Wormholes and Entanglement

20 January, 2014

 

An apparent contradiction in what most physicists believe about black holes—the ‘firewall problem’—is making some very good physicists reach for some very crazy-sounding ideas to find a way out. In particular, Maldacena and Susskind have come up with the idea that any pair of quantum-entangled particles is actually connected by a wormhole.

Entanglement is a spooky way for far-away particles to be correlated, but you can’t use it communicate faster than light. It’s been seen in the lab, but it’s only possible thanks to quantum mechanics. The first people to make a fuss over entanglement were Einstein, Podolsky and Rosen, back in 1935.

A wormhole is a spooky way for far-away regions of space to be connected by a kind of ‘tunnel’—but you probably can’t use it to communicate faster than light. Nobody has ever seen one, but they’re theoretically possible thanks to general relativity. The first people to make a fuss over wormholes were Einstein and Rosen, back in 1935.

So, superficially, it makes sense that there should be a connection between wormholes and entanglement. But when you learn enough physics, you’ll see that Maldacena and Susskind’s proposal sounds completely hare-brained.

But when you learn more physics—maybe more than enough?—you might decide there’s some merit to this idea after all. At the Centre for Quantum Technologies last summer, Jamie Vicary and I noticed some interesting connections between wormholes and quantum entanglement. We now have a paper out!

In it, we study quantum gravity in a universe where space is just 2-dimensional, not 3-dimensional like ours. It’s not realistic, but it has one huge advantage: there’s a working theory of what quantum gravity could be like when space is 2-dimensional, so you can calculate stuff!

So, we calculate what happens when a wormhole forms, and we show the ends look like a particle and its antiparticle (this was already known), and we note that this particle-antiparticle pair is entangled. In fact it’s completely entangled: any piece of information you might want to know about one can also be found in the other.

However, in a sense that Jamie and I make precise, this entanglement is ‘fake’. The reason is that the two ends of the wormhole are not independent things. They’re just two views of the same thing… and, technically, it doesn’t count as entanglement when something is ‘entangled with itself’. This fact is crucial to how Maldacena and Susskind want to get around the firewall problem.

For more details, try this:

Wormholes and entanglement, The n-Category Café.

This has links to other stuff, including our paper, but also some blog articles explaining the firewall problem, the paper by Maldacena and Susskind, and the original Einstein–Podolsky–Rosen and Einstein–Rosen papers (in English).

Since this quantum gravity stuff is more suited to the n-Category Café than here, I won’t enable comments here. If you want to talk, please go there. Sorry!


Quantropy (Part 4)

11 November, 2013

There’s a new paper on the arXiv:

• John Baez and Blake Pollard, Quantropy.

Blake is a physics grad student at U. C. Riverside who plans to do his thesis with me.

If you have carefully read all my previous posts on quantropy (Part 1, Part 2 and Part 3), there’s only a little new stuff here. But still, it’s better organized, and less chatty.

And in fact, Blake came up with a lot of new stuff for this paper! He studied the quantropy of the harmonic oscillator, and tweaked the analogy between statistical mechanics and quantum mechanics in an interesting way. Unfortunately, we needed to put a version of this paper on the arXiv by a deadline, and our writeup of this new work wasn’t quite ready (my fault). So, we’ll put that other stuff in a new version—or, I’m thinking now, a separate paper.

But here are two new things.

First, putting this paper on the arXiv had the usual good effect of revealing some existing work on the same topic. Joakim Munkhammar emailed me and pointed out this paper, which is free online:

• Joakim Munkhammar, Canonical relational quantum mechanics from information theory, Electronic Journal of Theoretical Physics 8 (2011), 93–108.

You’ll see it cites Garrett Lisi’s paper and pushes forward in various directions. There seems to be a typo where he writes the path integral

Z = \displaystyle{ \int e^{-\alpha S(q) } D q}

and says

In order to fit the purpose Lisi concluded that the Lagrange multiplier value \alpha \equiv 1/i \hbar. In similarity with Lisi’s approach we shall also assume that the arbitrary scaling-part of the constant \alpha is in fact 1/\hbar.

I’m pretty sure he means 1/i\hbar, given what he writes later. However, he speaks of ‘maximizing entropy’, which is not quite right for a complex-valued quantity; Blake and I prefer to give this new quantity a new name, and speak of ‘finding a stationary point of quantropy’.

But in a way these are small issues; being a mathematician, I’m more quick to spot tiny technical defects than to absorb significant new ideas, and it will take a while to really understand Munkhammar’s paper.

Second, while writing our paper, Blake and I noticed another similarity between the partition function of a classical ideal gas and the partition function of a quantum free particle. Both are given by an integral like this:

Z = \displaystyle{\int e^{-\alpha S(q) } D q }

where S is a quadratic function of q \in \mathbb{R}^n. Here n is the number of degrees of freedom for the particles in the ideal gas, or the number of time steps for a free particle on a line (where we are discretizing time). The only big difference is that

\alpha = 1/kT

for the ideal gas, but

\alpha = 1/i \hbar

for the free particle.

In both cases there’s an ambiguity in the answer! The reason is that to do this integral, we need to pick a measure D q. The obvious guess is Lebesgue measure

dq = dq_1 \cdots dq_n

on \mathbb{R}^n. But this can’t be right, on physical grounds!

The reason is that the partition function Z needs to be dimensionless, but d q has units. To correct this, we need to divide dq by some dimensionful quantity to get D q.

In the case of the ideal gas, this dimensionful quantity involves the ‘thermal de Broglie wavelength’ of the particles in the gas. And this brings Planck’s constant into the game, even though we’re not doing quantum mechanics: we’re studying the statistical mechanics of a classical ideal gas!

That’s weird and interesting. It’s not the only place where we see that classical statistical mechanics is incomplete or inconsistent, and we need to introduce some ideas from quantum physics to get sensible answers. The most famous one is the ultraviolet catastrophe. What are all rest?

In the case of the free particle, we need to divide by a quantity with dimensions of lengthn to make

dq = dq_1 \cdots dq_n

dimensionless, since each dq_i has dimensions of length. The easiest way is to introduce a length scale \Delta x and divide each dq_i by that. This is commonly done when people study the free particle. This length scale drops out of the final answer for the questions people usually care about… but not for the quantropy.

Similarly, Planck’s constant drops out of the final answer for some questions about the classical ideal gas, but not for its entropy!

So there’s an interesting question here, about what this new length scale \Delta x means, if anything. One might argue that quantropy is a bad idea, and the need for this new length scale to make it unambiguous is just proof of that. However, the mathematical analogy to quantum mechanics is so precise that I think it’s worth going a bit further out on this limb, and thinking a bit more about what’s going on.

Some weird sort of déjà vu phenomenon seems to be going on. Once upon a time, people tried to calculate the partition functions of classical systems. They discovered they were infinite or ambiguous until they introduced Planck’s constant, and eventually quantum mechanics. Then Feynman introduced the path integral approach to quantum mechanics. In this approach one is again computing partition functions, but now with a new meaning, and with complex rather than real exponentials. But these partition functions are again infinite or ambiguous… for very similar mathematical reasons! And at least in some cases, we can remove the ambiguity using the same trick as before: introducing a new constant. But then… what?

Are we stuck in an infinite loop here? What, if anything, is the meaning of this ‘second Planck’s constant’? Does this have anything to do with second quantization? (I don’t see how, but I can’t resist asking.)


The Elitzur–Vaidman Bomb-Testing Method

24 August, 2013

Quantum mechanics forces us to refine our attitude to counterfactual conditionals: questions about what would have happened if we had done something, even though we didn’t.

“What would the position of the particle be if I’d measured that… when actually I measured its momentum?” Here you’ll usually get no definite answer.

But sometimes you can use quantum mechanics to find out what would have happened if you’d done something… when classically it seems impossible!

Suppose you have a bunch of bombs. Some have a sensor that will absorb a photon you shine on it, and make the bomb explode! Others have a broken sensor, that won’t interact with the photon at all.

Can you choose some working bombs? You can tell if a bomb works by shining a photon on it. But if it works, it blows up—and then it doesn’t work anymore!

