Efficient semiclassical approach for time delays

In this section we derive a semiclassical approximation for the Wigner time delay that is based on representation (13) of the Wigner–Smith matrix Q. It is the third semiclassical approximation after (5) and (10). The starting point is the Green function $\mathcal{G}({\boldsymbol{r}} ,{{{\boldsymbol{r}} }^{\prime }},E)$ of the open cavity with an arbitrary number of leads. Its semiclassical approximation is a sum over all trajectories from ${{{\boldsymbol{r}} }^{\prime }}$ to ${\boldsymbol{r}}$ [16, 38]

$$\begin{eqnarray}&&\mathcal{G}({\boldsymbol{r}} ,{{{\boldsymbol{r}} }^{\prime }},E)\approx \frac{1}{{\rm i}\hbar \sqrt{2\pi {\rm i}\hbar }}\mathop{\sum }\limits_{\gamma }\frac{1}{\sqrt{{{v}_{\gamma }}\;{{v}_{\gamma }}^{\prime} \;|{{({{M}_{\gamma }})}_{12}}|}}{\rm exp} \left( \frac{{\rm i}}{\hbar }{{\mathcal{S}}_{\gamma }}-\frac{{\rm i}\;\pi }{2}{{\nu }_{\gamma }} \right).\end{eqnarray} \tag{ 19 }$$

Here, ${{\mathcal{S}}_{\gamma }}={{\int }_{\gamma }}{\boldsymbol{p}} \;{\rm d}{\boldsymbol{q}}$ is the action along the trajectory γ, ${{\nu }_{\gamma }}$ the number of conjugate points, and $v_{\gamma }^{\prime }$ (${{v}_{\gamma }}$) is the speed at initial (final) point (for a cavity without potential $v_{\gamma }^{\prime }={{v}_{\gamma }}$). Furthermore, ${{M}_{\gamma }}$ denotes the stability matrix that describes linearized motion near the trajectory. It connects perpendicular deviations from the trajectory at the end point to those at the initial point

$$\begin{eqnarray}&&\left( \begin{array}{ccccccccccccccc} {\rm d}{{q}_{\bot }} \\ {\rm d}{{p}_{\bot }} \\ \end{array} \right)={{M}_{\gamma }}\left( \begin{array}{ccccccccccccccc} {\rm d}q_{\bot }^{\prime } \\ {\rm d}p_{\bot }^{\prime } \\ \end{array} \right)=\left( \begin{array}{ccccccccccccccc} {{({{M}_{\gamma }})}_{11}} & {{({{M}_{\gamma }})}_{12}} \\ {{({{M}_{\gamma }})}_{21}} & {{({{M}_{\gamma }})}_{22}} \\ \end{array} \right)\left( \begin{array}{ccccccccccccccc} {\rm d}q_{\bot }^{\prime } \\ {\rm d}p_{\bot }^{\prime } \\ \end{array} \right).\end{eqnarray} \tag{ 20 }$$

Formally, the effective Hamiltonian ${{\mathcal{H}}_{{\rm eff}}}$ of an open cavity is an infinite dimensional operator [78–80], corresponding to $N\to \infty$ of the matrix truncation in (12) and (13). The position representation of the resolvent ${{(E-{{\mathcal{H}}_{{\rm eff}}})}^{-1}}$ can then be identified with the Green function $\mathcal{G}({\boldsymbol{r}} ,{{{\boldsymbol{r}} }^{\prime }},E)$, whereas V corresponds to a projection onto the transverse wavefunctions in the leads such that [81]

$$\begin{eqnarray}&&\langle {\boldsymbol{r}} |\;{{b}_{c}}=\langle {\boldsymbol{r}} |\frac{1}{E-{{\mathcal{H}}_{{\rm eff}}}}{{V}_{c}}=\sqrt{\hbar v_{\parallel }^{\prime }}\int _{0}^{W}{\rm d}y^{\prime} \;\mathcal{G}({\boldsymbol{r}} ,(x^{\prime} ,y^{\prime} ),E)\;{{\Phi }_{n(c)}}(y^{\prime} ).\end{eqnarray} \tag{ 21 }$$

The integration is over the cross section at the beginning of the lead that contains the cth incoming mode, $v_{\parallel }^{\prime }=\frac{\hbar }{m}{{k}_{\parallel }}=\frac{\hbar }{m}\sqrt{{{k}^{2}}-{{(n\pi /W)}^{2}}}$ is the longitudinal velocity (m is the mass) and W is the lead width. The corresponding transverse wavefunction is

$$\begin{eqnarray}&&{{\Phi }_{n}}(y)=\sqrt{\frac{2}{W}}{\rm sin} \left( \frac{n\pi y}{W} \right).\end{eqnarray} \tag{ 22 }$$

Note that $1\leqslant c\leqslant M$ labels the modes in all leads, whereas n labels the modes in one particular lead, so the choice of the lead and n depend on c.

The semiclassical approximation for the internal part $\langle {\boldsymbol{r}} |\;{{b}_{c}}$ follows by evaluating the integral in (21) in stationary phase approximation. After writing the sine in (22) as sum of two complex exponentials, the stationary phase condition reads

$$\begin{eqnarray}&&\frac{\partial \mathcal{S}}{\partial y^{\prime} }=-{{p}_{y}}^{\prime} =-\frac{\bar{n}\hbar \pi }{W}\qquad \Longrightarrow \qquad {\rm sin} {{\theta }_{{\bar{n}}}}=\frac{\bar{n}\pi }{kW},\end{eqnarray} \tag{ 23 }$$

where $\bar{n}=\pm n$. This fixes the starting angle of the trajectories (${\rm sin} \theta ={{p}_{y}}^{\prime} /p^{\prime}$) entering the cavity. Performing the stationary phase approximation results in

$$\begin{eqnarray}&&\langle {\boldsymbol{r}} |{{b}_{c}}\approx \frac{1}{\sqrt{\hbar }}\mathop{\sum }\limits_{\gamma (c\to {\boldsymbol{r}} )}{{A}_{\gamma }}{{{\rm e}}^{\frac{{\rm i}}{\hbar }{{\mathcal{S}}_{\gamma }}}},\end{eqnarray} \tag{ 24 }$$

where the sum runs over all trajectories that enter the cavity with the angle fixed by (23) and end at ${\boldsymbol{r}}$. The amplitudes are given by

$$\begin{eqnarray}&&{{A}_{\gamma }}=\frac{-{\rm sign}(\bar{n})}{\sqrt{2vW{\rm cos} {{\theta }_{{\bar{n}}}}\;|{{({{M}_{\gamma }})}_{11}}|}}\;{\rm exp} \left( \frac{{\rm i}\bar{n}\pi y^{\prime} }{W}-\frac{{\rm i}\pi }{2}{{\mu }_{\gamma }} \right),\end{eqnarray} \tag{ 25 }$$

where ${{\mu }_{\gamma }}$ is the number of conjugate points for neighbouring trajectories with the same entrance angle.

The expressions (24) and (25) allow us to represent the elements of the time-delay matrix (13) in terms of the trajectories specified above. In particular, the semiclassical approximation for the Wigner time delay follows from (2) and (13) as

$$\begin{eqnarray}&&{{\tau }_{{\rm W}}}\approx \frac{1}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \mathop{\sum }\limits_{\gamma ,\gamma ^{\prime} (c\to {\boldsymbol{r}} )}{{A}_{\gamma }}A_{\gamma ^{\prime} }^{*}\;{{{\rm e}}^{\frac{{\rm i}}{\hbar }({{\mathcal{S}}_{\gamma }}-{{\mathcal{S}}_{\gamma ^{\prime} }})}},\end{eqnarray} \tag{ 26 }$$

where the integral is over the interior of the cavity. This is the new representation for ${{\tau }_{{\rm W}}}$ that serves as the starting point for the semiclassical calculations in this article.

We first apply (26) to calculate the mean time delay. The approximation sums over pairs of trajectories that contribute with highly oscillatory terms. After spectral averaging most terms can be neglected and the only remaining terms are from pairs of trajectories that are correlated. These pairs will be discussed in the following.

The trajectories involved in (26) are similar to those that occur in the semiclassics of the current density [82] which in turn is related to the survival probability [83, 84].

The leading contribution to the average time delay comes from the diagonal approximation where $\gamma ^{\prime} =\gamma$:

$$\begin{eqnarray}&&\langle {{\tau }_{{\rm W}}}\rangle \approx \left\langle \frac{1}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \mathop{\sum }\limits_{\gamma (c\to {\boldsymbol{r}} )}|{{A}_{\gamma }}{{|}^{2}} \right\rangle .\end{eqnarray} \tag{ 27 }$$

The evaluation of this expression requires a sum rule for the type of trajectories in (27); see also [86] for related sum rules. To obtain the sum rule, we fix one of the leads and consider the probability density that trajectories starting in the opening with angle θ and energy E will arrive after time T at point ${\boldsymbol{r}}$,

$$\begin{eqnarray}&&P({\boldsymbol{r}} ,T,\theta ,E)=\frac{1}{W}\int _{0}^{W}\delta ({\boldsymbol{r}} (T)-{\boldsymbol{r}} )\;{\rm d}y^{\prime} ,\end{eqnarray} \tag{ 28 }$$

where $y^{\prime}$ denotes the position in the opening, and the initial conditions of ${\boldsymbol{r}} (T)$ are determined by $y^{\prime}$, θ and E. The integral can be evaluated in local coordinates that are parallel and perpendicular to the trajectory,

$$\begin{eqnarray}&&{{P}_{\varepsilon }}({\boldsymbol{r}} ,T,\theta ,E)=\mathop{\sum }\limits_{\gamma }\frac{1}{vW{\rm cos} \theta \;|{{({{M}_{\gamma }})}_{11}}|}{{\delta }_{\varepsilon }}(T-{{T}_{\gamma }}).\end{eqnarray} \tag{ 29 }$$

This sum runs over all trajectories and has wild oscillations which can be damped by a conventional smoothing of the delta-function $\delta \to {{\delta }_{\varepsilon }}$.

In an open chaotic cavity with area A the asymptotic form of the probability density is ${{P}_{\varepsilon }}({\boldsymbol{r}} ,T,\theta ,E)\sim {{{\rm e}}^{-\mu T}}/A$ as $T\to \infty$. Here the exponential term describes the asymptotic escape of trajectories and the $1/A$ reflects the fact that each end point in the cavity is equally likely. We hence obtain the following sum rule

$$\begin{eqnarray}&&\mathop{\sum }\limits_{\gamma (c\to {\boldsymbol{r}} )}|{{A}_{\gamma }}{{|}^{2}}\;{{\delta }_{\varepsilon }}(T-{{T}_{\gamma }})\sim \frac{1}{A}{{{\rm e}}^{-\mu T}}.\end{eqnarray} \tag{ 30 }$$

Note that channel c corresponds to two angles θ which cancels a factor $1/2$ coming from (25). With this sum rule we can evaluate the diagonal approximation and obtain

$$\begin{eqnarray}&&\langle {{\tau }_{{\rm W}}}\rangle \approx \frac{1}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \;\frac{1}{A}\int {{{\rm e}}^{-\mu T}}{\rm d}T=\frac{1}{\mu }.\end{eqnarray} \tag{ 31 }$$

This is already the correct expression for the mean time delay. We will now show that off-diagonal contributions leave this result intact.

Off-diagonal contributions come from trajectories that have close self-encounters in which two or more stretches of the trajectory are almost parallel or anti-parallel [48–50, 58–60]. These trajectories have close neighbours that differ in the way in which the remaining longer parts of the trajectory, the links, are connected in the encounter regions.

The simplest example is a trajectory with one encounter as shown in figure 1(a). The neighbouring trajectory (dashed line) starts with the same angle θ and arrives at the same end point ${\boldsymbol{r}}$, but it traverses the loop in the opposite direction. For this reason these pairs exist only in systems with TRS. For the evaluation of the semiclassical contribution of these orbits we follow [59, 60].

