Space-time propagation of photon pulses in dielectric media,

illustrations with beam splitters

M. Federico, V. Dorier, S. Guérin, H.R. Jauslin∗

Laboratoire Interdisciplinaire Carnot de Bourgogne - ICB, UMR 6303 CNRS

Université Bourgogne Franche-Comté, BP 47870, 21078 Dijon, France

∗ e-mail<EMAIL_ADDRESS>August 27, 2024

Published in: J. Phys. B 55 (2022) 174002
https://doi.org/10.1088/1361-6455/ac7e0e

Coherent Control: Photons, Atoms and Molecules

\- a special issue honouring the life and work of Bruce W. Shore -

###### Contents

  1. 1 Introduction
  2. 2 Passive inhomogeneous dielectric media
    1. 2.1 Hamiltonian structure of Maxwell’s equations for a passive inhomogeneous dielectric medium
    2. 2.2 Quantization of the electromagnetic field in passive inhomogeneous dielectric media
  3. 3 Evolution of the photon states determined from the dynamics of the classical modes
  4. 4 One-photon pulse propagation through a beam splitter
    1. 4.1 Description of the initial classical mode and its time evolution
    2. 4.2 Quantum dynamics of the one-photon state
    3. 4.3 Remarks on the literature on the beam splitter, involving a fourth port representing “incoming fluctuations of the vacuum”
  5. 5 The Hong-Ou-Mandel effect
    1. 5.1 Description of the initial classical modes and their time evolution
    2. 5.2 Quantum dynamics of the two photon state
    3. 5.3 Photons are not like classical waves nor like classical particles
  6. 6 Remarks on the concept of photons
  7. A Relation between the position Fock space and the reciprocal Fock space
  8. B Proof of the time evolution formulas of N-photon states

###### Abstract

Photons are the elementary quantum excitations of the electromagnetic field.
Quantization is usually constructed on the basis of an expansion in
eigenmodes, in the form of plane waves. Since they form a basis, other
electromagnetic configurations can be constructed by linear combinations. In
this presentation we discuss a formulation constructed in the general
formalism of bosonic Fock space, in which the quantum excitation can be
constructed directly on localized pulses of arbitrary shape. Although the two
formulations are essentially equivalent, the direct formulation in terms of
pulses has some conceptual and practical advantages, which we illustrate with
some examples. The first one is the passage of a single photon pulse through a
beam splitter. The analysis of this formulation in terms of pulses in Fock
space shows that there is no need to introduce “vacuum fluctuations entering
through the unused port”, as is often done in the literature. Another example
is the Hong-Ou-Mandel effect. It is described as a time dependent process in
the Schrödinger representation in Fock space. The analysis shows explicitly
how the two essential ingredients of the Hong-Ou-Mandel effect are the same
shape of the pulses and the bosonic nature of photons. This formulation shows
that all the phenomena involving linear quantum optical devices can be
described and calculated on the basis of the time dependent solution of the
corresponding classical Maxwell’s equations for pulses, from which the quantum
dynamics in Fock space can be immediately constructed.

##  1 Introduction

Recent progress in the techniques for the production and in the detection of
single photons has allowed to create photons at demand and heralded with pulse
shapes that can be engineered to a large extent [1, 2, 3, 4, 5, 6, 7, 8, 9].
These photon pulses can be made to propagate through different optical devices
[10] [11], like semi-transparent mirrors, beam splitters, lenses, and optical
fibers, and then detected with space-time resolution. In this article we
describe a framework for the propagation of single- or few-photon pulses
through linear optical media for which dissipation and dispersion can be
neglected in the range of considered frequencies. The presented approach is
based on the pioneering work of Glauber and Lewenstein [12] [13], combined
with Fock space tools of quantum field theory [14] [15] [16]. A key role is
played by the frequency operator, which contains the main information for the
dynamics, and makes the link between the classical Maxwell equations and the
quantum theory. The interaction of propagating photon pulses with matter has
been addressed in the literature from different perspectives, as described
e.g. in [17] [18] [19][20] and in the references therein.

##  2 Passive inhomogeneous dielectric media

A passive dielectric described by a dielectric coefficient
$\epsilon(\vec{x})>0$, is a linear medium for which the dissipation and the
dispersion can be neglected in the considered frequency range. The
corresponding wave equation, which is equivalent to the macroscopic
phenomenological Maxwell equations, can be written as

$\displaystyle\epsilon(\vec{x})\frac{\partial^{2}\vec{A}}{\partial t^{2}}$
$\displaystyle=$ $\displaystyle-c^{2}\nabla\times\nabla\times\vec{A}$ (2.1)
$\displaystyle\phantom{xxxxx}\vec{E}$ $\displaystyle=$
$\displaystyle-\frac{\partial\vec{A}}{\partial
t},\qquad\vec{B}=\nabla\times\vec{A},$ (2.2)

where $\vec{A}$ is the vector potential in the generalized Coulomb gauge [12]
[13], which satisfies the generalized transversality constraint
$\nabla\cdot(\epsilon\vec{A})=0$. In order to put this equation in Hamiltonian
form with a self-adjoint operator [12] [13], we make the following change of
variables:

$\vec{A}^{\prime}:=\sqrt{\epsilon}\vec{A},$ (2.3)

which leads to

$\frac{\partial^{2}\vec{A}^{\prime}}{\partial
t^{2}}=-\frac{1}{\sqrt{\epsilon}}c^{2}\nabla\times\nabla\times\frac{1}{\sqrt{\epsilon}}\leavevmode\nobreak\
\vec{A}^{\prime}.$ (2.4)

The operator

$\frac{{c}}{\sqrt{\epsilon}}\nabla\times\nabla\times\frac{{c}}{\sqrt{\epsilon}}=\Xi^{\dagger}\Xi,\qquad{\rm
with\leavevmode\nobreak\ }\Xi=\nabla\times\frac{{c}}{\sqrt{\epsilon}}$ (2.5)

is positive and selfadjoint, and thus there is a unique positive operator
$\Omega$, such that

$\Xi^{\dagger}\Xi=\Omega^{2},\qquad{\rm i.e.\leavevmode\nobreak\
}\Omega=\left(\frac{{c}}{\sqrt{\epsilon}}\nabla\times\nabla\times\frac{{c}}{\sqrt{\epsilon}}\right)^{1/2}.$
(2.6)

Thus Maxwell’s equations in a passive dielectric medium can be written as

$\frac{\partial^{2}\vec{A}^{\prime}}{\partial
t^{2}}=-\Omega^{2}\vec{A}^{\prime}.$ (2.7)

which has the same structure as the ones in empty space, where
$\Omega_{0}^{2}=-c^{2}\Delta$, but with the frequency operator $\Omega$
defined in (2.6) and the generalized transversality constraint. Thus we can
formulate the Hamiltonian structure for the passive dielectric medium as
follows.

###  2.1 Hamiltonian structure of Maxwell’s equations for a passive
inhomogeneous dielectric medium

By defining

$\vec{\Pi}^{\prime}:=\epsilon_{0}\frac{\partial\vec{A}^{\prime}}{\partial t}$
(2.8)

as the momentum canonically conjugate to $\vec{A}^{\prime}$, the wave equation
can be written in Hamiltonian form

$\displaystyle\frac{\partial\vec{A}^{\prime}}{\partial t}$ $\displaystyle=$
$\displaystyle\frac{\delta
H}{\delta\vec{\Pi}^{\prime}(\vec{x})}=\frac{1}{\epsilon_{0}}\vec{\Pi}^{\prime}$
(2.9) $\displaystyle\frac{\partial\vec{\Pi}^{\prime}}{\partial t}$
$\displaystyle=$ $\displaystyle{-}\frac{\delta
H}{\delta\vec{A}^{\prime}(\vec{x})}=-\epsilon_{0}\Omega^{2}\vec{A}^{\prime}.$
(2.10)

with the Hamilton functional

$\displaystyle H$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{3}}d^{3}x\leavevmode\nobreak\
\biggl{(}\frac{1}{2\epsilon_{0}}\vec{\Pi}^{\prime}(\vec{x})\cdot\vec{\Pi}^{\prime}(\vec{x})+\frac{\epsilon_{0}}{2}\vec{A}^{\prime}(\vec{x})\cdot\Omega^{2}\vec{A}^{\prime}(\vec{x})\biggr{)}.$
(2.11)

and the constraints

$\displaystyle\nabla\cdot\left(\sqrt{\epsilon}\vec{A}^{\prime}\right)=0,\quad\nabla\cdot\left(\sqrt{\epsilon}\vec{\Pi}^{\prime}\right)=0.$
(2.12)

###  2.2 Quantization of the electromagnetic field in passive inhomogeneous
dielectric media

The electromagnetic field in a passive inhomogeneous dielectric medium can be
quantized in the same way as in free space [12] [13] [14] [15] [16]. The only
difference is the form of the operator $\Omega^{2}$, and the generalized
transversality constraints (2.12). Our presentation is based on the general
formulation of [15] [16], adapted to the case of the electromagnetic field,
which requires to take into account the vector character and the
transversality constraints. This formulation is particularly well-adapted for
the space-time description of the propagation of the photon states. The main
steps can be summarized as follows. A) By introducing the following complex
representation of the fields

$\displaystyle\vec{\psi}:=\frac{1}{\sqrt{2\hbar}}\biggl{(}(\epsilon_{0}\Omega)^{1/2}\vec{A}^{\prime}+i(\epsilon_{0}\Omega)^{-1/2}\vec{\Pi}^{\prime}\biggr{)},$
(2.13)

the classical Maxwell equations can be written equivalently as

$\displaystyle i\frac{\partial\vec{\psi}}{\partial t}=\Omega\vec{\psi}.$
(2.14)

B) The classical phase space $\vec{\Pi}^{\prime},\vec{A}^{\prime}$ has a
natural Hilbert space structure, which in the complex representation takes the
form

$\displaystyle\mathcal{H}_{cl}:=\\{\vec{\psi}(\vec{x})\leavevmode\nobreak\
|\leavevmode\nobreak\ \langle\vec{\psi}|\vec{\psi}\rangle<\infty\\},$ (2.15)

with the scalar product

$\displaystyle\langle\vec{\psi}_{A}|\vec{\psi}_{B}\rangle:=\int_{\mathbb{R}^{3}}d^{3}x\leavevmode\nobreak\
\vec{\psi}_{A}^{*}(\vec{x})\cdot\vec{\psi}_{B}(\vec{x}).$ (2.16)

The fields in this space are essentially fields with a finite electromagnetic
energy. Nomenclature: We will use the following nomenclature, based on the one
of Refs. [17][18]: A classical mode is a configuration of the classical
electromagnetic field described equivalently by
$[\vec{E}(\vec{x}),\vec{A}(\vec{x})]$, $[\vec{E}(\vec{x}),\vec{B}(\vec{x})]$
or the complex representation $\vec{\psi}(\vec{x})$. The classical modes can
be normalized using the scalar product (2.16). A classical mode, also called
non-monochromatic mode can be viewed as an equivalence class (i.e. up to
normalization) of possible initial conditions for Maxwell’s equations.
Classical eigenmodes $\vec{\varphi}_{\kappa}$ , called also classical normal
modes, are a particular type of modes that are eigenfunctions of the frequency
operator, i.e. satisfying

$\displaystyle\Omega^{2}\vec{\varphi}_{\kappa}=\omega_{\kappa}^{2}\vec{\varphi}_{\kappa}.$
(2.17)

C) Based on this classical Hilbert space one can construct the associated
bosonic Fock space ${\mathcal{F}}^{\mathfrak{B}}(\mathcal{H}_{cl})$, defined
as follows [14] [15] [16] [21, Sect. 18.5]:

$\displaystyle{\mathcal{F}}^{\mathfrak{B}}(\mathcal{H}_{cl})$
$\displaystyle:=$
$\displaystyle\oplus_{m=0}^{\infty}{\mathcal{F}}^{\mathfrak{B}}_{m},$ (2.18)
$\displaystyle{\mathcal{F}}^{\mathfrak{B}}_{0}$ $\displaystyle:=$
$\displaystyle\mathbb{C}$ (2.19)
$\displaystyle{\mathcal{F}}^{\mathfrak{B}}_{1}$ $\displaystyle:=$
$\displaystyle\mathcal{H}_{cl}$ (2.20)
$\displaystyle{\mathcal{F}}^{\mathfrak{B}}_{2}$ $\displaystyle:=$
$\displaystyle\hat{P}_{2}\left(\mathcal{H}_{cl}\otimes\mathcal{H}_{cl}\right)\equiv\mathcal{H}_{cl}\otimes_{S}\mathcal{H}_{cl}$
$\displaystyle\vdots$ $\displaystyle{\mathcal{F}}^{\mathfrak{B}}_{m}$
$\displaystyle:=$
$\displaystyle\hat{P}_{m}\left(\mathcal{H}_{cl}\otimes\mathcal{H}_{cl}\otimes\ldots\otimes\mathcal{H}_{cl}\otimes\mathcal{H}_{cl}\right)$
$\displaystyle\vdots$

where $\hat{P}_{m}$ are the symmetrization operators, for instance
$\hat{P}_{2}|\vec{\psi}_{1}\otimes\vec{\psi}_{2}\rangle\equiv|\vec{\psi}_{1}\otimes_{S}\vec{\psi}_{2}\rangle:=\frac{1}{2}(|\vec{\psi}_{1}\otimes\vec{\psi}_{2}\rangle+|\vec{\psi}_{2}\otimes\vec{\psi}_{1}\rangle)$.

