# Phorgy Phynance

## Network Theory and Discrete Calculus – Noether’s Theorem

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This post is part of a series

As stated in the Introduction, one of the motivations for this series is to work in parallel with John Baez’ series on network theory to highlight some applications of discrete calculus. In this post, I reformulate some of the material in Part 11 pertaining to 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.

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Written by Eric

December 25, 2011 at 9:09 am