So, it sounds impossible. But with quantum mechanics you can do it. You can find some bombs that would have exploded if you had shone photons at them!

Here’s how:

Put a light that emits a single photon at A. Have the photon hit the half-silvered mirror at lower left, so it has a 50% chance of going through to the right, and a 50% chance of reflecting and going up. But in quantum mechanics, it sort of does both!

Put a bomb at B. Recombine the photon’s paths using two more mirrors. Have the two paths meet at a second half-silvered mirror at upper right. You can make it so that if the bomb doesn’t work, the photon interferes with itself and definitely goes to C, not D.

But if the bomb works, it absorbs the photon and explodes unless the photon takes the top route… in which case, when it hits the second half-silvered mirror, it has a 50% chance of going to C and a 50% chance of going to D.

So:

• If the bomb doesn’t work, the photon has a 100% chance of going to C.

• If the bomb works, there’s a 50% chance that it absorbs the photon and explodes. There’s also a 50% chance that the bomb does not explode—and then the photon is equally likely to go to either C or D. So, the photon has a 25% chance of reaching C and a 25% chance of reaching D.

So: if you see a photon at D, you know you have a working bomb… but the bomb has not exploded!

For each working bomb there’s:

• a 50% chance that it explodes,
• a 25% chance that it doesn’t explode but you can’t tell if it works,
• a 25% chance that it doesn’t explode but you can tell that it works.

This is the Elitzur–Vaidman bomb-testing method. It was invented by Avshalom Elitzur and Lev Vaidman in 1993. One year later, physicists actually did an experiment to show this idea works… but alas, not using actual bombs!

In 1996, Kwiat showed that using more clever methods, you can reduce the percentage of wasted working bombs as close to zero as you like. And pushing the idea even further, Graeme Mitchison and Richard Jozsa showed in 1999 that you can get a quantum computer to do a calculation for you without even turning it on!

This sounds amazing, but it’s really no more amazing than the bomb-testing method I’ve already described.

References

For details, read these:

• A. Elitzur and L. Vaidman, Quantum mechanical interaction-free measurements, Found. Phys. 23 (1993), 987–997.

• Paul G. Kwiat, H. Weinfurter, T. Herzog, A. Zeilinger, and M. Kasevich, Experimental realization of “interaction-free” measurements.

• Paul G. Kwiat, Interaction-free measurements.

• Graeme Mitchison and Richard Jozsa, Counterfactual computation, Proc. Roy. Soc. Lond. A457 (2001), 1175–1194.

The picture is from the Wikipedia article, which also has other references:

Elitzur–Vaidman bomb tester, Wikipedia.

Bas Spitters pointed out this category-theoretic analysis of the issue:

• Robert Furber and Bart Jacobs, Towards a categorical account of conditional probability.


Quantum Network Theory (Part 2)

13 August, 2013

guest post by Tomi Johnson

Last time I told you how a random walk called the ‘uniform escape walk’ could be used to analyze a network. In particular, Google uses it to rank nodes. For the case of an undirected network, the steady state of this random walk tells us the degrees of the nodes—that is, how many edges come out of each node.

Now I’m going to prove this to you. I’ll also exploit the connection between this random walk and a quantum walk, also introduced last time. In particular, I’ll connect the properties of this quantum walk to the degrees of a network by exploiting its relationship with the random walk.

This is pretty useful, considering how tricky these quantum walks can be. As the parts of the world that we model using quantum mechanics get bigger and have more complicated structures, like biological network, we need all the help in understanding quantum walks that we can get. So I’d better start!

Flashback

Starting with any (simple, connected) graph, we can get an old-fashioned ‘stochastic’ random walk on this graph, but also a quantum walk. The first is the uniform escape stochastic walk, where the walker has an equal probability per time of walking along any edge leaving the node they are standing at. The second is the related quantum walk we’re going to study now. These two walks are generated by two matrices, which we called S and Q. The good thing is that these matrices are similar, in the technical sense.

We studied this last time, and everything we learned is summarized here:

Diagram outlining the main concepts (again)

where:

G is a simple graph that specifies

A the adjacency matrix (the generator of a quantum walk) with elements A_{i j} equal to unity if nodes i and j are connected, and zero otherwise (A_{i i} = 0), which subtracted from

D the diagonal matrix of degrees D_{i i} = \sum_j A_{i j} gives

L = D - A the symmetric Laplacian (generator of stochastic and quantum walks), which when normalized by D returns both

S = L D^{-1} the generator of the uniform escape stochastic walk and

Q = D^{-1/2} L D^{-1/2} the quantum walk generator to which it is similar!

Now I hope you remember where we are. Next I’ll talk you through the mathematics of the uniform escape stochastic walk S and how it connects to the degrees of the nodes in the large-time limit. Then I’ll show you how this helps us solve aspects of the quantum walk generated by Q.

Stochastic walk

The uniform escape stochastic walk generated by S is popular because it has a really useful stationary state.

To recap from Part 20 of the network theory series, a stationary state of a stochastic walk is one that does not change in time. By the master equation

\displaystyle{ \frac{d}{d t} \psi(t) = -S \psi(t)}

the stationary state must be an eigenvector of S with eigenvalue 0.

A fantastic pair of theorems hold:

• There is always a unique (up to multiplication by a positive number) positive eigenvector \pi of S with eigenvalue 0. That is, there is a unique stationary state \pi.

• Regardless of the initial state \psi(0), any solution of the master equation approaches this stationary state \pi in the large-time limit:

\displaystyle{ \lim_{t \rightarrow \infty} \psi(t) = \pi }

To find this unique stationary state, consider the Laplacian L, which is both infinitesimal stochastic and symmetric. Among other things, this means the rows of L sum to zero:

\displaystyle{ \sum_j L_{i j} = 0 }

Thus, the ‘all ones’ vector \mathbf{1} is an eigenvector of L with zero eigenvalue:

L \mathbf{1} = 0

Inserting the identity I = D^{-1} D into this equation we then find D \mathbf{1} is a zero eigenvector of S:

L \mathbf{1} =  ( L D^{-1} ) ( D \mathbf{1} ) = S ( D \mathbf{1} ) = 0

Therefore we just need to normalize this to get the large-time stationary state of the walk:

\displaystyle{ \pi = \frac{D \mathbf{1}}{\sum_i D_{i i}} }

If we write i for the basis vector that is zero except at the ith node of our graph, and 1 at that node, the inner product \langle i , \pi \rangle is large-time probability of finding a walker at that node. The equation above implies this is proportional to the degree D_{i i} of node i.

We can check this for the following graph:

Illustration of a simple graph

We find that \pi is

\displaystyle{ \left( \begin{matrix} 1/6 \\ 1/6 \\ 1/4 \\ 1/4 \\ 1/6 \end{matrix} \right) }

which implies large-time probability 1/6 for nodes 1, 2 and 5, and 1/4 for nodes 3 and 4. Comparing this to the original graph, this exactly reflects the arrangement of degrees, as we knew it must.

Math works!

The quantum walk

Next up is the quantum walk generated by Q. Not a lot is known about quantum walks on networks of arbitrary geometry, but below we’ll see some analytical results are obtained by exploiting the similarity of S and Q.

Where to start? Well, let’s start at the bottom, what quantum physicists call the ground state. In contrast to stochastic walks, for a quantum walk every eigenvector \phi_k of Q is a stationary state of the quantum walk. (In Puzzle 5, at the bottom of this page, I ask you to prove this). The stationary state \phi_0 is of particular interest physically and mathematically. Physically, since eigenvectors of the Q correspond to states of well-defined energy equal to the associated eigenvalue, \phi_0 is the state of lowest energy, energy zero, hence the name ‘ground state’. (In Puzzle 3, I ask you to prove that all eigenvalues of Q are non-negative, so zero really does correspond to the ground state.)