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The encounter is described in a Poincaré surface of section in the encounter region. The relative distance of the two piercings of the original trajectory through the Poincaré surface is specified by coordinates s and u along the stable and unstable manifolds. This information is sufficient to determine the neighbouring trajectory and the action difference that is given by ${{\mathcal{S}}_{\gamma }}-{{\mathcal{S}}_{\gamma ^{\prime} }}=su$ in the linearized approximation. The duration of the encounter region is specified by requiring that the distance of the two stretches of the trajectory along the stable and unstable directions remain smaller than some small constant c, leading to

$$\begin{eqnarray}&&{{t}_{{\rm enc}}}(s,u)=\frac{1}{\lambda }{\rm ln} \frac{{{c}^{2}}}{|su|},\end{eqnarray} \tag{ 32 }$$

where λ is the Lyapunov exponent. The summation over the trajectory pairs is done by applying probabilistic arguments. Let ${{w}_{T}}(s,u)$ be the probability density in a chaotic system that a trajectory of long duration T has a close self-encounter that is specified by coordinates s and u. The contribution of the trajectory pairs to the average time delay can then be expressed as

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{1a}\rangle =\frac{1}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \int {\rm d}s\;{\rm d}u\mathop{\sum }\limits_{\gamma ,\gamma ^{\prime} (c\to {\boldsymbol{r}} )}|{{A}_{\gamma }}{{|}^{2}}\;{{w}_{T}}(s,u)\;{{{\rm e}}^{\frac{{\rm i}}{\hbar }su}}.\end{eqnarray} \tag{ 33 }$$

The density ${{w}_{T}}(s,u)$ is given by an integral over the first two link durations

$$\begin{eqnarray}&&{{w}_{T}}(s,u)=\int _{0}^{T-2{{t}_{{\rm enc}}}}{\rm d}{{t}_{1}}\int _{0}^{T-2{{t}_{{\rm enc}}}-{{t}_{1}}}{\rm d}{{t}_{2}}\ \frac{1}{\Omega \;{{t}_{{\rm enc}}}(s,u)},\end{eqnarray} \tag{ 34 }$$

where Ω is the volume of the surface of constant energy in phase space. Applying the sum rule (30) and changing the integral over the orbit length T to an integral over the third link duration results in

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{1a}\rangle =\frac{1}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \frac{1}{A}\int _{0}^{\infty }{\rm d}{{t}_{1}}\;{\rm d}{{t}_{2}}\;{\rm d}{{t}_{3}}\int {\rm d}s\;{\rm d}u\;\frac{{{{\rm e}}^{\frac{{\rm i}}{\hbar }su}}\;{{{\rm e}}^{-\mu ({{t}_{1}}+{{t}_{2}}+{{t}_{3}}+{{t}_{{\rm enc}}}(s,u))}}}{\Omega \;{{t}_{{\rm enc}}}(s,u)}.\end{eqnarray} \tag{ 35 }$$

Equation (35) contains a small correction to the sum rule (30) that is necessary when applied to trajectories with self-encounters. Namely, the trajectory time T in the exponent has to be replaced by the exposure time that counts each encounter duration only once. The reason is that if a trajectory does not escape during the first traversal of an encounter region, it will not escape during subsequent traversals of this region.

The integral over s and u can now be evaluated by noting that the only contribution that survives in the semiclassical limit is one where the encounter duration ${{t}_{{\rm enc}}}$ in the denominator is exactly cancelled by an encounter duration in the numerator. In other words, after expanding ${\rm exp} (-\mu {{t}_{{\rm enc}}})$ in a Taylor series the only term that survives is the linear term in ${{t}_{{\rm enc}}}$. With $\int {\rm d}s\;{\rm d}u{\rm exp} ({\rm i}su/\hbar )=2\pi \hbar$, we finally obtain

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{1a}\rangle ={{\left( \frac{1}{\mu } \right)}^{3}}\left( -\frac{\mu }{{{T}_{{\rm H}}}} \right)=-\frac{{{T}_{{\rm H}}}}{{{M}^{2}}},\end{eqnarray} \tag{ 36 }$$

where we have used ${{T}_{{\rm H}}}=\Omega /2\pi \hbar$ and $\mu =M/{{T}_{{\rm H}}}$.

The contribution (36) would lead to a deviation from the correct result (27). There is, however, a further contribution of the same order. It arises from the so-called one-leg loops, which are correlated trajectories that have an end point in an encounter region as in figure 1(b). These type of correlations do not occur in transport problems, but they arise when trajectories have one or both their end points in the cavity as for example in problems involving the survival probability, the current density or the fidelity [82–85].

The encounter regions of one-leg loops require a different treatment than the usual encounters. It can be shown, however, that this difference can effectively be taken into account by adding a further factor of ${{t}_{{\rm enc}}}$ in the integral over s and u [82]. So the contribution of the one-leg loop trajectories differs from (36) by a missing integration over the third link time t₃ and an additional factor of ${{t}_{{\rm enc}}}$,

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{1b}\rangle =\frac{1}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \frac{1}{A}\int _{0}^{\infty }{\rm d}{{t}_{1}}\;{\rm d}{{t}_{2}}\int {\rm d}s\;{\rm d}u\;\frac{{{{\rm e}}^{\frac{{\rm i}}{\hbar }su}}\;{{{\rm e}}^{-\mu ({{t}_{1}}+{{t}_{2}}+{{t}_{{\rm enc}}}(s,u))}}}{\Omega }.\end{eqnarray} \tag{ 37 }$$

The integrals are then evaluated similarly as before and result in

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{1b}\rangle ={{\left( \frac{1}{\mu } \right)}^{2}}\left( \frac{1}{{{T}_{{\rm H}}}} \right)=\frac{{{T}_{{\rm H}}}}{{{M}^{2}}}.\end{eqnarray} \tag{ 38 }$$

This contribution cancels exactly the contribution (36).

One could further consider shrinking the first link in figure 1(b) so that the encounter moves into the lead creating a 'coherent back-scattering' type of diagram. However the freedom of how much of the encounter box overlaps with the lead provides a further factor of the encounter time. In calculating the semiclassical contribution, the integrand then becomes at least linear in ${{t}_{{\rm enc}}}$ and the integral vanishes in the semiclassical limit. In general our attention may simply be restricted to diagrams with at most one end point in each encounter [82–85].

For higher-order corrections one considers trajectories with arbitrarily many self-encounters. These self-encounters can involve two or more stretches of a trajectory that are almost parallel or anti-parallel, where the latter case requires TRS. One speaks of an l-encounter if it involves l stretches of a trajectory. The types of a trajectoryʼs encounters are detailed in a vector ${\boldsymbol{v}}$ whose lth component v_l specifies the number of l-encounters. The total number of encounters is thus $V={{\sum }_{l}}{{v}_{l}}$, and the total number of stretches in all encounter regions is $L={{\sum }_{l}}l\;{{v}_{l}}$.

Trajectories with self-encounters have close neighbouring trajectories that differ in the way in which the links are connected in the encounter regions. For a given vector ${\boldsymbol{v}}$ there are many different configurations in which the encounters and the reconnections can be arranged along a trajectory pair. The number of these structures or families is denoted by $\mathcal{N}({\boldsymbol{v}} )$. The action difference of a trajectory pair can again be determined in terms of the separation of the trajectory stretches in the encounter regions. There are now altogether $L-V$ pairs of coordinates in the stable and unstable directions ($l-1$ pairs for each l-encounter). These coordinates are combined into vectors ${\boldsymbol{s}}$ and ${\boldsymbol{u}}$, and in the linearized approximation one has ${{\mathcal{S}}_{\gamma }}-{{\mathcal{S}}_{\gamma ^{\prime} }}={\boldsymbol{su}}$.

The definition of the encounter duration (32) is generalized to arbitrary l-encounters by requiring that the separations of the l trajectory stretches remain smaller than some constant c in the stable and unstable directions,

$$\begin{eqnarray}&&t_{{\rm enc}}^{\alpha }({\boldsymbol{s}} ,{\boldsymbol{u}} )=\frac{1}{\lambda }{\rm ln} \frac{{{c}^{2}}}{{{{\rm max} }_{j}}|{{s}_{\alpha ,j}}|\times {{{\rm max} }_{j}}|{{u}_{\alpha ,j}}|},\end{eqnarray} \tag{ 39 }$$

where α labels the encounters, $1\leqslant \alpha \leqslant V$. One applies again probabilistic arguments to replace the summation over trajectory pairs in (26) by one over self-encounters,

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{{\boldsymbol{v}} a}\rangle =\frac{\mathcal{N}({\boldsymbol{v}} )}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \int {\rm d}{\boldsymbol{s}} \;{\rm d}{\boldsymbol{u}} \mathop{\sum }\limits_{\gamma ,\gamma ^{\prime} (c\to {\boldsymbol{r}} )}|{{A}_{\gamma }}{{|}^{2}}\;{{w}_{T}}({\boldsymbol{s}} ,{\boldsymbol{u}} )\;{{{\rm e}}^{\frac{{\rm i}}{\hbar }{\boldsymbol{su}} }},\end{eqnarray} \tag{ 40 }$$

where ${{w}_{T}}({\boldsymbol{s}} ,{\boldsymbol{u}} )$ is the probability density that a trajectory of long duration T has self-encounters that are specified by the vector ${\boldsymbol{v}}$ and the separations ${\boldsymbol{s}}$ and ${\boldsymbol{u}}$. This density can be expressed by an integral over the first L link durations

$$\begin{eqnarray}&&{{w}_{T}}({\boldsymbol{s}} ,{\boldsymbol{u}} )=\int _{0}^{T-{{t}_{{\rm enc}}}}{\rm d}{{t}_{1}}\;\ldots \int _{0}^{T-{{t}_{{\rm enc}}}-{{t}_{1}}-\ldots -{{t}_{L-1}}}{\rm d}{{t}_{L}}\;\frac{1}{{{\Omega }^{L-V}}\;\mathop{\prod }\limits_{\alpha }t_{{\rm enc}}^{\alpha }({\boldsymbol{s}} ,{\boldsymbol{u}} )}.\quad \end{eqnarray} \tag{ 41 }$$

At a final step, the sum rule (30) is applied. As earlier in (35), it requires replacing the time in the exponent by the exposure time counting all encounter times only once. After replacing the integral over the trajectory time T by an integral over the final link duration, one obtains

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{{\boldsymbol{v}} a}\rangle =\frac{\mathcal{N}({\boldsymbol{v}} )}{M}\mathop{\sum }\limits_{c=1}^{M}\ \int {{{\rm d}}^{2}}{\boldsymbol{r}} \frac{1}{A}\ \ {{\left( \int _{0}^{\infty }{\rm d}t\;{{{\rm e}}^{-\mu t}} \right)}^{L+1}}\int {\rm d}{\boldsymbol{s}} \;{\rm d}{\boldsymbol{u}} \;\frac{{{{\rm e}}^{\frac{{\rm i}}{\hbar }{\boldsymbol{su}} }}\;{{{\rm e}}^{-\mu \mathop{\sum }\limits_{\alpha }t_{{\rm enc}}^{\alpha }({\boldsymbol{s}} ,{\boldsymbol{u}} )}}}{{{\Omega }^{L-V}}\;\mathop{\prod }\limits_{\alpha }t_{{\rm enc}}^{\alpha }({\boldsymbol{s}} ,{\boldsymbol{u}} )}.\end{eqnarray} \tag{ 42 }$$

In the integrals over the ${\boldsymbol{s}}$ and ${\boldsymbol{u}}$ coordinates the only terms that survive the semiclassical limit are those where all the encounter times in the denominator are exactly cancelled by corresponding encounter times in the numerator. The leading contribution thus again arises from the linear terms in the Taylor expansion of ${\rm exp} (-\mu {{\sum }_{\alpha }}t_{{\rm enc}}^{\alpha }({\boldsymbol{s}} ,{\boldsymbol{u}} ))$. Using $\int {\rm d}{\boldsymbol{s}} \;{\rm d}{\boldsymbol{u}} {\rm exp} ({\rm i}{\boldsymbol{su}} /\hbar )=2\pi \hbar$, the final result reads

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{{\boldsymbol{v}} a}\rangle =\mathcal{N}({\boldsymbol{v}} )\ \ \frac{1}{{{\mu }^{L+1}}}\;\frac{{{(-\mu )}^{V}}}{T_{{\rm H}}^{L-V}}=\mathcal{N}({\boldsymbol{v}} )\;{{(-1)}^{V}}\;\frac{{{T}_{{\rm H}}}}{{{M}^{L-V+1}}}.\end{eqnarray} \tag{ 43 }$$

As in the transport problem [59, 60], one can identify simple diagrammatic rules from this result. Each link contributes by a factor $1/M$, and each encounter contributes a factor $(-M)$. The Heisenberg times cancel up to one. Note further that the sum over the channels gives a factor of M that cancels the prefactor $1/M$ in (2).

As we have seen in section 2.3, there are additional trajectory correlations for the new semiclassical representation (26) of the Wigner time delay due to the one-leg loops that do not occur in the transport problem. In fact, for every trajectory configuration or structure in the above calculation there is a corresponding configuration where the end point is now inside the last encounter. The semiclassical calculation can be easily modified to obtain these additional contributions. The difference to (42) is that the integral over the final link duration is missing, and the integrals over the ${\boldsymbol{s}}$ and ${\boldsymbol{u}}$ coordinates contain an additional factor of the last encounter time [82]. These modifications lead to

$$\begin{eqnarray}&&\langle \tau _{{\rm W}}^{{\boldsymbol{v}} b}\rangle =\mu \left( -\frac{1}{\mu } \right)\langle \tau _{{\rm W}}^{{\boldsymbol{v}} a}\rangle =-\langle \tau _{{\rm W}}^{{\boldsymbol{v}} a}\rangle .\end{eqnarray} \tag{ 44 }$$

As expected all off-diagonal terms cancel. This calculation allows us to extend the diagrammatic rules: encounters that include an end point simply contribute a factor of 1.

