PS: [nanotech] toasted sentience

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• ... You ve exceeded my physics training, so I can t reply to this statement. I can t say that I feel this conversation has achieved closure, since the parts I
Message 1 of 1 , Dec 9, 2000
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Mark Gubrud wrote:
>
> Spin-1/2 states such as [ 1+i , i-1 ] (normalization assumed) are pure
> states. They have entropy zero. This is very easy to show:
>
> S = - tr( R log(R) ); R = density operator
>
> Suppose U is a unitary operator, and R = U P U+. Then
>
> S = - tr( U P U+ log( U P U+ ) ).
>
> Now, A = log( B ) means B = exp(A) = 1 + A + A^2/2 + ...
>
> So, if A = log(R) = log( U P U+) then
>
> P = U+ R U = U+ exp(A) U = U+ U + U+ A U + U+ A^2/2 U + ...
>
> = 1 + U+ A U + (U+ A U)^2/2 + ... = exp( U+ A U )
>
> therefore log(P) = log( U+ R U) = U+ log(R) U and
>
> S = - tr( U P log(P) U+) = -tr( P log(P) ).
>
> So entropy is invariant under a unitary transformation (basis rotation).
>
> Now, given any normalized state vector, we can construct (Gram-Schmidt) a
> complete basis which has this state vector as one of its basis vectors.
> In this basis, the density matrix of the state will have only one nonzero
> element, so the entropy is zero. If we first wrote the density matrix in
> some other basis, we could find a unitary matrix which would bring us to
> the basis in which the density matrix has only one nonzero element.
>
> You constructed 2-particle states; if the particles are distinguishable,
> the appropriate density operator will be a 4x4 matrix which, because you
> specified a pure state, will have zero entropy at the start of your
> unspecified process. If the particles are indistinguishable, you can
> treat them as a statistical mixture described by a 2x2 density matrix of
> nonzero entropy. In either case, any type of unitary evolution, such as
> would be described by an interaction hamiltonian, will preserve the
> entropy. Measurement of the final state, which involves interaction with
> uncontrolled environmental degrees of freedom, may increase the
> entropy, but will never decrease it.

You've exceeded my physics training, so I can't reply to this statement.
I can't say that I feel this conversation has achieved closure, since the
parts I did understand did seemed to say only that unitary evolution
preserves entropy, which I already knew. But perhaps you provided a
stunning refutation that simply exceeded my physical-sciences
understanding. I'm afraid that I don't know enough physics to say that
you've won, but it certainly does seem I've lost.

-- -- -- -- --
Eliezer S. Yudkowsky http://singinst.org/
Research Fellow, Singularity Institute for Artificial Intelligence
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