The vectors in this Fock space can be represented as
$(\phi_{0},\phi_{1},\phi_{2},\ldots)$, where
$\phi_{m}\in{\mathcal{F}}^{\mathfrak{B}}_{m}$ is the $m-$photon component. The
normalized vacuum state is defined up to a phase factor as
$|\varnothing\rangle:=(1,0,0,\ldots).$ On this symmetric Fock space one can
construct, for each $\vec{\psi}\in\mathcal{H}_{cl}$, the following creation-
annihilation operators

${\hat{B}}_{\vec{\psi}}:\mathcal{F}^{\mathfrak{B}}_{m}\to\mathcal{F}^{\mathfrak{B}}_{m-1}\qquad{\hat{B}}^{\dagger}_{\vec{\psi}}:\mathcal{F}^{\mathfrak{B}}_{m}\to\mathcal{F}^{\mathfrak{B}}_{m+1}$
(2.23)

defined by their action

$\displaystyle\kern-50.0pt{\hat{B}}^{\dagger}_{\vec{\psi}}|\vec{\psi}_{1}\otimes\ldots\otimes\vec{\psi}_{m}\rangle$
$\displaystyle:=$ $\displaystyle\sqrt{m+1}\leavevmode\nobreak\
\hat{P}_{m+1}|\vec{\psi}\otimes\vec{\psi}_{1}\otimes\ldots\otimes\vec{\psi}_{m}\rangle$
(2.24)
$\displaystyle\kern-50.0pt{\hat{B}}_{\vec{\psi}}|\vec{\psi}_{1}\otimes\ldots\otimes\vec{\psi}_{m}\rangle$
$\displaystyle:=$ $\displaystyle\leavevmode\nobreak\
\frac{1}{\sqrt{m}}\sum_{j=1}^{m}\langle\vec{\psi}|\vec{\psi}_{j}\rangle\leavevmode\nobreak\
\hat{P}_{m-1}\leavevmode\nobreak\
|\vec{\psi}_{1}\otimes\ldots\otimes\check{\vec{\psi}}_{j}\ldots\otimes\vec{\psi}_{m}\rangle,$
(2.25)

where the notation $\check{\vec{\psi}}_{j}$ indicates that this term is
missing. Examples of one- and two-photon states: If $\vec{\psi}(\vec{x})$ is a
pulse-shaped field, then

$\displaystyle{\hat{B}}^{\dagger}_{\vec{\psi}}|\varnothing\rangle=|\vec{\psi}\rangle=\vec{\psi}(\vec{x})$
(2.26)

is a one-photon state carried by the classical mode $\vec{\psi}(\vec{x})$,
described in the position space representation.

$\displaystyle{\hat{B}}^{\dagger}_{\vec{\psi}_{A}}{\hat{B}}^{\dagger}_{\vec{\psi}_{B}}|\varnothing\rangle=\sqrt{2}\leavevmode\nobreak\
|\vec{\psi}_{A}\otimes_{S}\vec{\psi}_{B}\rangle=\frac{1}{\sqrt{2}}(|\vec{\psi}_{A}\otimes\vec{\psi}_{B}\rangle+|\vec{\psi}_{B}\otimes\vec{\psi}_{A}\rangle)$
(2.27)

is a two-photon state carried by the classical modes $\vec{\psi}_{A}$ and
$\vec{\psi}_{B}$.
The annihilation operators act on 0, 1 and 2-photon states by

$\displaystyle{\hat{B}}_{\vec{\psi}}|\varnothing\rangle$ $\displaystyle=0$
(2.28) $\displaystyle{\hat{B}}_{\vec{\psi}_{A}}|\vec{\psi}_{B}\rangle$
$\displaystyle=\langle\vec{\psi}_{A}|\vec{\psi}_{B}\rangle\leavevmode\nobreak\
|\varnothing\rangle$ (2.29)
$\displaystyle{\hat{B}}_{\vec{\psi}_{A}}|\vec{\psi}_{B}\otimes_{S}\vec{\psi}_{C}\rangle$
$\displaystyle=\frac{1}{\sqrt{2}}\left(\langle\vec{\psi}_{A}|\vec{\psi}_{B}\rangle\leavevmode\nobreak\
|\vec{\psi}_{C}\rangle+\langle\vec{\psi}_{A}|\vec{\psi}_{C}\rangle\leavevmode\nobreak\
|\vec{\psi}_{B}\rangle\right).$ (2.30)

The bosonic creation-annihilation operators satisfy the following commutation
relations:

$\displaystyle\left[{\hat{B}}_{\vec{\psi}_{A}},{\hat{B}}^{\dagger}_{\vec{\psi}_{B}}\right]$
$\displaystyle=$
$\displaystyle\langle{\vec{\psi}_{A}}|{\vec{\psi}_{B}}\rangle$ (2.31)
$\displaystyle\left[{\hat{B}}_{\vec{\psi}_{A}},{\hat{B}}_{\vec{\psi}_{B}}\right]=$
$\displaystyle 0$
$\displaystyle=\left[{\hat{B}}^{\dagger}_{\vec{\psi}_{A}},{\hat{B}}^{\dagger}_{\vec{\psi}_{B}}\right].$
(2.32)

We remark that in the particular case where $\vec{\varphi}_{A}$ and
$\vec{\varphi}_{B}$ are normalized vectors orthogonal to each other, for
instance if $\vec{\varphi}_{A}$ and $\vec{\varphi}_{B}$ are orthonormal
eigenmodes of a cavity, the commutation relation (2.31) takes the form

$\displaystyle\left[{\hat{B}}_{\vec{\varphi}_{A}},{\hat{B}}^{\dagger}_{\vec{\varphi}_{B}}\right]$
$\displaystyle=$ $\displaystyle\delta_{A,B},$ (2.33)

which is the commutation relation that is traditionally used in the treatment
of systems of harmonic oscillators and in cavity quantum electrodynamics. The
commutation relations (2.31) are a slight generalization [22, p.105,
Eq.(3.192)], [18, Eq. (38)], in which the classical carrying modes of the
photons are not necessarily normal modes nor orthonormal. As we will see in
the applications to beam splitters this generalization is very useful for the
description of the propagation of pulse-shaped photon states.

Remark: In Eqs. (2.24), (2.25) (2.24) (2.25) the creation-annihilation
operators $\hat{B}^{\dagger}_{\vec{\psi}},\hat{B}_{\vec{\psi}}$ are labeled by
fields $\vec{\psi}$ that belong to the single-photon Hilbert space
$\mathcal{H}_{cl}$, i.e. which are square integrable and have a finite
classical energy. The definition can be extended formally to include the
continuum eigenfunctions $\vec{\varphi}_{\kappa}$ of the frequency operator
$\Omega$, that are not in the Hilbert space, (e.g. plane waves in the case
without a medium). When applied to the vacuum,
$\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}|\varnothing\rangle=|\vec{\varphi}_{\kappa}\rangle$,
can be interpreted formally as a single-photon state carried by a
monochromatic plane wave. These objects are useful e.g. for the mathematical
representation of the electromagnetic field observables (2.36)(2.37)(2.38),
since the plane waves are a “generalized basis” of the Hilbert space. They are
also useful to represent the pulse states as linear combinations of plane
waves,

$\displaystyle\kern-50.0pt|\psi\rangle=\hat{B}^{\dagger}_{\vec{\psi}}|\varnothing\rangle=\hat{B}^{\dagger}_{\int
d\kappa\leavevmode\nobreak\
z(\kappa)\vec{\varphi}_{\kappa}}|\varnothing\rangle=\int
d\kappa\leavevmode\nobreak\
z(\kappa)\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}|\varnothing\rangle=\int
d\kappa\leavevmode\nobreak\ z(\kappa)|\vec{\varphi}_{\kappa}\rangle,$ (2.34)

where we have used the linearity property (B.2) of the creation operators.
However, $|\vec{\varphi}_{\kappa}\rangle$ is not a single-photon state that
can be created in the laboratory, since it would contain an infinite energy,
and it would occupy uniformly the whole physical space $\mathbb{R}^{3}$. Thus,
purely monochromatic fields, although mathematically useful, cannot be
physical single-photon states. This is reflected in the theory by the fact
that they do not belong to the Hilbert space. In experiments what is often
created are almost-monochromatic photon states, that can be represented by a
monochromatic plane wave multiplied by a pulse-shaped envelope, which is what
we use below for the illustrations with beam splitters. They have a finite
energy and they belong to the Hilbert space of square integrable fields
$\mathcal{H}_{cl}$. We remark that this issue is different in free space than
in a cavity, where purely monochromatic single-photon states can in principle
be created, since the volume and thus the energy are finite. D) The basic
electromagnetic field observables are represented by the following operators
acting on the bosonic Fock space. We start with a complete orthonormal set of
continuum eigenfunctions $\vec{\varphi}_{\kappa}$ of the frequency operator,

$\displaystyle\Omega^{2}\vec{\varphi}_{\kappa}=\omega^{2}_{\kappa}\vec{\varphi}_{\kappa},\qquad\omega_{\kappa}>0.$
(2.35)

The electromagnetic field operators can be expressed as

$\displaystyle\sqrt{\epsilon}\vec{\hat{A}}=\vec{\hat{A}^{\prime}}(\vec{x})$
$\displaystyle=$ $\displaystyle\sqrt{\frac{\hbar}{2\epsilon_{0}}}\int
d\kappa\leavevmode\nobreak\
\omega_{\kappa}^{-1/2}\bigl{(}\vec{\varphi}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{B}_{\vec{\varphi}_{\kappa}}+\vec{\varphi}^{\kern
0.90417pt*}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{B}_{\vec{\varphi}_{\kappa}}^{\dagger}\bigr{)}$ (2.36)
$\displaystyle-\epsilon_{0}\sqrt{\epsilon}\vec{\hat{E}}=\vec{{\hat{\Pi^{\prime}}}}(\vec{x})$
$\displaystyle=$ $\displaystyle-i\sqrt{\frac{\hbar\epsilon_{0}}{2}}\int
d\kappa\leavevmode\nobreak\
\omega^{1/2}_{\kappa}\bigl{(}\vec{\varphi}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{B}_{\vec{\varphi}_{\kappa}}-\vec{\varphi}^{\kern
0.90417pt*}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{B}_{\vec{\varphi}_{\kappa}}^{\dagger}\bigr{)}.$ (2.37)

and the total energy operator (after normal ordering) as

$\displaystyle\hat{H}$ $\displaystyle=$ $\displaystyle\int
d\kappa\leavevmode\nobreak\
\hbar\omega_{\kappa}\hat{B}_{\vec{\varphi}_{\kappa}}^{\dagger}\hat{B}_{\vec{\varphi}_{\kappa}}.$
(2.38)

The notation $\int d\kappa$ is to be interpreted as an abbreviation of a
combination of integrals and discrete sums on the labels of the
eigenfunctions. For instance in the case of the free field
$\vec{\varphi}_{\kappa}(\vec{x})=(2\pi)^{-3/2}\vec{\varepsilon}_{s}(\vec{k})e^{i\vec{k}\cdot\vec{x}}$,
$\kappa=(\vec{k},s)$, where $\vec{k}\in\mathbb{R}^{3}$ is the wave vector,
$s=\pm$ is an index for the two polarizations, and $\int
d\kappa\leavevmode\nobreak\ \equiv\int_{\mathbb{R}^{3}}d^{3}k\sum_{s=\pm}$.
The explicit link with the usual formulas of the observables for the free
field is described in Appendix A.

