If you want, I can provide proofs that an infinitesimal stochastic leads to a stochastic

No i trust you on that one, and i think that intuitively it sort of makes sense given that exp(0)=1.

Also the other missing piece that i didn’t see before is that if all the columns of a stochastic Hamiltonian adds up to one, then **any given convex combination of these columns also must add up to one**. As a consequence, if is in the simplex so must be , and the simplex is invariant.

Thanks for reading and commenting!

I think you understand what’s going on very well.

We are indeed considering only vectors in the simplex (and ).

The evolution equations do actually have the property that if is in the simplex then so is for all .

This property is ensured whenever the classical Hamiltonian is infinitesimal stochastic (its columns sum to zero, its off-diagonals are non-positive, and diagonals are non-negative). You can then show that the evolution operator is stochastic (its elements are non-negative and columns add to unity). The stochastic map then preserves the normalization and non-negativity of the .

If you want, I can provide proofs that an infinitesimal stochastic leads to a stochastic .

Also you can check that is infinitesimal stochastic.

I think this was explained more in Part 1 of the series, in case you missed that.

Let me know if anything is still unclear!

P.S. Yes, that factor of i makes a big difference doesn’t it!? So does the fact that in the quantum case the Hamiltonian is Hermitian rather than infinitesimal stochastic – again this is required to preserve probability, which as you point out are different for quantum.

]]>Given the definition of (its elements represent the probability of the walker being found on that node) it seems we are implicitly considering only those vectors in the simplex .

If that is true then both the initial condition as well as the stationary state must lie on this surface, but this constraint does not seem (to me) to be enforced by the evolution equations. Perhaps is this the reason why you say that you “need” to normalize the zero eigenvector *D***1** ? I am puzzled by why you do this normalization otherwise.

in general if we could chose any independent of any constraint, then i would expect the “arrival” state to be some kind of projection of along the direction spanned by the unitary vector , so in that case the initial condition would still matter.

I don’t know, I am sure i am getting something wrong, (sorry if this was somewhat covered previously, i haven’t been able to find anything about it).

Other than that i think that the big difference in the behavior between stochastic and quantum walk must be due to the imaginary number that multiplies the Hamiltonian in the evolution equation, which completely changes the game. That, and the fact that now square amplitudes, not just amplitudes, represent probabilities, so the constraint now has a different shape, that is a sphere. Hopefully i am getting at least this part somewhat right :-).

]]>The squaring is weird. Taking what I think you’re saying at face value, it seems that a quantum walker must be times as likely to walk during its 2nd millisecond spent at a given node than during its 1st millisecond, so its walking amplitude must be sqrt(3) times larger for the 2nd ms than for the 1st ms. But this makes the concept ‘amplitude [to walk] per time’ seem incoherent or at least poorly named.

…so I must be confused.

Let’s look at it a bit more carefully.

Say there are just two nodes. Say the amplitude to stay at the first node is while the amplitude to hop to the second one is Since we take the absolute value squared to get from an amplitude to a probability, and probabilities sum to 1, we must have

So, and are the coordinates of a point on a circle! If the ‘amplitude to walk per time’ equals 1, this point goes around the circle at speed 1. The simplest option is this:

This works just fine:

Now suppose we wait just a short amount of time, so that is small. The amplitude to be at the first node at is

The amplitude to be at the second node is:

This increases linearly with time (for small times), so I feel justified in talking about the ‘amplitude to walk per time’: it’s 1.

However, the *probability* to be at the second node at time is

So yes, the probability to be there after 2 milliseconds is 4 times the probability to be there after 1 millisecond.

I think maybe you were mixing up the change of the square of something and the square of the change.

Philosophically, part of the point is you can only *tell* if the quantum walker has stepped from one node to another if you *look*. If you look often enough, you can reduce the probability of it doing this as small as you like. “The watched pot never boils.” But if you don’t look, it doesn’t make much sense to ask whether the walker made the step in the first millisecond or the second—since there is no way to answer this question!

This is the kind of thing that makes some people think quantum mechanics is strange. But these people are stuck in an old worldview, which ignores the danger of counterfactual questions like “what *would* I have seen *if* I’d looked, even though I didn’t?”

PS Happy Earth Overshoot Day.

]]>As John said, one implication of this is the quantum Zeno effect. If you add some on-site dephasing (equivalent to spy observing which node the walker is on but not telling you what he sees – just destroying the inter-site coherence), this will never turn the quantum walk into a classical walk. Rather it will just, for strong enough dephasing (frequent enough observation by the spy) cause the walker to stop.

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