<![CDATA[Quantum Network Theory (Part 2)]]>

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.

• $A$ 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

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

• $Q = D^{-1/2} L D^{-1/2}$ 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:

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

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

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

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 $i.$

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

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.

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

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

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

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

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

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:

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

• 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,

• 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:

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

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

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 $\pi.$

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 $2 \eta.$

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:

• 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:

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

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:

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

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.