##  3 Evolution of the photon states determined from the dynamics of the
classical modes

We are going to show that the time evolution of photon states in Fock space,
determined by the equation

$\displaystyle i\hbar\frac{\partial|\Phi\rangle}{\partial t}$ $\displaystyle=$
$\displaystyle\hat{H}|\Phi\rangle,\qquad|\Phi\rangle\in\mathcal{F}^{\mathfrak{B}}$
(3.1)

can be expressed in terms of the time evolution of the classical modes,
determined by the classical wave equation, i.e. by Maxwell’s equations.
(a) In particular, for a one-photon initial condition
$|\Phi(t=0)\rangle=\hat{B}_{\vec{\psi}(t=0)}^{\dagger}|\varnothing\rangle$ the
time evolution is

$\displaystyle|\Phi(t)\rangle$ $\displaystyle=$
$\displaystyle\hat{B}_{\vec{\psi}(t)}^{\dagger}|\varnothing\rangle,$ (3.2)

where $\vec{\psi}(t)$ is the solution of the classical equation (2.14) with
initial condition $\vec{\psi}(t=0)$. (b) For a two-photon initial condition
$|\Phi(t=0)\rangle=\hat{B}_{\vec{\psi}_{B}(t=0)}^{\dagger}\hat{B}_{\vec{\psi}_{A}(t=0)}^{\dagger}|\varnothing\rangle$
the time evolution is

$\displaystyle|\Phi(t)\rangle=\hat{B}_{\vec{\psi}_{B}(t)}^{\dagger}\hat{B}_{\vec{\psi}_{A}(t)}^{\dagger}|\varnothing\rangle,$
(3.3)

where $\vec{\psi}_{A}(t)$ and $\vec{\psi}_{B}(t)$ are the solutions of the
classical equation (2.14) for the corresponding initial conditions. (c) In the
general case of an N-photon initial condition
$|\Phi(t=0)\rangle=\prod_{j=1}^{N}\hat{B}_{\vec{\psi}_{j}(t=0)}^{\dagger}|\varnothing\rangle$
the time evolution is

$\displaystyle|\Phi(t)\rangle=\prod_{j=1}^{N}\hat{B}_{\vec{\psi}_{j}(t)}^{\dagger}|\varnothing\rangle,$
(3.4)

where $\vec{\psi}_{j}(t)$ are the solutions of the classical equation (2.14)
for the corresponding initial conditions. A proof of the time evolution
formulas (3.2),(3.3) and (3.4) is given in Appendix B. Remark: The family of
operators $\hat{B}_{\vec{\psi}(t)}^{\dagger}$ parametrized by time can be
thought of as a time-dependent operator which can be written as

$\displaystyle\hat{B}_{\vec{\psi}(t)}^{\dagger}=\hat{U}(t)\leavevmode\nobreak\
\hat{B}_{\vec{\psi}(t=0)}^{\dagger}\leavevmode\nobreak\ \hat{U}^{\dagger}(t)$
(3.5)

where $\hat{U}(t)=e^{-\frac{i}{\hbar}\hat{H}\leavevmode\nobreak\ t}$ is the
propagator in Fock space. This can be verified e.g. by applying both sides to
the vacuum state: Since

$\displaystyle\hat{U}^{\dagger}(t)|\varnothing\rangle$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\left(\frac{i}{\hbar}t\right)^{n}\left(\sum_{\kappa}\hbar\omega_{\kappa}\hat{B}_{\vec{\varphi}_{\kappa}}^{\dagger}\hat{B}_{\vec{\varphi}_{\kappa}}\right)^{n}|\varnothing\rangle=|\varnothing\rangle$
(3.6)

we can write

$\displaystyle\hat{U}(t)\leavevmode\nobreak\
\hat{B}_{\vec{\psi}(t=0)}^{\dagger}\hat{U}^{\dagger}(t)|\varnothing\rangle$
$\displaystyle=$ $\displaystyle\hat{U}(t)\leavevmode\nobreak\
\hat{B}_{\vec{\psi}(t=0)}^{\dagger}|\varnothing\rangle=\hat{U}(t)\leavevmode\nobreak\
|\vec{\psi}(t=0)\rangle$ (3.7) $\displaystyle=$
$\displaystyle|\vec{\psi}(t)\rangle=\hat{B}_{\vec{\psi}(t)}^{\dagger}|\varnothing\rangle.$
(3.8)

We remark however that it is not the time evolution in the Heisenberg picture,
which is given by a different expression [22, p.84]:

$\displaystyle\left(\hat{B}_{\vec{\psi}(t=0)}^{\dagger}\right)^{{\rm
Heisenberg}}(t)=\hat{U}^{\dagger}(t)\leavevmode\nobreak\
\hat{B}_{\vec{\psi}(t=0)}^{\dagger}\leavevmode\nobreak\
\hat{U}(t)=\hat{B}_{\vec{\psi}(-t)}^{\dagger}\neq\hat{B}_{\vec{\psi}(t)}^{\dagger}.$
(3.9)

##  4 One-photon pulse propagation through a beam splitter

We consider a single pulse arriving on a 50-50% beam splitter at $45^{o}$.

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{i}}$
$y$$x$$D_{Y}$$D_{X}$(a)

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{T}}$
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{R}}$
$y$$x$$D_{Y}$$D_{X}$(b)

Figure 1. Schematic illustration of the partial transmission and reflection of
a single-photon state through a 50%-50% beam splitter. (a) Incoming pulse, (b)
reflected and transmitted pulses. The dynamics of the single-photon state is
the same as the classical dynamics of pulses through beam splitters according
to Eq. (3.2). $D_{X}$ and $D_{Y}$ represent detectors.

###  4.1 Description of the initial classical mode and its time evolution

The pulse at the initial time $t_{i}$, before the pulse arrives at the beam
splitter, is a classical mode.
We define the following pulse shape function
$\vec{S}(r_{1},r_{2},r_{3},t;\leavevmode\nobreak\ k)$, depending on three
spatial arguments, $r_{1},r_{2},r_{3}$, a temporal variable $t$, and a wave
number $k$,

$\displaystyle\vec{S}(r_{1},r_{2},r_{3},k,t)$ $\displaystyle:=$
$\displaystyle\mathcal{N}\vec{e}_{z}\mathcal{E}(r_{1}-ct)e^{i(kr_{1}-\omega
t)}g(r_{2},r_{3})$ (4.1)

where $\vec{e}_{z}=(0,0,1)$ is the linear polarization vector, $\omega=ck>0$
is the carrier frequency, and $\mathcal{E}(r_{1})$ is the pulse envelope in
the direction of propagation. $g(r_{2},r_{3})$ is the transverse profile,
which we assume to be smaller than the beam splitter, and in the usual
circumstances does not change significantly with the propagation. For the
pulse envelope $\mathcal{E}(r_{1})$ one can take e.g. a $\sin^{2}$ with a
finite support, so that there is no ambiguity about when the process starts,
or a Gaussian multiplied by a characteristic function to give it a finite
support. $\mathcal{N}$ is a normalization constant, chosen such that
$\int_{\mathbb{R}^{3}}d^{3}r\leavevmode\nobreak\ \vec{S}^{*}\cdot\vec{S}=1$.
With these assumptions the propagation of the photon pulse can be reduced to a
one dimensional propagation. The initial mode is chosen as

$\displaystyle\vec{\psi}_{i}(\vec{x})$ $\displaystyle=$
$\displaystyle\vec{S}(x,y,z;\leavevmode\nobreak\ k,t=0)=:\vec{S}(x,t=0),$
(4.2)

where we introduce an abridged notation $\vec{S}(x,t)$, indicating only the
first spatial argument of its direction of propagation and the time argument.
After crossing the beam splitter, the classical mode evolves into a reflected
and a transmitted pulse (see Figure 1)

$\displaystyle\vec{\psi}_{i}\rightarrow\vec{\psi}_{R}+\vec{\psi}_{T}$ (4.3)
$\displaystyle\vec{\psi}_{R}(\vec{x},t)$ $\displaystyle=$
$\displaystyle\tilde{r}\leavevmode\nobreak\
\vec{S}(y-ct,t)=:\tilde{r}\leavevmode\nobreak\ \vec{Y}(y)$ (4.4)
$\displaystyle\vec{\psi}_{T}(\vec{x},t)$ $\displaystyle=$
$\displaystyle\tilde{t}\leavevmode\nobreak\
\vec{S}(x-ct,t)=:\tilde{t}\leavevmode\nobreak\ \vec{X}(x),$ (4.5)

where $\tilde{r},\tilde{t}$ are the reflection and transmission coefficients,
which in general satisfy the relations $|\tilde{r}|^{2}+|\tilde{t}|^{2}=1$ and
$\tilde{r}^{*}\tilde{t}+\tilde{r}\tilde{t}^{*}=0$. We have introduced the
abridged notations $\vec{Y}(y),\vec{X}(x)$ to improve the readability of the
construction below. ($\vec{X}$: propagation along the $x$-axis; $\vec{Y}$:
propagation along the $y$-axis).

###  4.2 Quantum dynamics of the one-photon state

We assume that the system is prepared with one photon on the initial mode,
described in the Fock space by

$\displaystyle|\Phi_{i}\rangle$ $\displaystyle=$
$\displaystyle\hat{B}^{\dagger}_{\vec{\psi}_{i}}|\varnothing\rangle=|\vec{\psi}_{i}\rangle.$
(4.6)

The mode function must be normalized:

$\displaystyle\int d^{3}x|\vec{\psi}_{i}(\vec{x})|^{2}=1,$ (4.7)

which entails

$\displaystyle\left[\hat{B}_{\vec{\psi}_{i}}\leavevmode\nobreak\
,\leavevmode\nobreak\ \hat{B}^{\dagger}_{\vec{\psi}_{i}}\right]$
$\displaystyle=$ $\displaystyle 1\kern-4.0pt1.$ (4.8)

This implies that the initial state (4.6) is normalized,
$\langle\Phi_{i}|\Phi_{i}\rangle=1$. After crossing the beam splitter, this
one-photon initial state evolves to

$\displaystyle|\vec{\psi}_{i}\rangle\rightarrow$
$\displaystyle|\vec{\psi}_{R}+\vec{\psi}_{T}\rangle\equiv\hat{B}^{\dagger}_{\vec{\psi}_{R}+\vec{\psi}_{T}}|\varnothing\rangle$
(4.9) $\displaystyle=$ $\displaystyle|\tilde{r}\leavevmode\nobreak\
\vec{Y}(y)+\tilde{t}\leavevmode\nobreak\ \vec{X}(x)\rangle.$ (4.10)

The physical interpretation of the state at time $t$ after the crossing of the
beam splitter is as follows: $|\vec{\psi}_{R}+\vec{\psi}_{T}\rangle$ is a one-
photon state on the single classical mode $\vec{\psi}_{R}+\vec{\psi}_{T}$,
which has two spatially disjoint components, one propagating in the $x$
direction and the other one in the $y$ direction (see Figure 1).

