Comment author: Luke_A_Somers 26 April 2012 03:16:57PM *  0 points [-]

You have already asked these 3 questions and had them answered: http://lesswrong.com/lw/btv/seq_rerun_on_being_decoherent/6f5f

To clarify the answer at point 3, if you phase shift by half a cycle and add, well, that's called 'subtraction'.

Comment author: DaveInNYC 23 May 2012 06:44:44PM 0 points [-]

Thanks; sorry about the duplicate question post, I had not been able to find the "replay" version of this particular article.

Comment author: DavidPlumpton 19 April 2012 09:11:13AM 0 points [-]

Does anybody else not like the general phrasing "The system is in the superposition STATE1 + STATE2" ?

The way I'm thinking of it there is no such thing as a superposition. There is simply more than one configuration in the (very recent) past that contributes a significant amount of amplitude to the "current" configuration.

Have I got this wrong?

Comment author: DaveInNYC 19 April 2012 04:00:57PM 0 points [-]

I think it is a good way to map what people have commonly called "superposition," but the sentence should probably be "The system is in the superposition STATE1 + STATE2, relative to STATE 3, where STATE 3 roughly factors out". STATE 3 in this case is usually an observer. I mean, if I flip a "quantum coin" and I have not told you if it is heads or tails, then the coin (and I) is in a superposition of "HEADS + TAILS" relative to you, but due to decoherence on my end, it is not in a superposition relative to me. For me this was an important concept to learn, as it helped me see that "many worlds" is a local and non-discrete phenomenon.

Comment author: DaveInNYC 19 April 2012 01:03:58AM 1 point [-]

And another quantum-related question. - In The Fabric of the Cosmos by Brian Greene (p. 196), he describes a setup of the two slit experiment where half of the particles have their "which way" information recorded, thus causing decoherence and not showing an interference pattern, and the other half of the particles are not measured, and thus do show an interference pattern. After the fact one can look at which photons were not measured, and these do indeed form the interference pattern.

However, he then goes on to explain an identical setup, with the difference that the decision as to whether to measure the 1/2 of the particles can be made many (light) years after the photons register on the screen, and only later, when the person making this decision light years away comes and tells you whether they measured or not, do you see if the unmeasured photons make an interference pattern.

This would all make sense to me IF there was no way to distinguish a totally non-interfering pattern, and a non-interfering pattern overlaid with an interfering one. Intuitively it seems like one WOULD be able to distinguish this, with a pretty high degree of confidence, by subtracting an "average" non-interfering pattern from the total pattern. Is this not the case?

BTW, I have been re-reading the QM sequence every 6 months or so since it was first posted, and get a bit more out of it each time. I am AMAZED at how it has explained things that, before reading it, seemed so freaky and inexplicable to me that it bordered on the supernatural.

Comment author: DaveInNYC 19 April 2012 12:51:36AM 0 points [-]

So this is sorta off-topic for this thread, but I cannot see where one can start a new one. I posted the following questions at http://lesswrong.com/lw/q2/spooky_action_at_a_distance_the_nocommunication/, as I cannot find the "rerun" version of it. Anyway, here goes. FWIW, the topic was about EPR experiments.

For all these types of experiments, how do they "aim" the particle so it hits its target from far away? It would seem that the experimenters would know pretty much where the particle is when it shoots out of the gun (or whatever), so would not the velocity be all over the place? In the post on the Heisenberg principle, there was an example of letting the sun shine through a hole in a piece of paper, which caused the photons to spread pretty widely, pretty quickly.

Does the polarization vector change as the photon moves along? It seems to be very similar to a photon's "main" wave function, as it can be represented as a complex number (and is even displayed as an arrow, like Feynman uses). But I know those Feynman arrows spin according to the photon's wavelength.

Finally - and this is really tripping me up - why can we put in the minus sign in the equation that you say "we will need" later, instead of a + sign? If you have two blobs of amplitude, you need to add them to get the wave function, yes? If that is not the case, I have SEVERELY misunderstood the most basic posts of this sequence.

Comment author: DaveInNYC 15 April 2012 06:15:23PM 2 points [-]

(I can't find the "rerun" version of this page, so am posting my questions here).

  1. For all these types of experiments, how do they "aim" the particle so it hits its target from far away? It would seem that the experimenters would know pretty much where the particle is when it shoots out of the gun (or whatever), so would not the velocity be all over the place? In the post on the Heisenberg principle, there was an example of letting the sun shine through a hole in a piece of paper, which caused the photons to spread pretty widely, pretty quickly.

  2. Does the polarization vector change as the photon moves along? It seems to be very similar to a photon's "main" wave function, as it can be represented as a complex number (and is even displayed as an arrow, like Feynman uses). But I know those Feynman arrows spin according to the photon's wavelength.

