The 1 on the right hand side of these equations follows since a box with no wires in or out is equal to 1.  The equation on the right is the normalisation condition for states.   In standard Quantum Theory the normalisation condition is $\text{trace}(\rho)=\text{trace}(\hat{\mathds{1}}\rho)=1$.  However, this form for the normalisation condition assumes we have adopted the forward gauge, wherein $\alpha_\mathsfb{a}\sqrt{N_\mathsfb{a}}=1$ for all system types (see Sec. <ref>).  In the forward gauge we represent the ignore result, $\hat{I}_{\mathsfb{b}_2}$ by $\hat{\mathds{1}}_\mathsfb{b_2}$ and the ignore preparation, $\hat{I}^{\mathsfb{a}_1}$ by $\hat{\mathds{1}}^{\mathsfb{a}_1}/N_{\mathsfb{a}_1}$.  However, if we use a different gauge then the normalisation will be different.  For example, in the time symmetric gauge, we represent $\hat{I}_{\mathsfb{b}_2}$ by $\hat{\mathds{1}}_{\mathsfb{b}_2}/\sqrt{N_\mathsfb{b}}$.  In this gauge, the equation on the right of (<ref>) becomes $\hat{\mathbb{B}}^{\mathsfb{b}_2} \hat{\mathds{1}}_{\mathsfb{b}_2}/\sqrt{N_\mathsfb{b}} = 1$.  These considerations mean that the sum of the entries along the diagonal of $\hat{\mathbb{B}}^{\mathsfb{b}_2}$ must equal $\sqrt{N_\mathsfb{b}}$.  In fact, from row 8 of Table <ref> we know that $\hat{\mathsfb{B}}^{\mathsfb{a}_1}=\hat{\mathsf{I}}^{\mathsfb{a}_1}$.  In the time symmetric gauge this means that $\hat{\mathbb{B}}^{\mathsfb{a}_1}=\hat{\mathds{1}}^{\mathsfb{a}_1}/\sqrt{N_\mathsfb{b}}$.  This does, indeed, have the property that the diagonal entries sum to $\sqrt{N_\mathsfb{b}}$.    These considerations reveal that there is an extra layer of representation when we go from abstract operator tensors to explicit representation of these operator tensors wherein we must specify $\alpha_\mathsfb{a}$.

§ MATHEMATICAL AXIOM FOR QUANTUM THEORY

We say that an operator is physical if (i) it is $T$-positive and (ii) it satisfies double causality.  We can state the following simple mathematical axiom for Time Symmetric Operational Quantum Theory.

Quantum Axiom.
Every physical operator has an operation corresponding to it.

That's it.  To support the reasonableness of this axiom we will provide an extension theorem (basically this the analogue of Stinespring's theorem adjusted to apply to this time symmetric situation) and also show this is equivalent to standard (time asymmetric) operational quantum theory.

Note that this is not a reconstruction in terms of reasonable axioms in the spirit of [43, 16, 26, 92, 79, 80, 37, 24, 67, 38, 59, 34, 32, 12, 84].  This axiom needs, up front, Hilbert spaces and the usual mathematical machinery of Quantum Theory.

§ EXTENSION THEOREM

We will prove the following

Extension theorem.  All physical operators can be written in the (extended) form

\begin{equation}\label{extensiontheoremequation}
\begin{Compose}{0}{0}\setdefaultfont{} \setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledouble{U}{2}{2}{0,0} \csymbol{B} \thispoint{AL}{-4,0} \csymbolalt[-20,0]{x} \thispoint{AR}{4,0} \csymbolalt[20,0]{y}
\joinlrnoarrowthick{U}{0}{AL}{0} \joinrlnoarrowthick{U}{0}{AR}{0}
\thispoint{a}{0,-4} \csymbolthird[0,-20]{a} \jointbnoarrow{a}{0}{U}{0}
\thispoint{d}{0,4} \csymbolthird[0,20]{b} \joinbtnoarrow{d}{0}{U}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{}\setsecondfont{\mathsfb}\setthirdfont{\mathtt}
\crectangledouble{U}{3}{2}{0,0} \csymbol{\hat{U}_\mathsf{B}}
\relpoint{U}{-0.3,4.5}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt{z}
\relpoint{U}{0.3,-4.5}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[left]{Id}{0}{U}{0.3} \csymbolalt[-3,0]{w}
\relpoint{U}{2.3,4.5}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos} \csymbol{\hat{Y}}  \jointbnoarrow[left]{U}{2.3}{Y}{0}  \csymbolalt{y}
\relpoint{Y}{4,0}{r} \joinlrnoarrowthick[above]{r}{0}{Y}{0} \csymbolthird[0,3]{y}
\relpoint{U}{-2.3,-4.5}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos} \csymbol{\hat{X}}  \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt{x} \relpoint{X}{-4,0}{l} \joinrlnoarrowthick[above]{l}{0}{X}{0} \csymbolthird{x}
\thispoint{down}{0,-7} \thispoint{up}{0,7}
\jointbnoarrow[right]{down}{2.3}{U}{2.3} \csymbolalt{a}
\jointbnoarrow[left]{U}{-2.3}{up}{-2.3} \csymbolalt{b}
\end{Compose}
\end{equation}

where $\hat{U}$ is unitary, and $\hat{X}$ and $\hat{Y}$ are maximal.

The proof of this theorem is as follows.  First, we note that we can write

\begin{equation}
\begin{Compose}{0}{0}\setdefaultfont{\hat}\setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledouble{U}{2}{2}{0,0} \csymbol{B} \thispoint{AL}{-4,0} \csymbolalt[-20,0]{x} \thispoint{AR}{4,0} \csymbolalt[20,0]{y}
\joinlrnoarrowthick{U}{0}{AL}{0} \joinrlnoarrowthick{U}{0}{AR}{0}
\thispoint{a}{0,-4} \csymbolthird[0,-20]{a} \jointbnoarrow{a}{0}{U}{0}
\thispoint{d}{0,4} \csymbolthird[0,20]{b} \joinbtnoarrow{d}{0}{U}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{\hat} \setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledouble{B}{2}{2}{0,0} \csymbol{B}
\crectangledouble{X}{0.9}{0.9}{-6,0} \csymbol{X} \crectangledouble{Y}{0.9}{0.9}{6,0} \csymbol{Y}
\crectangledouble{XD}{0.9}{0.9}{-6,-6} \csymbol{X} \crectangledouble{YU}{0.9}{0.9}{6,6} \csymbol{Y}
\relpoint{XD}{-3,0}{l} \relpoint{YU}{3,0}{r}
\joinrlnoarrowthick[above]{X}{0}{B}{0} \csymbolalt{x}  \joinrlnoarrowthick[above]{B}{0}{Y}{0} \csymbolalt{y}
\jointbnoarrow[left]{XD}{0}{X}{0} \csymbolthird{x}  \jointbnoarrow[left]{Y}{0}{YU}{0} \csymbolthird{y}
\joinrlnoarrowthick[above]{l}{0}{XD}{0} \csymbolalt{x}  \joinrlnoarrowthick[above]{YU}{0}{r}{0} \csymbolalt{y}
\thispoint{Bd}{0,-7} \thispoint{Bu}{0,7}
\jointbnoarrow[left]{Bd}{0}{B}{0} \csymbolthird{a}  \jointbnoarrow[left]{B}{0}{Bu}{0} \csymbolthird{b}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{\hat} \setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledmark[-1]{B}{3}{2}{0,0} \csymbol{V}
%\crectangledouble{X}{0.9}{0.9}{-6,0} \csymbol{X} \crectangledouble{Y}{0.9}{0.9}{6,0} \csymbol{Y}
\crectangledouble{XD}{0.9}{0.9}{-2,-6} \csymbol{X} \crectangledouble{YU}{0.9}{0.9}{2,6} \csymbol{Y}
\relpoint{XD}{-3,0}{l} \relpoint{YU}{3,0}{r}
%\joinrlnoarrow[above]{X}{0}{B}{0} \csymbolalt{x}  \joinrlnoarrow[above]{B}{0}{Y}{0} \csymbolalt{y}
\jointbnoarrow[left]{XD}{0}{B}{-2} \csymbolthird{x}  \jointbnoarrow[right]{B}{2}{YU}{0} \csymbolthird{y}
\joinrlnoarrowthick[above]{l}{0}{XD}{0} \csymbolalt{x}  \joinrlnoarrowthick[above]{YU}{0}{r}{0} \csymbolalt{y}
\thispoint{Bd}{0,-7} \thispoint{Bu}{0,7}
\jointbnoarrow[right]{Bd}{0}{B}{0} \csymbolthird{a}  \jointbnoarrow[left]{B}{0}{Bu}{0} \csymbolthird{b}
\end{Compose}
\end{equation}

using double maximality.  Now look at $\hat{V}$. Physicality requires $T$-positivity. Hence, following the steps in Sec. <ref>,  we can write

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{\hat} \setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledmark[-1]{B}{3}{2.5}{0,0} \csymbol{V}
\thispoint{Bd}{0,-7} \thispoint{Bu}{0,7}
\jointbnoarrow[right]{Bd}{0}{B}{0} \csymbolthird{a}  \jointbnoarrow[left]{B}{0}{Bu}{0} \csymbolthird{b}
\jointbnoarrow[right]{Bd}{-2}{B}{-2} \csymbolthird{x}  \jointbnoarrow[left]{B}{2}{Bu}{2} \csymbolthird{y}
\end{Compose}
\begin{Compose}{0}{0}\setdefaultfont{} \setsecondfont{\mathsfb}
\crectangledleft[-1]{A}{1.5}{2.5}{-4,0} \csymbol{V}
\relpoint{A}{0,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{0} \csymbolalt[-8,0]{xa}
%\relpoint{A}{-1,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{-1} \csymbolalt{x}
\relpoint{A}{0,7}{Cd} \jointbleft[left]{A}{0}{Cd}{0} \csymbolalt[-8,0]{by}
%\relpoint{A}{-1,7}{Ed} \jointbleft[left]{A}{-1}{Ed}{0} \csymbolalt{b}
\crectangledright[-1]{AA}{1.5}{2.5}{4,0} \csymbol{V}
\relpoint{AA}{0,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{0} \csymbolalt[8,0]{xa}
%\relpoint{AA}{-1,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{-1} \csymbolalt{a}
\relpoint{AA}{0,7}{Cd} \jointbright[right]{AA}{0}{Cd}{0} \csymbolalt[8,0]{y}
%\relpoint{AA}{-1,7}{Ed} \jointbright[right]{AA}{-1}{Ed}{0} \csymbolalt{b}
\joinrlnoarrow[above]{A}{1.7}{AA}{1.7}\csymbol{xa}
\joinrlnoarrow[above]{A}{-1.7}{AA}{-1.7}\csymbol[0,3]{by}
\end{Compose}
\begin{Compose}{0}{0}\setdefaultfont{} \setsecondfont{\mathsfb}
\crectangledouble{Iup}{1.5}{1.5}{0,8}\csymbolthird{\hat{\mathds{1}}} \crectangledouble{Idown}{1.5}{1.5}{0,-8} \csymbolthird{\hat{\mathds{1}}}
\crectangledleft[-1]{A}{1.5}{2.5}{-6,0} \csymbol{V}
\relpoint{A}{0,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{0} \csymbolalt[-8,0]{xa}
%\relpoint{A}{-1,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{-1} \csymbolalt{x}
\relpoint{A}{0,7}{Cd} \jointbleft[left]{A}{0}{Cd}{0} \csymbolalt[-8,0]{by}
%\relpoint{A}{-1,7}{Ed} \jointbleft[left]{A}{-1}{Ed}{0} \csymbolalt{b}
\crectangledleft{XA}{0.8}{0.8}{-2,3} \crectangledleft{BY}{0.8}{0.8}{-2,-3}
\relpoint{XA}{0.8,-1.8}{XAA} \joinrrnoarrow{XA}{0}{XAA}{0}
\joinrlnoarrow[below]{A}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{0.8,1.8}{BYY} \joinrrnoarrow{BY}{0}{BYY}{0}
\joinrlnoarrow[above]{A}{-1.2}{BYY}{0} \csymbol{by}
\jointbleft[left]{XA}{0}{Iup}{-1} \csymbolalt[-8,6]{xa} \jointbleft[left]{Idown}{-1}{BY}{0} \csymbolalt[-8,-6]{by}
\crectangledright[-1]{AA}{1.5}{2.5}{6,0} \csymbol{V}
\relpoint{AA}{0,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{0} \csymbolalt[8,0]{xa}
%\relpoint{AA}{-1,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{-1} \csymbolalt{a}
\relpoint{AA}{0,7}{Cd} \jointbright[right]{AA}{0}{Cd}{0} \csymbolalt[8,0]{by}
%\relpoint{AA}{-1,7}{Ed} \jointbright[right]{AA}{-1}{Ed}{0} \csymbolalt{b}
\crectangledright{XA}{0.8}{0.8}{2,3} \crectangledleft{BY}{0.8}{0.8}{2,-3}
\relpoint{XA}{-0.8,-1.8}{XAA} \joinllnoarrow{XA}{0}{XAA}{0}
\joinlrnoarrow[below]{AA}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{-0.8,1.8}{BYY} \joinllnoarrow{BY}{0}{BYY}{0}
\joinlrnoarrow[above]{AA}{-1.2}{BYY}{0} \csymbol{by}
\jointbright[right]{XA}{0}{Iup}{1} \csymbolalt[8,6]{xa} \jointbright[right]{Idown}{1}{BY}{0} \csymbolalt[8,-6]{by}
\end{Compose}
\end{equation}