The previous section has shown one main advantage of the new semiclassical approach. It results in the simple diagrammatic rules that are similar to the ones in transport problems [60], thus strongly simplifying the calculations. As in transport, one can generalize the diagrammatic rules to higher moments. We sketch this by considering the moments of the proper time-delays defined as

$$\begin{eqnarray}&&{{m}_{n}}=\frac{1}{M}\left\langle {\rm Tr}\left[ {{{\rm Q}}^{n}} \right] \right\rangle .\end{eqnarray} \tag{ 45 }$$

The representation (13) and the semiclassical approximation (24) lead to

$$\begin{eqnarray}&&{{m}_{n}}=\frac{1}{M}\sum 1,\ldots ,{{i}_{n}}=1Mi\int {{{\rm d}}^{2}}{{{\boldsymbol{r}} }_{1}}\ldots {{{\rm d}}^{2}}{{{\boldsymbol{r}} }_{n}}\mathop{\sum }\limits_{{\boldsymbol{ \gamma }} ,{\boldsymbol{ \gamma }} ^{\prime} }\left( \mathop{\prod }\limits_{j=1}^{n}{{A}_{{{\gamma }_{j}}}}A_{\gamma _{j}^{\prime }}^{*} \right){{{\rm e}}^{{\rm i}({{\mathcal{S}}_{{\boldsymbol{ \gamma }} }}-{{\mathcal{S}}_{{\boldsymbol{ \gamma }} ^{\prime} }})/\hbar }}.\end{eqnarray} \tag{ 46 }$$

Here ${{i}_{1}},\ldots ,{{i}_{n}}$ label the n incoming channels and ${{{\boldsymbol{r}} }_{1}},\ldots ,{{{\boldsymbol{r}} }_{n}}$ denote the n final points. The symbols ${\boldsymbol{ \gamma }} =\{{{\gamma }_{1}},\ldots ,{{\gamma }_{n}}\}$ and ${\boldsymbol{ \gamma }} ^{\prime} =\{\gamma _{1}^{\prime },\ldots ,\gamma _{n}^{\prime }\}$ stand for two sets of n trajectories, where ${{\gamma }_{j}}$ goes from channel i_j to the point ${{{\boldsymbol{r}} }_{j}}$, and $\gamma _{j}^{\prime }$ from ${{i}_{j+1}}$ to ${{{\boldsymbol{r}} }_{j}}$ (we identify ${{i}_{n+1}}$ with i₁). The total actions of the two sets are ${{\mathcal{S}}_{{\boldsymbol{ \gamma }} }}={{\sum }_{j}}{{\mathcal{S}}_{{{\gamma }_{j}}}}$ and ${{\mathcal{S}}_{{\boldsymbol{ \gamma }} ^{\prime} }}={{\sum }_{j}}{{\mathcal{S}}_{\gamma _{j}^{\prime }}}$, respectively.

Dealing with the correlated trajectories that survive the spectral averaging in (46) now involves two trajectory sets, ${\boldsymbol{ \gamma }}$ and ${\boldsymbol{ \gamma }} ^{\prime}$. The trajectories in the set ${\boldsymbol{ \gamma }}$ have encounters in which two or more trajectory stretches are almost parallel or anti-parallel. The set ${\boldsymbol{ \gamma }} ^{\prime}$ follows the set ${\boldsymbol{ \gamma }}$ very closely along the links, but differs in the way those are connected in the encounter regions. One can then replace the double sum over ${\boldsymbol{ \gamma }}$ and ${\boldsymbol{ \gamma }} ^{\prime}$ by a single sum over ${\boldsymbol{ \gamma }}$ plus an integral over a probability density for the self-encounters. The result can be split into contributions from links and encounters according to the following diagrammatic rules:

• The summation over each incoming channel gives a factor of M.
• Each link contributes a factor of $1/M$.
• Each encounter gives a factor of $(-M)$, unless it contains an end point.
• An encounters contributes a factor of 1 if it contains one end point, and a factor of 0 if it contains more than one end point.

There is furthermore an overall factor of T_Hⁿ, and a factor of $1/M$ from (45). The rules for the encounters follow from the fact that each end point inside an encounter provides an additional factor of the encounter time.

As an example, we discuss the leading order contribution to the second moment m₂. In analogy to transport problems we denote the jth end point by o_j instead of ${{{\boldsymbol{r}} }_{j}}$. The simplest trajectory configuration with an encounter is the one that is schematically shown in figure 2(a). The two trajectories belonging to ${\boldsymbol{ \gamma }}$ are shown by the full lines and go from channel i₁ to end point o₁ and from i₂ to o₂, respectively. They have one encounter that is indicated by the circle. The neighbouring trajectories belonging to ${\boldsymbol{ \gamma }} ^{\prime}$ (dashed lines) go from i₁ to o₂ and from i₂ to o₁, respectively. According to the above rules we obtain the following contribution of this configuration to m₂

$$\begin{eqnarray}&&\frac{-{{M}^{3}}}{{{M}^{4}}}\;\frac{T_{{\rm H}}^{2}}{M}=-\frac{1}{{{\mu }^{2}}}.\end{eqnarray} \tag{ 47 }$$

This contains $(-M)$ from the encounter, M² from the incoming channels, $1/{{M}^{4}}$ from the links, and the overall factor $T_{{\rm H}}^{2}/M$.

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The simpler configuration in figure 2(b) does not usually contribute to m₂ since the trajectories of ${\boldsymbol{ \gamma }} ^{\prime}$ (dashed lines) donʼt connect the correct initial and final points. Note, however, that this configuration is possible if the incoming channels coincide. Then one obtains the configuration in figure 3(a). It contributes at the same order as (47) since it has two links, one encounter and one incoming channel less. There is the further possibility that one of the end points is in the encounter as in figures 3(b) and (c), which again contribute at the same order. All in all the result is

$$\begin{eqnarray}&&\left( \frac{-{{M}^{3}}}{{{M}^{4}}}+\frac{M}{{{M}^{2}}}+2\frac{{{M}^{2}}}{{{M}^{3}}} \right)\;\frac{T_{{\rm H}}^{2}}{M}=2\frac{T_{{\rm H}}^{2}}{{{M}^{2}}}=\frac{2}{{{\mu }^{2}}}.\end{eqnarray} \tag{ 48 }$$

This is indeed the leading order term of m₂ in systems with or without TRS.

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We have obtained it by considering figures 2(a) and 3(a)–(c) as different trajectory configurations that all contribute to the moment. There is an even simpler and alternative point of view in which one considers all the contributions in (48) to come from figure 2(a). The diagrams in figures 3(a)–(c) are considered to be limiting cases of figure 2(a) where either the encounter moves into the incoming lead or one of the end points moves into the encounter. These limiting cases can be included by changing the contribution of the encounter.

For the description we adopt the language from transport and call the initial points of ${\boldsymbol{ \gamma }}$ i-leaves and the final points o-leaves. For the calculation of the moments m_n we then have to take into account the following cases:

• Encounters of size l can move into the incoming lead when connected directly to l i-leaves.
• The o-leaves can be moved into the encounter they are connected to, but each encounter can only take one at a time.

In terms of the semiclassical contributions, in both of these situations we change the rule for the affected encounter to include these cases. In the first case we lose l links, $(l-1)$ channel summations and the $(-M)$ from the encounter, and in the second case we lose one link and the $(-M)$ from the encounter. The contribution of encounter α is hence changed to $M(-1+\delta +{{s}_{\alpha }})$ with δ being 1 when the encounter can move into the lead (and 0 otherwise) and ${{s}_{\alpha }}$ the number of o-leaves attached.

In our example, if we change the encounter contribution for figure 2(a) in (47) from $(-M)$ to $M(-1+1+2)=2M$ then we obtain the same result as in (48). One can also easily see that the off-diagonal contributions for the average time delay all vanish, because the last encounter must have $\delta =0$ (l is at least 2 and there is only 1 i-leaf) and s_α = 1 since it is connected to the outgoing channel. The product of contributions is then 0.

We will show below that these rules can be employed systematically to calculate the second moments of the proper time-delays and the variance of the Wigner time delay, and they can as well be incorporated into the graphical framework [62, 63, 77, 87–90] to give moment generating functions. Previous approaches to the time delay involved including an energy dependence so that calculating the semiclassical contribution of any diagram required knowledge of its complete structure. Here instead the semiclassical contributions are much closer to standard transport moments where only the difference in the number of links and encounters matters. The only additional complication is that now some information about the position of the o-leaves is necessary. Determining this is however much less demanding than treating full energy dependence.

If a periodic orbit pair is cut along one of its links, it can be deformed into a pair of scattering trajectories which contributes to the first transport moment, the average conductance [59]. The cut link becomes two so that conductance diagrams have $L+1$ links. The number of structures $N({\boldsymbol{v}} )$ of the scattering trajectories is the same as the one for periodic orbits. Since encounters contribute with a minus sign, terms in a ${{M}^{-1}}$ expansion of the conductance can be related to the sum [59]

$$\begin{eqnarray}&&{{C}_{K}}=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}N({\boldsymbol{v}} ),\qquad K\geqslant 1.\end{eqnarray} \tag{ 49 }$$

This is the relevant quantity to evaluate. Without TRS, one can show C_K = 0 for $K\geqslant 1$ so only the diagonal pair is important [59]. It can be included in the formalism by defining $L=V=0$ and ${{C}_{0}}=1$ for it.

For the second transport moments (the shot noise and the conductance variance), the semiclassical diagrams involve four trajectories. These trajectory quadruplets can be divided into two groups: d-quadruplets and x-quadruplets. Consider a pair of trajectories ${{\gamma }_{1}}$ and ${{\gamma }_{2}}$ connecting channel i₁ to o₁ and i₂ to o₂ respectively. If the partner trajectory ${{\gamma }_{1}}^{\prime}$ also connects channel i₁ to o₁ then we have a d-quadruplet. Otherwise if ${{\gamma }_{1}}^{\prime}$ connects i₁ to o₂ we have an x-quadruplet. The remaining partner trajectory ${{\gamma }_{2}}^{\prime}$ must connect i₂ to the other outgoing channel. For example, the diagram in figure 2(a) is an x-quadruplet of trajectories meeting at a single 2-encounter while the diagram in figure 2(b) is the simplest d-quadruplet. It is made up of two independent links.

Denote the number of x-quadruplet diagrams corresponding to a vector ${\boldsymbol{v}}$ by ${{N}_{x}}({\boldsymbol{v}} )$, and the number of d-quadruplet diagrams by ${{N}_{d}}({\boldsymbol{v}} )$, then the conductance variance and the shot noise can be related to the sums

$$\begin{eqnarray}&&{{D}_{K}}=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}{{N}_{d}}({\boldsymbol{v}} ),\end{eqnarray} \tag{ 50 }$$

$$\begin{eqnarray}&&{{X}_{K}}=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}{{N}_{x}}({\boldsymbol{v}} ).\end{eqnarray} \tag{ 51 }$$

These sums can again be related to periodic orbit pairs. Without TRS, as detailed in [60], cutting periodic orbit pairs twice along links leads to a d-type quadruplet. In general this can be done in $L(L+1)$ ways since we may cut the same link twice (and the order matters). Note that the final d-quadruplet will have $L+2$ links. On the other hand, an x-quadruplet can be created by cutting out an entire 2-encounter from a periodic orbit pair. In general this can be done in 2v₂ ways and the final x-quadruplet will have one 2-encounter fewer and the same number of links as the periodic orbit pair.

With TRS, the cutting is more complicated, but in both symmetry cases the quantities D_K and X_K can be related to the following sums

$$\begin{eqnarray}&&{{A}_{K}}=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}(L({\boldsymbol{v}} )+1)N({\boldsymbol{v}} ),\end{eqnarray} \tag{ 52 }$$

$$\begin{eqnarray}&&{{B}_{K}}=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}\frac{2{{v}_{2}}}{L({\boldsymbol{v}} )}N({\boldsymbol{v}} ).\end{eqnarray} \tag{ 53 }$$

Without TRS, both are equal to 1 for even K and 0 otherwise; and we have D_K = A_K while ${{X}_{K}}=-{{B}_{K+1}}$ [60]. We further have ${{D}_{0}}=1$ and ${{X}_{0}}=0$.

The sums X_K are known, but for the time delays we need to make a further distinction because of the different rules for the o-leaves. We divide the x-quadruplets into two groups: we denote those where both final points are connected to the same encounter by $\tilde{x}$, and those where they are connected to different encounters by $x^{\prime}$. For each vector ${\boldsymbol{v}}$ we count the structures in each group as ${{N}_{\tilde{x}}}({\boldsymbol{v}} )$ and ${{N}_{x^{\prime} }}({\boldsymbol{v}} )$, with

$$\begin{eqnarray}&&{{N}_{x}}({\boldsymbol{v}} )={{N}_{\tilde{x}}}({\boldsymbol{v}} )+{{N}_{x^{\prime} }}({\boldsymbol{v}} ),\end{eqnarray} \tag{ 54 }$$

and we define the sums

$$\begin{eqnarray}&&{{\tilde{X}}_{K}}=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}{{N}_{\tilde{x}}}({\boldsymbol{v}} ),\end{eqnarray} \tag{ 55 }$$

$$\begin{eqnarray}&&X_{K}^{\prime }=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}{{N}_{x^{\prime} }}({\boldsymbol{v}} )={{X}_{K}}-{{\tilde{X}}_{K}}.\end{eqnarray} \tag{ 56 }$$

Only the ${{\tilde{X}}_{K}}$ part will contribute to the second moment of the time delays. This follows from the rules in section 2.5. If the two final points are connected to different encounters then at least one of these encounters cannot be moved into the incoming lead and contributes with a factor of $M(-1+\delta +{{s}_{\alpha }})=0$.