Mathematically, the relationship between eigenvectors implied by the similarity of S and Q means

\phi_0 \propto D^{-1/2} \pi \propto  D^{1/2} \mathbf{1}

So in the ground state, the probability of our quantum walker being found at node i is

| \langle i , \phi_0 \rangle |^2 \propto | \langle i , D^{1/2} \rangle \mathbf{1} |^2 = D_{i i}

Amazingly, this probability is proportional to the degree and so is exactly the same as \langle i , \pi \rangle, the probability in the stationary state \pi of the stochastic walk!

In short: a zero energy quantum walk Q leads to exactly the same distribution of the walker over the nodes as in the large-time limit of the uniform escape stochastic walk S. The classically important notion of degree distribution also plays a role in quantum walks!

This is already pretty exciting. What else can we say? If you are someone who feels faint at the sight of quantum mechanics, well done for getting this far, but watch out for what’s coming next.

What if the walker starts in some other initial state? Is there some quantum walk analogue of the unique large-time state of a stochastic walk?

In fact, the quantum walk in general does not converge to a stationary state. But there is a probability distribution that can be thought to characterize the quantum walk in the same way as the large-time state characterizes the stochastic walk. It’s the large-time average probability vector P.

If you didn’t know the time that had passed since the beginning of a quantum walk, then the best estimate for the probability of your measuring the walker to be at node i would be the large-time average probability

\displaystyle{ \langle i , P \rangle = \lim_{T \rightarrow \infty} \frac{1}{T} \int_0^T | \psi_i (t) |^2 d t }

There’s a bit that we can do to simplify this expression. As usual in quantum mechanics, let’s start with the trick of diagonalizing Q. This amounts to writing

\displaystyle{  Q= \sum_k \epsilon_k \Phi_k }

where \Phi_k are projectors onto the eigenvectors $\phi_k$ of Q, and \epsilon_k are the corresponding eigenvalues of Q. If we insert this equation into

\psi(t)  = e^{-Q t} \psi(0)

we get

\displaystyle{  \psi(t)  = \sum_k e^{-\epsilon_k t} \Phi_k \psi(0) }

and thus

\displaystyle{ \langle i , P \rangle = \lim_{T \rightarrow \infty} \frac{1}{T} \int_0^T | \sum_k e^{-i \epsilon_k t} \langle i, \Phi_k \psi (0) \rangle |^2 d t }

Due to the integral over all time, the interference between terms corresponding to different eigenvalues averages to zero, leaving:

\displaystyle{ \langle i , P \rangle = \sum_k | \langle i, \Phi_k \psi(0) \rangle |^2 }

The large-time average probability is then the sum of terms contributed by the projections of the initial state onto each eigenspace.

So we have a distribution that characterizes a quantum walk for a general initial state, but it’s a complicated beast. What can we say about it?

Our best hope of understanding the large-time average probability is through the term | \langle i, \Phi_0 \psi (0) \rangle |^2 associated with the zero energy eigenspace, since we know everything about this space.

For example, we know the zero energy eigenspace is one-dimensional and spanned by the eigenvector \phi_0. This means that the projector is just the usual outer product

\Phi_0 = | \phi_0 \rangle \langle \phi_0 | = \phi_0 \phi_0^\dagger

where we have normalized \phi_0 according to the inner product \langle \phi_0, \phi_0\rangle = 1. (If you’re wondering why I’m using all these angled brackets, well, they’re a notation named after Dirac that is adored by quantum physicists.)

The zero eigenspace contribution to the large-time average probability then breaks nicely into two:

\begin{array}{ccl} | \langle i, \Phi_0 \psi (0) \rangle |^2 &=& | \langle i, \phi_0\rangle \; \langle \phi_0,  \psi (0) \rangle |^2  \\  \\  &=& | \langle i, \phi_0\rangle |^2 \; | \langle \phi_0 , \psi (0) \rangle |^2 \\   \\  &=& \langle i ,  \pi \rangle \; | \langle \phi_0 ,  \psi (0) \rangle |^2 \end{array}

This is just the product of two probabilities:

• first, the probability \langle i , \pi \rangle for a quantum state in the zero energy eigenspace to be at node i, as we found above,

and

• second, the probability | \langle \phi_0, \psi (0)\rangle |^2 of being in this eigenspace to begin with. (Remember, in quantum mechanics the probability of measuring the system to have an energy is the modulus squared of the projection of the state onto the associated eigenspace, which for the one-dimensional zero energy eigenspace means just the inner product with the ground state.)

This is all we need to say something interesting about the large-time average probability for all states. We’ve basically shown that we can break the large-time probability vector P into a sum of two normalized probability vectors:

P = (1- \eta) \pi + \eta \Omega

the first \pi being the stochastic stationary state associated with the zero energy eigenspace, and the second $\Omega$ associated with the higher energy eigenspaces, with

\displaystyle{ \langle i , \Omega \rangle = \frac{ \sum_{k\neq 0} | \langle i, \Phi_k  \psi (0) \rangle |^2  }{ \eta} }

The weight of each term is governed by the parameter

\eta =  1 - | \langle \phi_0, \psi (0)\rangle |^2

which you could think of as the quantumness of the result. This is one minus the probability of the walker being in the zero energy eigenspace, or equivalently the probability of the walker being outside the zero energy eigenspace.

So even if we don’t know \Omega, we know its importance is controlled by a parameter \eta that governs how close the large-time average distribution P of the quantum walk is to the corresponding stochastic stationary distribution \pi.

What do we mean by ‘close’? Find out for yourself:

Puzzle 1. Show, using a triangle inequality, that the trace distance between the two characteristic stochastic and quantum distributions \{ \langle i , P \rangle \}_i and \{ \langle i , \pi \rangle \}_i is upper-bounded by 2 \eta.

Can we say anything physical about when the quantumness \eta is big or small?

Because the eigenvalues of Q have a physical interpretation in terms of energy, the answer is yes. The quantumness \eta is the probability of being outside the zero energy state. Call the next lowest eigenvalue \Delta = \min_{k \neq 0} \epsilon_k the energy gap. If the quantum walk is not in the zero energy eigenspace then it must be in an eigenspace of energy greater or equal to \Delta. Therefore the expected energy E of the quantum walker must bound the quantumness E \ge \eta \Delta.

This tells us that a quantum walk with a low energy is similar to a stochastic walk in the large-time limit. We already knew this was exactly true in the zero energy limit, but this result goes further.

So little is known about quantum walks on networks of arbitrary geometry that we were very pleased to find this result. It says there is a special case in which the walk is characterized by the degree distribution of the network, and a clear physical parameter that bounds how far the walk is from this special case.

Also, in finding it we learned that the difficulties of the initial state dependence, enhanced by the lack of convergence to a stationary state, could be overcome for a quantum walk, and that the relationships between quantum and stochastic walks extend beyond those with shared generators.

What next?

That’s all for the latest bit of idea sharing at the interface between stochastic and quantum systems.

I hope I’ve piqued your interest about quantum walks. There’s so much still left to work out about this topic, and your help is needed!

Other questions we have include: What holds analytically about the form of the quantum correction? Numerically it is known that the so-called quantum correction \Omega tends to enhance the probability of being found on nodes of low degree compared to \pi. Can someone explain why? What happens if a small amount of stochastic noise is added to a quantum walk? Or a lot of noise?

It’s difficult to know who is best placed to answer these questions: experts in quantum physics, graph theory, complex networks or stochastic processes? I suspect it’ll take a bit of help from everyone.

Background reading

A couple of textbooks with comprehensive sections on non-negative matrices and continuous-time stochastic processes are:

• Peter Lancaster and Miron Tismenetsky, The Theory of Matrices: with Applications, 2nd edition, Academic Press, San Diego, 1985.

• James R. Norris, Markov Chains, Cambridge University Press, Cambridge, 1997.

There is, of course, the book that arose from the Azimuth network theory series, which considers several relationships between quantum and stochastic processes on networks:

• John Baez and Jacob Biamonte, A Course on Quantum Techniques for Stochastic Mechanics, 2012.

Another couple of books on complex networks are:

• Mark Newman, Networks: An Introduction, Oxford University Press, Oxford, 2010.

• Ernesto Estrada, The Structure of Complex Networks: Theory and Applications, Oxford University Press, Oxford, 2011. Note that the first chapter is available free online.