We have grouped $x$ and $a$ together into a single wire so the diagram does not become too messy (and similarly with $b$ and $y$).  We have included the small black squares so we can track taking the adjoint in the following.   Using the relationships between $\hat{I}$ and $\mathds{1}$ in (<ref>), we obtain

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{\hat} \setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledmark[-1]{B}{3}{2.5}{0,0} \csymbol{V}
\thispoint{Bd}{0,-7} \thispoint{Bu}{0,7}
\jointbnoarrow[right]{Bd}{0}{B}{0} \csymbolthird{a}  \jointbnoarrow[left]{B}{0}{Bu}{0} \csymbolthird{b}
\jointbnoarrow[right]{Bd}{-2}{B}{-2} \csymbolthird{x}  \jointbnoarrow[left]{B}{2}{Bu}{2} \csymbolthird{y}
\end{Compose}
\begin{Compose}{0}{0}\setdefaultfont{} \setsecondfont{\mathsfb}
\crectangle[thin]{coef}{3.2}{2}{-12,0} \csymbol{\frac{\alpha_\mathsfb{xa}\sqrt{N_\mathsfb{xa}N_\mathsfb{by}}}{\alpha_\mathsfb{by}} }
\crectangledouble{Iup}{1.5}{1.5}{0,8}\csymbolthird{\hat{I}} \crectangledouble{Idown}{1.5}{1.5}{0,-8} \csymbolthird{\hat{I}}
\crectangledleft[-1]{A}{1.5}{2.5}{-6,0} \csymbol{V}
\relpoint{A}{0,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{0} \csymbolalt[-8,0]{xa}
%\relpoint{A}{-1,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{-1} \csymbolalt{x}
\relpoint{A}{0,7}{Cd} \jointbleft[left]{A}{0}{Cd}{0} \csymbolalt[-8,0]{by}
%\relpoint{A}{-1,7}{Ed} \jointbleft[left]{A}{-1}{Ed}{0} \csymbolalt{b}
\crectangledleft{XA}{0.8}{0.8}{-2,3} \crectangledleft{BY}{0.8}{0.8}{-2,-3}
\relpoint{XA}{0.8,-1.8}{XAA} \joinrrnoarrow{XA}{0}{XAA}{0}
\joinrlnoarrow[below]{A}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{0.8,1.8}{BYY} \joinrrnoarrow{BY}{0}{BYY}{0}
\joinrlnoarrow[above]{A}{-1.2}{BYY}{0} \csymbol{by}
\jointbleft[left]{XA}{0}{Iup}{-1} \csymbolalt[-8,6]{xa} \jointbleft[left]{Idown}{-1}{BY}{0} \csymbolalt[-8,-6]{by}
\crectangledright[-1]{AA}{1.5}{2.5}{6,0} \csymbol{V}
\relpoint{AA}{0,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{0} \csymbolalt[8,0]{xa}
%\relpoint{AA}{-1,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{-1} \csymbolalt{a}
\relpoint{AA}{0,7}{Cd} \jointbright[right]{AA}{0}{Cd}{0} \csymbolalt[8,0]{by}
%\relpoint{AA}{-1,7}{Ed} \jointbright[right]{AA}{-1}{Ed}{0} \csymbolalt{b}
\crectangledright{XA}{0.8}{0.8}{2,3} \crectangledleft{BY}{0.8}{0.8}{2,-3}
\relpoint{XA}{-0.8,-1.8}{XAA} \joinllnoarrow{XA}{0}{XAA}{0}
\joinlrnoarrow[below]{AA}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{-0.8,1.8}{BYY} \joinllnoarrow{BY}{0}{BYY}{0}
\joinlrnoarrow[above]{AA}{-1.2}{BYY}{0} \csymbol{by}
\jointbright[right]{XA}{0}{Iup}{1} \csymbolalt[8,6]{xa} \jointbright[right]{Idown}{1}{BY}{0} \csymbolalt[8,-6]{by}
\end{Compose}
\end{equation}

where the expression in the box is a multiplicative factor.   Hence, if we can show that

\begin{equation}\label{candidateunitary}
\begin{Compose}{0}{0}\setdefaultfont{} \setsecondfont{\mathsfb}
\crectangle[thin]{coef}{3.2}{2}{-12,0} \csymbol{\frac{\alpha_\mathsfb{xa}\sqrt{N_\mathsfb{xa}N_\mathsfb{by}}}{\alpha_\mathsfb{by}} }\thispoint{Iup}{0,7} \thispoint{Idown}{0,-7}
\crectangledleft[-1]{A}{1.5}{2.5}{-6,0} \csymbol{V}
\relpoint{A}{0,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{0} \csymbolalt[-8,0]{xa}
%\relpoint{A}{-1,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{-1} \csymbolalt{x}
\relpoint{A}{0,7}{Cd} \jointbleft[left]{A}{0}{Cd}{0} \csymbolalt[-8,0]{by}
%\relpoint{A}{-1,7}{Ed} \jointbleft[left]{A}{-1}{Ed}{0} \csymbolalt{b}
\crectangledleft{XA}{0.8}{0.8}{-2,3} \crectangledleft{BY}{0.8}{0.8}{-2,-3}
\relpoint{XA}{0.8,-1.8}{XAA} \joinrrnoarrow{XA}{0}{XAA}{0}
\joinrlnoarrow[below]{A}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{0.8,1.8}{BYY} \joinrrnoarrow{BY}{0}{BYY}{0}
\joinrlnoarrow[above]{A}{-1.2}{BYY}{0} \csymbol{by}
\jointbleft[left]{XA}{0}{Iup}{-2} \csymbolalt[-8,6]{xa} \jointbleft[left]{Idown}{-2}{BY}{0} \csymbolalt[-8,-6]{by}
\crectangledright[-1]{AA}{1.5}{2.5}{6,0} \csymbol{V}
\relpoint{AA}{0,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{0} \csymbolalt[8,0]{xa}
%\relpoint{AA}{-1,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{-1} \csymbolalt{a}
\relpoint{AA}{0,7}{Cd} \jointbright[right]{AA}{0}{Cd}{0} \csymbolalt[8,0]{by}
%\relpoint{AA}{-1,7}{Ed} \jointbright[right]{AA}{-1}{Ed}{0} \csymbolalt{b}
\crectangledright{XA}{0.8}{0.8}{2,3} \crectangledleft{BY}{0.8}{0.8}{2,-3}
\relpoint{XA}{-0.8,-1.8}{XAA} \joinllnoarrow{XA}{0}{XAA}{0}
\joinlrnoarrow[below]{AA}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{-0.8,1.8}{BYY} \joinllnoarrow{BY}{0}{BYY}{0}
\joinlrnoarrow[above]{AA}{-1.2}{BYY}{0} \csymbol{by}
\jointbright[right]{XA}{0}{Iup}{2} \csymbolalt[8,6]{xa} \jointbright[right]{Idown}{2}{BY}{0} \csymbolalt[8,-6]{by}
\end{Compose}
\end{equation}

is a unitary operator tensor then we have shown we can write $\hat{B}$ in the form in (<ref>). To prove this unitary property we use double causality (as required by physicality).   Applying backward causality to (<ref>) gives

\begin{equation}
\begin{Compose}{0}{0}\setdefaultfont{} \setsecondfont{\mathsfb}
\crectangle[thin]{coef}{3.2}{2}{-12,0} \csymbol{\frac{\alpha_\mathsfb{xa}\sqrt{N_\mathsfb{xa}N_\mathsfb{by}}}{\alpha_\mathsfb{by}} } \crectangledouble{Iup}{1.5}{1.5}{0,8}\csymbolthird{\hat{I}}\crectangledouble{Iupper}{1.5}{1.5}{0,14}\csymbolthird{\hat{I}}  \thispoint{Idown}{0,-7}
\crectangledleft[-1]{A}{1.5}{2.5}{-6,0} \csymbol{V}
\relpoint{A}{0,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{0} \csymbolalt[-8,0]{xa}
%\relpoint{A}{-1,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{-1} \csymbolalt{x}
\relpoint{A}{0,7}{Cd} \jointbleft[left]{A}{0}{Iupper}{-1} \csymbolalt[-8,0]{by}
%\relpoint{A}{-1,7}{Ed} \jointbleft[left]{A}{-1}{Ed}{0} \csymbolalt{b}
\crectangledleft{XA}{0.8}{0.8}{-2,3} \crectangledleft{BY}{0.8}{0.8}{-2,-3}
\relpoint{XA}{0.8,-1.8}{XAA} \joinrrnoarrow{XA}{0}{XAA}{0}
\joinrlnoarrow[below]{A}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{0.8,1.8}{BYY} \joinrrnoarrow{BY}{0}{BYY}{0}
\joinrlnoarrow[above]{A}{-1.2}{BYY}{0} \csymbol{by}
\jointbleft[left]{XA}{0}{Iup}{-1} \csymbolalt[-8,6]{xa} \jointbleft[left]{Idown}{-2}{BY}{0} \csymbolalt[-8,-6]{by}
\crectangledright[-1]{AA}{1.5}{2.5}{6,0} \csymbol{V}
\relpoint{AA}{0,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{0} \csymbolalt[8,0]{xa}
%\relpoint{AA}{-1,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{-1} \csymbolalt{a}
\relpoint{AA}{0,7}{Cd} \jointbright[right]{AA}{0}{Iupper}{1} \csymbolalt[8,0]{by}
%\relpoint{AA}{-1,7}{Ed} \jointbright[right]{AA}{-1}{Ed}{0} \csymbolalt{b}
\crectangledright{XA}{0.8}{0.8}{2,3} \crectangledleft{BY}{0.8}{0.8}{2,-3}
\relpoint{XA}{-0.8,-1.8}{XAA} \joinllnoarrow{XA}{0}{XAA}{0}
\joinlrnoarrow[below]{AA}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{-0.8,1.8}{BYY} \joinllnoarrow{BY}{0}{BYY}{0}
\joinlrnoarrow[above]{AA}{-1.2}{BYY}{0} \csymbol{by}
\jointbright[right]{XA}{0}{Iup}{1} \csymbolalt[8,6]{xa} \jointbright[right]{Idown}{2}{BY}{0} \csymbolalt[8,-6]{by}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\crectangledouble{Iup}{1.5}{1.5}{0,0}\csymbolthird{\hat{I}}
\relpoint{Iup}{0,-7}{Ad}
\jointbleft[left]{Ad}{-1}{Iup}{-1} \csymbolalt[-8,0]{xa}
\jointbright[right]{Ad}{1}{Iup}{1} \csymbolalt[8,0]{xa}
\crectangledouble{Iup}{1.5}{1.5}{7,0}\csymbolthird{\hat{I}}
\relpoint{Iup}{0,-7}{Ad}
\jointbleft[left]{Ad}{-1}{Iup}{-1} \csymbolalt[-8,0]{by}
\jointbright[right]{Ad}{1}{Iup}{1} \csymbolalt[8,0]{by}
\end{Compose}
\end{equation}

