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Quantum techniques for stochastic mechanics (course) lecture 2

Quantum Techniques for Stochastic Mechanics

Lecture Content

The master equation vs the rate equation

A stochastic petri net has

we want to know

ddtx i=??? \frac{d}{d t} x_i = ???

one for each 1ik 1\leq i \leq k.

For each transition we end up with a term like this

ddtx i=r(n im i)x 1 m 1x k m k \frac{d}{d t} x_i = r (n_i - m_i)x_1^{m_1}\cdots x_k^{m_k}

We will use index free notation.

x=(x 1,,x k)[0,) x = (x_1,\ldots, x_k) \in [0, \infinity)

is the concentration vector and


ddt=r(nm)x m \frac{d}{d t} = r(n - m)x^m
ddtx= τTr(τ)[n(τ)m(τ)]x m(τ) \frac{d}{d t} x = \sum_{\tau \in T} r(\tau) [n(\tau) - m(\tau)]x^m(\tau)


The rate equations of motion are

ddtR(t)=β(21)R(t)+γ(01)R(t)W(t) \frac{d}{d t} R(t) = \beta (2-1)R(t) + \gamma (0-1) R(t) W(t)
ddtW(t)=δ(01)W(t)+γ(21)R(t)W(t) \frac{d} {d t} W(t) = \delta (0-1) W(t) + \gamma (2-1) R(t) W(t)

The master equation

Let n=(n 1,,n k)N k n = (n_1, \ldots, n_k) \in N^k and ψ n=ψ n 1,,ψ n k \psi_n = \psi_{n_1}, \ldots, \psi_{n_k} and then write a monomial

z n=z 1 n 1z k n k z^n = z^{n_1}_1\cdots z^{n_k}_k

and express any stochastic state as

Ψ= nN kψ nz n \Psi = \sum_{n\in N^k}\psi_n z^n

A state has

Ψ| z=1= nψ n=1 \Psi |_{z=1} = \sum_n\psi_n = 1

The master equation provides a description of how Ψ \Psi changes with time.

ddtψ(t)=Hψ(t) \frac{d}{d t} \psi(t) = H \psi(t)


H= τTr(τ)[a n(τ)a m(τ)]a m(τ) H = \sum_{\tau\in T} r(\tau) [ a^{\dagger n(\tau)} - a^{\dagger m(\tau)}]a^{m(\tau)}


a m(τ)=a 1 m 1(τ)a k m k(τ) a^{m(\tau)} = a_1^{m_1(\tau)}\cdots a_k^{m_k(\tau)} a m(τ)=a 1 m 1(τ)a k m k(τ) a^{\dagger m(\tau)} = a_1^{\dagger m_1(\tau)}\cdots a_k^{\dagger m_k(\tau)}

How can we understand each term?

Consider again


ψ= nψ nz n \psi = \sum_n \psi_n z^n


z n=z 1 n 1z 2 n 2 z^n = z_1^{n_1} z_2^{n_2}

and so

ψ n=ψ n 1,n 2 \psi_n = \psi_{n_1, n_2}

is the probability of having n 1 n_1 rabbits and n 2 n_2 wolves. These probabilities evolve according to

ddtψ(t)=Hψ(t) \frac{d}{d t} \psi(t) = H \psi(t)
H=βB+γC+δD H = \beta B + \gamma C + \delta D
H= τTr(τ)[a n(τ)a m(τ)]a m(τ) H = \sum_{\tau\in T} r(\tau) [ a^{\dagger n(\tau)} - a^{\dagger m(\tau)}]a^{m(\tau)}

order the input and output vectors as (R,W) (R, W) then

B=r 1[a 1 a 1 a 1 ]a 1 B = r_1[ a_1^\dagger a_1^\dagger - a_1^\dagger] a_1
C=r 2[a 2 a 2 a 1 a 2 ]a 1a 2 C = r_2[ a_2^\dagger a_2^\dagger - a_1^\dagger a_2^\dagger]a_1a_2
D=r 3[1a 2 ]a 2 D = r_3[1-a_2^\dagger]a_2

Basic properties of stochastic mechanics

In the lecture, we briefly recalled some of the basic properties of stochastic mechanics, which can be found in the book, as well as several of the blog articles on azimuth.

Amoeba field theory

It shows a world with one state, amoeba with concentration A(t) A(t), and two transitions:

reproduction, where one amoeba turns into two. Let’s call the rate constant for this transition α\alpha.

competition, where two amoebas battle for resources and only one survives. Let’s call the rate constant for this transition β\beta.


ddtA(t)=αAβA 2 \frac{d}{d t} A(t) = \alpha A - \beta A^2

which has solutions

equilibrium. The horizontal blue line corresponds to the case where the initial population P 0P_0 exactly equals the carrying capacity. In this case the population is constant.

dieoff. The three decaying curves above the horizontal blue line correspond to cases where initial population is higher than the carrying capacity. The population dies off over time and approaches the carrying capacity.

growth. The four increasing curves below the horizontal blue line represent cases where the initial population is lower than the carrying capacity. Now the population grows over time and approaches the carrying capacity.

Quick intro to the deficiency zero theorem

The deficiency zero theorem gives conditions on the network that allow one to state that the master equation has a unique equilibrium solution.

For a harmonic oscillator,

ΔpΔq/2 \Delta p \Delta q \geq \hbar/2

A coherence state of the harmonic oscillator minimizes ΔpΔq \Delta p \Delta q with Δp=Δq \Delta p = \Delta q

We can write down

Ψ= n=0 ψ nz n \Psi = \sum_{n=0}^{\infinity} \psi_n z^n
Ψ=e cz= n=0 c nn!z n \Psi = e^{c z} = \sum_{n=0}^{\infinity} \frac{c^n}{n!}z^n

For amoebas

Ψ=e cze c=e c n=0 c nn!z n \Psi = \frac{e^{c z}}{e^c} = e^{-c} \sum_{n=0}^{\infinity} \frac{c^n}{n!}z^n

now the probability distribution ψ n=e cc nn!\psi_n = e^{-c} \frac{c^n}{n!} is called a poisson distribution. The expected number of amoebas is

NΨ=a aΨ=a ae cze c \sum N \Psi = \sum a^\dagger a \Psi = \sum a^\dagger a \frac{e^{c z}}{e^c}
=zddze cze c=cze cze c=ca Ψ=cΨ = \sum z \frac{d}{d z} \frac{e^{c z}}{e^c} = c\sum z \frac{e^{c z}}{e^c} = c \sum a^\dagger \Psi = c \sum \Psi