###  4.3 Remarks on the literature on the beam splitter, involving a fourth
port representing “incoming fluctuations of the vacuum”

First we remark that the creation operator in (4.9) can be decomposed as the
sum of two terms

$\displaystyle\hat{B}^{\dagger}_{\vec{\psi}_{R}+\vec{\psi}_{T}}=\hat{B}^{\dagger}_{\vec{\psi}_{R}}+\hat{B}^{\dagger}_{\vec{\psi}_{T}}.$
(4.11)

This property is a direct consequence of the definition (2.24), due to the
linearity of the tensor product in each of its arguments. Since the time
evolution of the classical modes is unitary, it implies that the norm of the
states in Fock space is preserved, and also

$\displaystyle\left[\hat{B}_{\vec{\psi}_{R}+\vec{\psi}_{T}}\leavevmode\nobreak\
,\leavevmode\nobreak\
\hat{B}^{\dagger}_{\vec{\psi}_{R}+\vec{\psi}_{T}}\right]$ $\displaystyle=$
$\displaystyle 1\kern-4.0pt1.$ (4.12)

However, the separate terms in (4.11) satisfy the commutation relations

$\displaystyle\left[\hat{B}_{\vec{\psi}_{R}}\leavevmode\nobreak\
,\leavevmode\nobreak\ \hat{B}^{\dagger}_{\vec{\psi}_{R}}\right]$
$\displaystyle=$ $\displaystyle\langle\vec{\psi}_{R}\leavevmode\nobreak\
|\leavevmode\nobreak\ \vec{\psi}_{R}\rangle=|\tilde{r}|^{2}\neq 1\kern-4.0pt1$
$\displaystyle\left[\hat{B}_{\vec{\psi}_{T}}\leavevmode\nobreak\
,\leavevmode\nobreak\ \hat{B}^{\dagger}_{\vec{\psi}_{T}}\right]$
$\displaystyle=$ $\displaystyle\langle\vec{\psi}_{T}\leavevmode\nobreak\
|\leavevmode\nobreak\ \vec{\psi}_{T}\rangle=|\tilde{t}|^{2}\neq
1\kern-4.0pt1.$ (4.13)

Thus, the separate creation operators cannot be “bosonic creation operators”.
This is presented in the literature as a major problem that needs to be
corrected. In order to analyze this question we first make the link with the
notation that is used conventionally in most of the literature, like e.g. in
[23] p.511:

$\displaystyle\hat{B}_{\vec{\psi}_{i}}\rightarrow\hat{a}_{1},\qquad\hat{B}_{\vec{\psi}_{R}}\rightarrow\hat{a}_{2},\qquad\hat{B}_{\vec{\psi}_{T}}\rightarrow\hat{a}_{3}.$
(4.14)

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{i}}$
${\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}\vec{\psi}_{vac,i}}$
$y$$x$$D_{Y}$$D_{X}$(a)

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{T}}$
${\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}\vec{\psi}_{vac,R}}$
${\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}\vec{\psi}_{vac,T}}$
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{R}}$
$y$$x$$D_{Y}$$D_{X}$(b)

Figure 2. Schematic illustration of the description of beam splitters which
uses a fourth port representing the “incoming fluctuations of the vacuum”. (a)
incoming pulses representing the initial state (in blue) and the vacuum
fluctuations (dashed green curve) (b) reflection and transmission of the
initial pulse and the fluctuation pulse. Fluctuation pulses are drawn in
dashed green lines since they don’t have any meaning neither in the classical
nor in the quantum theory.

The proposed correction is to add a fourth port represented by an operator
$\hat{a}_{0}$ that should correspond to “incoming fluctuations of the vacuum”,
represented by green dashed lines in Figure 2 (see e.g. [23] p.511 and p.640,
[24] p.8, 229, [25] p.70, [26] p.96, 123, [27] p.102, [28] p.138, [29] p.432,
[22] p.249 and footnote 1). The addition of this port is supposed to solve the
problem, since instead of

$\displaystyle\hat{a}_{2}$ $\displaystyle=$
$\displaystyle\tilde{t}\leavevmode\nobreak\ \hat{a}_{1}$ (4.15)
$\displaystyle\hat{a}_{3}$ $\displaystyle=$
$\displaystyle\tilde{r}\leavevmode\nobreak\ \hat{a}_{1},$ (4.16)

its inclusion allows to construct modified operators

$\displaystyle\hat{a}^{\prime}_{2}$ $\displaystyle:=$
$\displaystyle\tilde{t}\leavevmode\nobreak\
\hat{a}_{1}+\tilde{r}\leavevmode\nobreak\ \hat{a}_{0}$ (4.17)
$\displaystyle\hat{a}^{\prime}_{3}$ $\displaystyle:=$
$\displaystyle\tilde{r}\leavevmode\nobreak\
\hat{a}_{1}+\tilde{t}\leavevmode\nobreak\ \hat{a}_{0}$ (4.18)

that satisfy the bosonic commutation relations

$\displaystyle\left[\hat{a}^{\prime}_{2}\leavevmode\nobreak\
,\leavevmode\nobreak\ \hat{a}^{\prime{\dagger}}_{2}\right]$ $\displaystyle=$
$\displaystyle 1\kern-4.0pt1$ (4.19)
$\displaystyle\left[\hat{a}^{\prime}_{3}\leavevmode\nobreak\
,\leavevmode\nobreak\ \hat{a}^{\prime{\dagger}}_{3}\right]$ $\displaystyle=$
$\displaystyle 1\kern-4.0pt1$ (4.20)

since $|\tilde{r}|^{2}+|\tilde{t}|^{2}=1$.

Although formally this seems to solve the “problem”, yielding operators with
apparently bosonic commutation relations, we have two criticisms of this
construction:

(a) It is not a problem that the operators in (4.13) do not satisfy the
commutation relation for bosons. The photon is not an excitation of the
individual transmitted pulse nor of the reflected one. The photon is an
excitation of the global mode $\vec{\psi}_{R}+\vec{\psi}_{T}$ and thus the
bosonic nature of the photon is expressed by the global commutation relation
(4.12). So in fact the commutation relations (4.13) are correct, and they do
not pose any problem, since creation operators in Fock space are also well-
defined for unnormalized modes $\vec{\psi}$.

(b) The formal expression $\hat{a}^{\dagger}_{0}$ that would have to
correspond to $\hat{B}^{\dagger}_{vac}$ does not have any well defined sense
in the quantum field theory of light. There is no creation operator that could
create “fluctuations of the vacuum”. In Figure 2, the pulses drawn in dashed
lines, which are supposed to refer to vacuum fluctuations, don’t have any
meaning. Creation operators can only add photons carried by classical modes,
to the vacuum or to other states.

Our main conclusion from the analysis presented above is that vacuum
fluctuations don’t play any role in the transmission of a one-photon pulse
through a beam splitter.
Our explanation for the discrepancy with much of the literature is that the
notation $\hat{a}_{1},\hat{a}_{2},\hat{a}_{3},\hat{a}_{0}$ is misleading and
clearly insufficient for the description of the transmission of a one-photon
pulse through a beam splitter. One difficulty is that the operator (4.11),
$\hat{B}^{\dagger}_{\vec{\psi}_{R}+\vec{\psi}_{T}}$ has no natural
representation in the conventional notation: it could maybe be noted as
$\hat{a}^{\dagger}_{2+3}$, but it is not usually done. The notation also
suggests that all operators that are denoted with $\hat{a}_{j}$ should satisfy
the standard bosonic commutation relations, which is not well suited for the
beam splitter, as shown by Eqs. (4.13). Another shortcoming of this notation
is that it does not include the propagation in time: When one writes
$\hat{a}_{3}=\tilde{t}\hat{a}_{1}$, the operator $\hat{a}_{1}$ is not the
incoming one, but implicitly it must be interpreted as the time evolution of
the incoming one (which is not the evolution in the Heisenberg representation,
as we remarked in (3.5) (3.9)). All this difficulties disappear if one uses
the more precise notation $\hat{B}^{\dagger}_{\vec{\psi}}$, which indicates on
which classical mode $\vec{\psi}$ a photon is created. With this notation it
is immediately clear that it does not make sense to define a creation operator
like $\hat{a}^{\dagger}_{0}$ that would create “fluctuations of the vacuum”,
since there is no $\vec{\psi}$ that can do that.

The above remarks do not mean that there are no quantum fluctuations in the
one-photon pulse transmission through a beam splitter. The detector readings
have quantum fluctuations that can be traced back to the fact that the state
of the optical quantum field is not an eigenstate of the observable measured
by the detector. Our point is only that the statement that “vacuum
fluctuations enter through the unused port” is not well-defined within the
theory and that it is unnecessary, since, without any such addition, the
theory already gives a complete and consistent description of the process,
including the fluctuations that will be manifest at the detectors. The
fluctuations are a global feature of the quantum system. They are everywhere
and delocalized. They do not enter through any particular port. The photon
states and in particular the vacuum are global entities, they are not
localized in any specific port. If the electromagnetic field is in a one-
photon state, one cannot claim that it is in the vacuum state anywhere, in
particular not in an “unused port”. The fluctuations that may be measured in a
particular detector, will be the ones determined by the one-photon state, and
not by some partially localized vacuum.

As a final remark, in fact the model with four ports can describe a concrete
physical situation, but only when an actual classical mode is injected into
the fourth port, as we will see in the next section on the Hong-Ou-Mandel
effect (HOM) [30] [32].

##  5 The Hong-Ou-Mandel effect

We consider two pulses arriving on a 50-50% beam splitter at $45^{o}$.
$\vec{\psi}_{A,i}$ moves horizontally in the $x$ direction and
$\vec{\psi}_{B,i}$ moves vertically in the $y$.

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{A,i}}$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\vec{\psi}_{B,i}}$
$y$$x$$D_{Y}$$D_{X}$$D_{Y}$$D_{X}$(a)

${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{A,T}}$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\vec{\psi}_{B,R}}$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\vec{\psi}_{B,T}}$
${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\vec{\psi}_{A,R}}$
$y$$x$$D_{Y}$$D_{X}$$D_{Y}$$D_{X}$(b)

Figure 3. Schematic illustration of the Hong-Ou-Mandel effect. (a) Two
disjoint identical pulses impinge on the beam splitter at the same time (b)
each of them produces reflected and transmitted pulses. $D_{X}$ and $D_{Y}$
represent detectors, located at the same distance from the beam splitter.

