\nonumber \\ \end{aligned}$$

(5.18)

A complete set of states in the one-soliton sector is generated by functions $\psi (\phi _0)$ times products of $b^\dag _k$ acting on $|0\rangle _0$. Inserting the identity written in terms of such a complete set of states in between $\phi _C(x_n)$ and $\phi _B(x_1)$, one sees that all terms with $b^\dag$ operators vanish because the right matrix element contains no other $b^\dag$ operators. Similarly, if one considers an orthogonal basis of functions of $\phi _0$, then only the trivial function 1 will contribute as for all other functions the left matrix element will vanish by orthogonality, as none of the operators on the left contain $\phi _0$ or $\pi _0$ and they all commute with $\phi _0$. Thus we have argued

$$\begin{aligned} F(x_1\ldots x_n)= & {} \sum _{S\subset [1,n]}F_C(x_S)F_B(x_{[1,n]{\setminus } S})\nonumber \\ F_M(x_1\ldots x_n)= & {} {}_0\langle 0|:\phi _M(x_1)\ldots \phi _M(x_n):_a|0\rangle _0 \end{aligned}$$

(5.19)

where for any set $S={i_1,..,i_n}$ we have adopted the shorthand that $x_S$ is $x_{i_1}\ldots x_{i_n}$.

These products of fields can be rewritten in terms of our symbols N as in (4.10) but with no symmetrization

$$\begin{aligned} F_M(x_1\ldots x_n)= & {} \int \frac{d^np}{(2\pi )^n}\frac{e^{-i\sum _i x_ip_i}}{\sqrt{2^n\omega _{p_1}\ldots \omega _{p_n}}}\nonumber \\&\times {}_0\langle 0|N^M_n(p_1\ldots p_n)|0\rangle _0. \end{aligned}$$

(5.20)

As all matrix elements of normal mode normal ordered operators vanish, only the completely contracted operators $(n=2m)$ contribute. These depend on the coefficients $\alpha _{n,n/2}$. Immediately we see that the n-point functions vanish unless n is even.

Let us begin with the bound state components. Solving the recursion relation (4.5) in a few cases we find

$$\begin{aligned}&\alpha _{n,0}=1,\quad \nonumber \\&\alpha _{n,1}=-\sum _{j=1}^{n-1} \frac{j}{2\omega _{p_{j+1}}},\quad \nonumber \\&\alpha _{4,2}=\frac{1}{4\omega _{p_2}\omega _{p_4}}+\frac{1}{2\omega _{p_3}\omega _{p_4}}. \end{aligned}$$

(5.21)

The completely contracted terms

$$\begin{aligned} F_B(x_1\ldots x_n)=\int \frac{d^np}{(2\pi )^n} e^{-i\sum _i x_ip_i}\left( \prod _{j=1}^n{\tilde{g}}_B(p_j)\right) \alpha _{n,n/2}\nonumber \\ \end{aligned}$$

(5.22)

can be written immediately in the first few cases

$$\begin{aligned} F_B(x_1,x_2)= & {} g_B(x_1){\hat{g}}_B(x_2) \nonumber \\ F_B(x_1,x_2,x_3,x_4)= & {} g_B(x_1){\hat{g}}_B(x_4)\nonumber \\&\times \left( {\hat{g}}_B(x_2)g_B(x_3)+2g_B(x_2){\hat{g}}_B(x_3)\right) .\nonumber \\ \end{aligned}$$

(5.23)

It is remarkably asymmetrical in the $x_i$. This is to be expected, as we made an arbitrary choice in our convention for the bound state normal ordering. The correlator $F_B$ depends on that choice. Only the product F is independent.

