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# Total number operator conservation in chemical reaction networks

There has been a lot of nice work recently related to chemical reaction networks here on Azimuth!

The two papers have been blogged about on Azimuth recently, check out this!

After helping write the networks book with John, I dove into complex network theory and tried to connect it to quantum mechanics. It worked! We have a paper on this online and one more on the way. Now that this has worked, it's time for me to get back into the reaction networks stuff, and hopefully talk a lot more with David Tanzer as well as a few people I started talking with over email from Harvard (Felipe Herrera, Dmitrij Rappoport, Ryan Babbush, Dmitry Zubarev) and others to get some ideas going. I want to write some follow up papers to those listed above on the topic of reaction networks.

It's a highly interesting topic.

Today I'm going to post a calculation I did in November 2011, stored here.

## The reaction network Hamiltonian

$$H = \sum_{\tau\in T} r(\tau)(a^\dagger^{n(\tau)} - a^\dagger^{m(\tau)})a^{m(\tau)}$$ The number operator for a single species is

$$N_i = a_i^\dagger a_i$$ and the total number operator for all the species is a sum over the single species

$$N = \sum_i N_i$$

• (Remark added) We know from Stochastic Noether's Theorem (as done by John and Brendan) that if this commutes with the Hamiltonian, then it relates to a conserved quantity. This has been talked about recently on the forum here.

• (Remark added) The point of the following calculation is take an alegraic approach and see if we can determine a bit of the properties that a stochastic Hamiltonian must have for $N$ to be a symmetry. To do that, we are faced with simplyfying (if possible) the expression $$[H, N] = 0$$ to determine conditions on $H$ which ensures this.

## Commutation relations

It can be shown (using induction) that

$$[a, a^\dagger^k] = k a^\dagger^{k-1}$$ $$[a^k,a^\dagger] = k a^{k-1}$$

• exercise. Let $[a, a^\dagger] = 1$ be the base case and assume $[a, a^\dagger^k] = k a^\dagger^{k-1}$ and show that these assumptions imply the formula for $k+1$ and hence using induction or otherwise, prove the first commutation relation (or both if you wish) listed above.

Here we will use the following notation.

$$a^{m(\tau)}a^\dagger_i = a^{m_i(\tau)}_i a^\dagger_i a^{m'(\tau)}$$ $$a_i a^\dagger^{n(\tau)} = a^\dagger^{n'(\tau)}a_i a_i^\dagger^{n_i(\tau)}$$ where the vector $m'(\tau)$ has its $i$th component zero which is why we are able to move term(s) $a^{m'(\tau)}$ to the right. This enables us to express the more general commutation relations,

$$[a^{m(\tau)},a^\dagger_i] = m_i(\tau)a_i^{m_i(\tau)-1}a^{m'(\tau)}$$ $$[a_i, a^\dagger^{n(\tau)}] = n_i(\tau) (a_i^\dagger)^{n_i(\tau)-1}a^\dagger^{n'(\tau)}$$ Using these relations, it follows that (is better notation possible?)

$$[H, a_i^\dagger] = \sum_{\tau\in T} r(\tau)(a^\dagger^{n(\tau)} - a^\dagger^{m(\tau)})a_i^{m_i(\tau)-1} a^{m'(\tau)} m_i(\tau)$$ and also

$$[a_i, H] = \sum_{\tau\in T} r(\tau)[n_i(\tau)a^\dagger^{n_i(\tau)-1}a^\dagger^{n'(\tau)} - m_i(\tau)a^\dagger^{m_i(\tau)-1}a^\dagger^{m'(\tau)}]a^{m(\tau)}$$ These will simplify the calculation of the commutation of the Hamiltonian and the number operator.

$$[\sum_i a_i^\dagger a_i, H] = \sum_i [a_i^\dagger a_i, H] = \sum_i(a_i^\dagger [a_i, H] - [H, a_i^\dagger]a_i)$$ For this to vanish, the following quantity must vanish identically.