We likewise partition the d-quadruplets and define the sums

$$\begin{eqnarray}&&{{\tilde{D}}_{K}}=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}{{N}_{\tilde{d}}}({\boldsymbol{v}} ),\end{eqnarray} \tag{ 57 }$$

$$\begin{eqnarray}&&D_{K}^{\prime }=\mathop{\sum }\limits_{{\boldsymbol{v}} }^{L-V=K}{{(-1)}^{V}}{{N}_{d^{\prime} }}({\boldsymbol{v}} ),\end{eqnarray} \tag{ 58 }$$

where again we are only interested in the ${{\tilde{D}}_{K}}$ part. However, along with ${{N}_{\tilde{d}}}({\boldsymbol{v}} )$ and ${{N}_{d^{\prime} }}({\boldsymbol{v}} )$, we also need to take into account a third group where one or both end points are not connected to an encounter at all. These diagrams involve a direct link between an incoming channel and an end point, while the other trajectory pair can be any arbitrary conductance (first moment) diagram. We then have

$$\begin{eqnarray}&&{{D}_{K}}={{\tilde{D}}_{K}}+D_{K}^{\prime }+2{{C}_{K}}-{{\delta }_{K,0}},\end{eqnarray} \tag{ 59 }$$

where the last term is a correction to avoid overcounting the diagram in figure 2(b) made up of two direct links.

To obtain some information about ${{\tilde{D}}_{K}}$ and ${{\tilde{X}}_{K}}$ we need to build a recursion relation between them. We start by considering ways in which an $\tilde{d}$-quadruplet can be obtained from other diagrams. For a $\tilde{d}$-quadruplet both end points are connected to the same encounter (of size l say), and we can add a 2-encounter at the end. Now we have an x-quadruplet whose final 2-encounter has two links connected to the same l-encounter. Next we shrink the connecting links and merge the two encounters. If l = 2 both 2-encounters vanish and we have an arbitrary x-quadruplet with one encounter and two links fewer than the original $\tilde{d}$-quadruplet. This process is depicted in figure 4. If $l\gt 2$ there are two possibilities: A link could separate from the encounter which becomes one smaller ($l\to l-1$) to leave an arbitrary x-quadruplet with the same number of encounters but one link fewer than before. An example is given in figure 5. Alternatively the encounter could break into two separate encounters each connected to an outgoing channel hence giving a $x^{\prime}$-quadruplet. This has one more encounter than the original $\tilde{d}$-quadruplet and the same number of links.

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Zoom In Zoom Out Reset image size

Reversing the three processes, we can obtain any $\tilde{d}$-quadruplet in exactly one of the following ways. We may add a 2-encounter to an arbitrary x-quadruplet (figure 4). Alternatively for any x-quadruplet, we may join the link before an end point into the last encounter before the other end point (figure 5). We may also join the final distinct encounters of any $x^{\prime}$-quadruplet by pulling out a 2-encounter to create a $\tilde{d}$-quadruplet. Accounting for the minus sign from each encounter and the change in L and V we arrive at the relation

$$\begin{eqnarray}\begin{array}{rcl} {{\tilde{D}}_{K}} & = & -{{X}_{K-1}}+2{{X}_{K-1}}-X_{K-1}^{\prime } \\ {} & = & {{\tilde{X}}_{K-1}}, \\ \end{array}\end{eqnarray} \tag{ 60 }$$

since $X_{K}^{\prime }={{X}_{K}}-{{\tilde{X}}_{K}}$. Another way to look at this is the following: one can describe an l-encounter as a cyclic permutation $({{a}_{1}}\;\ldots {{a}_{l}})$ where the a_j are labels corresponding to the order encounter stretches are visited in the entire diagram [50]. If stretches a_i and a_j are connected to the outgoing channel, then adding a 2-encounter and shrinking the intervening links corresponds to multiplying (on the left) by $({{a}_{i}}\;{{a}_{j}})$. This breaks the l-cycle into a k and $l-k$ cycle. If k or $l-k$ are 1 then a link is separated from the encounter (leaving only a link if l = 2) otherwise the encounter breaks into two.

We can repeat the same process of adding a 2-encounter and shrinking the adjoining links but starting with an $\tilde{x}$-quadruplet. This again leads to three cases of d-quadruplets with a lower value of $L-V$: one with a 2-encounter removed, one with an $l\gt 2$ encounter reduced by 1 and one where the l encounter breaks into two separate ones. Reversing the steps, for any $d^{\prime}$-quadruplet we can connect the two distinct encounters before the two end points by pulling out an 2-encounter (which we then remove) to give a contribution of $-D_{K-1}^{\prime }$ to ${{\tilde{X}}_{K}}$. The minus sign derives from the final diagram having one encounter less. Then, for any d-quadruplet where o₂ is connected to an encounter, we can join the link to o₁ into the encounter giving a contribution of ${{D}_{K-1}}-{{C}_{K-1}}$. Here we remove the cases where o₂ connects directly to i₂. Swapping the roles of o₂ and o₁ gives a factor of 2. Finally we may connect the final links of any d-quadruplet with a 2-encounter creating an $\tilde{x}$-quadruplet with an additional encounter (and two extra links) and hence a contribution of $-{{D}_{K-1}}$ to ${{\tilde{X}}_{K}}$. Putting it all together,

$$\begin{eqnarray}\begin{array}{rcl} {{\tilde{X}}_{K}} & = & -{{D}_{K-1}}+2{{D}_{K-1}}-2{{C}_{K-1}}-D_{K-1}^{\prime } \\ {} & = & {{\tilde{D}}_{K-1}}-{{\delta }_{K,1}}, \\ \end{array}\end{eqnarray} \tag{ 61 }$$

using (59).

Since the only diagram for K = 1 is in figure 2(a), we have ${{\tilde{X}}_{1}}=-1$ and ${{\tilde{D}}_{1}}=0$, so that ${{\tilde{X}}_{K}}=-1$ for odd K while ${{\tilde{D}}_{K}}=-1$ for even K and both are 0 otherwise. (Both are also 0 for K = 0).

We also need to consider the cases when an encounter of a diagram can be moved into an incoming lead. This is only possible if both incoming channels are connected to the same 2-encounter. If, for example, an $\tilde{x}$-quadruplet starts with such a 2-encounter, then this encounter can be cut out (by moving the incoming channels to after the encounter) leaving a $\tilde{d}$-quadruplet with a value of $L-V$ which is one smaller. (The only exception is figure 2(a) where the removal of the 2-encounter leads to the d-quadruplet in figure 2(b)). Reversing the process, all such $\tilde{x}$-quadruplets can be built by adding a 2-encounter to the front of any $\tilde{d}$-quadruplet (or figure 2(b)). Similarly, one can interchange the roles of the $\tilde{x}$- and $\tilde{d}$-quadruplets and create any $\tilde{d}$-quadruplet starting with a 2-encounter by adding a 2-encounter to the front of any $\tilde{x}-$ quadruplet with a value of $L-V$ which is smaller by one. If we denote by an undertilde quadruplets with an encounter that can be moved into a lead then we have

$$\begin{eqnarray}&&{\underset{\sim}{\tilde{X}}_{K}}=-{{\tilde{D}}_{K-1}},\quad {\underset{\sim}{\tilde{D}}_{K}}=-{{\tilde{X}}_{K-1}},\quad K\gt 1,\end{eqnarray} \tag{ 62 }$$

and ${\underset{\sim}{\tilde{X}}_{1}}=-{{D}_{0}}$.

We now return to the calculation of m₂. It is based on the $\tilde{x}$-quadruplets. For each quadruplet we have $L+2$ links, V encounters and a factor of M for each channel. Both end points can also be moved into the last encounter, and one encounter can possibly be moved into an incoming lead, giving a combined contribution of

$$\begin{eqnarray}&&\frac{{{M}^{V+2}}}{{{M}^{L+2}}}{{(-1)}^{V-2}}(-1+\delta )\;(-1+2)=-\frac{{{(-1)}^{V}}}{{{M}^{L-V}}}(1-\delta ),\quad {\rm if}\ \ \ L-V\gt 1,\end{eqnarray} \tag{ 63 }$$

where $\delta =1$ if an encounter can move into a lead, and 0 otherwise. The case $L-V=1$ is different, because then we have the diagram in figure 2(a) where the same encounter can receive the end points and move into a lead. Then the brackets $(-1+\delta )(-1+2)$ are replaced by $(-2)$. When we sum over all diagrams using the results of section 3.3, using (62) for the $\delta =1$ case (and further multiply by $T_{{\rm H}}^{2}/M=M/{{\mu }^{2}}$), we have

$$\begin{eqnarray}\begin{array}{rcl} {{\mu }^{2}}{{m}_{2}} & = & -2{{\tilde{X}}_{1}}-\mathop{\sum }\limits_{K=2}^{\infty }\frac{{{\tilde{X}}_{K}}}{{{M}^{K-1}}}-\mathop{\sum }\limits_{K=2}^{\infty }\frac{{{\tilde{D}}_{K-1}}}{{{M}^{K-1}}} \\ {} & = & 2\mathop{\sum }\limits_{k=0}^{\infty }\frac{1}{{{M}^{2k}}}=\frac{2{{M}^{2}}}{{{M}^{2}}-1}, \\ \end{array}\end{eqnarray} \tag{ 64 }$$

where the term $-2{{\tilde{X}}_{1}}$ gives the leading order result in (48). The final expression in (64) is exactly the RMT result (18) at $\beta =2$.

This task requires the second moment of a different type, $\langle \tau _{{\rm W}}^{2}\rangle =\frac{1}{{{M}^{2}}}\langle {{[{\rm Tr}Q]}^{2}}\rangle$. The trajectories belonging to ${\boldsymbol{ \gamma }} ^{\prime}$ connect now i₁ to o₁ and i₂ to o₂ and hence the d-quadruplets are the relevant diagrams. We note that for computing the variance we only need to consider connected diagrams, since the remaining diagrams merely cancel the average time delay squared (in fact, the first d-quadruplet in figure 2(b) does this). Compared to the calculation for the moment m₂ in the previous section, the role of the $\tilde{x}$- and $\tilde{d}$-quadruplets are simply reversed, the case K = 1 does not contribute, and due to the normalization of the variance we donʼt have to multiply by $M/{{\mu }^{2}}$. The variance of the Wigner time delay is therefore given by the sum

$$\begin{eqnarray}\begin{array}{rcl} \operatorname{var}({{\tau }_{{\rm W}}})=\frac{1}{\bar{\tau }_{W}^{2}}\left\langle {{({{\tau }_{W}}-{{{\bar{\tau }}}_{W}})}^{2}} \right\rangle & = & -\mathop{\sum }\limits_{K=2}^{\infty }\frac{{{\tilde{D}}_{K}}}{{{M}^{K}}}-\mathop{\sum }\limits_{K=2}^{\infty }\frac{{{\tilde{X}}_{K-1}}}{{{M}^{K}}} \\ {} & = & 2\mathop{\sum }\limits_{k=1}^{\infty }\frac{1}{{{M}^{2k}}}=\frac{2}{{{M}^{2}}-1}. \\ \end{array}\end{eqnarray} \tag{ 65 }$$

This result fully agrees with RMT, as can be seen by setting $\beta =2$ in (4).

As already discussed in the beginning of section 2, for systems without TRS the statistics of the partial time-delays turns out to be equivalent (at perfect coupling) to those of the diagonal elements, q_c. Hence, we can consider $\operatorname{var}({{q}_{c}})$ which involves pairs of trajectories going to different end points but all starting in the same channel. Since the incoming channels coincide, the $\tilde{d}$- and $\tilde{x}$-quadruplets now all have the same number of free channels and contribute at the same order. The leading order d-quadruplet cancels the mean part squared (as for $\operatorname{var}({{\tau }_{{\rm W}}})$), leaving

$$\begin{eqnarray}\begin{array}{rcl} \operatorname{var}({{q}_{c}}) & = & -\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{X}}_{K}}}{{{M}^{K}}}-\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{D}}_{K}}}{{{M}^{K}}} \\ {} & = & \mathop{\sum }\limits_{k=1}^{\infty }\frac{1}{{{M}^{k}}}=\frac{1}{M-1}. \\ \end{array}\end{eqnarray} \tag{ 66 }$$

This is in line with the RMT result (15) at $\beta =2$, see also appendix A for an alternative calculation of $\operatorname{var}({{t}_{c}})$ explicitly.

By creating the relations above between the $\tilde{x}$- and $\tilde{d}$-diagrams, we could derive the second moments without any recourse to cutting periodic orbit diagrams. This is notably simpler than the results for the standard transport second moments (e.g. shot noise and the conductance variance) [60]. Treating the case with TRS requires, however, a more involved calculation which we pursue in the appendices. First, we map the problem without TRS to periodic orbits in appendix B and then evaluate the semiclassical sums which arise in appendix C. The corresponding sums for systems invariant under time-reversal are worked out in appendix D. This allows us to finally obtain, at any M, semiclassical results for the second moments of time-delays in the case of orthogonal symmetry in appendix E. We find identical results to RMT for the variance of the Wigner time delay and the second moment of the proper time-delays, and derive a new result for the variance of the diagonal elements (which in the case of orthogonal symmetry is not the same as the variance of the partial time-delays). We also demonstrate the equivalence of the semiclassical approach for the time-delay problem developed here to previous treatments in appendix F.