There are plenty more useful references in our article on this topic:

• Mauro Faccin, Tomi Johnson, Jacob Biamonte, Sabre Kais and Piotr Migdał, Degree distribution in quantum walks on complex networks.

Puzzles for the enthusiastic

Sadly I didn’t have space to show proofs of all the theorems I used. So here are a few puzzles that guide you to doing the proofs for yourself:

Stochastic walks and stationary states

Puzzle 2. (For the hard core.) Prove there is always a unique positive eigenvector for a stochastic walk generated by S. You’ll need the assumption that the graph G is connected. It’s not simple, and you’ll probably need help from a book, perhaps one of those above by Lancaster and Tismenetsky, and Norris.

Puzzle 3. Show that the eigenvalues of S (and therefore Q) are non-negative. A good way to start this proof is to apply the Perron-Frobenius theorem to the non-negative matrix M = - S + I \max_i S_{i i}. This implies that M has a positive eigenvalue r equal to its spectral radius

r = \max_k | \lambda_k |

where \lambda_k are the eigenvalues of M, and the associated eigenvector v is positive. Since S = - M + I \max_i S_{i i}, it follows that S shares the eigenvectors of M and the associated eigenvalues are related by inverted translation:

\epsilon_k = - \lambda_k + \max_i S_{i i}

Puzzle 4. Prove that regardless of the initial state \psi(0), the zero eigenvector \pi is obtained in the large-time limit \lim_{t \rightarrow \infty} \psi(t) = \pi of the walk generated by S. This breaks down into two parts:

(a) Using the approach from Puzzle 5, to show that S v  = \epsilon_v v, the positivity of v and the infinitesimal stochastic property \sum_i S_{i j} = 0 imply that \epsilon_v = \epsilon_0 = 0 and thus v = \pi is actually the unique zero eigenvector and stationary state of S (its uniqueness follows from puzzle 4, you don’t need to re-prove it).

(b) By inserting the decomposition S = \sum_k \epsilon_k \Pi_k into e^{-S t} and using the result of puzzle 5, complete the proof.

(Though I ask you to use the diagonalizability of S, the main results still hold if the generator is irreducible but not diagonalizable.)

Quantum walks

Here are a couple of extra puzzles for those of you interested in quantum mechanics:

Puzzle 5. In quantum mechanics, probabilities are given by the moduli squared of amplitudes, so multiplying a state by a number of modulus unity has no physical effect. By inserting

\displaystyle{ Q= \sum_k \epsilon_k \Phi_k }

into the quantum time evolution matrix e^{-Q t}, show that if

\psi(0) = \phi_k

then

\psi(t)  = e^{ - i \epsilon_k t} \psi(0)

hence \phi_k is a stationary state in the quantum sense, as probabilities don’t change in time.

Puzzle 6. By expanding the initial state \psi(0) in terms of the complete orthogonal basis vectors \phi_k show that for a quantum walk \psi(t) never converges to a stationary state unless it began in one.


Quantum Network Theory (Part 1)

5 August, 2013

guest post by Tomi Johnson

If you were to randomly click a hyperlink on this web page and keep doing so on each page that followed, where would you end up?

As an esteemed user of Azimuth, I’d like to think you browse more intelligently, but the above is the question Google asks when deciding how to rank the world’s web pages.

Recently, together with the team (Mauro Faccin, Jacob Biamonte and Piotr Migdał) at the ISI Foundation in Turin, we attended a workshop in which several of the attendees were asking a similar question with a twist. “What if you, the web surfer, behaved quantum mechanically?”

Now don’t panic! I have no reason to think you might enter a superposition of locations or tunnel through a wall. This merely forms part of a recent drive towards understanding the role that network science can play in quantum physics.

As we’ll find, playing with quantum networks is fun. It could also become a necessity. The size of natural systems in which quantum effects have been identified has grown steadily over the past few years. For example, attention has recently turned to explaining the remarkable efficiency of light-harvesting complexes, comprising tens of molecules and thousands of atoms, using quantum mechanics. If this expansion continues, perhaps quantum physicists will have to embrace the concepts of complex networks.

To begin studying quantum complex networks, we found a revealing toy model. Let me tell you about it. Like all good stories, it has a beginning, a middle and an end. In this part, I’ll tell you the beginning and the middle. I’ll introduce the stochastic walk describing the randomly clicking web surfer mentioned above and a corresponding quantum walk. In part 2 the story ends with the bounding of the difference between the two walks in terms of the energy of the walker.

But for now I’ll start by introducing you to a graph, this time representing the internet!

If this taster gets you interested, there are more details available here:

• Mauro Faccin, Tomi Johnson, Jacob Biamonte, Sabre Kais and Piotr Migdał, Degree distribution in quantum walks on complex networks, arXiv:1305.6078 (2013).

What does the internet look like from above?

As we all know, the idea of the internet is to connect computers to each other. What do these connections look like when abstracted as a network, with each computer a node and each connection an edge?

The internet on a local scale, such as in your house or office, might look something like this:

 

Local network

with several devices connected to a central hub. Each hub connects to other hubs, and so the internet on a slightly larger scale might look something like this:

 

Regional network

What about the full global, not local, structure of the internet? To answer this question, researchers have developed representations of the whole internet, such as this one:

 

Global network

While such representations might be awe inspiring, how can we make any sense of them? Or are they merely excellent desktop wallpapers and new-age artworks?

In terms of complex network theory, there’s actually a lot that can be said that is not immediately obvious from the above representation.

For example, we find something very interesting if we plot the number of web pages with different incoming links (called degree) on a log-log axis. What is found for the African web is the following:

Power law degree distribution

This shows that very few pages are linked to by a very large number others, while a very large number of pages receive very few links. More precisely, what this shows is a power law distribution, the signature of which is a straight line on a log-log axis.

In fact, power law distributions arise in a diverse number of real world networks, human-built networks such as the internet and naturally occurring networks. It is often discussed alongside the concept of the preferential attachment; highly connected nodes seem to accumulate connections more quickly. We all know of a successful blog whose success had led to an increased presence and more success. That’s an example of preferential attachment.

It’s clear then that degree is an important concept in network theory, and its distribution across the nodes a useful characteristic of a network. Degree gives one indication of how important a node is in a network.

And this is where stochastic walks come in. Google, who are in the business of ranking the importance of nodes (web pages) in a network (the web), use (up to a small modification) the idealized model of a stochastic walker (web surfer) who randomly hops to connected nodes (follows one of the links on a page). This is called the uniform escape model, since the total rate of leaving any node is set to be the same for all nodes. Leaving the walker to wander for a long while, Google then takes the probability of the walker being on a node to rank the importance of that node. In the case that the network is undirected (all links are reciprocated) this long-time probability, and therefore the rank of the node, is proportional to the degree of the node.

So node degrees and the uniform escape model play an important role in the fields of complex networks and stochastic walks. But can they tell us anything about the much more poorly understood topics of quantum networks and quantum walks? In fact, yes, and demonstrating that to you is the purpose of this pair of articles.

Before we move on to the interesting bit, the math, it’s worth just listing a few properties of quantum walks that make them hard to analyze, and explaining why they are poorly understood. These are the difficulties we will show how to overcome below.

No convergence. In a stochastic walk, if you leave the walker to wander for a long time, eventually the probability of finding a walker at a node converges to a constant value. In a quantum walk, this doesn’t happen, so the walk can’t be characterized so easily by its long-time properties.

Dependence on initial states. In some stochastic walks the long-time properties of the walk are independent of the initial state. It is possible to characterize the stochastic walk without referring to the initialization of the walker. Such a characterization is not so easy in quantum walks, since their evolution always depends on the initialization of the walker. Is it even possible then to say something useful that applies to all initializations?

Stochastic and quantum generators differ. Those of you familiar with the network theory series know that some generators produce both stochastic and quantum walks (see part 16 for more details). However, most stochastic walk generators, including that for the uniform escape model, do not generate quantum walks and vice versa. How do we then compare stochastic and quantum walks when their generators differ?

With the task outlined, let’s get started!

