Spud
Oct11-06, 03:07 PM
http://xxx.lanl.gov/abs/quant-ph/0505204
> David McAnally wrote:
> In fact, no matter what the state of the m+2 photons is, the probability
> that the message receiver will receive is 0 is independent of whether the
> sender sends a 0 or a 1. Suppose the density operator for the m+2 photons
> is
> \rho = \sum_{i_0,...,i_{m+1},j_1,...,j_{m+1}} p[i_0 i_1...i_{m+1},j_0 j_1
> ...j_{m+1}] |i_0 i_1...i_{m+1}><j_0 j_1...j_{m+1}|,
> for complex numbers p[i_0...i_{m+1},j_0...j_{m+1}], noting specifically
> that is a positive semidefinite self-adjoint operator of trace 1.
> If the sender sends a 0, then the state of the sender's photon after
> measurement is either |0> or |1>. The probability that the sender's
> photon is in state |0> and the receiver receives 0 is
> p[000...0,000...0] + p[011...1,011...1],
> and the probability that the sender's photon is in state |1> and the
> receiver receives 0 is
> p[100...0,100...0] + p[11...1,111...1],
> and so the nett probability that the receiver receives a 0, given that
> the sender sends 0, is
> p[000...0,000...0] + p[011...1,011...1] + p[100...0,100...0]
> + p[11...1,111...1].
> If the sender sends a 1, then the state of the sender's photon after
> measurement is either
> (1/sqrt(2)) (|0> + |1>)
> or
> (1/sqrt(2)) (|0> - |1>).
> The probability that the sender's photon is in state
> (1/sqrt(2)) (|0> + |1>)
> and the receiver receives 0 is
> (1/2) (p[000...0,000...0] + p[011...1,011...1]
> + p[000...0,100...0] + p[011...1,111...1]
> + p[100...0,000...0] + p[111...1,011...1]
> + p[100...0,100...0] + p[111...1,111...1]),
> and the probability that the sender's photon is in state
> (1/sqrt(2)) (|0> - |1>).
> and the receiver receives 0 is
> (1/2) (p[000...0,000...0] + p[011...1,011...1]
> - p[000...0,100...0] - p[011...1,111...1]
> - p[100...0,000...0] - p[111...1,011...1]
> + p[100...0,100...0] + p[111...1,111...1]),
> and so the nett probability that the receiver receives a 0, given that
> the sender sends 1, is
> p[000...0,000...0] + p[011...1,011...1] + p[100...0,100...0]
> + p[11...1,111...1].
> And so the probability of the receiver receiving 0 is independent of
> whether the sender sends 0 or 1.
> -----
No useful information can be received although there are vauge hints
the problem has been solved by reseting the receivers ? using, ? Wait
for the experiment in other words.
Is this hint just wishful thinking ?
Spud
> David McAnally wrote:
> In fact, no matter what the state of the m+2 photons is, the probability
> that the message receiver will receive is 0 is independent of whether the
> sender sends a 0 or a 1. Suppose the density operator for the m+2 photons
> is
> \rho = \sum_{i_0,...,i_{m+1},j_1,...,j_{m+1}} p[i_0 i_1...i_{m+1},j_0 j_1
> ...j_{m+1}] |i_0 i_1...i_{m+1}><j_0 j_1...j_{m+1}|,
> for complex numbers p[i_0...i_{m+1},j_0...j_{m+1}], noting specifically
> that is a positive semidefinite self-adjoint operator of trace 1.
> If the sender sends a 0, then the state of the sender's photon after
> measurement is either |0> or |1>. The probability that the sender's
> photon is in state |0> and the receiver receives 0 is
> p[000...0,000...0] + p[011...1,011...1],
> and the probability that the sender's photon is in state |1> and the
> receiver receives 0 is
> p[100...0,100...0] + p[11...1,111...1],
> and so the nett probability that the receiver receives a 0, given that
> the sender sends 0, is
> p[000...0,000...0] + p[011...1,011...1] + p[100...0,100...0]
> + p[11...1,111...1].
> If the sender sends a 1, then the state of the sender's photon after
> measurement is either
> (1/sqrt(2)) (|0> + |1>)
> or
> (1/sqrt(2)) (|0> - |1>).
> The probability that the sender's photon is in state
> (1/sqrt(2)) (|0> + |1>)
> and the receiver receives 0 is
> (1/2) (p[000...0,000...0] + p[011...1,011...1]
> + p[000...0,100...0] + p[011...1,111...1]
> + p[100...0,000...0] + p[111...1,011...1]
> + p[100...0,100...0] + p[111...1,111...1]),
> and the probability that the sender's photon is in state
> (1/sqrt(2)) (|0> - |1>).
> and the receiver receives 0 is
> (1/2) (p[000...0,000...0] + p[011...1,011...1]
> - p[000...0,100...0] - p[011...1,111...1]
> - p[100...0,000...0] - p[111...1,011...1]
> + p[100...0,100...0] + p[111...1,111...1]),
> and so the nett probability that the receiver receives a 0, given that
> the sender sends 1, is
> p[000...0,000...0] + p[011...1,011...1] + p[100...0,100...0]
> + p[11...1,111...1].
> And so the probability of the receiver receiving 0 is independent of
> whether the sender sends 0 or 1.
> -----
No useful information can be received although there are vauge hints
the problem has been solved by reseting the receivers ? using, ? Wait
for the experiment in other words.
Is this hint just wishful thinking ?
Spud