Starting with unrooted trees, there are two types: one with an excess of outgoing leaves whose generating function we call f and one with a excess of incoming leaves counted in $\hat{f}$. For example, removing the top i₁ channel from the tree in figure 6(a) to obtain an unrooted tree, an excess of o-leaves remains forming a subtree included in f. The generating variable will be r whose power counts the number of leaves, which is related to the order of the moment. Breaking the trees at the top encounter node gives the recursion

$$\begin{eqnarray}&&f=r-\mathop{\sum }\limits_{l=2}^{\infty }{{f}^{l}}{{\hat{f}}^{l-1}}+r\mathop{\sum }\limits_{l=2}^{\infty }l{{f}^{l-1}}{{\hat{f}}^{l-1}},\end{eqnarray} \tag{ 68 }$$

where the first term is an empty tree going straight to an outgoing leaf, the next term are trees with encounter nodes of size l with $l\geqslant 2$ and the last term is the correction for allowing one outgoing leaf to move into the encounter (and replacing the corresponding general tree f) where we have the factor of r to account for the lost leaf. Summing we get

$$\begin{eqnarray}&&\frac{f}{(1-h)}=r\frac{\partial }{\partial h}\frac{1}{(1-h)}=\frac{r}{{{(1-h)}^{2}}},\end{eqnarray} \tag{ 69 }$$

with $h=f\hat{f}$. This also gives us the useful relation

$$\begin{eqnarray}&&r=f(1-h).\end{eqnarray} \tag{ 70 }$$

The trees of type $\hat{f}$ with an excess of incoming leaves can actually also move into the incoming lead if all the $\hat{f}$ type trees after the top encounter end directly in an incoming channel. Then we have the recursion

$$\begin{eqnarray}&&\hat{f}=r-\mathop{\sum }\limits_{l=2}^{\infty }{{\hat{f}}^{l}}{{f}^{l-1}}+r\mathop{\sum }\limits_{l=2}^{\infty }(l-1){{\hat{f}}^{l}}{{f}^{l-2}}+\mathop{\sum }\limits_{l=2}^{\infty }{{r}^{l}}{{f}^{l-1}},\end{eqnarray} \tag{ 71 }$$

where the first three terms again correspond to an empty tree, trees with top encounter of size l and moving outgoing leaves into the encounter while the last term is the new possibility of moving the encounter into the incoming lead. Summing gives

$$\begin{eqnarray}&&\frac{{\hat{f}}}{(1-h)}=\frac{r{{{\hat{f}}}^{2}}}{{{(1-h)}^{2}}}+\frac{r}{(1-rf)}.\end{eqnarray} \tag{ 72 }$$

Substituting for r from (70) in just the first term on the right gives the simplification

$$\begin{eqnarray}&&\frac{{\hat{f}}}{(1-h)}=\frac{h\hat{f}}{(1-h)}+\frac{r}{(1-rf)},\qquad \hat{f}=\frac{r}{(1-rf)},\end{eqnarray} \tag{ 73 }$$

from which we can get a quadratic for f, $\hat{f}$ and more importantly h

$$\begin{eqnarray}&&{{h}^{2}}+(s-1)h+s=0,\end{eqnarray} \tag{ 74 }$$

with $s={{r}^{2}}$.

To get the full leading order moments with generating function G₀, we need to root our trees by adding an incoming channel to the f trees (and divide by M) or adding an outgoing leaf to the $\hat{f}$ trees at the top. The first option then allows the trees to also move into the incoming lead to give

$$\begin{eqnarray}&&{{G}_{0}}=rf+\mathop{\sum }\limits_{l=2}^{\infty }{{r}^{l}}{{f}^{l}}=\frac{rf}{1-rf}.\end{eqnarray} \tag{ 75 }$$

Using (73) this is just

$$\begin{eqnarray}&&{{G}_{0}}=h.\end{eqnarray} \tag{ 76 }$$

Alternatively, and more simply, we can place an outgoing leaf on top of the $\hat{f}$ type trees

$$\begin{eqnarray}&&{{G}_{0}}=r\hat{f}+r\mathop{\sum }\limits_{l=2}^{\infty }{{\hat{f}}^{l}}{{f}^{l-1}}=\frac{r\hat{f}}{1-h}=h,\end{eqnarray} \tag{ 77 }$$

where the sum is over the additional possibility of placing the new outgoing leaf into the encounter. Either way, the end result is that

$$\begin{eqnarray}&&{{G}_{0}}=\frac{1-s-\sqrt{1-6s+{{s}^{2}}}}{2},\end{eqnarray} \tag{ 78 }$$

from the correct solution of (74). This is exactly what was in [62] and equivalent to the previous RMT result [30]. Compared to the energy dependent cubic equations of [62] with corresponding energy derivatives and identity matrix corrections, the result (78) can now be obtained much more simply and directly with our new semiclassical approach for the proper time-delays.

Moving to the subleading order, we can continue looking for the dominant diagrams. The non-vanishing contribution exists only for systems with TRS, whereas for those without TRS the first correction occurs in the second subleading order (see below). As shown in [63], the relevant diagram in the former case has the topology of a Möbius strip that arises after merging the quadruplets with time-reversal partners. Around the Möbius strip there are two types of nodes, those with an even number of subtrees on each side and those with an odd number. We shall denote these as an even node and odd node respectively. For an l encounter there are $(l-1)$ subtrees of each type (two stretches are the Möbius loop itself) which can be arranged in l ways with an even number on each side and $(l-1)$ ways with an odd number on each side. These nodes cannot move into the incoming lead, but directly connecting odd leaves can be moved into the encounter node (one at a time). Using slightly different notation than [63] which is more useful for the higher orders in section 4.4 and appendix G, the contribution of an even node is

$$\begin{eqnarray}\begin{array}{rcl} \mathcal{A} & = & -\mathop{\sum }\limits_{l=2}^{\infty }l{{h}^{l-1}}+r\hat{f}\mathop{\sum }\limits_{l=2}^{\infty }l(l-1){{h}^{l-2}} \\ {} & = & \frac{h(h-2)}{{{(1-h)}^{2}}}+\frac{2r\hat{f}}{{{(1-h)}^{3}}}=\frac{{{h}^{2}}}{{{(1-h)}^{2}}}, \\ \end{array}\end{eqnarray} \tag{ 79 }$$

while an odd node contributes

$$\begin{eqnarray}&&\mathcal{B}=-\mathop{\sum }\limits_{l=2}^{\infty }(l-1){{h}^{l-1}}+r\hat{f}\mathop{\sum }\limits_{l=2}^{\infty }{{(l-1)}^{2}}{{h}^{l-2}}=-\frac{h}{{{(1-h)}^{2}}}+\frac{r\hat{f}(1+h)}{{{(1-h)}^{3}}}=\frac{{{h}^{2}}}{{{(1-h)}^{2}}}.\end{eqnarray} \tag{ 80 }$$

Both include the link leading up to the encounter node (but not the one leaving as this is included in the next node).

Around the Möbius loop we have an arbitrary number of encounter nodes, but we need to divide by their number because of the rotational symmetry. As such we define a generating function of an arbitrary arrangement of nodes around a loop

$$\begin{eqnarray}&&{{\tilde{\mathcal{K}}}_{1}}=\frac{1}{2}\mathop{\sum }\limits_{k}\frac{{{(\mathcal{A}+p\mathcal{B})}^{k}}}{k}=-\frac{1}{2}{\rm log} \left( 1-\mathcal{A}-p\mathcal{B} \right),\end{eqnarray} \tag{ 81 }$$

where we divide by 2 to account for the inside/outside symmetry of the loop. A factor p is included with the odd nodes since we actually need to have an odd number of odd nodes to close the loop properly. This is then achieved by comparing the values of ${{\tilde{\mathcal{K}}}_{1}}$ at $p=\pm 1$, giving

$$\begin{eqnarray}&&{{\mathcal{K}}_{1}}=\frac{{{\tilde{\mathcal{K}}}_{1}}(p=1)-{{\tilde{\mathcal{K}}}_{1}}(p=-1)}{2}=-\frac{1}{4}{\rm log} \left( 1-\frac{2{{h}^{2}}}{{{(1-h)}^{2}}} \right),\end{eqnarray} \tag{ 82 }$$

which is the integrated moment generating function. For the generating function itself, we differentiate

$$\begin{eqnarray}&&G_{1}^{{\rm O}}=2\;s\frac{{\rm d}{{\mathcal{K}}_{1}}}{{\rm d}s},\end{eqnarray} \tag{ 83 }$$

so that by using the solution for h from (74) we get the result

$$\begin{eqnarray}&&G_{1}^{{\rm O}}=\frac{1-3s-\sqrt{1-6s+{{s}^{2}}}}{2(1-6s+{{s}^{2}})},\end{eqnarray} \tag{ 84 }$$

which is the same as in [63] but obtained in much more straightforward way, without needing to add and then remove an energy dependence.

To treat higher-order corrections, we can use the algorithmic approach of [90] which makes use of combinatorial base structures built on permutations describing the diagramʼs edges and vertices. The possible permutations are generated via a computer search, while the permissible edge and vertex components are matched up according to the prescription of the permutation. All that is then needed by the algorithm are the general semiclassical contribution of the possible edge and vertex components which we list in appendix G.

Here instead we merely state the results, which for unitary symmetry are

$$\begin{eqnarray}&&G_{2}^{{\rm U}}=\frac{2{{s}^{2}}}{{{(1-6s+{{s}^{2}})}^{\frac{5}{2}}}},\end{eqnarray} \tag{ 85 }$$

confirming the guess in [63], and

$$\begin{eqnarray}&&G_{4}^{{\rm U}}=\frac{2{{s}^{2}}(1+30s+3{{s}^{2}}-12{{s}^{3}}+8{{s}^{4}})}{{{(1-6s+{{s}^{2}})}^{\frac{11}{2}}}}.\end{eqnarray} \tag{ 86 }$$

For the orthogonal case (with TRS), the results are

$$\begin{eqnarray}&&G_{2}^{{\rm O}}(s)=\frac{s(s-3)}{{{\left( 1-6s+{{s}^{2}} \right)}^{2}}}+\frac{3s{{(s-1)}^{2}}+2{{s}^{2}}}{{{\left( 1-6s+{{s}^{2}} \right)}^{\frac{5}{2}}}},\end{eqnarray} \tag{ 87 }$$

again confirming the guess in [63], while at higher order we have

$$\begin{eqnarray}&&G_{3}^{{\rm O}}(s)=-\frac{2s\left( 6{{s}^{4}}-5{{s}^{3}}+9{{s}^{2}}-15s-3 \right)}{{{\left( 1-6s+{{s}^{2}} \right)}^{4}}}-\frac{2s\left( 3+19s-9{{s}^{2}}+2{{s}^{3}} \right)}{{{\left( 1-6s+{{s}^{2}} \right)}^{\frac{7}{2}}}},\end{eqnarray} \tag{ 88 }$$

and

$$\begin{eqnarray}\begin{array}{rcl} G_{4}^{{\rm O}}(s) & = & \frac{4s\left( 6{{s}^{4}}-30{{s}^{4}}+123{{s}^{3}}-147{{s}^{2}}-85s-3 \right)}{{{\left( 1-6s+{{s}^{2}} \right)}^{5}}} \\ {} & {} & +\frac{2s\left( 6+163s+216{{s}^{2}}-219{{s}^{3}}+24{{s}^{4}}+20{{s}^{5}}+36{{s}^{6}} \right)}{{{(1-6s+{{s}^{2}})}^{\frac{11}{2}}}}. \\ \end{array}\end{eqnarray} \tag{ 89 }$$

At the highest two orders for both symmetry classes, the generating functions are new results not yet calculated semiclassically [63] or using RMT [32].

To obtain an arbitrary d-quadruplet, one can cut any pair of links of any correlated orbit pair (including the same link twice) [60]. If we cut links that leave different encounters however, then the resulting quadruplet will have one o-leaf on each encounter and the diagramʼs contribution will be 0.

To remain attached to the same l-encounter, we simply need to cut different links leaving that encounter (we also cannot cut the same link twice) which can de done in $l(l-1)$ ordered ways. The total number of $\tilde{d}$-quadruplets for each vector v is then

$$\begin{eqnarray}&&{{N}_{\tilde{d}}}(v)=\mathop{\sum }\limits_{l}\frac{l(l-1){{v}_{l}}}{L}N(v),\end{eqnarray} \tag{ B.1 }$$

and so

$$\begin{eqnarray}&&{{\tilde{D}}_{K}}=\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\mathop{\sum }\limits_{l}\frac{l(l-1){{v}_{l}}}{L}N(v).\end{eqnarray} \tag{ B.2 }$$

To obtain an arbitrary x-quadruplet, one can cut out any 2-encounter [60]. But again we only need to consider the case when both outgoing leaves are connected to the same encounter as all other cases cancel.

