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# Stephen comments on Decoherence is Pointless - Less Wrong

8 29 April 2008 06:38AM

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Comment author: 29 April 2008 01:53:18PM 4 points [-]

"The physicists imagine a matrix with rows like Sensor=0.0000 to Sensor=9.9999, and columns like Atom=0.0000 to Atom=9.9999; and they represent the final joint amplitude distribution over the Atom and Sensor, as a matrix where the amplitude density is nearly all in the diagonal elements. Joint states, like (Sensor=1.234 * Atom=1.234), get nearly all of the amplitude; and off-diagonal elements like (Sensor=1.234 * Atom=5.555) get an only infinitesimal amount."

This is not what physicists mean when they refer to off-diagonal matrix elements. They are talking about the off diagonal matrix elements of a density matrix. In a density matrix the rows and columns both refer to the same system. It is not a matrix with rows corresponding to states of one subsystem and columns corresponding to states of another. To put it differently, the density matrix is made by an outer product, whereas the matrix you have formulated is a tensor product. Notice if the atom and sensor were replaced by discrete systems, then if these systems didn't have an equal number of states then your matrix would not be square. In that case the notion of diagonal elements doesn't even make sense.

Comment author: 05 March 2011 07:48:29AM 0 points [-]

Perhaps it still works as long as he was saying not Sensor(0) but Sensor(Atom(0)) ie "Sensor says Atom is in state 0"

So it doesn't matter how many states the sensor has, it's how that state reflects on the state of the Atom that matters. Then the states could correspond one-to-one regardless of how many states the sensor has, and the probability would be concentrated in the diagonal.