$$\sum_i (a_i^\dagger K_2 - K_1 a_i)$$ where

$$a_i^\dagger K_2 = a_i^\dagger[a_i, H] = \sum_i \sum_{\tau\in T} r(\tau)[n_i(\tau)a^\dagger^{n(\tau)} - m_i(\tau)a^\dagger^{m(\tau)}]a^{m(\tau)}$$ $$K_1 a_i = [H, a_i^\dagger]a_i = \sum_{\tau\in T} r(\tau)(a^\dagger^{n(\tau)} - a^\dagger^{m(\tau)})a^{m(\tau)}m_i(\tau)$$ and so provided everything here is correct, we assert that

• (result). The following quantity must identically vanish for particle number conservation.

$$\sum_i \sum_{\tau\in T} r(\tau)[n_i(\tau) - m_i(\tau)]a^\dagger^{n(\tau)}a^{m(\tau)}$$

## Remarks

And provided this is correct, it is perhaps interesting since it allows us to say a bit more about systems with conserved total particle number.

• (Todo). Note, should rephrase as iff statement.

• (Todo). Redo the calculation for $$P = \sum_i k_i N_i$$ where $k_i \in Z_+$

So that's it. The point is that this calculation should be verified by someone else, since it could contain a slip up, but seems OK. The other point is what to do with this? I'm going to post an example next and hope to hear from others about their ideas. I think it's all a bit of fun, and if we press on with this, I think we could get an interesting result. Where I see this going is a better classification of reaction network Hamiltonians. Perhaps you, the reader have other ideas/comments/suggestions/corrections --- I'd be keen to hear them!

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1.

## Example Time!

In the last comment, I posted that

• (Total particle conservation). The following quantity must identically vanish for particle number conservation.

$$\sum_i \sum_{\tau\in T} r(\tau)[n_i(\tau) - m_i(\tau)]a^\dagger^{n(\tau)}a^{m(\tau)}$$ Now let's do an example of this. I think this example has appeared on Azimuth several times, but let's do it again! (For example, see Network Theory Part 10 where John considered this same reversible reaction Hamiltonian.)

$$H = (a^\dagger - b^\dagger)(\beta b - \alpha a)$$ We find its commutation relations with the creation (destruction) operators of both species to be

$$[H, a^\dagger] = \alpha (b^\dagger - a^\dagger)$$ $$[H, a] = \alpha a - \beta b$$ $$[b^\dagger, H] = \beta (b^\dagger - a^\dagger )$$ $$[b, H] = \alpha a - \beta b$$ Now we see that the particle number is conserved for this Hamiltonian by calculating

$$[N, H] = [a^\dagger a, H] + [b^\dagger b, H] = a^\dagger[a, H] + [a^\dagger, H]a + b^\dagger[b, H] + [b^\dagger, H]b = 0$$ All of the general equations above confirm the independently found results here. The example is rather simple. We need to try something a bit larger.

Comment Source:## Example Time! ## In the last comment, I posted that * (**Total particle conservation**). The following quantity must identically vanish for particle number conservation. $\sum_i \sum_{\tau\in T} r(\tau)[n_i(\tau) - m_i(\tau)]a^\dagger^{n(\tau)}a^{m(\tau)}$ Now let's do an example of this. I think this example has appeared on Azimuth several times, but let's do it again! (For example, see Network Theory [Part 10](http://johncarlosbaez.wordpress.com/2011/09/16/network-theory-part-10/) where John considered this same reversible reaction Hamiltonian.) $$H = (a^\dagger - b^\dagger)(\beta b - \alpha a)$$ We find its commutation relations with the creation (destruction) operators of both species to be $$[H, a^\dagger] = \alpha (b^\dagger - a^\dagger)$$ $$[H, a] = \alpha a - \beta b$$ $$[b^\dagger, H] = \beta (b^\dagger - a^\dagger )$$ $$[b, H] = \alpha a - \beta b$$ Now we see that the particle number is conserved for this Hamiltonian by calculating $$[N, H] = [a^\dagger a, H] + [b^\dagger b, H] = a^\dagger[a, H] + [a^\dagger, H]a + b^\dagger[b, H] + [b^\dagger, H]b = 0$$ All of the general equations above confirm the independently found results here. The example is rather simple. We need to try something a bit larger.
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2.