If we replace the $\hat{I}$'s by $\hat{\mathds{1}}$'s using (<ref>), we obtain

\begin{equation}
\begin{Compose}{0}{0}\setdefaultfont{} \setsecondfont{\mathsfb}
\crectangle[thin]{coef}{3.2}{2}{-12,0} \csymbol{\frac{\alpha_\mathsfb{xa}\sqrt{N_\mathsfb{xa}N_\mathsfb{by}}}{\alpha_\mathsfb{by}} }
\thispoint{Iup}{0,7} \thispoint{Idown}{0,-7}
\crectangledleft[-1]{A}{1.5}{2.5}{-6,0} \csymbol{V}
\relpoint{A}{0,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{0} \csymbolalt[-8,0]{xa}
%\relpoint{A}{-1,-7}{Ad} \jointbleft[left]{Ad}{0}{A}{-1} \csymbolalt{x}
\relpoint{A}{0,7}{Cd} \jointbleft[left]{A}{0}{Cd}{0} \csymbolalt[-8,0]{by}
%\relpoint{A}{-1,7}{Ed} \jointbleft[left]{A}{-1}{Ed}{0} \csymbolalt{b}
\crectangledleft{XA}{0.8}{0.8}{-2,3} \crectangledleft{BY}{0.8}{0.8}{-2,-3}
\relpoint{XA}{0.8,-1.8}{XAA} \joinrrnoarrow{XA}{0}{XAA}{0}
\joinrlnoarrow[below]{A}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{0.8,1.8}{BYY} \joinrrnoarrow{BY}{0}{BYY}{0}
\joinrlnoarrow[above]{A}{-1.2}{BYY}{0} \csymbol{by}
\jointbleft[left]{XA}{0}{Iup}{-2} \csymbolalt[-8,6]{xa} \jointbleft[left]{Idown}{-2}{BY}{0} \csymbolalt[-8,-6]{by}
\crectangledright[-1]{AA}{1.5}{2.5}{6,0} \csymbol{V}
\relpoint{AA}{0,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{0} \csymbolalt[8,0]{xa}
%\relpoint{AA}{-1,-7}{Ad} \jointbright[right]{Ad}{0}{AA}{-1} \csymbolalt{a}
\relpoint{AA}{0,7}{Cd} \jointbright[right]{AA}{0}{Cd}{0} \csymbolalt[8,0]{by}
%\relpoint{AA}{-1,7}{Ed} \jointbright[right]{AA}{-1}{Ed}{0} \csymbolalt{b}
\crectangledright{XA}{0.8}{0.8}{2,3} \crectangledleft{BY}{0.8}{0.8}{2,-3}
\relpoint{XA}{-0.8,-1.8}{XAA} \joinllnoarrow{XA}{0}{XAA}{0}
\joinlrnoarrow[below]{AA}{1.2}{XAA}{0} \csymbol{xa}
\relpoint{BY}{-0.8,1.8}{BYY} \joinllnoarrow{BY}{0}{BYY}{0}
\joinlrnoarrow[above]{AA}{-1.2}{BYY}{0} \csymbol{by}
\jointbright[right]{XA}{0}{Iup}{2} \csymbolalt[8,6]{xa} \jointbright[right]{Idown}{2}{BY}{0} \csymbolalt[8,-6]{by}
\jointtnoarrow{Iup}{-2}{Iup}{2} \jointtnoarrow{Iup}{-6}{Iup}{6}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb}
\thispoint{b}{0,-7}\thispoint{t}{0,0}
\jointbleft[left]{b}{-6}{t}{-6} \csymbolalt[-8,0]{xa} \jointbleft[left]{b}{-2}{t}{-2} \csymbolalt[-8,0]{by}
\jointbright[right]{b}{6}{t}{6} \csymbolalt[8,0]{xa} \jointbright[right]{b}{2}{t}{2} \csymbolalt[8,0]{by}
\jointtnoarrow{t}{-2}{t}{2} \jointtnoarrow{t}{-6}{t}{6}
\end{Compose}
\end{equation}

Note, we have cancelled out the common factor arising from converting from $\hat{I}$'s to $\hat{\mathds{1}}$ (this is clearly the same factor on both sides of the equation).  Also, note we now use the caps notation for the $\mathds{1}$'s from Sec. <ref>.  We see that this is exactly one of the two conditions for unitarity given in (<ref>) for the operator tensor given in (<ref>).  We obtain the other condition by applying backwards causality and following similar steps (flipped in the vertical direction).  This proves the extension theorem.

§ PROOF THAT FORMULATION IS TIME SYMMETRIC

The definition of a time symmetric operational theory was provided in Sec. <ref>.  Given the correspondence from operations to operators, this means (i) we need a map from any operator to the time reverse of that operator such that (ii) the time reversed operator circuit equates to the same value (the probability).  The obvious map between an operator and its time reverse is the adjoint map discussed in Sec. <ref> (see equation (<ref>)).  However, there is a subtlety here because of the gauge associated with the choice of the gauge parameters $\alpha_\mathsfb{a}$ (introduced in Sec. <ref>) and $\beta_\mathsfb{a}$ (introduced in Sec. <ref>).   If we choose the symmetric gauges (whereby $\alpha_\mathsfb{a}=\beta_\mathtt{a}=1$ for all systems) then the adjoint does the job.  However, it is more interesting to allow freedom in the choice of gauge (especially since the standard presentation of quantum theory is in the forward gauge).  We define the time reverse cups and caps which act as normal cups and caps except that they also introduce an overall factor.  Using these instead of the standard cups and caps we can define the time reverse operator which is just the adjoint with but with these extra factors.

Let us begin by defining time reverse cups and caps.  For systems these are

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\thispoint{a}{0,-1}
\ccapTR{t}{2.25}{1}{0,1}
\jointbnoarrow[left]{a}{-2.25}{t}{-2.25}  \csymbolalt{a}
\jointbnoarrow[right]{a}{2.25}{t}{2.25}  \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\crectangle[thin]{A}{1.2}{1.5}{-5,0}\csymbolthird{\frac{1}{\alpha_\mathsfb{a}^2}}
\thispoint{a}{0,-1}
\ccap{t}{2.25}{1}{0,1}
\jointbnoarrow[left]{a}{-2.25}{t}{-2.25}  \csymbolalt{a}
\jointbnoarrow[right]{a}{2.25}{t}{2.25}  \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\thispoint{a}{0,1}
\ccupTR{t}{2.25}{1}{0,-1}
\jointbnoarrow[left]{t}{-2.25}{a}{-2.25}  \csymbolalt{a}
\jointbnoarrow[right]{t}{2.25}{a}{2.25}  \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\crectangle[thin]{A}{1.2}{1.1}{-5,0}\csymbolthird{\alpha_\mathsfb{a}^2}
\thispoint{a}{0,1}
\ccup{t}{2.25}{1}{0,-1}
\jointbnoarrow[left]{t}{-2.25}{a}{-2.25}  \csymbolalt{a}
\jointbnoarrow[right]{t}{2.25}{a}{2.25}  \csymbolalt{a}
\end{Compose}
\end{equation}

These defined in terms of the caps and cups introduced in Sec. <ref> except that they introduce a factor (given inside the box).  It can be verified from Sec. <ref> that this is the correct factor to convert outputs to inputs and vice versa.  For pointers we have

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\thispoint{l}{-1,0} \thispoint{c}{0,0}
\chcapTR{t}{2.25}{0}{0,0}
\joinrlnoarrowthick[below]{l}{-2.25}{c}{-2.25}  \csymbolalt{a}
\joinrlnoarrowthick[above]{l}{2.25}{c}{2.25}  \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\crectangle[thin]{A}{1.2}{1.5}{-5,0}\csymbolthird{\frac{1}{\beta_\mathtt{a}^2}}
\thispoint{l}{-1,0} \thispoint{c}{0,0}
\chcap{t}{2.25}{0}{0,0}
\joinrlnoarrowthick[below]{l}{-2.25}{c}{-2.25}  \csymbolalt{a}
\joinrlnoarrowthick[above]{l}{2.25}{c}{2.25}  \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\thispoint{l}{1,0} \thispoint{c}{0,0}
\chcupTR{t}{2.25}{0}{0,0}
\joinlrnoarrowthick[below]{l}{-2.25}{c}{-2.25}  \csymbolalt{a}
\joinlrnoarrowthick[above]{l}{2.25}{c}{2.25}  \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\crectangle[thin]{A}{1.2}{1.1}{-5,0}\csymbolthird{\beta_\mathtt{a}^2}
\thispoint{l}{1,0} \thispoint{c}{0,0}
\chcup{t}{2.25}{0}{0,0}
\joinlrnoarrowthick[below]{l}{-2.25}{c}{-2.25}  \csymbolalt{a}
\joinlrnoarrowthick[above]{l}{2.25}{c}{2.25}  \csymbolalt{a}
\end{Compose}
\end{equation}

That these are correct can be seen by applying them to the fiducial elements in Sec. <ref>.

Given these time reverse cups and caps, we can define the time reverse operator as follows

\begin{equation}\label{timereverseofB}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\utilde{\hat{B}}}
\thispoint{l}{-5,0} \csymbolthird[-20,0]{y}  \thispoint{r}{5,0} \csymbolthird[20,0]{x}
\thispoint{d}{0,-5} \csymbolalt[0,-20]{b}  \thispoint{u}{0,5} \csymbolalt[0,20]{a}
\joinrlnoarrowthick{l}{0}{B}{0} \joinlrnoarrowthick{r}{0}{B}{0}
\jointbnoarrow{d}{0}{B}{0} \joinbtnoarrow{u}{0}{B}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble[white]{B}{2}{2}{0,0}
\thispoint{l}{-3,0} \thispoint{ll}{-12,0} \thispoint{lll}{-12,4}\csymbolthird[-20,0]{y}
\thispoint{r}{3,0} \thispoint{rr}{12,0} \thispoint{rrr}{12,-4} \csymbolthird[20,0]{x}
\chcupTR{a}{2}{0}{-3,-2} \joinrlnoarrowthick{l}{0}{B}{0} \joinrlnoarrowthick{l}{-4}{rr}{-4}
\chcapTR{a}{2}{0}{3,2} \joinlrnoarrowthick{r}{0}{B}{0} \joinlrnoarrowthick{r}{4}{ll}{4}
\thispoint{d}{0,-5} \thispoint{dd}{0,-12} \thispoint{ddd}{6,-12} \csymbolalt[0,-20]{b}
\thispoint{u}{0,5} \thispoint{uu}{0,12} \thispoint{uuu}{-6,12} \csymbolalt[0,20]{a}
\ccupTR{a}{3}{0}{-3,-5}
\jointbnoarrowthickwhite{d}{0}{B}{0} \jointbnoarrowthickwhite{d}{-6}{uu}{-6}
\jointbnoarrow{d}{0}{B}{0} \jointbnoarrow{d}{-6}{uu}{-6}
\ccapTR{a}{3}{0}{3,5}
\joinbtnoarrowthickwhite{u}{0}{B}{0} \joinbtnoarrowthickwhite{u}{6}{dd}{6}
\joinbtnoarrow{u}{0}{B}{0} \joinbtnoarrow{u}{6}{dd}{6}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\hat{B}}
\end{Compose}
\end{equation}

If we take the time reverse of the time reverse we get back to the original operation because the over all factors associated with the time reverse caps and cups are reciprocals of one another.   The adjoint is defined similarly - as

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\hat{B}^\dagger}
\thispoint{l}{-5,0} \csymbolthird[-20,0]{y}  \thispoint{r}{5,0} \csymbolthird[20,0]{x}
\thispoint{d}{0,-5} \csymbolalt[0,-20]{b}  \thispoint{u}{0,5} \csymbolalt[0,20]{a}
\joinrlnoarrowthick{l}{0}{B}{0} \joinlrnoarrowthick{r}{0}{B}{0}
\jointbnoarrow{d}{0}{B}{0} \joinbtnoarrow{u}{0}{B}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble[white]{B}{2}{2}{0,0}
\thispoint{l}{-3,0} \thispoint{ll}{-12,0} \thispoint{lll}{-12,4}\csymbolthird[-20,0]{y}
\thispoint{r}{3,0} \thispoint{rr}{12,0} \thispoint{rrr}{12,-4} \csymbolthird[20,0]{x}
\chcup{a}{2}{0}{-3,-2} \joinrlnoarrowthick{l}{0}{B}{0} \joinrlnoarrowthick{l}{-4}{rr}{-4}
\chcap{a}{2}{0}{3,2} \joinlrnoarrowthick{r}{0}{B}{0} \joinlrnoarrowthick{r}{4}{ll}{4}
\thispoint{d}{0,-5} \thispoint{dd}{0,-12} \thispoint{ddd}{6,-12} \csymbolalt[0,-20]{b}
\thispoint{u}{0,5} \thispoint{uu}{0,12} \thispoint{uuu}{-6,12} \csymbolalt[0,20]{a}
\ccup{a}{3}{0}{-3,-5}
\jointbnoarrowthickwhite{d}{0}{B}{0} \jointbnoarrowthickwhite{d}{-6}{uu}{-6}
\jointbnoarrow{d}{0}{B}{0} \jointbnoarrow{d}{-6}{uu}{-6}
\ccap{a}{3}{0}{3,5}
\joinbtnoarrowthickwhite{u}{0}{B}{0} \joinbtnoarrowthickwhite{u}{6}{dd}{6}
\joinbtnoarrow{u}{0}{B}{0} \joinbtnoarrow{u}{6}{dd}{6}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\hat{B}}
\end{Compose}
\end{equation}