Graphs and walks

In the next couple of sections I’m going to explain the diagram below to you. If you’ve been following the network theory series, in particular part 20, you’ll find parts of it familiar. But as it’s been a while since the last post covering this topic, let’s start with the basics.

Diagram outlining the main concepts

A simple graph G can be used to define both stochastic and quantum walks. A simple graph is something like this:

Illustration of a simple graph

where there is at most one edge between any two nodes, there are no edges from a node to itself and all edges are undirected. To avoid complications, let’s stick to simple graphs with a finite number n of nodes. Let’s also assume you can get from every node to every other node via some combination of edges i.e. the graph is connected.

In the particular example above the graph represents a network of n = 5 nodes, where nodes 3 and 4 have degree (number of edges) 3, and nodes 1, 2 and 5 have degree 2.

Every simple graph defines a matrix A, called the adjacency matrix. For a network with n nodes, this matrix is of size n \times n, and each element A_{i j} is unity if there is an edge between nodes i and j, and zero otherwise (let’s use this basis for the rest of this post). For the graph drawn above the adjacency matrix is

\left( \begin{matrix} 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 \end{matrix} \right)

By construction, every adjacency matrix is symmetric:

A =A^T

(the T means the transposition of the elements in the node basis) and further, because each A is real, it is self-adjoint:

A=A^\dagger

(the \dagger means conjugate transpose).

This is nice, since (as seen in parts 16 and 20) a self-adjoint matrix generates a continuous-time quantum walk.

To recap from the series, a quantum walk is an evolution arising from a quantum walker moving on a network.

A state of a quantum walk is represented by a size n complex column vector \psi. Each element \langle i , \psi \rangle of this vector is the so-called amplitude associated with node i and the probability of the walker being found on that node (if measured) is the modulus of the amplitude squared |\langle i , \psi \rangle|^2. Here i is the standard basis vector with a single non-zero ith entry equal to unity, and \langle u , v \rangle = u^\dagger v is the usual inner product.

A quantum walk evolves in time according to the Schrödinger equation

\displaystyle{ \frac{d}{d t} \psi(t)= - i H \psi(t) }

where H is called the Hamiltonian. If the initial state is \psi(0) then the solution is written as

\psi(t) = \exp(- i t H) \psi(0)

The probabilities | \langle i , \psi (t) \rangle |^2 are guaranteed to be correctly normalized when the Hamiltonian H is self-adjoint.

There are other matrices that are defined by the graph. Perhaps the most familiar is the Laplacian, which has recently been a topic on this blog (see parts 15, 16 and 20 of the series, and this recent post).

The Laplacian L is the n \times n matrix

L = D - A

where the degree matrix D is an n \times n diagonal matrix with elements given by the degrees

\displaystyle{ D_{i i}=\sum_{j} A_{i j} }

For the graph drawn above, the degree matrix and Laplacian are:

\left( \begin{matrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 2 \end{matrix} \right)  \qquad \mathrm{and} \qquad \left( \begin{matrix} 2 & -1 & 0 & -1 & 0 \\ -1 & 2 & -1 & 0 & 0 \\ 0 & -1 & 3 & -1 & -1 \\ -1 & 0 & -1 & 3 & -1 \\ 0 & 0 & -1 & -1 & 2 \end{matrix} \right)

The Laplacian is self-adjoint and generates a quantum walk.

The Laplacian has another property; it is infinitesimal stochastic. This means that its off diagonal elements are non-positive and its columns sum to zero. This is interesting because an infinitesimal stochastic matrix generates a continuous-time stochastic walk.

To recap from the series, a stochastic walk is an evolution arising from a stochastic walker moving on a network.

A state of a stochastic walk is represented by a size n non-negative column vector \psi. Each element \langle i , \psi \rangle of this vector is the probability of the walker being found on node i.

A stochastic walk evolves in time according to the master equation

\displaystyle{ \frac{d}{d t} \psi(t)= - H \psi(t) }

where H is called the stochastic Hamiltonian. If the initial state is \psi(0) then the solution is written

\psi(t) = \exp(- t H) \psi(0)

The probabilities \langle i , \psi (t) \rangle are guaranteed to be non-negative and correctly normalized when the stochastic Hamiltonian H is infinitesimal stochastic.

So far, I have just presented what has been covered on Azimuth previously. However, to analyze the important uniform escape model we need to go beyond the class of (Dirichlet) generators that produce both quantum and stochastic walks. Further, we have to somehow find a related quantum walk. We’ll see below that both tasks are achieved by considering the normalized Laplacians: one generating the uniform escape stochastic walk and the other a related quantum walk.

Normalized Laplacians

The two normalized Laplacians are:

• the asymmetric normalized Laplacian S = L D^{-1} (that generates the uniform escape Stochastic walk) and

• the symmetric normalized Laplacian Q = D^{-1/2} L D^{-1/2} (that generates a Quantum walk).

For the graph drawn above the asymmetric normalized Laplacian S is

\left( \begin{matrix} 1 & -1/2 & 0 & -1/3 & 0 \\ -1/2 & 1 & -1/3 & 0 & 0 \\ 0 & -1/2 & 1 & -1/3 & -1/2 \\ -1/2 & 0 & -1/3 & 1 & -1/2 \\ 0 & 0 & -1/3 & -1/3 & 1 \end{matrix} \right)

The identical diagonal elements indicates that the total rates of leaving each node are identical, and the equality within each column of the other non-zero elements indicates that the walker is equally likely to hop to any node connected to its current node. This is the uniform escape model!

For the same graph the symmetric normalized Laplacian Q is

\left( \begin{matrix} 1 & -1/2 & 0 & -1/\sqrt{6}  & 0 \\ -1/2 & 1 & -1/\sqrt{6}  & 0 & 0 \\ 0 & -1/\sqrt{6}  & 1 & -1/3 & -1/\sqrt{6}  \\ -1/\sqrt{6} & 0 & -1/3 & 1 & -1/\sqrt{6}  \\ 0 & 0 & -1/\sqrt{6}  & -1/\sqrt{6}  & 1 \end{matrix} \right)

That the diagonal elements are identical in the quantum case indicates that all nodes are of equal energy, this is type of quantum walk usually considered.

Puzzle 1. Show that in general S is infinitesimal stochastic but not self-adjoint.

Puzzle 2. Show that in general Q is self-adjoint but not infinitesimal stochastic.

So a graph defines two matrices: one S that generates a stochastic walk, and one Q that generates a quantum walk. The natural question to ask is whether these walks are related. The answer is that they are!

Underpinning this relationship is the mathematical property that S and Q are similar. They are related by the following similarity transformation

S = D^{1/2} Q D^{-1/2}

which means that any eigenvector \phi_k of Q associated to eigenvalue \epsilon_k gives a vector

\pi_k \propto D^{1/2} \phi_k

that is an eigenvector of S with the same eigenvalue! To show this, insert the identity I = D^{-1/2} D^{1/2} into

Q \phi_k = \epsilon_k \phi_k

and multiply from the left with D^{1/2} to obtain

\begin{aligned} (D^{1/2}  Q D^{-1/2} ) (D^{1/2} \phi_k) &= \epsilon_k ( D^{1/2} \phi_k ) \\ S \pi_k &= \epsilon_k \pi_k  \end{aligned}

The same works in the opposite direction. Any eigenvector \pi_k of S gives an eigenvector

\phi_k \propto D^{-1/2} \pi_k

of Q with the same eigenvalue \epsilon_k.

The mathematics is particularly nice because Q is self-adjoint. A self-adjoint matrix is diagonalizable, and has real eigenvalues and orthogonal eigenvectors.

As a result, the symmetric normalized Laplacian can be decomposed as

Q = \sum_k \epsilon_k \Phi_k

where \epsilon_k is real and \Phi_k are orthogonal projectors. Each \Phi_k acts as the identity only on vectors in the space spanned by \phi_k and as zero on all others, such that

\Phi_k \Phi_\ell = \delta_{k \ell} \Phi_k.

