I’m thinking about quantum computers today.

Quantum computers are supposed to be “better” than ordinary digital computers in that they’re able to solve, in polynomial time, many problems that an ordinary digital computer can only solve in exponential time. This has enormous practical implications: notably, many cryptographic methods are based on the fact that there are mathematical problems that can only be solved in exponential time, rendering it impractical to break an encryption key by computer using any “brute force” method. However, if a quantum computer could solve the same problem in polynomial time, a “brute force” method may be practical.

But the thing is, quantum computers are not exactly unique in this respect. Any good old analog computer from the 1950s can also solve the same problems in polynomial time. At least, in principle.

And that’s the operative phrase here: in principle. An analog computer, which represents data in the form of continuous quantities such as lengths, currents, voltages, angles, etc., is limited by its accuracy: even the best analog computer rarely has an accuracy better than one part in a thousand. Not exactly helpful when you’re trying to factorize 1000-digit numbers, for instance.

A quantum computer also represents data in the form of a continuous quantity: the (phase of the) wave function. Like an analog computer, a quantum computer is also limited in accuracy: this limitation is known as decoherence, when the wave function collapses into one of its eigenstates, as if a measurement had been performed.

So why bother with quantum computers, then? Simple: it is widely believed that it is possible to restore coherence in a quantum computer. If this is indeed possible, then a quantum computer is like an analog computer on steroids: any intermediate calculations could be carried out to arbitrary precision, only the final measurement (i.e., reading out the result) would be subject to a classical measurement error, which is not really a big issue when the final result, for instance, is a yes/no type result.

So that’s what quantum computing boils down to: “redundant qubits” that can ensure that coherence is maintained throughout a calculation. Many think that this can be done… I remain somewhat skeptical.

According the CNN, it is confirmed by the Pentagon: Iran successfully launched an orbital satellite.

This is a tremendous accomplishment for a nation that exists in economic isolation.

On the other hand, it is a cause for tremendous concern: the missile belongs to a nation that has been openly advocating the destruction of Israel, and is likely in the advanced stages of a nuclear weapons program.

I guess what it boils down to is two questions: 1) Are the ayatollahs crazy enough to try to nuke Israel or lob an ICBM over the Atlantic? 2) Are other parties worried enough to start a major war by launching a preemptive strike against Iran?

If the answer is a yes to either of these questions, lots of people will die and lots of unpleasant things will happen to lots of other people.

I’m reading Robert Wald’s book, Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, and I am puzzled. According to Wald, the black hole equivalent of the First Law reads (for a Kerr black hole):

(1/8π)κdA = dM – ΩdJ,

where κ is the surface gravity, A is the area of the event horizon, M is the mass, Ω is the angular velocity of the event horizon, and J is the black hole’s angular momentum.

The analogy with thermodynamics is obvious if one write the First Law as

TdS = dU + pdV,

where T is the temperature, S is the entropy, U is the internal energy, p is the pressure, and V is the volume. Further, as per the black hole area theorem, which Wald proves, A always increases, in analogy with the thermodynamical entropy.

But… if I am to take this analogy seriously, then I am reminded of the fact that in a thermodynamical system the temperature is determined as a function of pressure and volume, i.e., there is a function f such that T = f(p, V). Is there an analogue of this in black hole physics? Is the surface gravity κ fully determined as a function of Ω and J? It is not obvious to me that this is the case, and Wald doesn’t say. Yet without it, there is no zeroth law and no thermodynamics. He does mention the zeroth law in the context of a single black hole having uniform surface gravity, but that’s not good enough. It doesn’t tell me how the surface gravity can be calculated from Ω and J alone, nor does it tell me anything about more than one black hole being involved, whereas in thermodynamics, the zeroth law is about multiple thermodynamical systems being in thermal equilibrium.

Another puzzling aspect is that the area theorem has often been quoted as “proof” that a black hole cannot evaporate. Yet again, if I take the analogy with thermodynamics seriously, the Second Law applies only to closed systems that exchange neither matter nor energy with their environment; it is, in fact, quite possible to reduce S in an open system, otherwise your fridge would not work. So if a black hole can exchange energy and matter with its environment, perhaps it can evaporate after all.

Moreover, for the analogy to be complete, we’d also be required to have

8π∂M/dA = κ,
M/∂J = Ω,

just as in ordinary thermodynamics, we have T = ∂U/∂S and p = –∂U/∂V. So, do these relationships hold for black holes?

I guess I’ll go to ArXiv and read some recent papers on black hole thermodynamics.

These are not unusual pictures for us up here in the Great White North:

Trouble is, these pictures are not from Ottawa, Toronto, or London, Ontario. They are from London, England, where it hasn’t stopped snowing yet.