Network Theory and Discrete Calculus – Noether’s Theorem

The State-Time Graph

The directed graphs associated with discrete stochastic mechanics are described in the post The Discrete Master Equation, where the simple state-time graph example below was presented

Conceptually, the thing to keep in mind is that any transition from one state to another requires a time step. Therefore a transition from node $i$ to node $j$ is more precisely a transition from node $(i,t)$ to node $(j,t+1)$ .

On a state-time graph, a discrete 0-form can be written as

$\begin{aligned} \psi = \sum_{i,t} \psi^t_i \mathbf{e}^{(i,t)}.\end{aligned}$

and a discrete 1-form can be written as

$\begin{aligned} P = \sum_{i,j,t} \sum_{\epsilon\in[i,j]} P^{\epsilon,t}_{i,j} \mathbf{e}^{(i,t)(j,t+1)}_\epsilon.\end{aligned}$

The Master Equation

The master equation for discrete stochastic mechanics can be expressed simply as

$\partial(\psi P) = 0,$

where $\psi$ is a discrete 0-form representing the state at all times with

$\begin{aligned} 0\le \psi_{i}^t \le 1 \quad\text{and}\quad \sum_{i} \psi_{i}^t = 1 \end{aligned}$

and $P$ is a discrete 1-form representing transition probabilities with

$\begin{aligned} 0\le P_{i,j}^t \le 1 \quad\text{and}\quad \sum_{j} P_{i,j}^t = 1 \end{aligned}$

for all $t$ . When expanded into components, the master equation becomes

$\begin{aligned} \psi_j^{t+1} = \sum_i \psi_i^{t} P_{i,j}^{t}. \end{aligned}$

Observables and Expectations

A general discrete 0-form on a state-time graph is defined over all states and all time. However, occasionally, we would like to consider a discrete 0-form defined over all states at a specific point in time. To facilitate this in a component-free manner, denote

$\begin{aligned} 1^t = \sum_i \mathbf{e}^{(i,t)} \end{aligned}$

so the identity can be expressed as

$\begin{aligned} 1 = \sum_t 1^t.\end{aligned}$

The discrete 0-form $1^t$ is a projection that projects a general discrete 0-form to a discrete 0-form defined only at time $t$ . For instance, given a discrete 0-form $\psi$ , let

$\begin{aligned} \psi^t = 1^t \psi = \sum_i \psi_i^t \mathbf{e}^{(i,t)}\end{aligned}$

so that

$\begin{aligned} \psi = \sum_t \psi^t.\end{aligned}$

In discrete stochastic mechanics, an observable is nothing more than a discrete 0-form

$\begin{aligned} O = \sum_t O^t = \sum_{i,t} O_i^t \mathbf{e}^{(i,t)}.\end{aligned}$

The expectation of an observable $O^t$ with respect to a state $\psi$ is given by

$\langle O^t\rangle = tr_0(O^t \psi) = \sum_i O_i^t \psi_i^t$

where $tr_0$ was defined in a previous post. Note: $O^t \psi = O^t \psi^t.$

Some Commutators

In preparation for the discrete Noether’s theorem, note that

$\begin{aligned} { [P,O] = \sum_{i,j,t} \sum_{\epsilon\in[i,j]} (O_j^{t+1} - O_i^t) P_{i,j}^{\epsilon,t} \mathbf{e}^{(i,t)(j,t+1)}_\epsilon. } \end{aligned}$

and

$\begin{aligned} { [[P,O],O] = \sum_{i,j,t} \sum_{\epsilon\in[i,j]} (O_j^{t+1} - O_i^t)^2 P_{i,j}^{\epsilon,t} \mathbf{e}^{(i,t)(j,t+1)}_\epsilon. } \end{aligned}$

For these commutators to vanish, we must have

$P_{i,j}^{\epsilon,t} \ne 0 \implies O_j^{t+1} = O_i^t.$

This implies $[P,O] = 0$ if and only if $O$ is constant on each connected component of the state-time graph.

Constant Expectations

In this section, we determine the conditions under which the expectation of an observable $O$ is constant in time, i.e.

$\langle O^{t+1}\rangle = \langle O^{t} \rangle$

for all $t$ . This is a fairly straightforward application of the discrete master equation, i.e.

$\begin{aligned} \langle O^{t+1}\rangle &= \sum_{j} \psi_j^{t+1} O_j^{t+1} \\ &= \sum_{i} {\psi_i^{t} \sum_j {\sum_{\epsilon\in[i,j]} { P_{i,j}^{\epsilon,t} O_j^{t+1}}}}\end{aligned}$

indicating the condition we’re looking for is

$\begin{aligned} O_i^{t} = \sum_j {\sum_{\epsilon\in[i,j]} { P_{i,j}^{\epsilon,t} O_j^{t+1}. }}\end{aligned}$

Noether’s Theorem

In this section, we demonstrate that when both $\langle O^t\rangle$ and $\langle (O^t)^2\rangle$ are constant in time, this implies

$tr_1\left( [[P,O],O] \right) = 0$

which, in turn, implies $[P,O] = 0$ . To do this, we first expand

$\begin{aligned} tr_1([[P,O],O]) = \sum_{i,j,t} \sum_{\epsilon\in[i,j]} (O_j^{t+1} - O_i^t)^2 P_{i,j}^{\epsilon,t}. \end{aligned}$

The condition for this trace to vanish is the same as the condition for the commutators above to vanish, i.e.

$P_{i,j}^{\epsilon,t} \ne 0 \implies O_j^{t+1} = O_i^t.$

Expanding the trace further results in

$\begin{aligned} tr_1([[P,O],O]) = \sum_{i,j,t} \sum_{\epsilon\in[i,j]} P_{i,j}^{\epsilon,t} {(O_j^{t+1})}^2 - 2 O_i^t (P_{i,j}^{\epsilon,t} O_j^{t+1}) + (O_i^t)^2 P_{i,j}^{\epsilon,t}.\end{aligned}$

Summing over $j$ and $\epsilon$ when $\langle O^t\rangle$ and $\langle (O^t)^2\rangle$ are constants results in

$\begin{aligned} \text{1st Term + 2nd Term} = -\sum_{i,t} (O_i^t)^2,\end{aligned}$

while summing $j$ and $\epsilon$ in the third term results in

$\begin{aligned} \text{3rd Term} = \sum_{i,t} (O_i^t)^2 \end{aligned}$

by definition of the transition 1-form. Consequently, when $\langle O^t\rangle$ and $\langle (O^t)^2\rangle$ are constants, it follows that

$tr_1([[P,O],O]) =0.$

Finally, this implies $[P,O] = 0$ if and only if $\langle O^t\rangle$ and $\langle (O^t)^2\rangle$ are constant in time.

Written by Eric

December 25, 2011 at 9:09 am

Posted in Directed Graphs, Discrete Calculus, John Baez, Network Theory

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