Consequently, we have

\begin{equation}\label{TRtoAdjointwithconversionfactor}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\utilde{\hat{B}}}
\thispoint{l}{-5,0} \csymbolthird[-20,0]{y}  \thispoint{r}{5,0} \csymbolthird[20,0]{x}
\thispoint{d}{0,-5} \csymbolalt[0,-20]{b}  \thispoint{u}{0,5} \csymbolalt[0,20]{a}
\joinrlnoarrowthick{l}{0}{B}{0} \joinlrnoarrowthick{r}{0}{B}{0}
\jointbnoarrow{d}{0}{B}{0} \joinbtnoarrow{u}{0}{B}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangle[thin]{A}{2}{1.5}{-3,4}\csymbolthird{\frac{\alpha_\mathsfb{a}^2\beta_\mathtt{x}^2}{\alpha_\mathsfb{b}^2\beta_\mathtt{y}^2}}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\hat{B}^\dagger}
\thispoint{l}{-5,0} \csymbolthird[-20,0]{y}  \thispoint{r}{5,0} \csymbolthird[20,0]{x}
\thispoint{d}{0,-5} \csymbolalt[0,-20]{b}  \thispoint{u}{0,5} \csymbolalt[0,20]{a}
\joinrlnoarrowthick{l}{0}{B}{0} \joinlrnoarrowthick{r}{0}{B}{0}
\jointbnoarrow{d}{0}{B}{0} \joinbtnoarrow{u}{0}{B}{0}
\end{Compose}
\end{equation}

where the conversion factor is inside the rectangle.  Note that if we are in the time symmetric gauge (where $\alpha_\mathsfb{a}=\alpha_\mathsfb{b}=\beta_\mathtt{x}=\beta_\mathtt{y}=1$ then the conversion factor is equal to 1 so the time reverse is just equal to the adjoint.  Also, in the case where the input and output are the same type and also the income and outcome are the same type then the conversion factor is equal to 1.  Note that if an operator is physical then its time reverse is also physical.  This is clear since (a) $T$-positivity is a time symmetric notion that does not depend on any overall factors and (b) double causality is also a time symmetric notion but where we need to look after the overall factors as is done by the time reverse caps and cups.

If we have an operator circuit we can replace each operator by the expression involving its time reverse.  Consider the following example of this procedure:

\begin{equation}\label{ABCreversea}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{A}{1.5}{1.5}{0,-6}\csymbol{\hat{A}} \crectangledouble{B}{1.5}{1.5}{0,0}\csymbol{\hat{B}} \crectangledouble{C}{1.5}{1.5}{0,6}\csymbol{\hat{C}}
\jointbnoarrow[left]{A}{0}{B}{0} \csymbolalt{a} \jointbnoarrow[left]{B}{0}{C}{0} \csymbolalt{b}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{A}{1.5}{1.5}{0,-6}\csymbol{\hat{A}} \crectangledouble{B}{1.5}{1.5}{0,0}\csymbol{\hat{B}} \crectangledouble{C}{1.5}{1.5}{0,6}\csymbol{\hat{C}}
\jointbnoarrow[left]{A}{0}{B}{0} \csymbolalt{a} \jointbnoarrow[left]{B}{0}{C}{0} \csymbolalt{b}
\crectangledouble{Id}{0.9}{0.9}{0,-11}\csymbol{\hat{I}}  \crectangledouble{Iu}{0.9}{0.9}{0,11}\csymbol{\hat{I}}
\jointbnoarrow[left]{Id}{0}{A}{0} \csymbolalt{0} \jointbnoarrow[left]{C}{0}{Iu}{0} \csymbolalt{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{A}{1.5}{1.5}{5,-6}\csymbol{\utilde{\hat{A}}} \crectangledouble{B}{1.5}{1.5}{10,0}\csymbol{\utilde{\hat{B}}} \crectangledouble{C}{1.5}{1.5}{15,6}\csymbol{\utilde{\hat{C}}}
\crectangledouble{Id}{0.9}{0.9}{2.5,-11}\csymbol{\hat{I}}  \crectangledouble{Iu}{0.9}{0.9}{17.5,11}\csymbol{\hat{I}}
\relpoint{A}{-1.25,1.5}{Aul} \ccapTR{IA}{1.25}{0}{Aul} \jointbnoarrow[left]{Id}{0}{Aul}{-1.25} \csymbolalt{0}
\relpoint{A}{1.25,-1.5}{Adr}  \relpoint{B}{-1.25,1.5}{Bul} \ccupTR{AB}{1.25}{0}{Adr} \jointbnoarrow[left]{Adr}{1.25}{Bul}{-1.25} \csymbolalt{a} \ccapTR{BC}{1.25}{0}{Bul}
\relpoint{B}{1.25,-1.5}{Bdr} \relpoint{C}{-1.25,1.5}{Cul} \ccupTR{BC}{1.25}{0}{Bdr} \jointbnoarrow[left]{Bdr}{1.25}{Cul}{-1.25} \csymbolalt{b} \ccapTR{BC}{1.25}{0}{Cul}
\relpoint{C}{1.25,-1.5}{Cdr}  \ccupTR{AB}{1.25}{0}{Cdr} \jointbnoarrow[right]{Cdr}{1.25}{Iu}{0} \csymbolalt{0}
\end{Compose}
\end{equation}

Note that we introduce the ignore operators with the null system (this is associated with a Hilbert space of dimension 1 so this changes nothing).  Also note that each time reverse cup is matched by a time reverse cap. Hence, the overall factors cancel (since they are reciprocals).  Hence, (<ref>) continues as

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{A}{1.5}{1.5}{5,-6}\csymbol{\utilde{\hat{A}}} \crectangledouble{B}{1.5}{1.5}{10,0}\csymbol{\utilde{\hat{B}}} \crectangledouble{C}{1.5}{1.5}{15,6}\csymbol{\utilde{\hat{C}}}
\crectangledouble{Id}{0.9}{0.9}{2.5,-11}\csymbol{\hat{I}}  \crectangledouble{Iu}{0.9}{0.9}{17.5,11}\csymbol{\hat{I}}
\relpoint{A}{-1.25,1.5}{Aul} \ccap{IA}{1.25}{0}{Aul} \jointbnoarrow[left]{Id}{0}{Aul}{-1.25} \csymbolalt{0}
\relpoint{A}{1.25,-1.5}{Adr}  \relpoint{B}{-1.25,1.5}{Bul} \ccup{AB}{1.25}{0}{Adr} \jointbnoarrow[left]{Adr}{1.25}{Bul}{-1.25} \csymbolalt{a} \ccap{BC}{1.25}{0}{Bul}
\relpoint{B}{1.25,-1.5}{Bdr} \relpoint{C}{-1.25,1.5}{Cul} \ccup{BC}{1.25}{0}{Bdr} \jointbnoarrow[left]{Bdr}{1.25}{Cul}{-1.25} \csymbolalt{b} \ccap{BC}{1.25}{0}{Cul}
\relpoint{C}{1.25,-1.5}{Cdr}  \ccup{AB}{1.25}{0}{Cdr} \jointbnoarrow[right]{Cdr}{1.25}{Iu}{0} \csymbolalt{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{C}{1.5}{1.5}{0,-6}\csymbol{\utilde{\hat{C}}} \crectangledouble{B}{1.5}{1.5}{0,0}\csymbol{\utilde{\hat{B}}} \crectangledouble{A}{1.5}{1.5}{0,6}\csymbol{\utilde{\hat{A}}}
\jointbnoarrow[left]{C}{0}{B}{0} \csymbolalt{a} \jointbnoarrow[left]{B}{0}{A}{0} \csymbolalt{b}
\crectangledouble{Id}{0.9}{0.9}{-3.5,3}\csymbol{\hat{I}}  \crectangledouble{Iu}{0.9}{0.9}{3.5,-3}\csymbol{\hat{I}}
\relpoint{A}{-1.75,1.5}{Aul} \ccap{IA}{1.75}{0}{Aul} \jointbnoarrow[left]{Id}{0}{Aul}{-1.75} \csymbolalt{0}
\relpoint{C}{1.75,-1.5}{Cdr}  \ccup{AB}{1.75}{0}{Cdr} \jointbnoarrow[right]{Cdr}{1.75}{Iu}{0} \csymbolalt{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{C}{1.5}{1.5}{0,-6}\csymbol{\utilde{\hat{C}}} \crectangledouble{B}{1.5}{1.5}{0,0}\csymbol{\utilde{\hat{B}}} \crectangledouble{A}{1.5}{1.5}{0,6}\csymbol{\utilde{\hat{A}}}
\jointbnoarrow[left]{C}{0}{B}{0} \csymbolalt{a} \jointbnoarrow[left]{B}{0}{A}{0} \csymbolalt{b}
\end{Compose}
\end{equation}

Thus, in this example, the operator circuit formed by replacing elements by their time reverse and inverting the circuit has the same probability.  This works for any circuit - even if it has some boxes in parallel.  One way to see this is to imagine foliating the circuit and then sliding one layer at a time over a cap.  This deals with the case where we have system (not pointer) types.  It clearly works when there are pointer types as well.  One way to see this is to bend the pointer types so they point up and down and apply the same reasoning.  However, for the simple case where we have $R$ and $x$ boxes attached to each operator, we  can see we have the required behaviour by considering the following.

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\hat{B}}
\thispoint{l}{-5,0}   \thispoint{r}{5,0}
\thispoint{d}{0,-5} \csymbolalt[0,-20]{a}  \thispoint{u}{0,5} \csymbolalt[0,20]{b}
\RxBoxincomedouble{B}{1.5}{0}{x} \RxBoxoutcomedouble{B}{1.5}{0}{y}
%\joinrlnoarrow{l}{0}{B}{0} \joinlrnoarrow{r}{0}{B}{0}
\jointbnoarrow{d}{0}{B}{0} \joinbtnoarrow{u}{0}{B}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble[white]{B}{2}{2}{0,0}
\thispoint{l}{-3,0} \thispoint{ll}{-5,0} \thispoint{lll}{-12,4}
\thispoint{r}{3,0} \thispoint{rr}{5,0} \thispoint{rrr}{12,-4}
\chcupTR{a}{2}{0}{-3,-2} \joinrlnoarrowthick{l}{0}{B}{0} \joinrlnoarrowthick[below]{l}{-4}{rr}{-4} \csymbolthird[60,-4]{y}
\GenxdifBoxoutcomedouble{rr}{0}{-4}{}{y}{R} %\RxBoxoutcomedouble{rr}{0}{-4}{}
\chcapTR{a}{2}{0}{3,2} \joinlrnoarrowthick{r}{0}{B}{0}\joinlrnoarrowthick[above]{r}{4}{ll}{4} \csymbolthird[-60,0]{x}
\GenxdifBoxincomedouble{ll}{0}{4}{}{x}{R} %\RxBoxincomedouble{ll}{0}{4}{}
\thispoint{d}{0,-5} \thispoint{dd}{0,-12} \thispoint{ddd}{6,-12} \csymbolalt[0,-20]{a}
\thispoint{u}{0,5} \thispoint{uu}{0,12} \thispoint{uuu}{-6,12} \csymbolalt[0,20]{b}
\ccupTR{a}{3}{0}{-3,-5}
\jointbnoarrowthickwhite{d}{0}{B}{0} \jointbnoarrowthickwhite{d}{-6}{uu}{-6}
\jointbnoarrow{d}{0}{B}{0} \jointbnoarrow{d}{-6}{uu}{-6}
\ccapTR{a}{3}{0}{3,5}
\joinbtnoarrowthickwhite{u}{0}{B}{0} \joinbtnoarrowthickwhite{u}{6}{dd}{6}
\joinbtnoarrow{u}{0}{B}{0} \joinbtnoarrow{u}{6}{dd}{6}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\utilde{\hat{B}}}
\end{Compose}
\end{equation}

If we now go through the rest of the steps inserting cups and caps in the horizontal direction then we obtain the following:

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\hat{B}}
\thispoint{l}{-5,0}   \thispoint{r}{5,0}
\thispoint{d}{0,-5} \csymbolalt[0,-20]{a}  \thispoint{u}{0,5} \csymbolalt[0,20]{b}
\RxBoxincomedouble{B}{1.5}{0}{x} \RxBoxoutcomedouble{B}{1.5}{0}{y}
%\joinrlnoarrow{l}{0}{B}{0} \joinlrnoarrow{r}{0}{B}{0}
\jointbnoarrow{d}{0}{B}{0} \joinbtnoarrow{u}{0}{B}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangledouble[white]{B}{2}{2}{0,0}
\thispoint{l}{-5,0}   \thispoint{r}{5,0}
\thispoint{d}{0,-5} %\csymbolalt[0,-20]{a}
\thispoint{u}{0,5}  % \csymbolalt[0,20]{b}
\RxBoxincomedouble{B}{1.5}{0}{y} \RxBoxoutcomedouble{B}{1.5}{0}{x}
%\joinrlnoarrow{l}{0}{B}{0} \joinlrnoarrow{r}{0}{B}{0}
%\jointbnoarrow{d}{0}{B}{0} \joinbtnoarrow{u}{0}{B}{0}
\thispoint{d}{0,-5} \thispoint{dd}{0,-12} \thispoint{ddd}{6,-12}% \csymbolalt[0,-20]{a}
\thispoint{u}{0,5} \thispoint{uu}{0,12} \thispoint{uuu}{-6,12} %\csymbolalt[0,20]{b}
\ccupTR{a}{5}{0}{-5,-5}
\jointbnoarrowthickwhite{d}{0}{B}{0} % \jointbnoarrowthickwhite{d}{-6}{uu}{-6}
\jointbnoarrow{d}{0}{B}{0} \jointbnoarrow[right]{d}{-10}{uu}{-10} \csymbolalt{a}
\ccapTR{a}{5}{0}{5,5}
\joinbtnoarrowthickwhite{u}{0}{B}{0} % \joinbtnoarrowthickwhite{u}{6}{dd}{6}
\joinbtnoarrow{u}{0}{B}{0} \joinbtnoarrow[left]{u}{10}{dd}{10} \csymbolalt{b}
\crectangledouble{B}{2}{2}{0,0} \csymbol{\utilde{\hat{B}}}
\end{Compose}
\end{equation}

It is clear that, in general, we can invert circuits as required. This completes the proof that the we have a time symmetric formulation of Quantum Theory as defined in Sec. <ref>.