Let us first consider instead the opposite case where the outgoing leaves are connected to different encounters and we have an $x^{\prime}$-quadruplet. The parts of the periodic orbit must be arranged as in figure B1 . Say that an l₁-encounter and an l₂-encounter are connected to the 2-encounter with encounter links numbered by a_i and b_i respectively. When the links between the encounters and the 2-encounter are shrunk, a single l-encounter with $l={{l}_{1}}+{{l}_{2}}$ is created. If the reconnection of the original l₁-encounter is represented by the permutation $({{a}_{1}}\;\ldots \;{{a}_{{{l}_{1}}}})$ and of the l₂ encounter by $({{b}_{1}}\;\ldots \;{{b}_{{{l}_{2}}}})$ and the 2-encounter swaps links $({{a}_{i}}{{b}_{j}})$ then the l-encounter has the permutation

$$\begin{eqnarray}&&({{a}_{i}}{{b}_{j}})({{a}_{1}}\;\ldots \;{{a}_{{{l}_{1}}}})({{b}_{1}}\;\ldots \;{{b}_{{{l}_{2}}}})=({{a}_{1}}\;\ldots \;{{a}_{i-1}}\ {{b}_{j}}\ldots {{b}_{{{l}_{2}}}}{{b}_{1}}\;\ldots \;{{b}_{j-1}}\ {{a}_{i}}\;\ldots \;{{a}_{{{l}_{1}}}}).\end{eqnarray} \tag{ B.3 }$$

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Reversing the process, we may pull a 2-encounter out of an l-encounter (and break it into an l₁- and l₂-encounter) if we can turn the cycle $(1\;\ldots \;l)$ into a 2-cycle followed by a l₁ and l₂ cycle with ${{l}_{1}},{{l}_{2}}\geqslant 2$. For this we can easily check that multiplying $(1\;\ldots \;l)$ on the left by $(ij)$ leads to cycles of the correct size as long as i and j are not equal or adjacent (cyclically). There are then $l(l-3)$ ways of pulling a 2-encounter out of an l-encounter for $l\gt 3$. Note that this process adds two links and two encounters so the value of $L-V$ remains constant.

For any periodic orbit pair described by a vector v, an $x^{\prime}$-quadruplet can therefore be created in ${{\sum }_{l\gt 3}}l(l-3){{v}_{l}}$ ways while x-quadruplets could be created in the 2v₂ ways of cutting out any 2-encounter. Taking the difference to obtain $\tilde{x}$-quadruplets leaves

$$\begin{eqnarray}&&2{{v}_{2}}-\mathop{\sum }\limits_{l\gt 3}l(l-3){{v}_{l}}=-\mathop{\sum }\limits_{l\geqslant 2}l(l-3){{v}_{l}}\end{eqnarray} \tag{ B.4 }$$

possibilities.

Removing a 2-encounter from a periodic orbit leaves a x-quadruplet with L links, $V-1$ encounters and a vector with a 2-encounter removed. Recalling the minus contribution of encounters leads to

$$\begin{eqnarray}&&{{\tilde{X}}_{K}}=\mathop{\sum }\limits_{v}^{L-V=K+1}{{(-1)}^{V}}\mathop{\sum }\limits_{l}\frac{l(l-3){{v}_{l}}}{L}N(v).\end{eqnarray} \tag{ B.5 }$$

The terms in the recursion relations above have a simple interpretation in terms of orbits. If we take an arbitrary orbit with $L-V=K-2$, we can take any point in any of the L links, combine it with any point in the $(L-1)$ remaining links or any point either side of the original point and join them together to make a new 3-encounter. This adds 3 links and one encounter. Including the $(-1)$ for each encounter gives

$$\begin{eqnarray}&&-\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}\frac{(L+1)L}{L}N(v).\end{eqnarray} \tag{ C.14 }$$

Alternatively one may take a point from any of the L links and move it to (before or after) any of the L encounter stretches to increase the size of that encounter by 2. This adds two more links and no new encounters

$$\begin{eqnarray}&&2\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}\frac{{{L}^{2}}}{L}N(v).\end{eqnarray} \tag{ C.15 }$$

Finally one can pick stretches from different encounters, a k- and l-encounter say and combine them into a $k+l-1$ encounter. This adds one link and reduces the number of encounters by 1. To count them we can pick any two (ordered) encounter stretches in $L(L-1)$ ways and remove the ${{\sum }_{l}}l(l-1){{v}_{l}}$ which come from the same encounter

$$\begin{eqnarray}&&-\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}\frac{L(L-1)}{L}N(v)+\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}\mathop{\sum }\limits_{l}\frac{l(l-1){{v}_{l}}}{L}N(v).\end{eqnarray} \tag{ C.16 }$$

Later we will also need the following sum

$$\begin{eqnarray}&&{{J}_{K}}=\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\mathop{\sum }\limits_{l}{{l}^{2}}{{v}_{l}}N(v)=\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}L\mathop{\sum }\limits_{l}lN(v,l),\end{eqnarray} \tag{ C.17 }$$

which we now treat in the same way as H_K for systems without TRS. First for l = 2 we have

$$\begin{eqnarray}\begin{array}{rcl} \mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}2LN(v,2) & = & 2\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}L\mathop{\sum }\limits_{k\geqslant 2}N({{v}^{[k,2\to k+1]}},k+1) \\ {} & = & -2\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K}{{(-1)}^{V^{\prime} }}(L^{\prime} +1)\mathop{\sum }\limits_{k\geqslant 2}N(v^{\prime} ,k+1), \\ \end{array}\end{eqnarray} \tag{ C.18 }$$

following the steps in (C.12) and since merging the encounters reduces the number of links by 1. Relabelling the sums and substituting into (C.17) leads to

$$\begin{eqnarray}&&{{J}_{K}}=\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\mathop{\sum }\limits_{l\geqslant 3}\left[ (l-2)L-2 \right]N(v,l).\end{eqnarray} \tag{ C.19 }$$

The term for l = 3 is

$$\begin{eqnarray}\begin{array}{rcl} \mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}(L-2)N(v,3) & = & -\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K}{{(-1)}^{V^{\prime} }}(L^{\prime} -1)\mathop{\sum }\limits_{k\geqslant 2}N(v^{\prime} ,k+2) \\ {} & {} & -\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K-2}{{(-1)}^{V^{\prime} }}{{(L^{\prime} +1)}^{2}}N(v^{\prime} ), \\ \end{array}\end{eqnarray} \tag{ C.20 }$$

since when two encounters are merged the number of links decreases by 1 while when a 3-encounter breaks into links, 3 links are lost in the end. Substituting into (C.17) and relabelling the sum

$$\begin{eqnarray}&&{{J}_{K}}=\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\mathop{\sum }\limits_{l\geqslant 4}[(l-3)L-1]N(v,l)-\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}{{(L+1)}^{2}}N(v).\end{eqnarray} \tag{ C.21 }$$

For l = 4 we have $(L-1)N(v,4)$ inside the first sum. We can continue to replace terms using (C.4) and the same steps we used for H_K as long as we keep track of how the number of links changes for different terms in the recursions. We find that as l increases we continue to have this factor of $(L-1)$ and the sum reduces to

$$\begin{eqnarray}\begin{array}{rcl} {{J}_{K}} & = & -\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}{{(L+1)}^{2}}N(v)+2\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}L(L+1)N(v) \\ {} & {} & -\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}{{L}^{2}}N(v)+\mathop{\sum }\limits_{v}^{L-V=K-2}{{(-1)}^{V}}\mathop{\sum }\limits_{l}{{l}^{2}}{{v}_{l}}N(v). \\ \end{array}\end{eqnarray} \tag{ C.22 }$$

This result can be read off directly from the orbit interpretation if we remember to multiply by the new number of links minus two when making a new 3-encounter and instead by the new number of links minus one in the other cases. We can now simplify the results from the three cases since the L² factors cancel. This leaves

$$\begin{eqnarray}&&{{J}_{K}}=-{{C}_{K-2}}+{{J}_{K-2}}.\end{eqnarray} \tag{ C.23 }$$

With ${{C}_{K-2}}=0$ for $(K-2)\gt 0$ then ${{J}_{K}}={{J}_{K-2}}$ for $K\gt 2$. Since J₂ is −1 and J₁ is 0, then J_K is always −1 for even K. For K = 0 with ${{C}_{0}}=1$ we could also set ${{J}_{0}}=0$ in line with a sum over a 0 vector.

As for the unitary case, we can first consider adding a 2-encounter to the end of an $\tilde{d}$-quadruplet. Since the relevant encounter stretches are traversed in the same direction, when the links to the new 2-encounter are shrunk, the steps are identical to those in section 3.3 giving the same relation

$$\begin{eqnarray}\begin{array}{rcl} {{\tilde{D}}_{K}} & = & -{{X}_{K-1}}+2{{X}_{K-1}}-X_{K-1}^{\prime } \\ {} & = & {{\tilde{X}}_{K-1}}+{{{\hat{X}}}_{K-1}}, \\ \end{array}\end{eqnarray} \tag{ E.5 }$$

albeit with an extra term arising when we replace $X^{\prime}$ using (E.2). Also nothing changes when we start with an $\tilde{x}$-quadruplet giving

$$\begin{eqnarray}\begin{array}{rcl} {{\tilde{X}}_{K}} & = & -{{D}_{K-1}}+2{{D}_{K-1}}-2{{C}_{K-1}}-D_{K-1}^{\prime } \\ {} & = & {{\tilde{D}}_{K-1}}+{{{\hat{D}}}_{K-1}}-{{\delta }_{K,1}}, \\ \end{array}\end{eqnarray} \tag{ E.6 }$$

using (E.4).

Now we consider adding the same 2-encounter to the end of an $\hat{d}$-quadruplet. Since the relevant encounter stretches are traversed in opposite directions, the l-encounter at the end before the new 2-encounter must have size $l\gt 2$. To see what happens, it is simplest to express the l-encounter as a permutation, and we need to record that of both the encounter stretches and their time reversals. We use a bar to represent time reversal and hence stretches travelling in the opposite direction. Say that stretches a and $\bar{b}$ were originally connected to the outgoing leaves so that the l-encounter has the permutation

$$\begin{eqnarray}&&(ax\;\ldots \;y\bar{b}z\;\ldots \;w)(\bar{w}\;\ldots \;\bar{z}b\bar{y}\;\ldots \;\bar{x}\bar{a}),\end{eqnarray} \tag{ E.7 }$$

consisting of a single l cycle and its time reversal. The entries $w\;\ldots \;z$ could contain bars while the actual numbering of the elements would correspond to the order of traversal. Adding the 2-encounter between the stretches from a and b corresponds to multiplying later by the permutation $(ab)$, and for the time reversal before by the permutation $(\bar{a}\bar{b})$ leaving us with

$$\begin{eqnarray}&&\begin{array}{l} (ab)(ax\;\ldots \;y\bar{b}z\;\ldots \;w)(\bar{w}\;\ldots \;\bar{z}b\bar{y}\;\ldots \;\bar{x}\bar{a})(\bar{a}\bar{b}) \\ \quad =(ax\;\ldots \;y\bar{b}\bar{w}\;\ldots \;\bar{z})(z\;\ldots \;wb\bar{y}\;\ldots \;\bar{x}\bar{a}), \\ \end{array}\end{eqnarray} \tag{ E.8 }$$

again a single l-encounter with both outgoing leaves still attached to stretches travelling in opposite directions. Adding a 2-encounter however changes a d-type quadruplet into an x-type and once the connecting links are shrunk we still have the same number of links and encounters. We illustrate this process in figure E1 . Performing the same steps to a $\hat{x}$-quadruplet we just go back to a $\hat{d}$-one so this is an involution and

$$\begin{eqnarray}&&{{\hat{D}}_{K}}={{\hat{X}}_{K}}.\end{eqnarray} \tag{ E.9 }$$

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So far these relations are not sufficient to determine the four quantities $\tilde{D}$, $\hat{D}$, $\tilde{X}$, $\hat{X}$ but to proceed we can use TRS to our advantage. With TRS, we can connect the outgoing leaves o₁ to o₂ together, and also the incoming channels i₂ to i₁ and create a periodic orbit. If we start with a $\tilde{d}$-quadruplet the link formed from connecting the outgoing leaves must return to the same encounter, but now in the opposite direction. Starting the partner orbit in the same direction it will also follow the link made from connecting the incoming channels in the same direction. Performing the same steps to a $\tilde{x}$-quadruplet we again have a link connecting an encounter to itself (but returning to the encounter in the opposite direction), as well as one where the orbit and its partner travel in different directions.

Given all the periodic orbits with a vector v we can cut any link which connects an encounter to itself (traversed in the opposite direction) and place the outgoing leaves there. Then we can cut any of the remaining $L-1$ links, placing the incoming channels appropriately, to create ${{N}_{\tilde{x}}}(v)+{{N}_{\tilde{d}}}(v)$. Fortunately the number of links connecting encounters to themselves (in the opposite direction) corresponds to the third line in (D.1). When we multiply by the required $(L-1)$ (which cancels with the denominator) this leads to

$$\begin{eqnarray}\begin{array}{rcl} {{\tilde{D}}_{K}}+{{\tilde{X}}_{K}} & = & \mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\mathop{\sum }\limits_{l}{{(l-1)}^{2}}({{v}_{l-1}}+1)N({{v}^{[l\to l-1]}}) \\ {} & = & -\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K-1}{{(-1)}^{V^{\prime} }}(L^{\prime} +1)N(v^{\prime} ) \\ {} & {} & +\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K-1}{{(-1)}^{V^{\prime} }}\mathop{\sum }\limits_{l^{\prime} }{{(l^{\prime} )}^{2}}{{v}_{l^{\prime} }}N(v^{\prime} ) \\ {} & = & -{{A}_{K-1}}+{{J}_{K-1}}={{I}_{K-1}}, \\ \end{array}\end{eqnarray} \tag{ E.10 }$$

by relabelling sums and treating the 1-cycles that arise from the l = 2 term as in [50]. The l = 2 term in the first line gives the first sum on the second line which has a simple orbit interpretation. Given an orbit with a vector $v^{\prime}$ with $L^{\prime}$ links, we can replace any of those links by a 2-encounter that returns to itself, adding two more links and one more encounter. Cutting the new link returning to the 2-encounter, $L^{\prime} +1$ other links can be cut to create a $\tilde{x}$- or $\tilde{d}$-quadruplet.