###  5.1 Description of the initial classical modes and their time evolution

Each pulse at the initial time $t_{i}$ (before the pulses arrive on the beam
splitter) is a classical mode. We will use the pulse shape function
$\vec{S}(r_{1},r_{2},r_{3};k,t)$ defined in Eq. (4.1). For the pulse envelope
$\mathcal{E}(r_{1})$ we have taken a $\sin^{2}$ with a finite support, so that
there is no ambiguity about when the process starts. The initial modes $A$ and
$B$ are chosen as

$\displaystyle\vec{\psi}_{A,i}(\vec{x})$ $\displaystyle=$
$\displaystyle\vec{S}(x,y,z;\leavevmode\nobreak\
k_{A},t=0)=:\vec{S}_{A}(x,t=0)$ (5.1) $\displaystyle\vec{\psi}_{B,i}(\vec{x})$
$\displaystyle=$ $\displaystyle\vec{S}(y,x,z;\leavevmode\nobreak\
k_{B},t=0)=:\vec{S}_{B}(y,t=0),$ (5.2)

one is oriented in the $x$ direction and the other one in the $y$ direction.
In (5.1)(5.2) we introduce an abridged notation $\vec{S}_{A},\vec{S}_{B}$,
indicating only the first spatial argument and the time argument. After
crossing the beam splitter, the classical modes evolve into a reflected pulse
and a transmitted pulse:

$\displaystyle\vec{\psi}_{A,i}$ $\displaystyle\rightarrow$
$\displaystyle\vec{\psi}_{A,R}+\vec{\psi}_{A,T}$ (5.3)
$\displaystyle\vec{\psi}_{B,i}$ $\displaystyle\rightarrow$
$\displaystyle\vec{\psi}_{B,R}+\vec{\psi}_{B,T}$ (5.4)
$\displaystyle\vec{\psi}_{A,R}(\vec{x},t)$ $\displaystyle=$
$\displaystyle\tilde{r}\leavevmode\nobreak\
\vec{S}_{A}(y-ct,t)=:\tilde{r}\leavevmode\nobreak\ \vec{Y}_{A}(y)$ (5.5)
$\displaystyle\vec{\psi}_{A,T}(\vec{x},t)$ $\displaystyle=$
$\displaystyle\tilde{t}\leavevmode\nobreak\
\vec{S}_{A}(x-ct,t)=:\tilde{t}\leavevmode\nobreak\ \vec{X}_{A}(x)$ (5.6)
$\displaystyle\vec{\psi}_{B,R}(\vec{x},t)$ $\displaystyle=$
$\displaystyle\tilde{r}\leavevmode\nobreak\
\vec{S}_{B}(x-ct,t=:\tilde{r}\leavevmode\nobreak\ \vec{X}_{B}(x)$ (5.7)
$\displaystyle\vec{\psi}_{B,T}(\vec{x},t)$ $\displaystyle=$
$\displaystyle\tilde{t}\leavevmode\nobreak\
\vec{S}_{B}(y-ct,t)=:\tilde{t}\leavevmode\nobreak\ \vec{Y}_{B}(y),$ (5.8)

where $\tilde{r},\tilde{t}$ are the reflection and transmission coefficients,
which in general satisfy the relations $|\tilde{r}|^{2}+|\tilde{t}|^{2}=1$ and
$\tilde{r}^{*}\tilde{t}+\tilde{r}\tilde{t}^{*}=0$. For a 50-50% beam splitter
they satisfy furthermore

$\displaystyle\tilde{t}=i\tilde{r},\qquad{\rm
i.e}\quad\tilde{r}^{2}+\tilde{t}^{2}=0.$ (5.9)

We again introduce the abridged notations $\vec{Y}_{A}(y),\vec{X}_{A}(x)$
$\vec{Y}_{B}(y),\vec{X}_{B}(x)$ to improve the readability of the construction
below. ($\vec{X}$: propagation along the $x$-axis; $\vec{Y}$: propagation
along the $y$-axis).

###  5.2 Quantum dynamics of the two photon state

We assume that the system is prepared with one photon on each mode. The two-
photon state in the Fock space is

$\displaystyle|\Phi_{i}\rangle$ $\displaystyle=$
$\displaystyle\hat{B}^{\dagger}_{\vec{\psi}_{A,i}}\hat{B}^{\dagger}_{\vec{\psi}_{B,i}}|\varnothing\rangle$
(5.10) $\displaystyle=$
$\displaystyle\sqrt{2}|\vec{\psi}_{A,i}\otimes_{S}\vec{\psi}_{B,i}\rangle=|\vec{\psi}_{A,i}\otimes\vec{\psi}_{B,i}+\vec{\psi}_{B,i}\otimes\vec{\psi}_{A,i}\rangle/\sqrt{2},$

The mode functions must be normalized:

$\displaystyle\int d^{3}x|\vec{\psi}_{A,i}(\vec{x})|^{2}=1,\qquad\int
d^{3}x|\vec{\psi}_{B,i}(\vec{x})|^{2}=1,$ (5.11)

which entails

$\displaystyle\left[\hat{B}_{\vec{\psi}_{A,i}}\leavevmode\nobreak\
,\leavevmode\nobreak\ \hat{B}^{\dagger}_{\vec{\psi}_{A,i}}\right]$
$\displaystyle=$ $\displaystyle
1\kern-4.0pt1,\qquad\left[\hat{B}_{\vec{\psi}_{B,i}}\leavevmode\nobreak\
,\leavevmode\nobreak\
\hat{B}^{\dagger}_{\vec{\psi}_{B,i}}\right]=1\kern-4.0pt1,$
$\displaystyle\left[\hat{B}_{\vec{\psi}_{A,i}}\leavevmode\nobreak\
,\leavevmode\nobreak\ \hat{B}^{\dagger}_{\vec{\psi}_{B,i}}\right]$
$\displaystyle=$
$\displaystyle\langle\vec{\psi}_{A,i}|\vec{\psi}_{B,i}\rangle=\int
d^{3}x\leavevmode\nobreak\
\vec{\psi}_{A,i}^{*}(\vec{x})\cdot\vec{\psi}_{B,i}(\vec{x})=0,$ (5.12)

since the support on the two classical modes $\vec{\psi}_{A,i}(\vec{x})$ and
$\vec{\psi}_{B,i}(\vec{x})$ is disjoint. This implies that the initial state
(5.10) is normalized, $\langle\Phi_{i}|\Phi_{i}\rangle=1$. This two-photon
initial state evolves to

$\displaystyle\vec{\psi}_{A,i}\otimes_{S}\vec{\psi}_{B,i}\rightarrow\left(\vec{\psi}_{A,R}+\vec{\psi}_{A,T}\right)\otimes_{S}\left(\vec{\psi}_{B,R}+\vec{\psi}_{B,T}\right)$
(5.13) $\displaystyle=$
$\displaystyle\vec{\psi}_{A,R}\otimes_{S}\vec{\psi}_{B,R}+\vec{\psi}_{A,R}\otimes\vec{\psi}_{B,T}+\vec{\psi}_{A,T}\otimes_{S}\vec{\psi}_{B,R}+\vec{\psi}_{A,T}\otimes_{S}\vec{\psi}_{B,T}$
$\displaystyle=$
$\displaystyle\tilde{r}\vec{Y}_{A}\otimes_{S}\tilde{r}\vec{X}_{B}+\tilde{r}\vec{Y}_{A}\otimes_{S}\tilde{t}\vec{Y}_{B}+\tilde{t}\vec{X}_{A}\otimes_{S}\tilde{r}\vec{X}_{B}+\tilde{t}\vec{X}_{A}\otimes_{S}\tilde{t}\vec{Y}_{B}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\Bigl{[}\phantom{\leavevmode\nobreak\
}\tilde{r}^{2}\vec{Y}_{A}\otimes\vec{X}_{B}+\tilde{t}^{2}\vec{X}_{A}\otimes\vec{Y}_{B}+\tilde{r}\tilde{t}\left(\vec{Y}_{A}\otimes\vec{Y}_{B}+\vec{X}_{A}\otimes\vec{X}_{B}\right)$
$\displaystyle+\leavevmode\nobreak\
\tilde{r}^{2}\vec{X}_{B}\otimes\vec{Y}_{A}+\tilde{t}^{2}\vec{Y}_{B}\otimes\vec{X}_{A}+\tilde{r}\tilde{t}\left(\vec{Y}_{B}\otimes\vec{Y}_{A}+\vec{X}_{B}\otimes\vec{X}_{A}\right)\Bigr{]}.$
(5.15)

In the degenerate Hong-Ou-Mandel effect the shapes of the two incoming
classical modes are the same and they arrive at the same time at the beam
splitter and at the detectors, which means that
$\vec{Y}_{A}=\vec{Y}_{B}=:\vec{Y}$ and $\vec{X}_{A}=\vec{X}_{B}=:\vec{X}$.
Therefore the above expression can be rewritten as

$\displaystyle\Phi_{f}$ $\displaystyle=$
$\displaystyle\frac{\tilde{r}^{2}+\tilde{t}^{2}}{\sqrt{2}}\left(\vec{X}\otimes\vec{Y}+\vec{Y}\otimes\vec{X}\right)+\sqrt{2}\leavevmode\nobreak\
\tilde{r}\tilde{t}\left(\vec{Y}\otimes\vec{Y}+\vec{X}\otimes\vec{X}\right).$
(5.16)

Since for a 50-50% beam splitter $\tilde{r}^{2}+\tilde{t}^{2}=0$, the final
state for the degenerate Hong-Ou-Mandel effect can be written as

$\displaystyle\Phi_{i}:=\sqrt{2}\left(\vec{\psi}_{A,i}\otimes_{S}\vec{\psi}_{B,i}\right)\rightarrow\Phi_{f}=\sqrt{2}\leavevmode\nobreak\
\tilde{r}\tilde{t}\left(\vec{Y}\otimes\vec{Y}+\vec{X}\otimes\vec{X}\right).$
(5.17)

From the expression (5.17) one can conclude that there will be no simultaneous
detection of one photon in each detector, since the state does not contain
terms of the form $\left(\vec{X}\otimes\vec{Y}+\vec{Y}\otimes\vec{X}\right)$.
In order to make this statement more precise we have to construct a model for
the detectors, i.e. we have to write the observables that correspond to single
and double detections. A simple model for these observables can be constructed
using the classical modes $\vec{X}$ and $\vec{Y}$: The observable
corresponding to the detection of a single photon in the considered mode shape
in the detector $D_{X}$ is

$\displaystyle\hat{O}_{1X}:=|\vec{X}\rangle\langle\vec{X}|\otimes
1\kern-4.0pt1+1\kern-4.0pt1\otimes|\vec{X}\rangle\langle\vec{X}|,$ (5.18)

and correspondingly in the detector $D_{Y}$

$\displaystyle\hat{O}_{1Y}:=|\vec{Y}\rangle\langle\vec{Y}|\otimes
1\kern-4.0pt1+1\kern-4.0pt1\otimes|\vec{Y}\rangle\langle\vec{Y}|.$ (5.19)

The observable corresponding to the detection of two photons in $D_{X}$ is

$\displaystyle\hat{O}_{2XX}:=|\vec{X}\rangle\langle\vec{X}|\otimes|\vec{X}\rangle\langle\vec{X}|,$
(5.20)

and correspondingly in the detector $D_{Y}$

$\displaystyle\hat{O}_{2\vec{Y}\vec{Y}}:=|\vec{Y}\rangle\langle\vec{Y}|\otimes|\vec{Y}\rangle\langle\vec{Y}|.$
(5.21)

The observable corresponding to the detection of one photon in $D_{X}$ and
simultaneously one photon in $D_{Y}$ is

$\displaystyle\hat{O}_{2XY}:=|\vec{Y}\rangle\langle\vec{Y}|\otimes|\vec{X}\rangle\langle\vec{X}|+|\vec{X}\rangle\langle\vec{X}|\otimes|\vec{Y}\rangle\langle\vec{Y}|.$
(5.22)

Since $\hat{O}_{2XY}$ is a projector, the probability to observe
simultaneously one photon in each detector in the final state
$|\Phi_{f}\rangle$ is given by

$\displaystyle\hbox{P}(D_{X}\leavevmode\nobreak\ {\rm and\leavevmode\nobreak\
}D_{Y})=\langle\Phi_{f}|\hat{O}_{2XY}|\Phi_{f}\rangle.$ (5.23)

Inserting the expression we obtained for $|\Phi_{f}\rangle$ we get

$\displaystyle\hbox{P}(D_{X}\leavevmode\nobreak\ {\rm and\leavevmode\nobreak\
}D_{Y})=\langle\Phi_{f}|\hat{O}_{2XY}|\Phi_{f}\rangle$ (5.24)
$\displaystyle=2|\tilde{r}|^{2}|\tilde{t}|^{2}\Big{\langle}\left(\vec{Y}\otimes\vec{Y}+\vec{X}\otimes\vec{X}\right)\Big{|}\leavevmode\nobreak\
\left[|\vec{Y}\rangle\langle\vec{Y}|\otimes|\vec{X}\rangle\langle\vec{X}|\right]\leavevmode\nobreak\
\Big{|}\left(\vec{Y}\otimes\vec{Y}+\vec{X}\otimes\vec{X}\right)\Big{\rangle}$
(5.25)

and since

$\displaystyle\left[|\vec{Y}\rangle\langle\vec{Y}|\otimes|\vec{X}\rangle\langle\vec{X}|\right]\leavevmode\nobreak\
|\vec{Y}\otimes\vec{Y}\rangle$ $\displaystyle=$ $\displaystyle 0$ (5.26)
$\displaystyle\left[|\vec{Y}\rangle\langle\vec{Y}|\otimes|\vec{X}\rangle\langle\vec{X}|\right]\leavevmode\nobreak\
|\vec{X}\otimes\vec{X}\rangle$ $\displaystyle=$ $\displaystyle 0$ (5.27)

we conclude that

$\displaystyle\hbox{P}(D_{X}\leavevmode\nobreak\ {\rm and\leavevmode\nobreak\
}D_{Y})=0,$ (5.28)

i.e. the probability for simultaneous detection of one photon in each detector
is zero, which is the main characteristic of the Hong-Ou-Mandel effect.