Now let us turn our attention to the continuum modes $\phi _C$. Solving the recursion relation (4.41) in the first few cases we find

$$\begin{aligned} \alpha _{n,0}^{k_1\ldots k_{n}}= & {} \prod _{j=1}^n {\tilde{g}}_{k_j}(p_j),\quad \nonumber \\ \alpha _{2,1}= & {} \int \frac{dk^\prime }{2\pi }{\tilde{g}}_{-k^\prime }(p_2) \left( \frac{1}{2\omega _{k^\prime }}-\frac{1}{2\omega _{p_2}}\right) {\tilde{g}}_{k^\prime }(p_1)\nonumber \\ \alpha _{3,1}^{k_1}= & {} {\tilde{g}}_{k_1}(p_3)\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{-k^\prime }(p_2)\nonumber \\&\times \left( \frac{1}{2\omega _{k^\prime }}-\frac{1}{2\omega _{p_2}}\right) {\tilde{g}}_{k^\prime }(p_1)\nonumber \\&+\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{-k^\prime }(p_3)\left( \frac{1}{2\omega _{k^\prime }}-\frac{1}{2\omega _{p_3}}\right) \nonumber \\&\times \left( {\tilde{g}}_{k^\prime }(p_1){\tilde{g}}_{k_1}(p_2)+{\tilde{g}}_{k_1}(p_1){\tilde{g}}_{k^\prime }(p_2)\right) \nonumber \\ \alpha _{4,2}= & {} \int \frac{dk^\prime _2}{2\pi }{\tilde{g}}_{-k^\prime _2}(p_4)\left( \frac{1}{2\omega _{k^\prime _2}}-\frac{1}{2\omega _{p_4}}\right) \alpha _{3,1}^{k^\prime _2}\nonumber \\= & {} \int \frac{d^{2}k^\prime }{(2\pi )^{2}}{\tilde{g}}_{-k^\prime _2}(p_4)\left( \frac{1}{2\omega _{k^\prime _2}}-\frac{1}{2\omega _{p_4}}\right) \nonumber \\&\times {\tilde{g}}_{k^\prime _2}(p_3){\tilde{g}}_{-k^\prime _1}(p_2)\left( \frac{1}{2\omega _{k^\prime _1}}-\frac{1}{2\omega _{p_2}}\right) {\tilde{g}}_{k^\prime _1}(p_1)\nonumber \\&+\int \frac{d^{2}k^\prime }{(2\pi )^{2}}{\tilde{g}}_{-k^\prime _2}(p_4)\left( \frac{1}{2\omega _{k^\prime _2}}-\frac{1}{2\omega _{p_4}}\right) \nonumber \\&\times {\tilde{g}}_{k^\prime _2}(p_1){\tilde{g}}_{-k^\prime _1}(p_3)\left( \frac{1}{2\omega _{k^\prime _1}}-\frac{1}{2\omega _{p_3}}\right) {\tilde{g}}_{k^\prime _1}(p_2)\nonumber \\&+\int \frac{d^{2}k^\prime }{(2\pi )^{2}}{\tilde{g}}_{-k^\prime _2}(p_4)\left( \frac{1}{2\omega _{k^\prime _2}}-\frac{1}{2\omega _{p_4}}\right) \nonumber \\&\times {\tilde{g}}_{k^\prime _2}(p_2){\tilde{g}}_{-k^\prime _1}(p_3)\left( \frac{1}{2\omega _{k^\prime _1}}-\frac{1}{2\omega _{p_3}}\right) {\tilde{g}}_{k^\prime _1}(p_1).\nonumber \\ \end{aligned}$$

(5.24)

The first completely contracted term

$$\begin{aligned} F_C(x_1\ldots x_n)=\int \frac{d^np}{(2\pi )^n} e^{-i\sum _i x_ip_i}\alpha _{n,n/2} \end{aligned}$$

(5.25)

is

$$\begin{aligned} F_C(x_1,x_2)=\int \frac{dk^\prime }{2\pi }g_{k^\prime }(x_1){\hat{g}}_{-k^\prime }(x_2) \end{aligned}$$

(5.26)

and so, using the fact that for the empty set S, $F_M(x_S)=1$ we find

$$\begin{aligned} F(x_1,x_2)= & {} F_B(x_1,x_2)+F_C(x_1,x_2)=g_B(x_1){\hat{g}}_B(x_2)\nonumber \\&+\int \frac{dk^\prime }{2\pi }g_{k^\prime }(x_1){\hat{g}}_{-k^\prime }(x_2). \end{aligned}$$

(5.27)

This is looks like the completeness relation (2.14) but is different because of the hats.