When we take the time reverse of an operator we pick up a gauge dependent factor in the representation of this operator (as seen in (<ref>)).  There is a natural time symmetric choice for $\alpha_\mathsfb{a}$ and $\beta_\mathtt{x}$ so this factor is equal to 1. However, the proof that the formulation is time symmetric presented in this section goes through regardless of what values we choose for these gauge parameters.

The extension theorem provides a nice way to take the time reverse of an operation.  By the above techniques, it is easy to prove that the time reverse of

\begin{equation}%\label{extensiontheoremequation}
\begin{Compose}{0}{0}\setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledouble{U}{2}{2}{0,0} \csymbol{B} \thispoint{AL}{-4,0} \csymbolalt[-20,0]{x} \thispoint{AR}{4,0} \csymbolalt[20,0]{y}
\joinlrnoarrowthick{U}{0}{AL}{0} \joinrlnoarrowthick{U}{0}{AR}{0}
\thispoint{a}{0,-4} \csymbolthird[0,-20]{a} \jointbnoarrow{a}{0}{U}{0}
\thispoint{d}{0,4} \csymbolthird[0,20]{b} \joinbtnoarrow{d}{0}{U}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{}\setsecondfont{\mathsfb}\setthirdfont{\mathtt}
\crectangledmark[-1]{U}{3}{2}{0,0} \csymbol{\hat{U}_\mathsf{B}}
\relpoint{U}{-0.3,4.5}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt{z}
\relpoint{U}{0.3,-4.5}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[left]{Id}{0}{U}{0.3} \csymbolalt[-3,0]{w}
\relpoint{U}{2.3,4.5}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos} \csymbol{\hat{Y}}  \jointbnoarrow[left]{U}{2.3}{Y}{0}  \csymbolalt{y}
\relpoint{Y}{4,0}{r} \joinlrnoarrowthick[above]{r}{0}{Y}{0} \csymbolthird[0,3]{y}
\relpoint{U}{-2.3,-4.5}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos} \csymbol{\hat{X}}  \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt{x} \relpoint{X}{-4,0}{l} \joinrlnoarrowthick[above]{l}{0}{X}{0} \csymbolthird{x}
\thispoint{down}{0,-7} \thispoint{up}{0,7}
\jointbnoarrow[right]{down}{2.3}{U}{2.3} \csymbolalt{a}
\jointbnoarrow[left]{U}{-2.3}{up}{-2.3} \csymbolalt{b}
\end{Compose}
\end{equation}

\begin{equation}%\label{extensiontheoremequation}
\begin{Compose}{0}{0}\setdefaultfont{}\setsecondfont{\mathtt} \setthirdfont{\mathsfb}
\crectangledouble{U}{2}{2}{0,0} \csymbol{\utilde{B}} \thispoint{AL}{-4,0} \csymbolalt[-20,0]{y} \thispoint{AR}{4,0} \csymbolalt[20,0]{x}
\joinlrnoarrowthick{U}{0}{AL}{0} \joinrlnoarrowthick{U}{0}{AR}{0}
\thispoint{a}{0,-4} \csymbolthird[0,-20]{b} \jointbnoarrow{a}{0}{U}{0}
\thispoint{d}{0,4} \csymbolthird[0,20]{a} \joinbtnoarrow{d}{0}{U}{0}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{}\setsecondfont{\mathsfb}\setthirdfont{\mathtt}
\crectangle[thin]{coeff}{3}{2}{-7.5,0}
\csymbol{
\frac{\alpha_\mathsfb{a}^2\beta_\mathtt{x}^2}{\alpha_\mathsfb{b}^2\beta_\mathtt{y}^2}
\crectangledmark[1]{U}{3}{2}{0,0} \csymbol{\hat{U}_\mathsf{B}}
\relpoint{U}{-0.3,4.5}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt[-3,0]{w}
\relpoint{U}{0.3,-4.5}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[left]{Id}{0}{U}{0.3} \csymbolalt{z}
\relpoint{U}{2.3,4.5}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos} \csymbol{\hat{X}}  \jointbnoarrow[left]{U}{2.3}{Y}{0}  \csymbolalt{x}
\relpoint{Y}{4,0}{r} \joinlrnoarrowthick[above]{r}{0}{Y}{0} \csymbolthird{x}
\relpoint{U}{-2.3,-4.5}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos} \csymbol{\hat{Y}}  \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt{y} \relpoint{X}{-4,0}{l} \joinrlnoarrowthick[above]{l}{0}{X}{0} \csymbolthird[0,3]{y}
\thispoint{down}{0,-7} \thispoint{up}{0,7}
\jointbnoarrow[right]{down}{2.3}{U}{2.3} \csymbolalt{b}
\jointbnoarrow[left]{U}{-2.3}{up}{-2.3} \csymbolalt{a}
\end{Compose}
\end{equation}

The black dot position indicates the adjoint.

§ EQUIVALENCE TO STANDARD OPERATIONAL FORMULATION

We will now show that the time symmetric operational formulation of Quantum Theory presented here is equivalent to the standard (time asymmetric) operational formulation of Quantum Theory which is construed in a forward time direction.  First we note that, by the extension theorem presented in Sec. <ref>, we know we can model every time symmetric operation by a corresponding operator that can be expressed in the language of standard quantum theory.  This proves that the standard operational formulation can model the time-symmetric formulation.   To prove equivalence we also need to prove the converse - that the time symmetric formulation can model the standard (time asymmetric) operational formulation of Quantum Theory.  In fact, we prove this at the more general level of operational theories since the pertinent properties ($T$-positivity and double causality) can be discussed at that level.

In the standard (time forward) operational formulation all incomes are regarded as settings - we condition on particular values of them.  Consider the circuit

\begin{equation}\label{ABCxyxuvwcircuit}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\crectangle{A}{2}{2}{0,0} \csymbol{A}  \crectangle{B}{2}{2}{5,9} \csymbol{B} \crectangle{C}{2}{2}{2,18} \csymbol{C}
\thispoint{mid}{-5,9}\csymbolalt[-20,0]{a} \jointbnoarrow[left]{A}{-1}{mid}{0}  \jointbnoarrow[left]{mid}{0}{C}{-1}
\jointbnoarrow[below right]{A}{1}{B}{0}  \csymbolalt[5,-5]{b}
\jointbnoarrow[above right]{B}{0}{C}{1}  \csymbolalt{a}
\Rxboxincome{A}{0}{x} \Rxboxoutcome{A}{0}{u}
\Rxboxincome{B}{0}{y} \Rxboxoutcome{B}{0}{v}
\Rxboxincome{C}{0}{z} \Rxboxoutcome{C}{0}{w}
\end{Compose}
\end{equation}

We are interested in the probability

\begin{equation}
\text{prob}(uvw|xyz)~~ = ~~
\frac{ \text{prob}\left(
\begin{Compose}{0}{-2.3} \setsecondfont{\mathsfb}
\crectangle{A}{2}{2}{0,0} \csymbol{A}  \crectangle{B}{2}{2}{5,9} \csymbol{B} \crectangle{C}{2}{2}{2,18} \csymbol{C}
\thispoint{mid}{-5,9}\csymbolalt[-20,0]{a} \jointbnoarrow[left]{A}{-1}{mid}{0}  \jointbnoarrow[left]{mid}{0}{C}{-1}
\jointbnoarrow[below right]{A}{1}{B}{0}  \csymbolalt[5,-5]{b}
\jointbnoarrow[above right]{B}{0}{C}{1}  \csymbolalt{a}
\Rxboxincome{A}{0}{x} \Rxboxoutcome{A}{0}{u}
\Rxboxincome{B}{0}{y} \Rxboxoutcome{B}{0}{v}
\Rxboxincome{C}{0}{z} \Rxboxoutcome{C}{0}{w}
\end{Compose}\right)
}{ \text{prob} \left(~~
\begin{Compose}{0}{-2.3} \setsecondfont{\mathsfb}
\crectangle{A}{2}{2}{0,0} \csymbol{A}  \crectangle{B}{2}{2}{5,9} \csymbol{B} \crectangle{C}{2}{2}{2,18} \csymbol{C}
\thispoint{mid}{-5,9}\csymbolalt[-20,0]{a} \jointbnoarrow[left]{A}{-1}{mid}{0}  \jointbnoarrow[left]{mid}{0}{C}{-1}
\jointbnoarrow[below right]{A}{1}{B}{0}  \csymbolalt[5,-5]{b}
\jointbnoarrow[above right]{B}{0}{C}{1}  \csymbolalt{a}
\Rxboxincome{A}{0}{x} \Routcome{A}{0}{u}
\Rxboxincome{B}{0}{y} \Routcome{B}{0}{v}
\Rxboxincome{C}{0}{z} \Routcome{C}{0}{w}
\end{Compose} ~~ \right)
\end{equation}

We know from backwards summation (or backwards flatness) that the denominator is equal to $\frac{1}{N_\mathtt{x} N_\mathtt{y} N_\mathtt{z}}$.  This means we can write

\begin{equation}
\text{prob}(uvw|xyz)~~ = ~~
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\crectangle{A}{2}{2}{0,-9} \csymbol{A}  \crectangle{B}{2}{2}{5,0} \csymbol{B} \crectangle{C}{2}{2}{2,9} \csymbol{C}
\thispoint{mid}{-5,0}\csymbolalt[-20,0]{a} \jointbnoarrow[left]{A}{-1}{mid}{0}  \jointbnoarrow[left]{mid}{0}{C}{-1}
\jointbnoarrow[below right]{A}{1}{B}{0}  \csymbolalt[5,-5]{b}
\jointbnoarrow[above right]{B}{0}{C}{1}  \csymbolalt{a}
\cxconditionincome[1.3]{A}{0}{x} \cxconditionoutcome[1.3]{A}{0}{u}
\cxconditionincome[1.3]{B}{0}{y} \cxconditionoutcome[1.3]{B}{0}{v}
\cxconditionincome[1.3]{C}{0}{z} \cxconditionoutcome[1.3]{C}{0}{w}
\end{Compose}
\end{equation}

\begin{equation}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionincome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\crectangle[thin]{N}{1}{1}{-8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxincome[0.8]{p}{0}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionoutcome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
%\crectangle[thin]{N}{1}{1}{-8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxoutcome[0.8]{p}{0}{0}{x}
\end{Compose}
\end{equation}

The $N_\mathtt{x}$ factor in the box is a multiplicative factor.  The interpretation of this is that we prepare $x$.  The $N_\mathtt{x}$ can be thought of as canceling the $\frac{1}{N_\mathtt{x}}$ from the $\mathsf R$ box.