###  5.3 Photons are not like classical waves nor like classical particles

We emphasize that the last equality of (5.15), and thus (5.16) and (5.17), are
only true because the photons are bosons (i.e. indistinguishable quanta or
“particles”). The effect of the bosonic symmetrization is essential for the
HOM effect.

In particular, as it is well known, that the HOM effect is an exclusively
quantum effect, that does not happen with classical waves. Indeed the
classical waves arrive at the two detectors simultaneously, and thus it is
excluded that only one of the detectors is activated. The events that for
photons have probability zero, for classical waves would have probability one.

The behavior of photons in the HOM effect is also completely different from
that of classical particles. Indeed, for classical particles having $1/2$
probability of transmission and $1/2$ of reflection, the probability of
detecting two particles in the detector $D_{X}$ would be $1/4$, and the same
$1/4$ for detector $D_{Y}$. The probability to detect one particle in each
detector would be $1/2$. Indeed, denoting by $D_{X},D_{Y}$ the horizontal and
the vertical detectors, and labeling the particles moving initially
horizontally and vertically by $X_{i}$ and $Y_{i}$ respectively, the classical
probabilities are

$\displaystyle\hbox{P}({\rm 2\leavevmode\nobreak\
particles\leavevmode\nobreak\ in\leavevmode\nobreak\ }D_{X})$ $\displaystyle=$
$\displaystyle\hbox{P}({\rm 1\leavevmode\nobreak\ particle\leavevmode\nobreak\
in\leavevmode\nobreak\ }D_{X})\times\hbox{P}({\rm 1\leavevmode\nobreak\
particle\leavevmode\nobreak\ in\leavevmode\nobreak\
}D_{X})=\frac{1}{2}\frac{1}{2}=\frac{1}{4}$ (5.29) $\displaystyle\hbox{P}({\rm
2\leavevmode\nobreak\ particles\leavevmode\nobreak\ in\leavevmode\nobreak\
}D_{Y})$ $\displaystyle=$ $\displaystyle\hbox{P}({\rm 1\leavevmode\nobreak\
particle\leavevmode\nobreak\ in\leavevmode\nobreak\ }D_{Y})\times\hbox{P}({\rm
1\leavevmode\nobreak\ particle\leavevmode\nobreak\ in\leavevmode\nobreak\
}D_{Y})=\frac{1}{2}\frac{1}{2}=\frac{1}{4}\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ $ (5.30)

and

$\displaystyle\hbox{P}({\rm 1\leavevmode\nobreak\ particle\leavevmode\nobreak\
in\leavevmode\nobreak\ }D_{X}\leavevmode\nobreak\ $
$\displaystyle\kern-15.0pt{\rm and\leavevmode\nobreak\ 1\leavevmode\nobreak\
particle\leavevmode\nobreak\ in\leavevmode\nobreak\ }D_{Y})$ (5.31)
$\displaystyle=$ $\displaystyle\hbox{P}({\rm particle\leavevmode\nobreak\
Y}_{i}{\rm\leavevmode\nobreak\ in\leavevmode\nobreak\
}D_{Y}{\rm\leavevmode\nobreak\ and\leavevmode\nobreak\
particle\leavevmode\nobreak\ X}_{i}{\rm\leavevmode\nobreak\
in\leavevmode\nobreak\ }D_{X})$ $\displaystyle+\hbox{P}({\rm
particle\leavevmode\nobreak\ Y}_{i}{\rm\leavevmode\nobreak\
in\leavevmode\nobreak\ }D_{X}{\rm\leavevmode\nobreak\ and\leavevmode\nobreak\
particle\leavevmode\nobreak\ X}_{i}{\rm\leavevmode\nobreak\
in\leavevmode\nobreak\ }D_{Y})\phantom{xxxxx}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}=\frac{1}{2},$
(5.32)

i.e. the event that has zero probability for photons has probability $1/2$ for
classical particles.

This example gives an illustration for the question on whether photons are
particles or waves, or both. The answer is that they are neither particles nor
waves in the classical sense. It is not that photons “behave sometimes like
particles and sometimes like waves”, as it was often stated in the early
stages of the development of quantum mechanics. They are purely quantum
entities, that have properties that do not exist in classical objects.

##  6 Remarks on the concept of photons

In his recent book [33], Bruce W. Shore gave a detailed account of the
historical evolution of the concept of photons, as well as the different
interpretations that have been formulated in different contexts. In the
present article we apply tools of quantum field theory to some questions of
quantum optics involving passive dielectrics, in order to highlight some
aspects of the concept of photons. From the theoretical point of view, in the
framework of quantum field theory, photons are the quanta of the
electromagnetic field, that are created by the bosonic operators
$\hat{B}^{\dagger}_{\vec{\psi}}$ carried by a classical mode $\vec{\psi}$ (or
electromagnetic configuration), applied on the vacuum state in Fock space.
They are bosons, which means that they are intrinsically indistinguishable and
their states are symmetric with respect to permutations. In the context of the
Hong-Ou-Mandel effect one often uses the expression “two indistinguishable
photons” with a slightly different meaning, which is to be interpreted as “a
two-photon state, carried by two classical modes (or electromagnetic
configurations) that are disjoint and have the same shape (carrier frequency,
pulse shape and polarization) and arrive simultaneously on the beam splitter”.

We can further remark that “two photons” does not mean “one photon and another
photon”. Two photons means a two-photon state in the bosonic Fock space.
Because of the bosonic nature, expressed by the symmetrization of the states,
two photons are not the juxtaposition of one photon and another photon. A two-
photon state is a global entity that has properties that don’t follow from the
properties of single-photon states. The HOM effect is an illustrative example
of this fact. If two photons were the juxtaposition of one photon and another
photon the HOM effect would not exist.

We remark furthermore that in the case of a single-photon pulse going through
a beam splitter, discussed in Section 4, once the pulse has gone through the
beam splitter, the single-photon state is both in the transmitted and in the
reflected pulses. Neither of them can be called individually a photon.

##  Appendix A Relation between the position Fock space and the reciprocal
Fock space

The description of photon states that we have used is defined on a position
space representation of the Hilbert space $\mathcal{H}_{cl}$. In the
literature, in particular for the quantized free fields, a reciprocal space
representation is often used [35][36]. In this Appendix we describe the
relation between the two representations. We first remark that in quantum
mechanics one has to construct together the Hilbert space of states a
representation and the operators describing the observables in that
representation. The mathematical form of the state space alone is not enough
to provide the physical interpretation of the theory. As formulated by Dirac
in the theory of transformations [34], if two Hilbert spaces are related by a
unitary transformation and the observables are transformed accordingly, then
all the physical predictions are identical. The physically observable
quantities appear through expectation values, involving both the states and
the observables. The reciprocal space of the Hilbert and Fock spaces can be
defined as follows. We start with the position space representation
$\mathcal{H}_{cl}$, and an orthonormal basis $\\{\vec{\varphi}_{\kappa}\\}$ of
eigenvectors of the frequency operator $\Omega$. Then any
$\vec{\psi}(\vec{x})\in\mathcal{H}_{cl}$ can be developed in this basis as

$\displaystyle\vec{\psi}(\vec{x})$ $\displaystyle=\int
d\kappa\leavevmode\nobreak\ z(\kappa)\leavevmode\nobreak\
\vec{\varphi}_{\kappa}(\vec{x}),\qquad{\rm with\leavevmode\nobreak\
}z(\kappa)=\int_{\mathbb{R}^{3}}d^{3}x\leavevmode\nobreak\
\vec{\varphi}^{\kern
1.0pt*}_{\kappa}(\vec{x})\cdot\vec{\psi}(\vec{x})\in\mathbb{C}.$ (A.1)

The coefficients $z(\kappa)$ are in the Hilbert space

$\displaystyle\tilde{\mathcal{H}}_{cl}$
$\displaystyle:=\\{z(\kappa)\leavevmode\nobreak\ |\leavevmode\nobreak\ \langle
z|z\rangle_{\tilde{\mathcal{H}}_{cl}}<\infty\\},$ (A.2)

with the scalar product

$\displaystyle\langle z_{A}|z_{B}\rangle_{\tilde{\mathcal{H}}_{cl}}$
$\displaystyle:=\int d\kappa\leavevmode\nobreak\
z_{A}^{*}(\kappa)\leavevmode\nobreak\ z_{B}(\kappa).$ (A.3)

The map

$\displaystyle\mathcal{I}:\mathcal{H}_{cl}$ $\displaystyle\to$
$\displaystyle\tilde{\mathcal{H}}_{cl}$ (A.4)
$\displaystyle\vec{\psi}(\vec{x})$ $\displaystyle\mapsto$
$\displaystyle\left[\mathcal{I}\vec{\psi}\right](\kappa)=z(\kappa)=\int_{\mathbb{R}^{3}}d^{3}x\leavevmode\nobreak\
\vec{\varphi}^{\kern 1.0pt*}_{\kappa}(\vec{x})\cdot\vec{\psi}(\vec{x})$ (A.5)

is a unitary isomorphism, since it is bijective and it preserves the scalar
products,

$\displaystyle\langle\vec{\psi}_{A}|\vec{\psi}_{B}\rangle=\langle\mathcal{I}\vec{\psi}_{A}|\mathcal{I}\vec{\psi}_{B}\rangle_{\tilde{\mathcal{H}}_{cl}}.$
(A.6)

This is a direct consequence of the fact that the basis
$\\{\vec{\varphi}_{\kappa}\\}$ is orthonormal. The isomorphism of the Hilbert
spaces $\mathcal{H}_{cl}$ and $\tilde{\mathcal{H}}_{cl}$ extends to the
corresponding Fock spaces ${\mathcal{F}}^{\mathfrak{B}}(\mathcal{H}_{cl})$ and
$\tilde{{\mathcal{F}}}^{\mathfrak{B}}:={\mathcal{F}}^{\mathfrak{B}}(\tilde{\mathcal{H}}_{cl})$,
where

$\displaystyle\tilde{{\mathcal{F}}}^{\mathfrak{B}}$ $\displaystyle:=$
$\displaystyle\oplus_{m=0}^{\infty}{\tilde{\mathcal{F}}}^{\mathfrak{B}}_{m},$
(A.7) $\displaystyle\tilde{\mathcal{F}}^{\mathfrak{B}}_{0}$ $\displaystyle:=$
$\displaystyle\mathbb{C}$ (A.8)
$\displaystyle\tilde{\mathcal{F}}^{\mathfrak{B}}_{1}$ $\displaystyle:=$
$\displaystyle\tilde{\mathcal{H}}_{cl}$ (A.9)
$\displaystyle\tilde{\mathcal{F}}^{\mathfrak{B}}_{2}$ $\displaystyle:=$
$\displaystyle\tilde{\mathcal{H}}_{cl}\otimes_{S}\tilde{\mathcal{H}}_{cl}$
$\displaystyle\vdots$