As we are working in the Schrodinger picture, all operators are time-independent and so these are equal time correlators. When $x_1\ne x_2$ the points are spacelike separated. Nonetheless this trivial calculation shows that the correlator does not vanish for spacelike separated $x_1$ and $x_2$. Also it is not translation invariant. Neither of these properties should be a surprise. This is not a correlation function in the vacuum, but in the 1-loop kink state $|0\rangle _1$ which is not translation invariant. It is, to one-loop, a Hamiltonian eigenstate and this condition implies that it has nonlocal correlations, just as the phase in a superconductor varies slowly over macroscopic distances or as parton wave functions are correlated in a hadron.

What about causality? Causality demands that

$$\begin{aligned} {}_0\langle 0|[\phi (x_1),\phi (x_2)]|0\rangle _0=0 \end{aligned}$$

(5.28)

for spacelike separated $x_1$ and $x_2$. This expression is not normal ordered and one needs to be careful when normal ordering a commutator as commutation and normal ordering do not commute

$$\begin{aligned} 0= & {} a^\dag a-a^\dag a=:a a^\dag :-:a^\dag a:\nonumber \\\ne & {} 1=:1{:}{=}{:}(aa^\dag -a^\dag a){:}{=}{:}[a,a^\dag ]:. \end{aligned}$$

(5.29)

If we start by using the equal time canonical commutation relation $[\phi (x_1),\phi (x_2)]=0$ then (5.28) is trivially satisfied and we have not succeeded in testing our correlation function. So instead rewrite (5.28) as

$$\begin{aligned}&{}_0\langle 0|\phi (x_1)\phi (x_2)|0\rangle _0={}_0\langle 0|\phi (x_2)\phi (x_1)|0\rangle _0. \end{aligned}$$

(5.30)

Now we can normal order both sides. So long as we do not to use the commutation relations, equality before normal ordering will imply equality after normal ordering and thus we will obtain a weaker but still necessary condition for causality

$$\begin{aligned} F(x_1,x_2)=F(x_2,x_1). \end{aligned}$$

(5.31)

Is this satisfied by (5.27)?

$$\begin{aligned} F(x_1,x_2)= & {} \int \frac{d^2p}{(2\pi )^2} e^{-i(x_1p_1+x_2p_2)}\left( - \frac{{\tilde{g}}_B(p_1){\tilde{g}}_B(p_2)}{2\omega _{p_{2}}}\right. \nonumber \\&+\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{-k^\prime }(p_2)\left( \frac{1}{2\omega _{k^\prime }}-\frac{1}{2\omega _{p_2}}\right) \left. {\tilde{g}}_{k^\prime }(p_1) \right) \nonumber \\= & {} \int \frac{d^2p}{(2\pi )^2} e^{-i(x_1p_1+x_2p_2)}\nonumber \\&\times \left( - \frac{2\pi \delta (p_1+p_2)}{2\omega _{p_{2}}}+\int \frac{dk^\prime }{2\pi }{\tilde{g}}_{-k^\prime }(p_2)\nonumber \right. \\&\times \left. \left( \frac{1}{2\omega _{k^\prime }}\right) {\tilde{g}}_{k^\prime }(p_1) \right) \nonumber \\= & {} -\int \frac{dp}{2\pi }\frac{e^{-ip(x_1-x_2)}}{2\omega _p}\nonumber \\&+\int \frac{dk^\prime }{2\pi }\frac{g_{-k^\prime }(x_2)g_{k^\prime }(x_1)}{2\omega _{k^\prime }}. \end{aligned}$$

(5.32)

We used the completeness relation to arrive at the second line. The last line has two terms. The first is symmetric under $x_1\leftrightarrow x_2$ because $\omega _p$ is an even function of p and so the odd piece of the exponential does not contribute to the integral. To see that the second term is also symmetric, it suffices to relabel the dummy variable $k^\prime \rightarrow -k^\prime$. Therefore the causality condition (5.31) is satisfied.

