Posted on 02/23/2006 8:50:48 AM PST by edcoil
The cat, actually, being dead, voted democrat.
I must be missing something.
"I don't understand this. Did the cat turn the computer on or not?
parsy, who demands the possibilty of an answer."
I demand rigidly defined areas of doubt and uncertainty!
:)
Quantum computing gives me a headache.
But then the whole oil market would crash. The middle east would be sent to abject poverty. They would turn around and...hate us.
Hmmm...
I posted this reply before this thread was even started. Took a while to show up here though, quantum anomolies and all..
The cat, actually, being dead, voted democrat
Eureka! You have provided the clue to solve this conundrum. . .to wit:
Dead democrats make less mistakes.
parsy, who is actually now a democrat, but a live, good, one.
Ok. You sound like the kind of fellow who would know the answer to this:
How many surrealists does it take to change a light bulb?
parsy, the whimsical.
I do not read the story, yet I undertand it.
It could be retaining power from a charge, but the point is
is is actually running, just at a particular state. If it wasn't running at all, how would it know to answer? It doesn't have ESP built in...
You have to love Zero's theory. Take your pencil and drop it on the table. Now, remember that clunk and look at the pencil on the table - that, did not just occur.
The theory is that for it to fall, it has to fall half way and to fall it goes half way again and again never reaching the other surface.
Have fun with that one.
The "none of the above", while not having identifiable face, would turn out just as bad as the rest of them. "Office makes a man" as they used to say about the popes.
And the answer is "42."
No, you're not understanding, this has nothing to do with power states or whether or not the computer itself is running. Instead, the zeno effect is put to use where the photon isn't actually run through the program, but (for lack of a better description) only influenced by the program. We know that an unstable particle can never decay if it's being observed, but the observation does influence the particle and alters it. That's what's at work here.
No. The good bombs will influence your test particle differently than the bad bombs. Zeno principle at work.
No, because in the case where the photon appears at detector C, and the bomb remains unexploded, the photon never actually hit the bomb. The wavefunction takes both paths, but the wavefunction is not the photon. The wavefunction is a description of possible paths (or locations, if you will) for the photon. If the detector at B isn't working, then there's no way the photon could ever end up at C, because the wavefunction that describes its allowed paths would cancel out. The two paths to C end up with the opposite phase by construction.
So we see a photon at C, and we see no explosion. What do we know? Well, we know that the detector at B works, else we couldn't have seen the photon at C. (The photon would necessarily have taken both paths, you see, leading to the wave cancellation at C. But since in the exploding case it can't take both paths, the wave cancellation at C never occurs.) We also know that the photon took the path that didn't go past the bomb, or the bomb would have exploded. So we know that although the bomb didn't explode, it must be good.
[Geek alert: the "photon", as we use the term here to describe the thing that makes the bomb go boom, refers to the position eigenstate of the wavefunction. Careful, though: sometimes the word "photon" can refer to the momentum eigenstate of the wavefunction (as in, "what was the frequency of that photon"), and sometimes it can refer to the wavefunction itself. In this example, though, what matters to the bomb is where the photon goes.]
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