Multiplying from the left by D^{1/2} and the right by D^{-1/2} results in a similar decomposition for S:

S = \sum_k \epsilon_k \Pi_k

with orthogonal projectors

\Pi_k  = D^{1/2} \Phi_k D^{-1/2}

I promised above that I would explain the following diagram:

Diagram outlining the main concepts (again)

Let’s summarize what it represents now:

G is a simple graph that specifies

A the adjacency matrix (generator of a quantum walk), which subtracted from

D the diagonal matrix of the degrees gives

L the symmetric Laplacian (generator of stochastic and quantum walks), which when normalized by D returns both

S the generator of the uniform escape stochastic walk and

Q the quantum walk generator to which it is similar!

What next?

Sadly, this is where we’ll finish for now.

We have all the ingredients necessary to study the walks generated by the normalized Laplacians and exploit the relationship between them.

Next time, in part 2, I’ll talk you through the mathematics of the uniform escape stochastic walk S and how it connects to the degrees of the nodes in the long-time limit. Then I’ll show you how this helps us solve aspects of the quantum walk generated by Q.

In other news

Before I leave you, let me tell you about a workshop the ISI team recently attended (in fact helped organize) at the Institute of Quantum Computing, on the topic of quantum computation and complex networks. Needless to say, there were talks on papers related to quantum mechanics and networks!

Some researchers at the workshop gave exciting talks based on numerical examinations of what happens if a quantum walk is used instead of a stochastic walk to rank the nodes of a network:

• Giuseppe Davide Paparo and Miguel Angel Martín-Delgado, Google in a quantum network, Sci. Rep. 2 (2012), 444.

• Eduardo Sánchez-Burillo, Jordi Duch, Jesús Gómez-Gardenes and David Zueco, Quantum navigation and ranking in complex networks, Sci. Rep. 2 (2012), 605.

Others attending the workshop have numerically examined what happens when using quantum computers to represent the stationary state of a stochastic process:

• Silvano Garnerone, Paolo Zanardi and Daniel A. Lidar, Adiabatic quantum algorithm for search engine ranking, Phys. Rev. Lett. 108 (2012), 230506.

It was a fun workshop and we plan to organize/attend more in the future!


Coherence for Solutions of the Master Equation

10 July, 2013

guest post by Arjun Jain

I am a master’s student in the physics department of the Indian Institute of Technology Roorkee. I’m originally from Delhi. Since some time now, I’ve been wanting to go into Mathematical Physics. I hope to do a PhD in that. Apart from maths and physics, I am also quite passionate about art and music.

Right now I am visiting John Baez at the Centre for Quantum Technologies, and we’re working on chemical reaction networks. This post can be considered as an annotation to the last paragraph of John’s paper, Quantum Techniques for Reaction Networks, where he raises the question of when a solution to the master equation that starts as a coherent state will remain coherent for all times. Remember, the ‘master equation’ describes the random evolution of collections of classical particles, and a ‘coherent state’ is one where the probability distribution of particles of each type is a Poisson distribution.

If you’ve been following the network theory series on this blog, you’ll know these concepts, and you’ll know the Anderson-Craciun-Kurtz theorem gives many examples of coherent states that remain coherent. However, all these are equilibrium solutions of the master equation: they don’t change with time. Moreover they are complex balanced equilibria: the rate at which any complex is produced equals the rate at which it is consumed.

There are also non-equilibrium examples where coherent states remain coherent. But they seem rather rare, and I would like to explain why. So, I will give a necessary condition for it to happen. I’ll give the proof first, and then discuss some simple examples. We will see that while the condition is necessary, it is not sufficient.

First, recall the setup. If you’ve been following the network theory series, you can skip the next section.

Reaction networks

Definition. A reaction network consists of:

• a finite set S of species,

• a finite set K of complexes, where a complex is a finite sum of species, or in other words, an element of \mathbb{N}^S,

• a graph with K as its set of vertices and some set T of edges.

You should have in mind something like this:

where our set of species is S = \{A,B,C,D,E\}, the complexes are things like A + E, and the arrows are the elements of T, called transitions or reactions. So, we have functions

s , t : T \to K

saying the source and target of each transition.

Next:

Definition. A stochastic reaction network is a reaction network together with a function r: T \to (0,\infty) assigning a rate constant to each reaction.

From this we can write down the master equation, which describes how a stochastic state evolves in time:

\displaystyle{ \frac{d}{dt} \Psi(t) = H \Psi(t) }

Here \Psi(t) is a vector in the stochastic Fock space, which is the space of formal power series in a bunch of variables, one for each species, and H is an operator on this space, called the Hamiltonian.

From now on I’ll number the species with numbers from 1 to k, so

S = \{1, \dots, k\}

Then the stochastic Fock space consists of real formal power series in variables that I’ll call z_1, \dots, z_k. We can write any of these power series as

\displaystyle{\Psi = \sum_{\ell \in \mathbb{N}^k} \psi_\ell z^\ell }

where

z^\ell = z_1^{\ell_1} \cdots z_k^{\ell_k}

We have annihilation and creation operators on the stochastic Fock space:

\displaystyle{ a_i \Psi = \frac{\partial}{\partial z_i} \Psi }

\displaystyle{ a_i^\dagger \Psi = z_i \Psi }

and the Hamiltonian is built from these as follows:

\displaystyle{ H = \sum_{\tau \in T} r(\tau) \, ({a^\dagger}^{t(\tau)} - {a^\dagger}^{s(\tau)}) \, a^{s(\tau)} }

John explained this here (using slightly different notation), so I won’t go into much detail now, but I’ll say what all the symbols mean. Remember that the source of a transition \tau is a complex, or list of natural numbers:

s(\tau) = (s_1(\tau), \dots, s_k(\tau))

So, the power a^{s(\tau)} is really an abbreviation for a big product of annihilation operators, like this:

\displaystyle{ a^{s(\tau)} = a_1^{s_1(\tau)} \cdots a_k^{s_k(\tau)} }

This describes the annihilation of all the inputs to the transition \tau. Similarly, we define

\displaystyle{ {a^\dagger}^{s(\tau)} = {a_1^\dagger}^{s_1(\tau)} \cdots {a_k^\dagger}^{s_k(\tau)} }

and

\displaystyle{ {a^\dagger}^{t(\tau)} = {a_1^\dagger}^{t_1(\tau)} \cdots {a_k^\dagger}^{t_k(\tau)} }

The result

Here’s the result:

Theorem. If a solution \Psi(t) of the master equation is a coherent state for all times t \ge 0, then \Psi(0) must be complex balanced except for complexes of degree 0 or 1.

This requires some explanation.

First, saying that \Psi(t) is a coherent state means that it is an eigenvector of all the annihilation operators. Concretely this means

\Psi (t) = \displaystyle{\frac{e^{c(t) \cdot z}}{e^{c_1(t) + \cdots + c_k(t)}}}

where

c(t) = (c_1(t), \dots, c_k(t)) \in [0,\infty)^k

and

z = (z_1, \dots, z_k)

It will be helpful to write

\mathbf{1}= (1,1,1,...)

so we can write

\Psi (t) = \displaystyle{ e^{c(t) \cdot (z - \mathbf{1})} }

Second, we say that a complex has degree d if it is a sum of exactly d species. For example, in this reaction network:

the complexes A + C and B + E have degree 2, while the rest have degree 1. We use the word ‘degree’ because each complex \ell gives a monomial

z^\ell = z_1^{\ell_1} \cdots z_k^{\ell_k}

and the degree of the complex is the degree of this monomial, namely

\ell_1 + \cdots + \ell_k

Third and finally, we say a solution \Psi(t) of the master equation is complex balanced for a specific complex \ell if the total rate at which that complex is produced equals the total rate at which it’s destroyed.

