X-Message-Number: 18490 Date: Tue, 05 Feb 2002 14:53:22 +0100 From: Henri Kluytmans <> Subject: Re: Simulation, qubits Yvan Bozzonetti wrote : >Please note: We have a 2^813 states, each is a 813 bits word. Indeed, if we had only 2^813 possible states... >To simulate the Universe (ten billion light years in radius for ten billion >years) you need 10^244 Planck's cubes ( One Planck's unit for each dimension >in space and time). This could be run by a quantum computer with a 813 bits >word in a single computation cycle. I could see this to be the case if, for example, only ONE Planck cube out of all 10^244 could be in the "on" state simultaneously for each overall state (and all the other Planck cubes would then be in the "off" state). But this doesn't seem to be the case. >This define a four dimensional classical universe. There is no storage >implied. After one computing round, everything is destroyed if there is no >coupling to a more long lived quantum state. And what would be that state you will be looking for ? I.e. what would be the state(s) you want to let the quantum computer collapse into? >What you write about, 2^(10^244), would define a simulation where each >instant of each elementary domain would interact with each other. This is a >world with time travel and multi stories: I agree, there are less than 10^244 qubits required. But certainly more than a mere 813. For example, to factor a product of two primes with a combined size of 1000 bits. You will need at least a 1000 qubits. You will use the right quantum logic to let the qubits collapse into that single state that contains the two primes. There are 2^1000 possible combinations, but there is only 1 possible collapsed state. The quantum logic will determine the collapsed state. There are 2^(2^813) possible combinations when you have 2^813 Planck cubes. Of course, because of time travel and other restrictions (e.g. the amount of matter, etc..) many possible combinations will not be allowed. At any certain point in time you have 10^183 Planck cubes. And again, not all possible combinations will or can occur. But that's usually what you use the quantum logic for : to filter out only certain allowed combinations by letting the system collapse. ****** Please explain to me why precisely 2^813 out of all 2^(2^813) combinations remain ??? What kind of projection are you using ? What kind of transformation is linking a 813 bits word to the state of the universe (i.e. the states of all 2^813 Planck cubes). Rate This Message: http://www.cryonet.org/cgi-bin/rate.cgi?msg=18490