All of Joey Marcellino's Comments + Replies

Sure, one can always embed a game inside another one and so alter the overall expectation values how one likes. That said, we still only want to play the meta-game if it had positive expectation value, no?

1Nathaniel Monson2mo
Minor semantic quibble: I would say we always want positive expected utility, but how that translates into money/time/various intangibles can vary tremendously both situationally and from person to person.

The conclusion seems rather to be "human metabolism is less efficient than solar panels," which, while perhaps true, has limited bearing on the question of whether or not the brain is thermodynamically efficient as a computer when compared to current or future AI. The latter is the question that recent discussion has been focused on, and to which the "No - " in the title makes it seem like you're responding.

Moreover, while a quick Google search turns up 100W as the average resting power output of a person, another search suggests the brain is only responsi... (read more)

2Maxwell Clarke6mo
Well, yes, the point of my post is just to point out that the number that actually matters is the end-to-end energy efficiency — and it is completely comparable to humans. The per-flop efficiency is obviously worse. But, that's irrelevant if AI is already cheaper for a given task in real terms. I admit the title is a little clickbaity but i am responding to a real argument (that humans are still "superior" to AI because the brain is more thermodynamically efficient per-flop)

What does quantum entanglement mean for causality? Due to entanglement, there can be spacelike separated measurements such that there exists a reference frame > where it looks like measurement A precedes and has a causal influence on the outcomes of measurement B, and > also a reference frame where it looks like measurement B precedes and has a causal influence on the outcomes of measurement A.

"Causality" is already a somewhat fraught notion in fundamental physics irrespective of quantum mechanics; it's not clear that one needs to have some sort ... (read more)

Just to (hopefully) make the distinction a bit more clear:

A true copying operation would take |psi1>|0> to |psi1>|psi1>; that's to say, it would take as input one qubit in an arbitrary quantum state and a second qubit in |0>, and output two qubits in the same arbitrary quantum state that the first qubit was in. For our example, we'll take |psi1> to be an equal superposition of 0 and 1: |psi1> = |0> + |1> (ignoring normalization).

If CNOT is a copying operation, it should take (|0> + |1>)|0> to (|0> + |1>)(|0> + |... (read more)

Thank you! Some context: I am a "quantum autodidact", and I am currently reading a book Q is for Quantum, which is a very gentle, beginner-friendly introduction to quantum computing. I was thinking how it relates to the things I have read before, and then I noticed that I was confused. I looked at Wikipedia, which said that CNOT does not violate the no-cloning theorem... but I didn't understand the explanation why. I think I get it now. |00> + |11> is not a copy (looking at one qubit collapses the other), |00> + |01> + |10> + |11> would be a copy (looking at one qubit would still leave the other as |0> + |1>).