The operators acting on these two Hilbert spaces are related by

$\displaystyle\hat{\tilde{O}}$
$\displaystyle=\mathcal{I}\hat{O}\mathcal{I}^{-1}.$ (A.11)

In particular, the creation-annihilation operators are related by

$\displaystyle\hat{\tilde{B}}^{\dagger}_{z}=\hat{\tilde{B}}^{\dagger}_{\mathcal{I}\vec{\psi}}$
$\displaystyle=\mathcal{I}\hat{B}^{\dagger}_{\vec{\psi}}\mathcal{I}^{-1}.$
(A.12)

For instance,

$\displaystyle\hat{\tilde{B}}^{\dagger}_{z}|\tilde{\varnothing}\rangle$
$\displaystyle=\mathcal{I}\hat{B}^{\dagger}_{\vec{\psi}}\mathcal{I}^{-1}|\tilde{\varnothing}\rangle=\mathcal{I}\hat{B}^{\dagger}_{\vec{\psi}}|\varnothing\rangle=\mathcal{I}|\vec{\psi}\rangle=|z\rangle=|\mathcal{I}\vec{\psi}\rangle=\hat{\tilde{B}}^{\dagger}_{\mathcal{I}\vec{\psi}}|\tilde{\varnothing}\rangle.$
(A.13)

With this construction we can make the link between the position Fock space
representation of the electromagnetic observables and their reciprocal Fock
space representation. For instance for the vector potential we have in the
position space representation

$\displaystyle\sqrt{\epsilon}\vec{\hat{A}}=\vec{\hat{A}^{\prime}}(\vec{x})$
$\displaystyle=$ $\displaystyle\sqrt{\frac{\hbar}{2\epsilon_{0}}}\int
d\kappa\leavevmode\nobreak\
\omega_{\kappa}^{-1/2}\left(\vec{\varphi}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{B}_{\vec{\varphi}_{\kappa}}+\vec{\varphi}^{\kern
0.90417pt*}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{B}_{\vec{\varphi}_{\kappa}}^{\dagger}\right)$ (A.14)

and in the reciprocal space representation

$\displaystyle\sqrt{\epsilon}\vec{\hat{\tilde{A}}}=\vec{\hat{\tilde{A}^{\prime}}}(\vec{x})$
$\displaystyle=$ $\displaystyle\sqrt{\frac{\hbar}{2\epsilon_{0}}}\int
d\kappa\leavevmode\nobreak\
\omega_{\kappa}^{-1/2}\left(\vec{\varphi}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{{\tilde{B}}}_{\mathcal{I}\vec{\varphi}_{\kappa}}+\vec{\varphi}^{\kern
0.90417pt*}_{\kappa}(\vec{x})\leavevmode\nobreak\
\hat{\tilde{B}}_{\mathcal{I}\vec{\varphi}_{\kappa}}^{\dagger}\right).$ (A.15)

We remark that
$\left(\mathcal{I}\vec{\varphi}_{\kappa}\right)(\kappa^{\prime})=\delta(\kappa^{\prime}-\kappa)$,
where $\delta(\kappa^{\prime}-\kappa)$ is a notation for combination of Dirac
and Kronecker delta functions. This follows from the definition (A.5), which
applied to $\vec{\psi}=\vec{\varphi}_{\kappa^{\prime}}$ yields

$\displaystyle\left(\mathcal{I}\vec{\varphi}_{\kappa^{\prime}}\right)(\kappa)=\int_{\mathbb{R}^{3}}d^{3}x\leavevmode\nobreak\
\vec{\varphi}^{\kern
1.0pt*}_{\kappa}(\vec{x})\cdot\vec{\varphi}_{\kappa^{\prime}}(\vec{x})=\delta(\kappa-\kappa^{\prime}).$
(A.16)

In order to make the link with the formulas used most often in the literature,
we consider the free field case, i.e. $\epsilon(\vec{x})=1$. In this case

$\displaystyle\vec{\varphi}_{\kappa}(\vec{x})=\frac{1}{(2\pi)^{3/2}}\vec{\varepsilon}_{s}(\vec{k})e^{i\vec{k}\cdot\vec{x}},$
(A.17)

where $\vec{k}\in\mathbb{R}^{3}$ and $s=\pm$ is an index for the two
polarizations, i.e. $\kappa=(\vec{k},s)$. In the reciprocal space

$\displaystyle\left(\mathcal{I}\vec{\varphi}_{\kappa^{\prime}}\right)(\kappa)=\delta(\vec{k}-\vec{k}^{\prime})\delta_{s,s^{\prime}}$
(A.18)

and the corresponding creation-annihilation operators are denoted

$\displaystyle\hat{a}^{\dagger}_{\vec{k}\kern
1.0pt^{\prime},s^{\prime}}:=\hat{\tilde{B}}^{\dagger}_{\delta(\vec{k}-\vec{k}^{\prime})\delta_{s,s^{\prime}}}.$
(A.19)

With this notation we can write

$\phantom{xxxx}\vec{\hat{\tilde{A}}}=\frac{1}{(2\pi)^{3/2}}\sqrt{\frac{\hbar}{2\epsilon_{0}}}\int
d^{3}k\sum_{s=\pm}\omega_{\vec{k}}^{-1/2}\left(\vec{\varepsilon}_{s}(\vec{k})e^{i\vec{k}\cdot\vec{x}}\leavevmode\nobreak\
\hat{a}_{\vec{k},s}+\vec{\varepsilon}_{s}^{\kern
2.0pt*}(\vec{k})e^{-i\vec{k}\cdot\vec{x}}\leavevmode\nobreak\
\hat{a}^{\dagger}_{\vec{k},s}\right),$ (A.20)

with $\omega_{\vec{k}}=c|\vec{k}|$. This expression coincides with the usual
formulas of the literature, like e.g. in [23, p.483, (10.4-38)][35], in the
infinite volume limit. We remark in conclusion that the Fock space that is
mostly used in the free field case is the reciprocal one. For the space-time
representation of the photons, in particular in the presence of a medium, we
used the position space representation of the Fock space, which allows a
spatial representation of the photons states, as in Eq. (2.26).

##  Appendix B Proof of the time evolution formulas of N-photon states

We are going to prove the formulas (3.2),(3.3),(3.4) of Section 3, i.e. that
for an N-photon initial condition
$|\Phi(t=0)\rangle=\prod_{j=1}^{N}\hat{B}_{\vec{\psi}_{j}(t=0)}^{\dagger}|\varnothing\rangle$
the time evolution is

$\displaystyle|\Phi(t)\rangle=\prod_{j=1}^{N}\hat{B}_{\vec{\psi}_{j}(t)}^{\dagger}|\varnothing\rangle,$
(B.1)

where $\vec{\psi}_{j}(t)$ are the solutions of the classical equation (2.14)
for the corresponding initial conditions. Proof: Although the proof of the
general N-photon case given in (c) below implies of course the results for the
one- and two-photon cases, since it is notationally harder to read we will
first give the proofs for the simplest particular cases. We will use the
relation

$\displaystyle\hat{B}^{\dagger}_{\alpha\vec{\psi}_{A}+\beta\vec{\psi}_{B}}=\alpha\hat{B}^{\dagger}_{\vec{\psi}_{A}}+\beta\hat{B}^{\dagger}_{\vec{\psi}_{B}}\quad{\rm
for\leavevmode\nobreak\ }\alpha,\beta\in\mathbb{C}.$ (B.2)

We will write $\sum_{\kappa}$ instead of $\int d\kappa$ to simplify the
notation.

(a) We first prove the statement for one-photon states, i.e. that the state
defined by (3.2) satisfies (3.1). We start by expressing the time evolution of
the classical modes in terms of the eigenfunctions $\vec{\varphi}_{\kappa}$ of
$\Omega$:

$\displaystyle\vec{\psi}(t)$ $\displaystyle=$
$\displaystyle\sum_{\kappa}e^{-i\omega_{\kappa}t}\vec{\varphi}_{\kappa}\alpha_{\kappa},\qquad\alpha_{\kappa}=\langle\vec{\varphi}_{\kappa}|\vec{\psi}(t=0)\rangle,$
(B.3)

and

$\displaystyle|\Phi(t)\rangle=\hat{B}_{\vec{\psi}(t)}^{\dagger}|\varnothing\rangle=\hat{B}^{\dagger}_{\sum_{\kappa}e^{-i\omega_{\kappa}t}\vec{\varphi}_{\kappa}\alpha_{\kappa}}|\varnothing\rangle=\sum_{\kappa}e^{-i\omega_{\kappa}t}\alpha_{\kappa}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}|\varnothing\rangle.$
(B.4)

With this representation we can write

$\displaystyle i\hbar\frac{\partial}{\partial t}|\Phi(t)\rangle$
$\displaystyle=$
$\displaystyle\sum_{\kappa}\hbar\omega_{\kappa}\leavevmode\nobreak\
e^{-i\omega_{\kappa}t}\alpha_{\kappa}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}|\varnothing\rangle.$
(B.5)

Using the representation of the Hamiltonian as
$\hat{H}=\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}$
we can write

$\displaystyle\hat{H}|\Phi(t)\rangle$ $\displaystyle=$
$\displaystyle\sum_{\kappa^{\prime},\kappa}\hbar\omega_{\kappa^{\prime}}e^{-i\omega_{\kappa}t}\alpha_{\kappa}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}|\varnothing\rangle$
(B.6) $\displaystyle=$
$\displaystyle\sum_{\kappa^{\prime},\kappa}\hbar\omega_{\kappa^{\prime}}e^{-i\omega_{\kappa}t}\alpha_{\kappa}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\left(\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}+\delta_{\kappa,\kappa^{\prime}}\right)|\varnothing\rangle$
(B.7) $\displaystyle=$
$\displaystyle\sum_{\kappa}\hbar\omega_{\kappa}e^{-i\omega_{\kappa}t}\alpha_{\kappa}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}|\varnothing\rangle=i\hbar\frac{\partial}{\partial
t}|\Phi(t)\rangle,$ (B.8)

where we have used the commutation relations
$\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}=\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}+\delta_{\kappa,\kappa^{\prime}}$,
the fact that $\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}|\varnothing\rangle=0$
and (B.5), which completes the proof.