Likewise, if we connect the outgoing leaves of a $\hat{x}$- or $\hat{d}$-quadruplet we arrive at a periodic orbit where a link connects an encounter to itself, but now in the same direction. From the periodic orbits with a vector v we therefore cut any link which connects an encounter to itself in the same direction and then cut any of the remaining $L-1$ links to create ${{N}_{{\hat{x}}}}(v)+{{N}_{{\hat{d}}}}(v)$. The number of links connecting encounters to themselves (in the same direction) corresponds to the second line in (D.1). Again the factor of $(L-1)$ cancels and

$$\begin{eqnarray}\begin{array}{rcl} {{{\hat{D}}}_{K}}+{{{\hat{X}}}_{K}} & = & 2\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\mathop{\sum }\limits_{l}\mathop{\sum }\limits_{m=1}^{l-2}(l-m-1)({{v}_{l-m-1}}+1)N({{v}^{[l\to m,l-m-1]}}) \\ {} & = & -2\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K-2}{{(-1)}^{V^{\prime} }}(L^{\prime} +2)(L^{\prime} +1)N(v^{\prime} ) \\ {} & {} & +4\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K-2}{{(-1)}^{V^{\prime} }}(L^{\prime} +1)L^{\prime} N(v^{\prime} ) \\ {} & {} & -2\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K-2}{{(-1)}^{V^{\prime} }}{{(L^{\prime} )}^{2}}N(v^{\prime} )+2\mathop{\sum }\limits_{v^{\prime} }^{L^{\prime} -V^{\prime} =K-2}{{(-1)}^{V^{\prime} }}\mathop{\sum }\limits_{l^{\prime} }{{(l^{\prime} )}^{2}}{{v}_{l^{\prime} }}N(v^{\prime} ) \\ {} & = & -2{{A}_{K-2}}-2{{C}_{K-2}}+2{{J}_{K-2}}=2{{I}_{K-2}}-2{{C}_{K-2}}, \\ \end{array}\end{eqnarray} \tag{ E.11 }$$

following the same steps as in appendix D.

Since $\hat{D}=\hat{X}$ from (E.9) we therefore have

$$\begin{eqnarray}&&{{\hat{D}}_{K}}={{\hat{X}}_{K}}={{I}_{K-2}}-{{C}_{K-2}}={{I}_{K-2}}-{{(-1)}^{K}},\end{eqnarray} \tag{ E.12 }$$

an explicit result for these quantities using (D.13). For $\tilde{D}$ and $\tilde{X}$ we first consider the difference

$$\begin{eqnarray}&&{{\tilde{X}}_{K}}-{{\tilde{D}}_{K}}=({{\tilde{D}}_{K-1}}+{{\hat{D}}_{K-1}})-({{\tilde{X}}_{K-1}}+{{\hat{X}}_{K-1}}),\end{eqnarray} \tag{ E.13 }$$

using (E.5) and (E.6) for $K\gt 1$. Again since $\hat{D}=\hat{X}$, this simplifies to

$$\begin{eqnarray}&&{{\tilde{X}}_{K}}-{{\tilde{D}}_{K}}=-({{\tilde{X}}_{K-1}}-{{\tilde{D}}_{K-1}}).\end{eqnarray} \tag{ E.14 }$$

Next $({{\tilde{X}}_{1}}-{{\tilde{D}}_{1}})$ can be checked to equal −1 so that

$$\begin{eqnarray}&&{{\tilde{X}}_{K}}-{{\tilde{D}}_{K}}={{(-1)}^{K}},\end{eqnarray} \tag{ E.15 }$$

for $K\gt 0$. Note that for K = 0 both ${{\tilde{X}}_{0}}$ and ${{\tilde{D}}_{0}}$ are 0. Substituting into (E.10) gives the explicit formulae

$$\begin{eqnarray}\begin{array}{rcl} 2{{\tilde{D}}_{K}} & = & {{I}_{K-1}}-{{(-1)}^{K}} \\ 2{{\tilde{X}}_{K}} & = & {{I}_{K-1}}+{{(-1)}^{K}}, \\ \end{array}\end{eqnarray} \tag{ E.16 }$$

in terms of (D.13). Useful for the second moments are the sums

$$\begin{eqnarray}\begin{array}{rcl} {{\tilde{D}}_{K+1}}+{{{\hat{D}}}_{K+1}}+{{\tilde{X}}_{K}}+{{{\hat{X}}}_{K}} & = & -\frac{4}{3}\left[ {{2}^{K}}-{{(-1)}^{K}} \right] \\ {{\tilde{X}}_{K+1}}+{{{\hat{X}}}_{K+1}}+{{\tilde{D}}_{K}}+{{{\hat{D}}}_{K}} & = & -\frac{2}{3}\left[ 2\cdot {{2}^{K}}+{{(-1)}^{K}} \right]. \\ \end{array}\end{eqnarray} \tag{ E.17 }$$

Finally, we can again consider the case where both incoming channels are connected to the same 2-encounter which can then be moved into the lead so that both incoming channels coincide. As for systems without TRS in section 3.3 such diagrams can be generated by simply appending a 2-encounter to the start of appropriate quadruplets. We therefore again have the relation (62) while for the $\hat{x}$ and $\hat{d}$ cases we analogously have

$$\begin{eqnarray}&&{\underset{\sim}{\hat{X}}_{K}}=-{{\hat{D}}_{K-1}},\quad {\underset{\sim}{\hat{D}}_{K}}=-{{\hat{X}}_{K-1}},\quad K\gt 1.\end{eqnarray} \tag{ E.18 }$$

For the calculation of m₂ the contribution of each diagram is as in section 3.4. When we sum over all diagrams (and further divide by ${{\mu }^{2}}M$) we have to evaluate

$$\begin{eqnarray}\begin{array}{rcl} {{\mu }^{2}}{{m}_{2}} & = & 2-\mathop{\sum }\limits_{K=2}^{\infty }\frac{{{\tilde{X}}_{K}}+{{{\hat{X}}}_{K}}}{{{M}^{K-1}}}-\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{D}}_{K}}+{{{\hat{D}}}_{K}}}{{{M}^{K}}} \\ {} & = & 2-\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{X}}_{K+1}}+{{{\hat{X}}}_{K+1}}+{{\tilde{D}}_{K}}+{{{\hat{D}}}_{K}}}{{{M}^{K}}} \\ {} & = & \frac{2}{3}\mathop{\sum }\limits_{K=0}^{\infty }\frac{2\cdot {{2}^{K}}+{{(-1)}^{K}}}{{{M}^{K}}}=\frac{2{{M}^{2}}}{(M+1)(M-2)}, \\ \end{array}\end{eqnarray} \tag{ E.19 }$$

where the leading order term from (48) is the same as for systems without TRS, and is included as the K = 0 term in the final sum. As discussed in section 2.5 this may be viewed as either a combination of ${{D}_{0}}-{{\tilde{X}}_{1}}-{{\hat{X}}_{1}}$ or simply $-2{{\tilde{X}}_{1}}$. The sum over d-quadruplets in (E.19) is again derived from x-quadruplets where the initial 2-encounter can be moved into the incoming lead as in (62) and (E.18). The result in (E.19) is exactly the RMT result; see expression (18) at $\beta =1$.

The variance of the time delay (dividing by the overall factor of M²) is given by the related sum

$$\begin{eqnarray}\begin{array}{rcl} \operatorname{var}({{\tau }_{{\rm W}}}) & = & -\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{X}}_{K}}+{{{\hat{X}}}_{K}}}{{{M}^{K+1}}}-\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{D}}_{K}}+{{{\hat{D}}}_{K}}}{{{M}^{K}}} \\ {} & = & -\frac{1}{M}\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{D}}_{K+1}}+{{{\hat{D}}}_{K+1}}+{{\tilde{X}}_{K}}+{{{\hat{X}}}_{K}}}{{{M}^{K}}} \\ {} & = & \frac{4}{3M}\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{2}^{K}}-{{(-1)}^{K}}}{{{M}^{K}}}=\frac{4}{(M+1)(M-2)}, \\ \end{array}\end{eqnarray} \tag{ E.20 }$$

since ${{\tilde{D}}_{1}}={{\hat{D}}_{1}}=0$. This result again agrees with RMT; set $\beta =1$ in (4).

For the symplectic symmetry class of spin-$\frac{1}{2}$ particles, the semiclassical diagrams are identical to those for the orthogonal TRS case. Spin–orbit interactions are instead included as additional spin propagators along the classical trajectories [105, 106]. Each channel or leaf is also split into a spin up and spin down version though observables are appropriately rescaled so that the semiclassical contribution of the diagonal pair to the average time delay remains unchanged. At subleading order, each of the diagrams in figure 1 gain an additional factor of $-\frac{1}{2}$ [107] but they still cancel. Continuing to all orders [84] all off-diagonal contributions similarly cancel.

Treating the second moment of the proper delay times, the spin semiclassical contributions of each diagram become more complicated [107] but effectively the leading order term remains unchanged while each higher order gains a factor of $-\frac{1}{2}$. This can be simply incorporated by substituting $M\to -2M$ in (E.19) to give

$$\begin{eqnarray}&&{{\mu }^{2}}{{m}_{2}}=\frac{4{{M}^{2}}}{(M+1)(2M-1)},\end{eqnarray} \tag{ E.21 }$$

which is exactly the RMT result (18) at $\beta =4$. Likewise, one finds

$$\begin{eqnarray}&&\operatorname{var}({{\tau }_{{\rm W}}})=\frac{2}{(M+1)(2M-1)},\end{eqnarray} \tag{ E.22 }$$

for the variance of the Wigner time delay in full agreement with RMT, see (4).

Returning to the orthogonal case, finally we can consider

$$\begin{eqnarray}&&\operatorname{var}({{q}_{c}})=-\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{\tilde{X}}_{K}}+{{{\hat{X}}}_{K}}+{{\tilde{D}}_{K}}+{{{\hat{D}}}_{K}}}{{{M}^{K}}}=\frac{{{M}^{2}}+5M+2}{{{(M+1)}^{2}}(M-2)},\end{eqnarray} \tag{ E.23 }$$

which is notably different from the RMT result for a diagonal element of the symmetrized Wigner–Smith matrix (or a partial time-delay) in (15). In particular, semiclassics predicts $1/M$ at leading order unlike the $2/(\beta M)$ of (15). The leading order term can however also be derived from an energy dependent correlator along the lines of the calculation in [61]. Since none of the diagrams at leading order for $\operatorname{var}({{q}_{c}})$ rely on TRS and we again simply obtain the unitary result $1/M$.

Of course the difference is due to correlations between the diagonal elements of Q and the symmetrization process. To explore this in more detail, we return to the definition of a partial time-delay in (A.1) but now perform averages over U using the COE results [89, 92, 108]. For the first moment, one finds

$$\begin{eqnarray}\begin{array}{rcl} \left\langle {{t}_{c}} \right\rangle & = & \frac{1}{M+1}\mathop{\sum }\limits_{a}\left\langle {{Q}_{ab}} \right\rangle \left( {{\delta }_{a,b}}+{{\delta }_{a,b,c}} \right) \\ {} & = & \frac{\left\langle {{\sum }_{a}}{{Q}_{aa}} \right\rangle +\left\langle {{Q}_{cc}} \right\rangle }{M+1}=\frac{\left\langle {\rm Tr}Q \right\rangle +\left\langle {{q}_{c}} \right\rangle }{M+1}={{\tau }_{{\rm W}}}, \\ \end{array}\end{eqnarray} \tag{ E.24 }$$

since $M\left\langle {{q}_{c}} \right\rangle =\left\langle {\rm Tr}Q \right\rangle$. Now the step of treating Q and U independently is no longer justified for the orthogonal case, but for the first moment it is easy to show semiclassically using our new approach. Treating the matrix elements in (A.1) semiclassically means considering four trajectories with end points mostly determined by the channel labels. However, the pair of trajectories coming from Q_ab end together somewhere inside the cavity instead. Correlations between Q and U would involve a quadruplet of trajectories which cannot be separated into two independent pairs (these simply give the result in (E.24)). In each semiclassical diagram there must therefore be an encounter before the link ending inside the cavity. Moving the end point into the encounter provides a second diagram, with the opposite contribution, and all possibilities simply cancel like for the average time delay.