What happens to the axiom of Quantum Theory when we go into this income conditioned picture?  In particular, how is the notion of physicality affected?  Physicality has two parts (i) $T$-positivity and (ii) double causality.  It is clear that $T$-positivity goes through unaltered (since multiplication by $N_\mathtt{x}$ factors does not change this property).  However, double causality needs to be revisited.  Double causality consists of the following two properties.

\begin{equation}\label{causalityconditionsagain}
\Funitalxyab{1}{1}{1}{1}    ~~~~~~~~~~~~~~~  \Bunitalyxba{1}{1}{1}{1}
\end{equation}

The forward causality property on the left becomes unapplicable since we are always conditioning on an income - except in the special case where there is no income when we get

\begin{equation}
\Funitalxyab{0}{1}{1}{1}
\end{equation}

This corresponds preparing the maximally mixed state. In this case, the outcome gives no information.  The backwards causality property (on the right of (<ref>)) is applicable.  We can rewrite it assuming that we have conditioned on a particular value of $x$ in which case it becomes

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\Crectangle{B}{1.3}{1.3}{0,0}
\Crectangle{I}{0.8}{0.8}{0,4.4} \jointbnoarrow[left]{B}{0}{I}{0} \csymbolalt{b}
\Routcome{B}{0}{y}
\cxconditionincome{B}{0}{x}
\thispoint{b}{0,-4}\jointbnoarrow[left]{b}{0}{B}{0} \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\Crectangle{I}{0.8}{0.8}{0,0.3}
\thispoint{b}{0,-4}\jointbnoarrow[left]{b}{0}{I}{0} \csymbolalt{a}
\end{Compose}
\end{equation}

This corresponds to the causality condition from standard Quantum Theory [12](see diagrammatic notation in [51]) whereby the deterministic effect is unique.

In the context of Quantum Theory, these considerations mean that we have, in this time asymmetric picture, the usual physicality conditions (as discussed in [51] for example).  These are

\begin{equation}
0\underset{T}{\leq} ~~~
\begin{Compose}{0}{-0.07} \setdefaultfont{\hat} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\Crectangledouble{B}{1.3}{1.3}{0,0}
\thispoint{b}{0,-4}\thispoint{t}{0,4} \thispoint{l}{-4,0} \thispoint{r}{4,0}
\jointbnoarrow[left]{b}{0}{B}{0} \csymbolalt{a}
\jointbnoarrow[left]{B}{0}{t}{0} \csymbolalt{b}
\joinrlnoarrowthick[above]{l}{0}{B}{0} \csymbolthird{x}
\joinrlnoarrowthick[above]{B}{0}{r}{0} \csymbolthird{y}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb}
\crectangledouble{B}{1.3}{1.3}{0,0} \csymbol{\hat{B}}
\crectangledouble{I}{0.9}{0.9}{0,4.4} \csymbol{\hat{I}} \jointbnoarrow[left]{B}{0}{I}{0} \csymbolalt{b}
\Routcomedouble{B}{0}{y}
\cxconditionincomedouble{B}{0}{x}
\thispoint{b}{0,-4}\jointbnoarrow[left]{b}{0}{B}{0} \csymbolalt{a}
\end{Compose}
\begin{Compose}{0}{0} \setdefaultfont{\hat} \setsecondfont{\mathsfb}
\Crectangledouble{I}{0.9}{0.9}{0,0.3}
\thispoint{b}{0,-4}\jointbnoarrow[left]{b}{0}{I}{0} \csymbolalt{a}
\end{Compose}
\end{equation}

In the usual parlance, the condition on the left is called complete positivity and the condition on the right tells us that, the trace is preserved if we sum over the outcomes.

We can also prove equivalence to a backwards formulation wherein we calculate probabilities for incomes conditioned on outcomes.  To do this, we would set

\begin{equation}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionincome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
%\crectangle[thin]{N}{1}{1}{-8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxincome[0.8]{p}{0}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionoutcome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\crectangle[thin]{N}{1}{1}{8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxoutcome[0.8]{p}{0}{0}{x}
\end{Compose}
\end{equation}

This is discussed further in Sec. <ref>.

§ PROOF THAT QUANTUM THEORY IS DOUBLY SUMMING

Consider a circuit like that in (<ref>).  The corresponding operator circuit is

\begin{equation}\label{ABCxyxuvwcircuitoperator}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb}
\crectangledouble{A}{2}{2}{0,0} \csymbol{\hat{A}}  \crectangledouble{B}{2}{2}{5,9} \csymbol{\hat{B}} \crectangledouble{C}{2}{2}{2,18} \csymbol{\hat{C}}
\thispoint{mid}{-5,9}\csymbolalt[-20,0]{a} \jointbnoarrow[left]{A}{-1}{mid}{0}  \jointbnoarrow[left]{mid}{0}{C}{-1}
\jointbnoarrow[below right]{A}{1}{B}{0}  \csymbolalt[5,-5]{b}
\jointbnoarrow[above right]{B}{0}{C}{1}  \csymbolalt{a}
\Rxboxincomedouble{A}{0}{x} \Rxboxoutcomedouble{A}{0}{u}
\Rxboxincomedouble{B}{0}{y} \Rxboxoutcomedouble{B}{0}{v}
\Rxboxincomedouble{C}{0}{z} \Rxboxoutcomedouble{C}{0}{w}
\end{Compose}
\end{equation}

If we substitute each operation for its extended version (as in (<ref>)) we obtain

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb}
\crectangledouble{U}{3.5}{2}{0,-20} \csymbol{\hat{U}_\mathsf{A}}
\relpoint{U}{-0.3,4.5}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt{q}
\relpoint{U}{0.3,-4.5}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[left]{Id}{0}{U}{0.3} \csymbolalt{r}
\relpoint{U}{2.3,4.5}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos}   \jointbnoarrow[left]{U}{2.3}{Y}{0}  \csymbolalt{u} \RxBoxoutcomedouble[0.5]{Y}{0.9}{0}{u}
\relpoint{U}{-2.3,-4.5}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos}  \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt{x} \RxBoxincomedouble[0.5]{X}{0.9}{0}{x}
\relpoint{U}{-2.3,2.1}{outputA}
\relpoint{U}{2.3,-2.1}{inputA}
\crectangledouble{U}{3}{2}{7,0} \csymbol{\hat{U}_\mathsf{B}}
\relpoint{U}{-0.3,4.5}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt{s}
\relpoint{U}{0.3,-4.5}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[left]{Id}{0}{U}{0.3} \csymbolalt{t}
\relpoint{U}{2.3,4.5}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos}   \jointbnoarrow[left]{U}{2.3}{Y}{0}  \csymbolalt{v} \RxBoxoutcomedouble[0.5]{Y}{0.9}{0}{v}
\relpoint{U}{-2.3,-4.5}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos}  \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt{y} \RxBoxincomedouble[0.5]{X}{0.9}{0}{y}
\relpoint{U}{-2.3,2.1}{outputB}
\relpoint{U}{2.3,-2.1}{inputB}
\crectangledouble{U}{3.5}{2}{4,20} \csymbol{\hat{U}_\mathsf{C}}
\relpoint{U}{-0.3,4.5}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt[-7,0]{m}
\relpoint{U}{0.3,-4.5}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[left]{Id}{0}{U}{0.3} \csymbolalt{n}
\relpoint{U}{2.3,4.5}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos}   \jointbnoarrow[left]{U}{2.3}{Y}{0}  \csymbolalt{w} \RxBoxoutcomedouble[0.5]{Y}{0.9}{0}{w}
\relpoint{U}{-2.3,-4.5}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos}  \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt{z} \RxBoxincomedouble[0.5]{X}{0.9}{0}{z}
\relpoint{U}{-2.3,2.1}{outputC}
\relpoint{U}{2.3,-2.1}{inputC}
\thispoint{AB}{$0.5*(outputA)+0.5*(inputB)$} \csymbolalt[0,20]{b}
\jointlnoarrow{outputA}{0.1}{AB}{0} \joinrbnoarrow{AB}{0}{inputB}{0}
\jointbnoarrow[right]{outputB}{0}{inputC}{0.8}  \csymbolalt{a}
\thispoint{AC}{-5,0}\csymbolalt[-20,0]{a} \jointbnoarrow{outputA}{-0.6}{AC}{0} \jointbnoarrow{AC}{0}{inputC}{0.3}
\end{Compose}
\end{equation}

The blank boxes are maximal elements.  We can stretch wires and move boxes so this is the same circuit as

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb}
\crectangledouble{U}{3.5}{2}{0,-10} \csymbol{\hat{U}_\mathsf{A}}
\thispoint{Iupos}{-8.6,20}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-2.3}{Iu}{0}  \csymbolalt[0,-8]{q}
\thispoint{Idpos}{1,-20}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[left]{Id}{0}{U}{0} \csymbolalt{r}
\thispoint{Ypos}{-6,20}  \crectangledouble{Y}{0.9}{0.9}{Ypos}   \jointbnoarrow[right]{U}{-0.3}{Y}{0}  \csymbolalt[5,0]{u} \RxBoxoutcomedouble[0.5]{Y}{0.9}{0}{u}
\thispoint{Xpos}{-2,-20}  \crectangledouble{X}{0.9}{0.9}{Xpos}  \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt{x} \RxBoxincomedouble[0.5]{X}{0.9}{0}{x}
\relpoint{U}{2.3,2.1}{outputA}
\relpoint{U}{2.3,-2.1}{inputA}
\crectangledouble{U}{3}{2}{9,0} \csymbol{\hat{U}_\mathsf{B}}
\relpoint{U}{8,20}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt[0,7]{s}
\relpoint{U}{8.6,-20}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}} \jointbnoarrow[right]{Id}{0}{U}{2.3} \csymbolalt[7,0]{t}
\relpoint{U}{10.6,20}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos}   \jointbnoarrow[left]{U}{2.3}{Y}{0}  \csymbolalt{v} \RxBoxoutcomedouble[0.5]{Y}{0.9}{0}{v}
\relpoint{U}{6,-20}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos}  \jointbnoarrow[left]{X}{0}{U}{0} \csymbolalt[0,-7]{y} \RxBoxincomedouble[0.5]{X}{0.9}{0}{y}
\relpoint{U}{-2.3,2.1}{outputB}
\relpoint{U}{-2.3,-2.1}{inputB}
\crectangledouble{U}{3.5}{2}{4,10} \csymbol{\hat{U}_\mathsf{C}}
\relpoint{U}{-0.3,10}{Iupos}  \crectangledouble{Iu}{0.9}{0.9}{Iupos} \csymbol{\hat{I}}  \jointbnoarrow[left]{U}{-0.3}{Iu}{0}  \csymbolalt[-5,0]{m}
\relpoint{U}{-14.4,-30}{Idpos}  \crectangledouble{Id}{0.9}{0.9}{Idpos} \csymbol{\hat{I}}
\jointbnoarrowthickwhite[left]{Id}{0}{U}{0.3}\jointbnoarrow[left]{Id}{0}{U}{0.3} \csymbolalt[0,7]{n}
\relpoint{U}{2.3,10}{Ypos}  \crectangledouble{Y}{0.9}{0.9}{Ypos}   \jointbnoarrow[right]{U}{2.3}{Y}{0}  \csymbolalt[5,0]{w} \RxBoxoutcomedouble[0.5]{Y}{0.9}{0}{w}
\relpoint{U}{-17,-30}{Xpos}  \crectangledouble{X}{0.9}{0.9}{Xpos}
\jointbnoarrowthickwhite[left]{X}{0}{U}{-2.3} \jointbnoarrow[left]{X}{0}{U}{-2.3} \csymbolalt[0,8]{z} \RxBoxincomedouble[0.5]{X}{0.9}{0}{z}
\relpoint{U}{-2.3,2.1}{outputC}
\relpoint{U}{2.3,-2.1}{inputC}
\jointbnoarrow[left]{outputA}{0.1}{inputB}{0}\csymbolalt[0,10]{b}
\jointbnoarrow[right]{outputB}{0}{inputC}{0.8}  \csymbolalt{a}
\jointbnoarrow[left]{outputA}{-0.6}{inputC}{-0.3} \csymbolalt{a}% \jointbnoarrow{AC}{0}{inputC}{0.3}
\crectangledouble{U}{3.5}{2}{4,10} \crectangledouble{Id}{0.9}{0.9}{Idpos} \crectangledouble{X}{0.9}{0.9}{Xpos} % drawn again because of the thick white lines
\crectangle[thin, dashed]{dots}{18}{13}{3,0}
\end{Compose}
\end{equation}

(We have taken some liberties with the positions at which the wires join the boxes to minimise wires crossing).
Now we see that we have a big unitary transformation happening in the middle (a dashed box is drawn round this).   Thus, this is equivalent to

\begin{equation}
\begin{Compose}{0}{0} \setdefaultfont{} \setsecondfont{\mathsfb}
\crectangledouble{U}{3}{2}{0,0} \csymbol{\hat{U}}
\crectangledouble{Iu}{0.9}{0.9}{-2,7} \csymbol{\hat{I}} \crectangledouble{Yu}{0.9}{0.9}{2,7}
\crectangledouble{Xd}{0.9}{0.9}{-2,-7} \crectangledouble{Id}{0.9}{0.9}{2,-7} \csymbol{\hat{I}}
\jointbnoarrow[left]{U}{-2}{Iu}{0} \csymbolalt{e} \jointbnoarrow[right]{U}{2}{Yu}{0} \csymbolalt{f}
\jointbnoarrow[left]{Xd}{0}{U}{-2} \csymbolalt{c} \jointbnoarrow[right]{Id}{0}{U}{2} \csymbolalt{d}
\RxBoxincomedouble{Xd}{0.9}{0}{c} \RxBoxoutcomedouble{Yu}{0.9}{0}{f}
\end{Compose}
\end{equation}

where we have formed composite systems to combine the ignore elements and the maximal elements with $R$'s and readout boxes attached.  It is now clear that we have the double summation property.  If we sum over $c$ then we have two ignore preparations going into the unitary $\hat{U}$ and so we get the ignore preparation out.  Then we have a tensor product of an ignore result and a maximal measurement.  Thus, the circuit equates to $\frac{1}{N_\mathsfb{f}}$. This proves forwards summation. Backwards summation follows by similar reasoning.