  3. Finally - and this is really tripping me up - why can we put in the minus sign in the equation that you say "we will need" later, instead of a + sign? If you have two blobs of amplitude, you need to add them to get the wave function, yes? If that is not the case, I have SEVERELY misunderstood the most basic posts of this sequence.

Comment author: DaveInNYC 07 April 2012 01:48:54AM 0 points [-]

In a previous post in this series, it was stated that if you shot the particles towards the mirrors at different times, but that difference was vanishingly small, then you would still see the same results, except for there would be a correspondingly vanishingly small chance that you would see both detectors register a single particle, since configurations were "smudgy". Why would not the same apply to two electrons that were distinguishable, but their differences were vanishingly small?

Comment author: Kevin 28 March 2011 01:16:38AM 2 points [-]

Bots only win at 1v1 limit poker. No bot can play professional no-limit poker, especially at a full table.

Again, the best humans are much, much better at poker than the best bots. The idea that optimal NL poker is computationally solvable and not demanding is just wrong. No one has solved it yet.

Comment author: DaveInNYC 31 March 2011 10:03:15PM 0 points [-]

Bots only win at 1v1 limit poker. No bot can play professional no-limit poker, especially at a full table.

I believe that they can win LIMIT poker at a full table; does not have to be 1-1 in that case.

Comment author: Manfred 25 March 2011 06:12:23PM *  5 points [-]

The a priori argument that using money is important doesn't stand up under closer examination. If you are incapable of generalizing from in-game currency to dollars, you won't be capable of generalizing from poker to other activities. And player behavior does not seem to be grossly different - take for example the fact that prediction markets work the same with real money or fake money.

Comment author: DaveInNYC 31 March 2011 09:59:30PM 2 points [-]

The a priori argument that using money is important doesn't stand up under closer >examination. If you are incapable of generalizing from in-game currency to dollars, you >won't be capable of generalizing from poker to other activities. And player behavior >does not seem to be grossly different - take for example the fact that prediction markets >work the same with real money or fake money.

It may not be much different for prediction markets, but it is VERY different for online poker. Even if you play exactly the same with or without money, your opponents will not, and therefore you will be "training" on different data than you think. This applies especially to NL games; risking 1,000 points on one bet is a lot different than risking 1,000 dollars.

Comment author: Perplexed 28 March 2011 04:10:15PM -1 points [-]

The difference between game theory and decision theory is that in game theory you need to worry about not just what is rational for you to do, you also have to consider what is rational for the other players to do.

When you play online poker, you are placing complete trust in your analysis of the trustworthiness of the 'house'. The house could cheat you, if it desires, and probably not get caught. But you analyze that cheating would be an irrational thing for the house to do an a large scale - because large scale cheating would get caught and they would lose customers.

So, what is your analysis of what they would do to a small number of their customers who violate their rules by using real-time machine assistance? They can't take those people to court. People who, if they were allowed to get away with it, would destroy the online poker business. Would it be rational for the Poker houses to try to cheat the rule-breakers?

Would they actually do that?

I used to count cards at blackjack. And when I did it in Reno, at a certain stage a new dealer would be brought to the table (outside the normal shift schedule). And from that point on, I would lose money. If I watched closely, I could see them dealing seconds.

Comment author: DaveInNYC 31 March 2011 09:46:26PM 0 points [-]

Would it be rational for the Poker houses to try to cheat the rule-breakers?

While I can't vouch for every single poker site out there, the chances of them trying to cheat you are almost nil. The amount of marginal income they would make compared to the risk of a tarnished reputation makes it a foolish play (note that to cheat you, they would need to not only rig the game, but have a shill in there to divert the money to).

I used to count cards at blackjack. And when I did it in Reno, at a certain stage a >new dealer would be brought to the table (outside the normal shift schedule). And >from that point on, I would lose money. If I watched closely, I could see them dealing >seconds.

Unless this was 1950, I believe your eyes fooled you. They would ban you, yes, but the idea of a Reno casino trying to win their money "back" via slight of hand is a little silly.

Comment author: DaveInNYC 31 March 2011 03:04:00PM 0 points [-]

So in the following transformation:

X1Y1 -> X2Y1 X1Y2 -> X4Y1 X1Y3 -> X6Y1 X1Y4 -> X8Y1

You say that while true entropy has not increased (it stays at 2 bits), apparent entropy has, due to the observer not keeping track of X and just lumping its possible states into X2-X8. If this is the case, why doesn't observed entropy decrease as well, since phase space is preserved with the following?

X2Y1 -> X1Y1 X4Y1 -> X1Y2 X6Y1 -> X1Y3 X8Y1 -> X1Y4

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