Treating t_c² directly would involve eight trajectories, akin to a fourth moment in standard transport, and far more complicated than the second moments considered here. We therefore do not confirm the RMT result in (15) but instead semiclassics provides us with the complementary result for $\operatorname{var}({{q}_{c}})$ in (E.23) which has so far not been obtained from RMT. To gauge the level of correlations between Q and U we can however compute t_c² assuming their independence, for which we need

$$\begin{eqnarray}\begin{array}{rcl} \left\langle {{U}_{{{b}_{1}}c}}{{U}_{{{b}_{2}}c}}U_{{{a}_{1}}c}^{*}U_{{{a}_{2}}c}^{*} \right\rangle & = & \frac{{{\delta }_{{{a}_{1}},{{b}_{1}}}}{{\delta }_{{{a}_{2}},{{b}_{2}}}}+{{\delta }_{{{a}_{1}},{{b}_{2}}}}{{\delta }_{{{a}_{2}},{{b}_{1}}}}+2{{\delta }_{{{a}_{1}},{{b}_{1}},{{a}_{2}},{{b}_{2}},c}}}{M(M+3)} \\ {} & {} & +\frac{{{\delta }_{{{a}_{1}},{{b}_{1}}}}{{\delta }_{{{a}_{2}},{{b}_{2}},c}}+{{\delta }_{{{a}_{1}},{{b}_{2}}}}{{\delta }_{{{a}_{2}},{{b}_{1}},c}}+{{\delta }_{{{a}_{2}},{{b}_{2}}}}{{\delta }_{{{a}_{1}},{{b}_{1}},c}}+{{\delta }_{{{a}_{2}},{{b}_{1}}}}{{\delta }_{{{a}_{1}},{{b}_{2}},c}}}{{{(M-1)}^{-1}}M(M+1)(M+3)}, \\ \end{array}\end{eqnarray} \tag{ E.25 }$$

giving

$$\begin{eqnarray}\begin{array}{rcl} \left\langle t_{c}^{2} \right\rangle & = & \frac{\left\langle {{\sum }_{a,b}}{{Q}_{aa}}{{Q}_{bb}}+{{Q}_{ab}}{{Q}_{ba}} \right\rangle +2\left\langle Q_{cc}^{2} \right\rangle }{M(M+3)} \\ {} & {} & +\frac{2(M-1)\left\langle {{\sum }_{a}}{{Q}_{aa}}{{Q}_{cc}}+{{Q}_{ac}}{{Q}_{ca}} \right\rangle }{M(M+1)(M+3)}. \\ \end{array}\end{eqnarray} \tag{ E.26 }$$

Semiclassically, the result for each term b in the sums in the top line is the same so that the sums in the second line incur a factor of ${{M}^{-1}}$ and we can write the full result as

$$\begin{eqnarray}&&\left\langle t_{c}^{2} \right\rangle =\frac{\left( {{M}^{2}}+3M-2 \right)\left( \left\langle {{[{\rm Tr}Q]}^{2}} \right\rangle +\left\langle {\rm Tr}[{{Q}^{2}}] \right\rangle \right)}{{{M}^{2}}(M+1)(M+3)}+2\frac{\left\langle q_{c}^{2} \right\rangle }{M(M+3)}.\end{eqnarray} \tag{ E.27 }$$

Using (E.19), (E.20) and (E.23) leads to

$$\begin{eqnarray}&&\frac{\left\langle t_{c}^{2} \right\rangle }{{{\overline{{{\tau }_{{\rm W}}}}}^{2}}}=\frac{M({{M}^{2}}+M+2)}{{{(M+1)}^{2}}(M-2)},\end{eqnarray} \tag{ E.28 }$$

giving a (rescaled) variance of

$$\begin{eqnarray}&&\operatorname{var}({{t}_{c}})=\frac{{{M}^{2}}+5M+2}{{{(M+1)}^{2}}(M-2)},\end{eqnarray} \tag{ E.29 }$$

which is actually equal to $\operatorname{var}({{q}_{c}})$, see (E.23). The difference between (E.29) and (15) are thus due to the correlations between Q and U.

Previous approaches to the moments of the proper time-delays involved including an energy dependence during the intermediate semiclassical calculations which is differentiated out in a final step [61–63]. For example, by including an energy dependence in the scattering matrix and considering the correlation function

$$\begin{eqnarray}&&C(\epsilon )=\frac{1}{M}{\rm Tr}\left[ {{S}^{\dagger }}\left( E-\frac{\epsilon \hbar \mu }{2} \right)S\left( E+\frac{\epsilon \hbar \mu }{2} \right) \right],\end{eqnarray} \tag{ F.1 }$$

the first moment can be obtained as follows

$$\begin{eqnarray}&&{{m}_{1}}=\frac{1}{{\rm i}\mu }\frac{{\rm d}}{{\rm d}\epsilon }C(\epsilon ){{|}_{\epsilon =0}},\end{eqnarray} \tag{ F.2 }$$

in line with (1). By treating correlated pairs of semiclassical trajectories, this was shown to equal the average time delay in [61]. For second moments one should in general consider a correlator of four (energy dependent) scattering matrices. To avoid that here, we can actually use the unitarity of the scattering matrix to obtain [88]

$$\begin{eqnarray}&&{{m}_{2}}=\frac{1}{{{({\rm i}\mu )}^{2}}}\frac{{{{\rm d}}^{2}}}{{\rm d}{{\epsilon }^{2}}}C(\epsilon ){{|}_{\epsilon =0}},\end{eqnarray} \tag{ F.3 }$$

and enormously reduce the complexity of the problem. We state the semiclassical result for the correlator [61]

$$\begin{eqnarray}&&C(\epsilon )=\left( 1+\frac{2-\beta }{\beta M} \right)\mathop{\sum }\limits_{K=0}^{\infty }\frac{1}{{{M}^{K}}}\sum _{v}^{L-V=K}{{(-1)}^{V}}\frac{\prod _{\sigma =1}^{V}(1-{\rm i}\epsilon {{l}_{\sigma }})}{{{(1-{\rm i}\epsilon )}^{L+1}}}N(v),\end{eqnarray} \tag{ F.4 }$$

in terms of a sum over periodic orbits (since these are cut once to obtain the first moment diagrams) and we include the diagonal term as a vector with 0 entries. The prefactor accounts for the coherent backscattering diagrams: With TRS, when the incoming and outgoing channel coincide, the time reversal of the partner trajectory can also be paired with the original trajectory. Differentiating

$$\begin{eqnarray}&&\frac{{\rm d}}{{\rm d}\epsilon }\frac{\prod _{\sigma =1}^{V}(1-{\rm i}\epsilon {{l}_{\sigma }})}{{{(1-{\rm i}\epsilon )}^{L+1}}}={\rm i}\left[ \frac{L+1}{(1-{\rm i}\epsilon )}-\mathop{\sum }\limits_{\sigma }\frac{{{l}_{\sigma }}}{(1-{\rm i}\epsilon {{l}_{\sigma }})} \right]\frac{\prod _{\sigma =1}^{V}(1-{\rm i}\epsilon {{l}_{\sigma }})}{{{(1-{\rm i}\epsilon )}^{L+1}}}\;,\end{eqnarray} \tag{ F.5 }$$

$$\begin{eqnarray}&&\frac{{{{\rm d}}^{2}}}{{\rm d}{{\epsilon }^{2}}}\frac{\prod _{\sigma =1}^{V}(1-{\rm i}\epsilon {{l}_{\sigma }})}{{{(1-{\rm i}\epsilon )}^{L+1}}}{{|}_{\epsilon =0}}=-\left[ L+1-\mathop{\sum }\limits_{\sigma }l_{\sigma }^{2} \right]-{{\left[ L+1-\mathop{\sum }\limits_{\sigma }{{l}_{\sigma }} \right]}^{2}}.\end{eqnarray} \tag{ F.6 }$$

Since ${{\sum }_{\sigma }}{{l}_{\sigma }}={{\sum }_{l}}l{{v}_{l}}=L$ the second term is simply 1 and so

$$\begin{eqnarray}&&\frac{{{{\rm d}}^{2}}}{{\rm d}{{\epsilon }^{2}}}C(\epsilon ){{|}_{\epsilon =0}}=-\left( 1+\frac{2-\beta }{\beta M} \right)\mathop{\sum }\limits_{K=0}^{\infty }\frac{1}{{{M}^{K}}}\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\left[ L+2-\mathop{\sum }\limits_{l}{{l}^{2}}{{v}_{l}} \right]N(v).\end{eqnarray} \tag{ F.7 }$$

These are sums we have already evaluated and hence

$$\begin{eqnarray}&&{{\mu }^{2}}{{m}_{2}}=\left( 1+\frac{2-\beta }{\beta M} \right)\mathop{\sum }\limits_{K=0}^{\infty }\frac{{{A}_{K}}+{{C}_{K}}-{{J}_{K}}}{{{M}^{K}}}.\end{eqnarray} \tag{ F.8 }$$

Without TRS, $\beta =2$, the sum ${{A}_{K}}+{{C}_{K}}-{{J}_{K}}=2$ for all even K (including K = 0) and hence

$$\begin{eqnarray}&&{{\mu }^{2}}{{m}_{2}}=\mathop{\sum }\limits_{k=0}^{\infty }\frac{2}{{{M}^{2k}}}=\frac{2{{M}^{2}}}{{{M}^{2}}-1},\end{eqnarray} \tag{ F.9 }$$

in agreement with (64). With TRS, $\beta =1$, we have the sum

$$\begin{eqnarray}\begin{array}{rcl} {{\mu }^{2}}{{m}_{2}}=2-\frac{1}{M}\mathop{\sum }\limits_{K=0}^{\infty }\frac{{{I}_{K}}+{{I}_{K+1}}}{{{M}^{K}}} & = & 2+\frac{2}{3M}\mathop{\sum }\limits_{K=0}^{\infty }\frac{4\cdot {{2}^{K}}-{{(-1)}^{K}}}{{{M}^{K}}} \\ {} & = & \frac{2{{M}^{2}}}{(M+1)(M-2)}, \\ \end{array}\end{eqnarray} \tag{ F.10 }$$

in agreement with (E.19). Even for the unitary case we are forced to use the semiclassical sums from appendix C and the calculation is only tractable in this way because we used (F.3). Nonetheless, this shows the agreement between the new semiclassical method presented here and previous approaches.

For the variance of the Wigner time delay, to avoid treating energy dependent correlations between quadruplets of scattering trajectories, we turn to the expression in terms of periodic orbit correlations like (6) as discussed in section 1. In particular we may derive an expression for the two point correlator of the time delay at different energies as a second differential (see the appendix in the preprint version of [60]). This can only be done for the off-diagonal terms, while the diagonal term was given in (8). Here we merely state the semiclassical expression

$$\begin{eqnarray}&&\operatorname{var}({{\tau }_{{\rm W}}})=\frac{4}{\beta {{M}^{2}}}-\frac{2}{\beta {{M}^{2}}}\frac{{{{\rm d}}^{2}}}{{\rm d}{{\epsilon }^{2}}}\mathop{\sum }\limits_{K=1}^{\infty }\frac{1}{{{M}^{K}}}\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\frac{\prod _{\sigma =1}^{V}(1-{\rm i}\epsilon {{l}_{\sigma }})}{L{{(1-{\rm i}\epsilon )}^{L}}}N(v){{|}_{\epsilon =0}}.\end{eqnarray} \tag{ F.11 }$$

The main difference is that since periodic orbits are closed, we divide by the number of links L to avoid overcounting the same orbits while in total there are L links which is one fewer than when the periodic orbits are cut open to create the conductance diagrams used for (F.4). With TRS we may also always compare an orbit with its correlated partner and its time reversal, giving the global factor of $\frac{2}{\beta }$. Performing the differentials

$$\begin{eqnarray}&&\frac{{{{\rm d}}^{2}}}{{\rm d}{{\epsilon }^{2}}}\frac{\prod _{\sigma =1}^{V}(1-{\rm i}\epsilon {{l}_{\sigma }})}{{{(1-{\rm i}\epsilon )}^{L}}}{{|}_{\epsilon =0}}=-\left[ L-\mathop{\sum }\limits_{\sigma }l_{\sigma }^{2} \right]-{{\left[ L-\mathop{\sum }\limits_{\sigma }{{l}_{\sigma }} \right]}^{2}},\end{eqnarray} \tag{ F.12 }$$

the second term cancels completely and we are left with

$$\begin{eqnarray}&&\operatorname{var}({{\tau }_{{\rm W}}})=\frac{4}{\beta {{M}^{2}}}+\frac{4}{\beta {{M}^{2}}}\mathop{\sum }\limits_{K=1}^{\infty }\frac{1}{{{M}^{K}}}\mathop{\sum }\limits_{v}^{L-V=K}{{(-1)}^{V}}\left[ 1-\mathop{\sum }\limits_{l}\frac{{{l}^{2}}{{v}_{l}}}{L} \right]N(v),\end{eqnarray} \tag{ F.13 }$$

again in terms of sums we know

$$\begin{eqnarray}&&\operatorname{var}({{\tau }_{{\rm W}}})=\frac{4}{\beta {{M}^{2}}}+\frac{4}{\beta {{M}^{2}}}\mathop{\sum }\limits_{K=1}^{\infty }\frac{{{C}_{K}}-{{H}_{K}}}{{{M}^{K}}}.\end{eqnarray} \tag{ F.14 }$$

Without TRS, ${{C}_{K}}-{{H}_{K}}=1$ for all K (including 0) and hence

$$\begin{eqnarray}&&\operatorname{var}({{\tau }_{{\rm W}}})=\frac{2}{{{M}^{2}}}\mathop{\sum }\limits_{k=0}^{\infty }\frac{1}{{{M}^{2k}}}=\frac{2}{{{M}^{2}}-1},\end{eqnarray} \tag{ F.15 }$$

the same as (65). With TRS instead

$$\begin{eqnarray}&&\operatorname{var}({{\tau }_{{\rm W}}})=\frac{4}{3{{M}^{2}}}\mathop{\sum }\limits_{k=0}^{\infty }\frac{2\cdot {{2}^{K}}+{{(-1)}^{K}}}{{{M}^{2k}}}=\frac{4}{(M+1)(M-2)},\end{eqnarray} \tag{ F.16 }$$

the same as (E.20).