End remarks

§ CONCLUSIONS

In this paper we have presented a time symmetric approach to operational theories in general and applied this to the particular case of Quantum Theory.  The starting point was a time symmetric notion of the basic idea of an operation. In addition to having inputs and outputs (for physical systems like electrons for example) this also has incomes as well as the usual outcomes.  We have presented many double properties (all of which hold for Quantum Theory).  Most importantly, we obtained a double causality principle that goes beyond the causality principle of Chiribella, D'Ariano, and Perinotti.

We have shown how to formulate Quantum Theory in a time symmetric operational way.  To do this we set up the theory of operator tensors with additional structure corresponding to the incomes and outcomes.  We have provided an extension theorem showing how an arbitrary operator (to which an operation corresponds) can be modeled as a unitary behaviour on a larger system (similar to the Stinespring extension theorem).

There remains much work to be done on this project.  I will discuss some future directions in the next section.

§ FUTURE DIRECTIONS

§.§ Interpretation

I have not discussed the interpretation of this formalism.  In part this is because the mathematical results stand for themselves and interpretation is better seen as additional discussion.  Nevertheless, there are some important questions.

We know, as a contingent fact, that the world is not time symmetric.  It is possible, as a matter of principle, for a theory to have some symmetry while the stuff that makes up the universe does not. For example, having a fundamental theory that is Lorentz invariant does not preclude the existence of some given oak table even though the oak table itself picks out a preferred frame of reference.  However, in the present case, it is not so simple.  In Sec. <ref> we argued that distributions over incomes (regarded forward in time and outcomes (regarded backwards in time) must be flat.  However, this does not appear to be true of the actual distributions we see.  What should we make of this discrepancy?  One attitude is the following. The formalism presented here provides a way of calculating any probabilities (including conditional probabilities) in terms of probabilities for circuits.  In the case of conditional probabilities we divide one circuit probability by another.  Furthermore, when we use the mathematical machinery in this paper to calculate the probability for these circuits, we get the same answer whether we do a calculation forwards or backwards in time.  This is a formal time symmetry at the level of calculations.

How should we interpret double causality?  If we look at the right column of row 8 of Table <ref> we see that results with input only (for a given system type) are all equivalent.  This means that different choices of result do not change the probability of a circuit.  Consequently we cannot signal backward in time (this is the Pavia causality condition [12]).  Similarly, the left column of row 8 in the same table tells us that there is a unique preparation having an output only.  This tells us that we cannot signal forward in time.  Of course, we are quite used to communicating forward in time so this is rather odd.  In fact, we can steer information forwards (or backwards) in time in the sense that an earlier income can be correlated to a later outcome just as measurement outcomes on an entangled state can be correlated.   This still does not explain why, typically, we appear to be able to signal to the future but not the past.  A discussion of exactly this situation is provided by Di Biagio, Donà, and Rovelli [29] which appeals to the second law.  For the current project, a fully satisfactory resolution would involve showing how this apparent time asymmetry can appear within the framework described in this paper.  Some steps in this direction are discussed below in Sec. <ref>.   Such a project may also require modeling the agent themselves.  The discussion of time asymmetry from the perspective of a causal agent by Evans, Milburn, and Shrapnel in [30] is a good starting point for this project.

§.§ Generalised theory of conditional frames

In the time symmetric formalism we calculate joint probabilities

\begin{equation}
\end{equation}

In Sec. <ref> we defined the following object

\begin{equation}\label{transtoforwardpicture}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionincome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\crectangle[thin]{N}{1}{1}{-8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxincome[0.8]{p}{0}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionoutcome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
%\crectangle[thin]{N}{1}{1}{-8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxoutcome[0.8]{p}{0}{0}{x}
\end{Compose}
\end{equation}

We could regard this as a transformation between different conditional frames of reference.  In the original frame we have a time symmetric formulation.  In the new frame we have a time forward formulation. In this new conditional frame of reference we calculate probabilities of the form

\begin{equation}
\end{equation}

as illustrated by the example in Sec. <ref>.
We could also define a time backwards frame of reference as follows

\begin{equation}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionincome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
%\crectangle[thin]{N}{1}{1}{-8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxincome[0.8]{p}{0}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionoutcome{p}{0}{x}
\end{Compose}
\begin{Compose}{0}{0}
\crectangle[thin]{N}{1}{1}{8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxoutcome[0.8]{p}{0}{0}{x}
\end{Compose}
\end{equation}

In this frame of reference we would calculate probabilities of the form

\begin{equation}
\end{equation}

All three conditional frames of reference (the time symmetric one, the forward one, and the backwards one) are useful and afford equivalent formulations of operational Quantum Theory (by virtue of the arguments in Sec. <ref>).

An interesting question is whether we can usefully define general conditional frames of reference that go beyond the three examples just discussed.  One thing that makes these three cases special is that the normalisation factor $N_\mathtt{x}$ does not depend on the circuit it pertains to.  In the forward frame, this is because, by double flatness, the probability $p(\text{incomes},-)$ (where we sum over outcomes) is equal to $\frac{1}{N_\text{incomes}}$ (where $N_\text{incomes}$ is the product of the $N$'s for the income types.  And consequently, we can attribute the appropriate $N$ factor to each of the incomes as in (<ref>). Similar remarks apply to the backwards frame.  There may be other examples that are more contingent on the form of the circuit in which we can do this (for example, see the example in (<ref>) discussed in Sec. <ref> on double $Z$-flatness).  However, in general, we the normalisation factor will depend on the circuit.  There is no reason we cannot deal consistently with such general conditional frames.

This project could be driven by analogy with the role of general coordinate systems in General Relativity. In that case the special case is when we have an inertial frame of reference and there are linear transformations between different inertial frames of reference. More generally, the transformations are nonlinear.

This project would inform the interpretational questions discussed in Sec. <ref>.  Each observer would carry with them beliefs/knowledge which they condition on. They would want to choose a conditional frame of reference consistent with these beliefs/knowledge.

§.§ Porting income and outcome information around

We introduces incomes as providing information that (within the confines of the operation) is available before the operation takes place but not afterwards.  At first sight this appears problematic because we need to have income and outcome information in one place if we are to analyse it (and deduce joint probabilities for example).  However, away from the confines of the operation we can allow the income information and outcome information to be processed.  In fact, we could imagine doing this by using classical operations.   Let us consider how we might go about this.   First consider the example discussed in Sec. <ref> wherein operations are wired together by their systems and then we simplify this to the situation

\begin{equation}
\begin{Compose}{0}{0}
\crectangle{M}{3}{2}{0,0} \csymbol{M}
\thispoint{u}{-6,0} \csymbol[-20,0]{u}
\thispoint{v}{6,0} \csymbol[20,0]{v}
\joinrlnoarrowthick[above]{u}{0}{M}{0}
\joinrlnoarrowthick[above]{M}{0}{v}{0}
\end{Compose}
\end{equation}

Now imagine we want to port the income readouts and outcome readouts to a single location so we can deduce joint probabilities.  We could attempt to do this in the following way

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\Crectangle{C}{2}{4}{-9,-2}
\crectangle{M}{2}{2}{0,0} \csymbol{M}
\thispoint{u}{4,-4} % \csymbol[-20,0]{u}
\thispoint{v}{4,0} % \csymbol[20,0]{v}
\joinrlnoarrowthick[above]{C}{2}{M}{0} \csymbolalt{u}
\joinrlnoarrowthick[above]{C}{-2}{u}{0}
\joinrlnoarrowthick[above]{M}{0}{v}{0}
\RxBoxoutcome{u}{0}{0}{u}
\RxBoxoutcome{v}{0}{0}{v}
\end{Compose}
\end{equation}

The idea is that the $\mathsf C$ operation correlates its two outcomes so that they have the same $u$ value and then we can read off both $u$ and $v$ after $\mathsf M$.  However, forward flatness tells us that

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\Crectangle{C}{2}{4}{0,0}
\thispoint{u}{4,-2}  \csymbolalt[20,0]{u}
\thispoint{v}{4,2}  \csymbolalt[20,0]{u}
\joinrlnoarrow{C}{-2}{u}{0}
\joinrlnoarrow{C}{2}{v}{0}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\Crectangle{R}{0.8}{0.8}{0,-2} \crectangle{RR}{0.8}{0.8}{0,2} \csymbol{R}
\thispoint{u}{4,-2}  \csymbolalt[20,0]{u}
\thispoint{v}{4,2}  \csymbolalt[20,0]{u}
\joinrlnoarrow{R}{0}{u}{0}
\joinrlnoarrow{RR}{0}{v}{0}
\end{Compose}
\end{equation}

and consequently it is impossible to arrange that $\mathsf C$ correlates its outcomes.  This strategy fails.  It seems that the only way we can bring information forward is if we have some information about the past. Thus, if we substitute $\mathsf C$ in the above for $\mathsf D$ which has an income

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathtt}
\crectangle{C}{2}{4}{0,0} \csymbol{D}
\thispoint{u}{4,-2}  \csymbolalt[20,0]{u}
\thispoint{v}{4,2}  \csymbolalt[20,0]{u}
\joinrlnoarrow{C}{-2}{u}{0}
\joinrlnoarrow{C}{2}{v}{0}
\RxBoxincome{C}{2}{0}{z}
\end{Compose}
\end{equation}

then forward flatness does apply as it did with $\mathsf C$.  However, this still leaves the problem of how to get the information $z$ to the future.  We could attempt to copy it but that would give rise to the same problem we had with copying $u$.  Formally, we could choose to go into a forward conditional frame of reference just with respect to this instance of $\mathtt z$ such that

\begin{equation}%\label{transtoforwardpicture}
\begin{Compose}{0}{0}
\thispoint{p}{0,0} \cxconditionincome{p}{0}{z}
\end{Compose}
\begin{Compose}{0}{0}
\crectangle[thin]{N}{1}{1}{-8,0} \csymbolfourth{N_\mathtt{x}}
\thispoint{p}{0,0} \RxBoxincome[0.8]{p}{0}{0}{z}
\end{Compose}
\end{equation}

In this case, we would be calculating the probability $p(u,v|z)$.  This enables us to set up the mathematics so we can port information around.  However, it does not resolve the interpretational issues of how we do it.

§.§ Implications of gauge parameters $\alpha_\mathsfb{x}$ and $\beta_\mathtt{x}$

In Sec. <ref> and Sec. <ref> we discussed the gauge parameters $\alpha_\mathsfb{x}$ and $\beta_\mathtt{x}$. Different choices of these gauge parameters go naturally with different points of view (forward, backward, and time symmetric).  We saw, further, in Sec. <ref> that these parameters enter when we transform an operation to its time reverse.  What are the implications of having such gauge parameters in the theory?  Do they represent a fundamental symmetry?  In the context of the Lagrangian framework, gauge symmetries are associated with conserved quantities according to Noether's theorem.  We are not in the Lagrangian framework here but it is interesting to ask whether there are conserved quantities associated with these parameters.

Operator tensors are tensor-like objects. One might ask how they transform.  In [52] I provided a brief discussion of this topic. General transformations of operator tensors may provide the appropriate framework in which to study $\alpha_\mathsfb{x}$ and $\beta_\mathtt{x}$.

§.§ Reasonable axioms for Quantum Theory

It was twenty years ago that I wrote the paper Can we obtain Quantum Theory from reasonable axioms?  Since then there have been many papers on this subject (see Sec. <ref>).  All this work has been in the forward time picture (a notable exception being the reconstruction due to Ding Jia [60]).  However, the time symmetric picture is arguably much more natural.  Consequently it is a better setting for such reconstruction efforts.  We might hold out the hope that, in this time symmetric setting, a really compelling set of axioms is possible (as has been advocated for by Fuchs [35]).