Now we are ready to prove the theorem:

Proof. Consider the master equation

\displaystyle { \frac{d \Psi (t)}{d t} = H \psi (t) }

Assume that \Psi(t) is a coherent state for all t \ge 0. This means

\Psi (t) = \displaystyle{ e^{c(t) \cdot (z - \mathbf{1})} }

For convenience, we write c(t) simply as c, and similarly for the components c_i. Then we have

\displaystyle{ \frac{d\Psi(t)}{dt}  = (\dot{c} \cdot (z - \mathbf{1})) \, e^{c \cdot (z - \mathbf{1})}   }

On the other hand, the master equation gives

\begin{array}{ccl} \displaystyle {\frac{d\Psi(t)}{dt}} &=&  \displaystyle{ \sum_{\tau \in T} r(\tau) \, ({a^\dagger}^{t(\tau)} - {a^\dagger}^{s(\tau)}) \, a^{s(\tau)} e^{c \cdot (z - \mathbf{1})} } \\  \\  &=& \displaystyle{\sum_{\tau \in T} c^{t(\tau)} r(\tau) \, ({z}^{t(\tau)} - {z}^{s(\tau)}) e^{c \cdot (z - \mathbf{1})} } \end{array}

So,

\displaystyle{ (\dot{c} \cdot (z - \mathbf{1})) \, e^{c \cdot (z - \mathbf{1})}  =\sum_{\tau \in T} c^{t(\tau)} r(\tau) \, ({z}^{t(\tau)} - {z}^{s(\tau)}) e^{c \cdot (z - \mathbf{1})} }

As a result, we get

\displaystyle{  \dot{c}\cdot z -\dot{c}\cdot\mathbf{1} =  \sum_{\tau \in T} c^{s(\tau)} r(\tau) \, ({z}^{t(\tau)} - {z}^{s(\tau)})  }.

Comparing the coefficients of all z^\ell, we obtain the following. For \ell = 0, which is the only complex of degree zero, we get

\displaystyle { \sum_{\tau: t(\tau)=0} r(\tau) c^{s(\tau)} - \sum_{\tau\;:\; s(\tau)= 0} r(\tau) c^{s(\tau)} = -\dot{c}\cdot\mathbf{1} }

For the complexes \ell of degree one, we get these equations:

\displaystyle { \sum_{\tau\;:\; t(\tau)=(1,0,0,\dots)} r(\tau) c^{s(\tau)} - \sum_{\tau \;:\;s(\tau)=(1,0,0,\dots)} r(\tau) c^{s(\tau)}= \dot{c_1} }

\displaystyle { \sum_{\tau\; :\; t(\tau)=(0,1,0,\dots)} r(\tau) c^{s(\tau)} - \sum_{\tau\;:\; s(\tau)=(0,1,0,\dots)} r(\tau) c^{s(\tau)} = \dot{c_2} }

and so on. For all the remaining complexes \ell we have

\displaystyle { \sum_{\tau\;:\; t(\tau)=\ell} r(\tau) c^{s(\tau)} = \sum_{\tau \;:\; s(\tau)=\ell} r(\tau) c^{s(\tau)} }.

This says that the total rate at which this complex is produced equals the total rate at which it’s destroyed. So, our solution of the master equation is complex balanced for all complexes \ell of degree greater than one. This is our necessary condition.                                                                                   █

To illustrate the theorem, I’ll consider three simple examples. The third example shows that the condition in the theorem, though necessary, is not sufficient. Note that our proof also gives a necessary and sufficient condition for a coherent state to remain coherent: namely, that all the equations we listed hold, not just initially but for all times. But this condition seems a bit complicated.

Introducing amoebae into a Petri dish

Suppose that there is an inexhaustible supply of amoebae, randomly floating around in a huge pond. Each time an amoeba comes into our collection area, we catch it and add it to the population of amoebae in the Petri dish. Suppose that the rate constant for this process is 3.

So, the Hamiltonian is 3(a^\dagger -1). If we start with a coherent state, say

\displaystyle { \Psi(0)=\frac{e^{cz}}{e^c} }

then

\displaystyle { \Psi(t) = e^{3(a^\dagger -1)t} \; \frac{e^{cz}}{e^c}  = \frac{e^{(c+3t)z}}{e^{c+3t}} }

which is coherent at all times.

We can see that the condition of the theorem is satisfied, as all the complexes in the reaction network have degree 0 or 1.

Amoebae reproducing and competing

This example shows a Petri dish with one species, amoebae, and two transitions: fission and competition. We suppose that the rate constant for fission is 2, while that for competition is 1. The Hamiltonian is then

H= 2({a^\dagger}^2-a^\dagger)a + (a^\dagger-{a^\dagger}^2)a^2

If we start off with the coherent state

\displaystyle{\Psi(0) = \frac{e^{2z}}{e^2}}

we find that

\displaystyle {\Psi(t)=e^{2(z^2-z)2+(z-z^2)4} \; \Psi(0)}=\Psi(0)

which is coherent. It should be noted that the chosen initial state

\displaystyle{ \frac{e^{2z}}{e^2}}

was a complex balanced equilibrium solution. So, the Anderson–Craciun–Kurtz Theorem applies to this case.

Amoebae reproducing, competing, and being introduced

This is a combination of the previous two examples, where apart from ongoing reproduction and competition, amoebae are being introduced into the dish with a rate constant 3.

As in the above examples, we might think that coherent states could remain coherent forever here too. Let’s check that.

Assuming that this was true, if

\displaystyle{\Psi(t) = \frac{e^{c(t)z}}{e^{c(t)}} }

then c(t) would have to satisfy the following:

\dot{c}(t) = c(t)^2 + 3 -2c(t)

and

c(t)^2=2c(t)

Using the second equation, we get

\dot{c}(t) = 3 \Rightarrow c = 3t+ c_0

But this is certainly not a solution of the second equation. So, here we find that initially coherent states do not remain remain coherent for all times.

However, if we choose

\displaystyle{\Psi(0) = \frac{e^{2z}}{e^2}}

then this coherent state is complex balanced except for complexes of degree 1, since it was in the previous example, and the only new feature of this example, at time zero, is that single amoebas are being introduced—and these are complexes of degree 1. So, the condition of the theorem does hold.

So, the condition in the theorem is necessary but not sufficient. However, it is easy to check, and we can use it to show that in many cases, coherent states must cease to be coherent.


The Large-Number Limit for Reaction Networks (Part 2)

6 July, 2013

I’ve been talking a lot about ‘stochastic mechanics’, which is like quantum mechanics but with probabilities replacing amplitudes. In Part 1 of this mini-series I started telling you about the ‘large-number limit’ in stochastic mechanics. It turns out this is mathematically analogous to the ‘classical limit’ of quantum mechanics, where Planck’s constant \hbar goes to zero.

There’s a lot more I need to say about this, and lots more I need to figure out. But here’s one rather easy thing.

In quantum mechanics, ‘coherent states’ are a special class of quantum states that are very easy to calculate with. In a certain precise sense they are the best quantum approximations to classical states. This makes them good tools for studying the classical limit of quantum mechanics. As \hbar \to 0, they reduce to classical states where, for example, a particle has a definite position and momentum.

We can borrow this strategy to study the large-number limit of stochastic mechanics. We’ve run into coherent states before in our discussions here. Now let’s see how they work in the large-number limit!

Coherent states

For starters, let’s recall what coherent states are. We’ve got k different kinds of particles, and we call each kind a species. We describe the probability that we have some number of particles of each kind using a ‘stochastic state’. For starters, this is a formal power series in variables z_1, \dots, z_k. We write it as

\displaystyle{\Psi = \sum_{\ell \in \mathbb{N}^k} \psi_\ell z^\ell }

where z^\ell is an abbreviation for

z_1^{\ell_1} \cdots z_k^{\ell_k}

But for \Psi to be a stochastic state the numbers \psi_\ell need to be probabilities, so we require that

\psi_\ell \ge 0

and

\displaystyle{ \sum_{\ell \in \mathbb{N}^k} \psi_\ell = 1}

Sums of coefficients like this show up so often that it’s good to have an abbreviation for them:

\displaystyle{ \langle \Psi \rangle =  \sum_{\ell \in \mathbb{N}^k} \psi_\ell}

Now, a coherent state is a stochastic state where the numbers of particles of each species are independent random variables, and the number of the ith species is distributed according to a Poisson distribution.