## Enter quantum physics

Note in particular that just because total particle number is conserved, this does not always mean that the chemical reaction Hamiltonian is certainly a quantum generator.

$$H = H^\dagger$$ implies that $H$ is certainly a valid quantum generator. So for our example

$$H = \beta a^\dagger b - \alpha a^\dagger a - \beta b^\dagger b + \alpha b^\dagger a$$ and

$$H^\dagger = \beta b^\dagger a - \alpha a^\dagger a - \beta b^\dagger b + \alpha a^\dagger b$$ Note that in the quantum case, we could have Hermitian terms such as

$$e^{i \theta} a^\dagger b + e^{-i \theta} b^\dagger a$$ for real $\theta$. However, the coefficeints of $a^\dagger a$ and $b^\dagger b$ in our case must both agree under the dagger, which implies they're real here.

So we are left with the following expression, which must vanish for $H$ to be a valid quantum generator

$$\beta a^\dagger b + \alpha b^\dagger a - \beta b^\dagger a - \alpha a^\dagger b = 0$$ Or, in other words

$$\beta - \alpha = 0$$ over the reals for $H$ to generate a valid quantum process.

An idea would perhaps be to inforce the condition that the generator is also valid for quantum physics in the expression for total particle conservation, and see what happens.

Comment Source:## Enter quantum physics ## Note in particular that just because total particle number is conserved, this does not always mean that the chemical reaction Hamiltonian is certainly a quantum generator. $$H = H^\dagger$$ implies that $H$ is certainly a valid quantum generator. So for our example $$H = \beta a^\dagger b - \alpha a^\dagger a - \beta b^\dagger b + \alpha b^\dagger a$$ and $$H^\dagger = \beta b^\dagger a - \alpha a^\dagger a - \beta b^\dagger b + \alpha a^\dagger b$$ Note that in the quantum case, we could have Hermitian terms such as $$e^{i \theta} a^\dagger b + e^{-i \theta} b^\dagger a$$ for real $\theta$. However, the coefficeints of $a^\dagger a$ and $b^\dagger b$ in our case must both agree under the dagger, which implies they're real here. So we are left with the following expression, which must vanish for $H$ to be a valid quantum generator $$\beta a^\dagger b + \alpha b^\dagger a - \beta b^\dagger a - \alpha a^\dagger b = 0$$ Or, in other words $$\beta - \alpha = 0$$ over the reals for $H$ to generate a valid quantum process. An idea would perhaps be to inforce the condition that the generator is also valid for quantum physics in the expression for total particle conservation, and see what happens.
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3.
edited June 2013

Did I miss anything? If so, please post it in the comments!

Maybe you would be interested in this discussion: http://forum.azimuthproject.org/discussion/1121/a-question-about-petri-nets/.

Comment Source:> Did I miss anything? If so, please post it in the comments! Maybe you would be interested in this discussion: [http://forum.azimuthproject.org/discussion/1121/a-question-about-petri-nets/](http://forum.azimuthproject.org/discussion/1121/a-question-about-petri-nets/).
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4.
edited June 2013

By the way, Jacob, I keep correcting your spelling of Brendan Fong's name...

More interesting comments later, I hope!

Comment Source:By the way, Jacob, I keep correcting your spelling of Brend**a**n Fong's name... More interesting comments later, I hope!
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5.

I'm very sorry John. How embarassing. I'm always trying to be more careful and won't make that mistake again. How embarassing. Sorry also to Brendan.

Comment Source:I'm very sorry John. How embarassing. I'm always trying to be more careful and won't make that mistake again. How embarassing. Sorry also to Brendan.
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6.

Hi Jacob, it would be great to talk about ideas for continued work on reaction networks!

Thanks

Comment Source:Hi Jacob, it would be great to talk about ideas for continued work on reaction networks! Thanks
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7.

Hi David! Today I'm going to plow through the rest of the Jordan normal form stuff you guys started on the other thread. I want to make sure I understand what everyone has been saying, in my quest to learn more math. Graham Jones, please help me with that!

I might summarize some thoughts if it seems like I have a different way to think about any part of this than what's already said. I might also summarize my thoughts anyway. In particular, as we know, for generators in stochastic mechanics, they're not guaranteed to be diagonalizable. Anyway, your other thread got a bunch of random thoughts going through my mind about Jordan chains, stochastic generators, upper triangular graphs. Fun stuff thanks!