To do this we need suitable axioms and a reconstruction.  In this paper we have presented many double properties and some subset of these might provide a suitable set of axioms.  The mathematical axiom provided in Sec. <ref> is not suitable as it stands since it refers to operators. It assumes the operator structure we would like to reconstruct.  However, the idea of correspondence from operations to some sort of mathematical object and physicality (so we have positive probabilities and causality) are reasonable.  Thus, one approach might be to start with the mathematical axiom and back fill so we do not assume the operator structure up front.  The approach in this paper is diagrammatic.  Other diagrammatic reconstructions may provide clues.  The reconstruction of Chiribella, D'Ariano, and Perinotti [12] invokes diagrams as does my reconstruction [51].   In the reconstruction of Selby, Scandolo, and Coecke [84] the diagrammatics are more central to the proof.  Interestingly, notions of time reversal play an important role.  Further, Postulate 5 (symmetric purification) in that work clearly has a connection with the extension theorem presented in Sec. <ref>.

Di Biagio, Dona, and Rovelli also have some speculations on directions to take for a reconstruction program [29] (see also [28]).

§.§ Conditional probabilities

To formulate these principles we have introduced diagrammatic notation for the operational theory, for duotensors, and for operator tensors (as well as a version of the diagrammatic notation of Coecke and Kissinger for Hilbert space objects).  In so doing, we have added classical structure corresponding to the incomes and outcomes.  Interestingly, the duotensor notation can be thought of as implementing the chain rule for conditional probabilities in the classical (square dot) case.  There is a similar looking structure for the round dot duotensors. However, the inverse of the hopping matrix can be negative so there is not a simple interpretation of this structure as implementing the chain rule for conditional probabilities.  However, this may relate to attempts to provide objects like a conditional density matrix [61, 63].

§.§ Combs and other generalisations

The quantum combs approach considers more general objects than operations called quantum combs.  These have inputs and outputs at multiple times (such that we could imagine, for example, feeding an output into an input on the same comb).  Associated with quantum combs is a generalization of the causality condition.  This generalisation is iterative.  However, it is in the time forward picture.  It would be useful to find the generalization of these conditions to the time symmetric picture presented in this work.

The approach to duotensors and operator tensors presented here goes beyond my previous work on these subjects.  This is because we add classical structure associated with incomes and outcomes.  In the duotensor formalism, for example, these are associated with square dots.  The work of Schmid, Selby, and Spekkens [82] does something similar adding inferential structure to a process framework.  In fact, this work is more general since it considers objects that might be regarded as inferential combs. In the time symmetric setting there are constraints on what can usefully be implemented (see the discussion of the attempt to construct a classical operation, $\mathsf C$, that provides two copies of an income in Sec. <ref> for an example).  Thus there is an interesting project (suggested, in fact, by Di Biagio in [28]) in developing the interplay between the inferential structure of Schmid et al. and formulating operational  Quantum Theory in a time symmetric fashion.

§.§ Time reverse experiments

There are many experiments discussed in the foundations of Quantum Theory and Quantum Information Theory having conceptual import.  We can take any of these and consider the time reverse experiment.  We must get the same predictions for this time reverse experiment.  Consequently, any conceptual implications that apply in one time direction will apply in the reverse direction as well.  Obvious examples to look at are Bell's theorem [7] and its variants (such as [39, 42]), quantum teleportation [8], and the Pusey, Barrett, and Rudolph result [78].  Wood and Spekkens's investigation of Bell's theorem [94] using causal discovery algorithms included retrocausal explanations on the same footing as forward in time causal explanations so this is of the right flavour.

To get a taste of how this might work. Here is the standard quantum teleportation experiment in the diagrammatic notation of this paper.

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangle{B}{2}{1.3}{0,0} \csymbol{B}
\crectangle{BB}{2}{1.3}{-6,8} \csymbol{\utilde{\mathsf{B}}} \Crectangle{C}{1.5}{1.5}{7,12}
\jointbnoarrow[above right]{B}{-1}{BB}{1} \csymbolalt{a} \jointbnoarrow[above left]{B}{1}{C}{0} \csymbolalt{a}
\relpoint{BB}{-4,-4}{pB} \jointbnoarrow[below right]{pB}{0}{BB}{-1} \csymbolalt{a}
\joinrlnoarrowthick[above left]{BB}{0}{C}{0} \csymbolthird{x}
\relpoint{C}{4,4}{pC} \jointbnoarrow[above left]{C}{0}{pC}{0} \csymbolalt{a}
\RxBoxincome[0.7]{B}{2}{0}{x}
\end{Compose}
\end{equation}

Note that we would normally condition on a given value of $x$. Then this implements the identity channel on $\mathsfb a$.  We could write $\mathtt{x}=\mathtt{aa}$.  The time reverse is the following

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangle{B}{2}{1.3}{0,0} \csymbol{\utilde{\mathsf{B}}}
\crectangle{BB}{2}{1.3}{6,-8} \csymbol{B} \crectangle{C}{1.5}{1.5}{-7,-12} \csymbol{\utilde{\mathsf{C}}}
\joinbtnoarrow[above right]{B}{1}{BB}{-1} \csymbolalt{a} \joinbtnoarrow[above left]{B}{-1}{C}{0} \csymbolalt{a}
\relpoint{BB}{4,4}{pB} \joinbtnoarrow[below right]{pB}{0}{BB}{1} \csymbolalt{a}
\joinlrnoarrowthick[above left]{BB}{0}{C}{0} \csymbolthird{x}
\relpoint{C}{-4,-4}{pC} \joinbtnoarrow[above left]{C}{0}{pC}{0} \csymbolalt{a}
\RxBoxoutcome[0.7]{B}{2}{0}{x}
\end{Compose}
\end{equation}

In the case where we post-select on the same given value of $x$, this will implement the identity channel on $\mathsfb a$.
We can model $\mathsf C$ in extended form as a controlled unitary implementing the standard operations in the quantum teleportation protocol.

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangle{C}{1.5}{1.5}{0,0} \csymbol{C}
\thispoint{d}{0,-5} \thispoint{u}{0,5} \thispoint{h}{-5,0}
\jointbnoarrow[left]{d}{0}{C}{0} \csymbolalt{a}
\jointbnoarrow[left]{C}{0}{u}{0} \csymbolalt{a}
\joinrlnoarrowthick[above]{h}{0}{C}{0} \csymbolthird{x}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangle{U}{2}{1.5}{0,0} \csymbol{U}
\crectangle{X}{0.8}{0.8}{-1.5,-4}  \crectangle{I}{0.8}{0.8}{-1.5,4} \csymbol{I}
\thispoint{d}{1.5,-5} \thispoint{u}{1.5,5}
\jointbnoarrow[left]{X}{0}{U}{-1.5} \csymbolalt{x}
\jointbnoarrow[left]{U}{-1.5}{I}{0} \csymbolalt{x}
\jointbnoarrow[right]{d}{0}{U}{1.5} \csymbolalt{a}
\jointbnoarrow[right]{U}{1.5}{u}{0} \csymbolalt{a}
\relpoint{X}{-3,0}{Xp} \joinrlnoarrowthick[above]{Xp}{0}{X}{0} \csymbolthird{x}
\end{Compose}
\end{equation}

where the blank box is a maximal preparation.  The time reverse of this is

\begin{equation}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangle{C}{1.5}{1.5}{0,0} \csymbol{C}
\thispoint{d}{0,5} \thispoint{u}{0,-5} \thispoint{h}{5,0}
\joinbtnoarrow[left]{d}{0}{C}{0} \csymbolalt{a}
\joinbtnoarrow[left]{C}{0}{u}{0} \csymbolalt{a}
\joinlrnoarrowthick[above]{h}{0}{C}{0} \csymbolthird{x}
\end{Compose}
\begin{Compose}{0}{0} \setsecondfont{\mathsfb} \setthirdfont{\mathtt}
\crectangle{U}{2}{1.5}{0,0} \csymbol{\utilde{\mathsf{U}}}
\crectangle{X}{0.8}{0.8}{1.5,4}  \crectangle{I}{0.8}{0.8}{1.5,-4} \csymbol{I}
\thispoint{d}{-1.5,5} \thispoint{u}{-1.5,-5}
\joinbtnoarrow[right]{X}{0}{U}{1.5} \csymbolalt{x}
\joinbtnoarrow[right]{U}{1.5}{I}{0} \csymbolalt{x}
\joinbtnoarrow[left]{d}{0}{U}{-1.5} \csymbolalt{a}
\joinbtnoarrow[left]{U}{-1.5}{u}{0} \csymbolalt{a}
\relpoint{X}{3,0}{Xp} \joinlrnoarrowthick[above]{Xp}{0}{X}{0} \csymbolthird{x}
\end{Compose}
\end{equation}

This is still a controlled unitary but where noise is sent in, then an outcome is read out at the end. The time reverse teleportation protocol works as follows.  The state, on system $\mathsfb a$, to be teleported is sent into $\mathsf C$.  A classical outcome is extracted and system $\mathsfb a$ continues.  The outcome from $\mathsf C$ is used to select a Bell state.  Meanwhile, a Bell measurement is made on the system and one of the two outputs from the Bell state.  If the outcome of this Bell measurement is the right one then the state of the system is transferred onto the outgoing $\mathsfb a$ system.

§.§ Connection to Operational Quantum Gravity program

While this may not be apparent, the real motivation for the present work is the need to solve the problem of Quantum Gravity (that is to find a theory that limits to Quantum Theory and General Relativity under appropriate circumstances).   I have pursued an operational approach to this problem [44, 56, 57] wherein the indefinite causal structure is regarded as representing the most basic challenge to making progress.  We expect to have indefinite causal structure in a theory of Quantum Gravity because causal structure (via the metric) is a dynamical variable in General Relativity while, in Quantum Theory, dynamical variables are typically subject to indefiniteness.  Indefinite causal structure is a problem because it makes it difficult to use standard concepts from physics.  In particular, how do we define causality if causal structure is indefinite.  It seems that indefinite causal structure needs to be tamed".  The best idea I have for taming it is the Quantum Equivalence Principle - this is the idea that it should be possible to find a quantum coordinate system such that, in the vicinity of any given point, we have definite causal structure [58] (see also [36, 66]).  Once we are in a quantum coordinate system where we have definite causal structure then we can attempt to impose causality.  However, the previous Pavia causality condition is time asymmetric which is at odds with the underlying symmetry of General Relativity.  This motivates the search for a time-symmetric approach which accounts for the present paper.

§.§ LaTeX/Tikz package for diagrams

In this paper I have drawn numerous diagrams. These were composed using the  packages that come with this arXiv submission. These packages define a number of commands built in terms of basic Tikz commands [89].  For example, the diagram

\[
\begin{Compose}{0}{0} \setsecondfont{\mathsfb}
\crectangle{A}{2}{2}{0,0} \csymbol{A}
\crectangle{B}{2}{2}{5,9} \csymbol{B}
\crectangle{C}{2}{2}{2,18} \csymbol{C}
\jointbnoarrow[left]{A}{-1}{C}{-1} \csymbolalt{a}
\jointbnoarrow[below right]{A}{1}{B}{0}  \csymbolalt[5,-5]{b}
\jointbnoarrow[above right]{B}{0}{C}{1}  \csymbolalt{a}
\Rxboxincome{A}{0}{x} \Rxboxoutcome{A}{0}{u}
\Rxboxoutcome{B}{0}{v}
\Rxboxincome{C}{0}{z} \Rxboxoutcome{C}{0}{w}
\end{Compose}
\]

is given by the code

However, the  packages, as they stand, have a lot of legacy structure in them having been used for multiple projects.  The ability to draw diagrams is basic to projects of this nature and some work on providing more user friendly packages is needed for the current project.

§ ACKNOWLEDGEMENTS

This project began in ernest following Zoom discussions that formed part of the QISS project (see below).   I am grateful to Flaminia Giacomini, Laurent Freidel, Doreen Fraser, Francesca Vidotto, Marios Christodoulou and especially Carlo Rovelli.  Further, I am grateful to Andrea Di Biagio who gave a presentation to this small QISS discussion group on an early version of his work (as outlined in Sec. <ref>).

I am further grateful to Bob Coecke, Ognyan Oreshkov, Mauro D'Ariano, Paulo Perinotti, Giulio Chribella, Robert Oeckl, David Schmid, Robert Spekkens, Ding Jia, Nitica Sakharwade and many others on topics related to this work.

I am grateful to my family (co-workers) during this stay-at-home covid era, especially to Zivy for frequently reminding me, with simple examples, that the world is not time symmetric.

Research at Perimeter Institute is supported in part by the Government
of Canada through the Department of Innovation, Science and Industry Canada and by the Province of Ontario through the Ministry of Colleges and Universities

As emphasised above and in Sec. <ref>, this work was made possible through the support of a grant from the John TempletonFoundation (ID# 61466) funding “The Quantum Information Structure of Spacetime (QISS)” project (qiss.fr).

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Wharton - and check his email
Leifer and Pussey
Kenneth references.
Carl Bender's stuff
CPT papers.