The continuum contribution to the four-point function is

$$\begin{aligned} F_C(x_1,x_2,x_3,x_4)= & {} \int \frac{d^4p}{(2\pi )^4} e^{-i\sum _i x_ip_i}\alpha _{4,2}\nonumber \\= & {} \int \frac{d^{2}k^\prime }{(2\pi )^{2}}\left( g_{k^\prime _1}(x_1){\hat{g}}_{-k^\prime _1}(x_2)g_{k^\prime _2}(x_3)\nonumber \right. \\&+g_{k^\prime _1}(x_2){\hat{g}}_{-k^\prime _1}(x_3)g_{k^\prime _2}(x_1)\nonumber \\&\left. +g_{k^\prime _1}(x_1){\hat{g}}_{-k^\prime _1}(x_3)g_{k^\prime _2}(x_2)\right) {\hat{g}}_{-k^\prime _2}(x_4).\nonumber \\ \end{aligned}$$

(5.33)

The total four-point function

$$\begin{aligned} F(x_1,x_2,x_3,x_4)= & {} F_C(x_1,x_2,x_3,x_4)\nonumber \\&+F_C(x_1,x_2)F_B(x_3,x_4)+F_C(x_1,x_3)F_B(x_2,x_4)\nonumber \\&+F_C(x_1,x_4)F_B(x_2,x_3)\nonumber \\&+F_C(x_2,x_3)F_B(x_1,x_4)+F_C(x_2,x_4)F_B(x_1,x_3)\nonumber \\&+F_C(x_3,x_4)F_B(x_1,x_2)+ F_B(x_1,x_2,x_3,x_4)\nonumber \\ \end{aligned}$$

(5.34)

is then given in terms of Eqs. (5.23), (5.26) and (5.33).

6 Remarks

We have found that plane wave normal ordering can be converted into normal mode normal ordering by following a simple rule, playing the role of Wick’s theorem. After decomposing a product of n fields into products of j field components, where each component corresponds to a set of normal modes, the components can be decomposed by summing over all possible contractions. For each contraction one replaces the pair of field components with the difference between the inverse plane wave energy $\omega _p$ and inverse normal mode energy, suitably normalized over the spectrum. Intuitively the first term arises from eliminating the plane wave normal ordering and the second from imposing the normal mode normal ordering. Of course with no normal ordering at all, one expects divergences. However the difference between these two energies is, when suitably averaged, quite small and thus all expressions are finite given either normal ordering scheme. Once we go beyond scalar theories and 1+1 dimensions there will be other divergences which must be regularized and renormalized.

In Ref. [9] the conversion between normal orderings was the most complicated part of the perturbation theory treatment of the one soliton sector. Now that we have treated this problem at all orders, and in a much more general class of theories, we expect that it will be easier to extend that calculation to two loops or beyond. In Ref. [20], we have used the Wick’s theorem here to find the two-loop ground state and mass of a scalar kink in a theory with an arbitrary potential. Previously the scalar kink mass was only found in the Sine-Gordon model [21,22,23], and our results agree in that case. However it is still not obvious that the solution to the zero mode problem in Ref. [9] also solves the problem at higher loops. If it does not, then it may be necessary to use other formalisms such as that of [7] and [8].

To go beyond perturbation theory, we will eventually need supersymmetry. In this context, coherent states have been constructed in Refs. [24, 25]. This will require a fermionic generalization of the Wick’s theorem found here. Perhaps the generalized Wick’s theorem of Ref. [18] can provide an efficient derivation.

Another application of our Wick’s theorem is the calculation of plane wave normal ordered correlation functions in soliton states. So far our calculations have been in the Schrodinger picture. As a result we can only evaluate equal time correlators. However we suspect that a generalization to the interaction picture would be straightforward, as interaction picture operators evolve according to the free Hamiltonian. This would allow us to compute correlators of operators at arbitrary time. We note that there are two interaction pictures, corresponding to the free Hamiltonians H and $H^\prime$ which describe perturbations in the zero-soliton and one-soliton sectors. For the calculation of one-soliton state correlation functions, one would use the latter as the one-loop one-soliton ground state is an eigenstate of the noninteracting part. As the one-soliton ground state is an eigenstate of $H^\prime$, it is even plausible that an LSZ reduction formula exists in the one-soliton sector. This would be the starting point for a treatment of scattering in this sector.

The recent discovery of spectral walls [26] caused by transitions between breather and continuum states has rekindled interest in kink scattering [27, 28]. The treatment of this phenomenon has so far been largely classical. While the current methodology is most straightforwardly applied to the one kink sector, it could nonetheless allow an understanding of the role played by breathers in fully quantum scattering. In particular the scattering of a kink with a plane wave or wave packet could be treated in the one kink sector.