Since we can pick ithe means of these Poisson distributions to be whatever we want, we get a coherent state \Psi_c for each list of numbers c \in [0,\infty)^k:

\displaystyle{ \Psi_c = \frac{e^{c \cdot z}}{e^c} }

Here I’m using another abbreviation:

e^{c} = e^{c_1 + \cdots + c_k}

If you calculate a bit, you’ll see

\displaystyle{  \Psi_c = e^{-(c_1 + \cdots + c_k)} \, \sum_{n \in \mathbb{N}^k} \frac{c_1^{n_1} \cdots c_k^{n_k}} {n_1! \, \cdots \, n_k! } \, z_1^{n_1} \cdots z_k^{n_k} }

Thus, the probability of having n_i things of the ith species is equal to

\displaystyle{  e^{-c_i} \, \frac{c_i^{n_i}}{n_i!} }

This is precisely the definition of a Poisson distribution with mean equal to c_i.

What are the main properties of coherent states? For starters, they are indeed states:

\langle \Psi_c \rangle = 1

More interestingly, they are eigenvectors of the annihilation operators

a_i = \displaystyle{ \frac{\partial}{\partial z_i} }

since when you differentiate an exponential you get back an exponential:

\begin{array}{ccl} a_i \Psi_c &=&  \displaystyle{ \frac{\partial}{\partial z_i} \frac{e^{c \cdot z}}{e^c} } \\ \\   &=& c_i \Psi_c \end{array}

We can use this fact to check that in this coherent state, the mean number of particles of the ith species really is c_i. For this, we introduce the number operator

N_i = a_i^\dagger a_i

where a_i^\dagger is the creation operator:

(a_i^\dagger \Psi)(z) = z_i \Psi(z)

The number operator has the property that

\langle N_i \Psi \rangle

is the mean number of particles of the ith species. If we calculate this for our coherent state \Psi_c, we get

\begin{array}{ccl} \langle a_i^\dagger a_i \Psi_c \rangle &=& c_i \langle a_i^\dagger \Psi_c \rangle \\  \\ &=& c_i \langle \Psi_c \rangle \\ \\ &=& c_i \end{array}

Here in the second step we used the general rule

\langle a_i^\dagger \Phi \rangle = \langle \Phi \rangle

which is easy to check.

Rescaling

Now let’s see how coherent states work in the large-numbers limit. For this, let’s use the rescaled annihilation, creation and number operators from Part 1. They look like this:

A_i = \hbar \, a_i

C_i = a_i^\dagger

\widetilde{N}_i = C_i A_i

Since

\widetilde{N}_i = \hbar N_i

the point is that the rescaled number operator counts particles not one at a time, but in bunches of size 1/\hbar. For example, if \hbar is the reciprocal of Avogadro’s number, we are counting particles in ‘moles’. So, \hbar \to 0 corresponds to a large-number limit.

To flesh out this idea some more, let’s define rescaled coherent states:

\widetilde{\Psi}_c = \Psi_{c/\hbar}

These are eigenvectors of the rescaled annihilation operators:

\begin{array}{ccl} A_i \widetilde{\Psi}_c &=& \hbar a_i \Psi_{c/\hbar}  \\  \\  &=& c_i \Psi_{c/\hbar} \\ \\  &=& c_i \widetilde{\Psi}_c  \end{array}

This in turn means that

\begin{array}{ccl} \langle \widetilde{N}_i \widetilde{\Psi}_c \rangle &=& \langle C_i A_i \widetilde{\Psi}_c \rangle \\  \\  &=& c_i \langle  C_i \widetilde{\Psi}_c \rangle \\  \\ &=& c_i \langle \widetilde{\Psi}_c \rangle \\ \\ &=& c_i \end{array}

Here we used the general rule

\langle C_i \Phi \rangle = \langle \Phi \rangle

which holds because the ‘rescaled’ creation operator C_i is really just the usual creation operator, which obeys this rule.

What’s the point of all this fiddling around? Simply this. The equation

\langle \widetilde{N}_i \widetilde{\Psi}_c \rangle = c_i

says the expected number of particles of the ith species in the state \widetilde{\Psi}_c is c_i, if we count these particles not one at a time, but in bunches of size 1/\hbar.

A simple test

As a simple test of this idea, let’s check that as \hbar \to 0, the standard deviation of the number of particles in the state \Psi_c goes to zero… where we count particle using the rescaled number operator.

The variance of the rescaled number operator is, by definition,

\langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle -   \langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle^2

and the standard deviation is the square root of the variance.

We already know the mean of the rescaled number operator:

\langle \widetilde{N}_i \widetilde{\Psi}_c \rangle = c_i

So, the main thing we need to calculate is the mean of its square:

\langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle

For this we will use the commutation relation derived last time:

[A_i , C_i] = \hbar

This implies

\begin{array}{ccl} \widetilde{N}_i^2 &=& C_i A_i C_i A_i \\  \\  &=&  C_i (C_i A_i + \hbar) A_i \\ \\  &=&  C_i^2 A_i^2 + \hbar C_i A_i \end{array}

so

\begin{array}{ccl} \langle \widetilde{N}_i^2\widetilde{\Psi}_c \rangle &=& \langle (C_i^2 A_i^2 + \hbar C_i A_i) \Psi_c \rangle \\   \\  &=&  c_i^2 + \hbar c_i  \end{array}

where we used our friends

A_i \Psi_c = c_i \Psi_c

and

\langle C_i \Phi \rangle = \langle \Phi \rangle

So, the variance of the rescaled number of particles is

\begin{array}{ccl} \langle \widetilde{N}_i^2 \widetilde{\Psi}_c \rangle  -   \langle \widetilde{N}_i \widetilde{\Psi}_c \rangle^2  &=& c_i^2 + \hbar c_i - c_i^2 \\  \\  &=& \hbar c_i \end{array}

and the standard deviation is

(\hbar c_i)^{1/2}

Good, it goes to zero as \hbar \to 0! And the square root is just what you’d expect if you’ve thought about stuff like random walks or the central limit theorem.

A puzzle

I feel sure that in any coherent state, not only the variance but also all the higher moments of the rescaled number operators go to zero as \hbar \to 0. Can you prove this?

Here I mean the moments after the mean has been subtracted. The pth moment is then

\langle (\widetilde{N}_i - c_i)^p \; \widetilde{\Psi}_c \rangle

I want this to go to zero as \hbar \to 0.

Here’s a clue that should help. First, there’s a textbook formula for the higher moments of Poisson distributions without the mean subtracted. If I understand it correctly, it gives this:

\displaystyle{ \langle N_i^m \; \Psi_c \rangle = \sum_{j = 1}^m {c_i}^j \; \left\{ \begin{array}{c} m \\ j \end{array} \right\} }

Here

\displaystyle{ \left\{ \begin{array}{c} m \\ j \end{array} \right\} }

is the number of ways to partition an m-element set into j nonempty subsets. This is called Stirling’s number of the second kind. This suggests that there’s some fascinating combinatorics involving coherent states. That’s exactly the kind of thing I enjoy, so I would like to understand this formula someday… but not today! I just want something to go to zero!

If I rescale the above formula, I seem to get

\begin{array}{ccl} \langle \widetilde{N}_i^m \; \widetilde{\Psi}_c \rangle &=& \hbar^m \langle N_i^m \Psi_{c/\hbar} \rangle \\ \\ &=& \hbar^m \; \displaystyle{ \sum_{j = 1}^m \left(\frac{c_i}{\hbar}\right)^j \left\{ \begin{array}{c} m \\ j \end{array} \right\} } \end{array}

We could plug this formula into

\langle (\widetilde{N}_i - c_i)^p \; \widetilde{\Psi}_c \rangle =  \displaystyle{ \sum_{m = 0}^p \, \binom{m}{p} \; \langle \widetilde{N}_i^m \;  \widetilde{\Psi}_c \rangle \, (-c_i)^{p - m} }

and then try to show the result goes to zero as \hbar \to 0. But I don’t have the energy to do that… not right now, anyway!

Maybe you do. Or maybe you can think of a better approach to solving this problem. The answer must be well-known, since the large-number limit of a Poisson distribution is a very important thing.


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