(b) Next we prove the relation for two-photon states, i.e. we prove that the
state defined by (3.3) satisfies (3.1). First we expand in terms of the
eigenfunction of $\Omega$:

$\displaystyle\vec{\psi}_{j}(t)$ $\displaystyle=$
$\displaystyle\sum_{\kappa_{j}}e^{-i\omega_{\kappa_{j}}t}\vec{\varphi}_{\kappa_{j}}\alpha_{\kappa_{j}},\qquad\alpha_{\kappa_{j}}=\langle\vec{\varphi}_{\kappa_{j}}|\vec{\psi}(t=0)\rangle,\qquad
j=A,B,$ (B.9)

which allows us to write

$\displaystyle|\Phi(t)\rangle$ $\displaystyle=$
$\displaystyle\hat{B}_{\vec{\psi}_{B}(t)}^{\dagger}\hat{B}_{\vec{\psi}_{A}(t)}^{\dagger}|\varnothing\rangle=\prod_{j=A,B}\left(\sum_{\kappa_{j}}e^{-i\omega_{\kappa_{j}}t}\alpha_{\kappa_{j}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\right)|\varnothing\rangle$
(B.10) $\displaystyle=$
$\displaystyle\left(\sum_{\kappa_{A}}e^{-i\omega_{\kappa_{A}}t}\alpha_{\kappa_{A}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\right)\left(\sum_{\kappa_{B}}e^{-i\omega_{\kappa_{B}}t}\alpha_{\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}\right)|\varnothing\rangle$
(B.11) $\displaystyle=$
$\displaystyle\sum_{\kappa_{A},\kappa_{B}}e^{-i(\omega_{\kappa_{A}}+\omega_{\kappa_{B}})t}\leavevmode\nobreak\
\alpha_{\kappa_{A}}\alpha_{\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle$
(B.12)

and

$\displaystyle i\hbar\frac{\partial}{\partial t}|\Phi(t)\rangle$
$\displaystyle=$
$\displaystyle\sum_{\kappa_{A},\kappa_{B}}\hbar(\omega_{\kappa_{A}}+\omega_{\kappa_{B}})\leavevmode\nobreak\
e^{-i(\omega_{\kappa_{A}}+\omega_{\kappa_{B}})t}\leavevmode\nobreak\
\alpha_{\kappa_{A}}\alpha_{\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle.$
(B.13)

Applying the Hamiltonian to (B.12) yields

$\displaystyle\hat{H}|\Phi(t)\rangle$ $\displaystyle=$
$\displaystyle\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}^{\dagger}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\leavevmode\nobreak\
\sum_{\kappa_{A},\kappa_{B}}e^{-i(\omega_{\kappa_{A}}+\omega_{\kappa_{B}})t}\leavevmode\nobreak\
\alpha_{\kappa_{A}}\alpha_{\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle$
(B.14) $\displaystyle=$
$\displaystyle\sum_{\kappa_{A},\kappa_{B}}\leavevmode\nobreak\
e^{-i(\omega_{\kappa_{A}}+\omega_{\kappa_{B}})t}\leavevmode\nobreak\
\alpha_{\kappa_{A}}\alpha_{\kappa_{B}}\leavevmode\nobreak\
\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle.$
(B.15)

Using the commutation relations
$\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}=\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}+\delta_{\kappa_{j},\kappa^{\prime}}$
and the fact that
$\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}|\varnothing\rangle=0$ we can write

$\displaystyle\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle$
$\displaystyle=$
$\displaystyle\delta_{\kappa^{\prime},\kappa_{A}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle+\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle$
(B.16) $\displaystyle=$
$\displaystyle\delta_{\kappa^{\prime},\kappa_{A}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle+\delta_{\kappa^{\prime},\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}|\varnothing\rangle+\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}|\varnothing\rangle$
(B.17) $\displaystyle=$
$\displaystyle\delta_{\kappa^{\prime},\kappa_{A}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle+\delta_{\kappa^{\prime},\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}|\varnothing\rangle,$
(B.18)

and thus

$\displaystyle\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle$
$\displaystyle=$
$\displaystyle\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\left(\delta_{\kappa^{\prime},\kappa_{A}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle+\delta_{\kappa^{\prime},\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}|\varnothing\rangle\right)\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ $ (B.19)
$\displaystyle=$
$\displaystyle\hbar\omega_{\kappa_{A}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle+\hbar\omega_{\kappa_{B}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}|\varnothing\rangle$
(B.20) $\displaystyle=$
$\displaystyle\left(\hbar\omega_{\kappa_{A}}+\hbar\omega_{\kappa_{B}}\right)\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle,$
(B.21)

which inserted into (B.15) yields

$\displaystyle\hat{H}|\Phi(t)\rangle$ $\displaystyle=$
$\displaystyle\sum_{\kappa_{A},\kappa_{B}}\leavevmode\nobreak\
e^{-i(\omega_{\kappa_{A}}+\omega_{\kappa_{B}})t}\leavevmode\nobreak\
\alpha_{\kappa_{A}}\alpha_{\kappa_{B}}\leavevmode\nobreak\
\left(\hbar\omega_{\kappa_{A}}+\hbar\omega_{\kappa_{B}}\right)\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{A}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{B}}}|\varnothing\rangle$
(B.22) $\displaystyle=$ $\displaystyle i\hbar\frac{\partial}{\partial
t}|\Phi(t)\rangle,$

where in the last equality we have used the relation (B.13), which completes
the proof.

(c) We can generalize the preceding proof for $N$-photon states as follows.
First we expand in terms of the eigenfunction of $\Omega$:

$\displaystyle\vec{\psi}_{j}(t)$ $\displaystyle=$
$\displaystyle\sum_{\kappa_{j}}e^{-i\omega_{\kappa_{j}}t}\vec{\varphi}_{\kappa_{j}}\alpha_{\kappa_{j}},\qquad\alpha_{\kappa_{j}}=\langle\vec{\varphi}_{\kappa_{j}}|\vec{\psi}(t=0)\rangle,\qquad
j=1,\ldots N,$ (B.23)

which allows us to write

$\displaystyle|\Phi(t)\rangle$ $\displaystyle=$
$\displaystyle\prod_{j=1}^{N}\hat{B}_{\vec{\psi}_{j}(t)}^{\dagger}|\varnothing\rangle=\prod_{j=1}^{N}\left(\sum_{\kappa_{j}}e^{-i\omega_{\kappa_{j}}t}\leavevmode\nobreak\
\alpha_{\kappa_{j}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\right)|\varnothing\rangle$
(B.24) $\displaystyle=$
$\displaystyle\left(\sum_{\kappa_{1}}e^{-i\omega_{\kappa_{1}}t}\leavevmode\nobreak\
\alpha_{\kappa_{1}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{1}}}\right)\ldots\left(\sum_{\kappa_{N}}e^{-i\omega_{\kappa_{N}}t}\leavevmode\nobreak\
\alpha_{\kappa_{N}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{N}}}\right)|\varnothing\rangle$
(B.25) $\displaystyle=$
$\displaystyle\sum_{\kappa_{1},\ldots,\kappa_{N}}e^{-i\sum_{j=1}^{N}\omega_{\kappa_{j}}t}\leavevmode\nobreak\
\left(\prod_{j^{\prime\prime}}\alpha_{\kappa_{j^{\prime\prime}}}\right)\leavevmode\nobreak\
\left(\prod_{j^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}\right)|\varnothing\rangle$
(B.26)

and

$\displaystyle i\hbar\frac{\partial}{\partial t}|\Phi(t)\rangle$
$\displaystyle=$
$\displaystyle\sum_{\kappa_{1},\ldots,\kappa_{N}}\left(\sum_{j}\hbar\omega_{\kappa_{j}}\right)e^{-i\sum_{j}\omega_{\kappa_{j}}t}\leavevmode\nobreak\
\prod_{j^{\prime\prime}}\alpha_{\kappa_{j^{\prime\prime}}}\leavevmode\nobreak\
\prod_{j^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle.$
(B.27)

Applying the Hamiltonian to (B.26) yields

$\displaystyle\hat{H}|\Phi(t)\rangle$
$\displaystyle=\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}^{\dagger}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\leavevmode\nobreak\
\sum_{\kappa_{1},\ldots,\kappa_{N}}e^{-i\sum_{j=1}^{N}\omega_{\kappa_{j}}t}\leavevmode\nobreak\
\left(\prod_{j^{\prime\prime}}\alpha_{\kappa_{j^{\prime\prime}}}\right)\leavevmode\nobreak\
\left(\prod_{j^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}\right)|\varnothing\rangle$
(B.28)
$\displaystyle=\sum_{\kappa_{1},\ldots,\kappa_{N}}e^{-i\sum_{j}\omega_{\kappa_{j}}t}\leavevmode\nobreak\
\prod_{j^{\prime\prime}}\alpha_{\kappa_{j}^{\prime\prime}}\leavevmode\nobreak\
\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\prod_{j^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle.\phantom{xxxxxxxx}$
(B.29)

Using the commutation relations
$\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}=\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}+\delta_{\kappa_{j},\kappa^{\prime}}$
one can show by recursion that

$\displaystyle\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\prod_{j=1}^{N}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{N}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}+\prod_{j=1}^{N}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}.$
(B.30)

Indeed, in Eq. (B.16) we have shown that it is true for $N=2$. If we assume
that it is true for $N$, the following relations show that it is true also for
$N+1$:

$\displaystyle\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\prod_{j=1}^{N+1}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}$
$\displaystyle=$
$\displaystyle\left(\sum_{j=1}^{N}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}+\prod_{j=1}^{N}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\right)\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{N+1}}}$
(B.31) $\displaystyle=$
$\displaystyle\sum_{j=1}^{N}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{N+1}}}+\prod_{j=1}^{N}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\left(\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{N+1}}}\right)$
(B.32) $\displaystyle=$
$\displaystyle\sum_{j=1}^{N}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{N+1}}}+\prod_{j=1}^{N}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\left(\delta_{\kappa^{\prime},\kappa_{N+1}}+\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{N+1}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\right)$
(B.33) $\displaystyle=$
$\displaystyle\sum_{j=1}^{N}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{N+1}}}+\prod_{j=1}^{N}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\delta_{\kappa^{\prime},\kappa_{N+1}}+\prod_{j=1}^{N+1}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ $ (B.34)
$\displaystyle=$
$\displaystyle\sum_{j=1}^{N+1}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}+\prod_{j=1}^{N+1}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}.$
(B.35)

Thus, using the fact that
$\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}|\varnothing\rangle=0$ we have

$\displaystyle\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\prod_{j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}|\varnothing\rangle$
$\displaystyle=$
$\displaystyle\sum_{j}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle$
(B.36)

and further

$\displaystyle\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\prod_{j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}|\varnothing\rangle$
$\displaystyle=$
$\displaystyle\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\sum_{j}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle$
(B.37) $\displaystyle=$
$\displaystyle\sum_{j}\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\delta_{\kappa^{\prime},\kappa_{j}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle$
(B.38) $\displaystyle=$
$\displaystyle\sum_{j}\hbar\omega_{\kappa_{j}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}\prod_{j^{\prime}\neq
j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle$
(B.39) $\displaystyle=$
$\displaystyle\left(\sum_{j}\hbar\omega_{\kappa_{j}}\right)\prod_{{\rm
all\leavevmode\nobreak\
}j^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle.$
(B.40)

Finally inserting (B.40) into (B.29) we obtain

$\displaystyle\hat{H}|\Phi(t)\rangle$ $\displaystyle=$
$\displaystyle\sum_{\kappa_{1},\ldots,\kappa_{N}}e^{-i\sum_{j}\omega_{\kappa_{j}}t}\leavevmode\nobreak\
\prod_{j^{\prime\prime}}\alpha_{\kappa_{j}^{\prime\prime}}\leavevmode\nobreak\
\sum_{\kappa^{\prime}}\hbar\omega_{\kappa^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa^{\prime}}}\hat{B}_{\vec{\varphi}_{\kappa^{\prime}}}\prod_{j}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j}}}|\varnothing\rangle$
(B.41) $\displaystyle=$
$\displaystyle\sum_{\kappa_{1},\ldots,\kappa_{N}}e^{-i\sum_{j}\omega_{\kappa_{j}}t}\leavevmode\nobreak\
\prod_{j^{\prime\prime}}\alpha_{\kappa_{j}^{\prime\prime}}\left(\sum_{j}\hbar\omega_{\kappa_{j}}\right)\prod_{j^{\prime}}\hat{B}^{\dagger}_{\vec{\varphi}_{\kappa_{j^{\prime}}}}|\varnothing\rangle$
$\displaystyle=$ $\displaystyle i\hbar\frac{\partial}{\partial
t}|\Phi(t)\rangle,$ (B.42)

where in the last equality we have used the relation (B.13), which completes
the proof.

Acknowledgments:  This work was supported by the “Investissements d’Avenir”
program, project ISITE-BFC / IQUINS (ANR-15-IDEX-03), QUACO-PRC
(ANR-17-CE40-0007-01) and the EUR-EIPHI Graduate School (17-EURE-0002). We
also acknowledge support from the European Union’s Horizon 2020 research and
innovation program under the Marie Sklodowska-Curie grant agreement No. 765075
(LIMQUET).

This article is dedicated to the memory of Bruce W. Shore,

who has been a great inspiration for us for so many years.

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