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As I understand it, faster than light communication is not possible, but I have a specific example which concludes that it is and I'm trying to find the mistake.

The scheme uses two things

1) An entangled Bell pair ## | \phi \rangle = | 0 0 \rangle + | 1 1 \rangle## ( neglecting normalization )

2) The fact that orthogonal states can be distinguished.

To begin with ## |\phi\rangle ## is given to Alice and Bob each having a qubit.

Bob has in his positions a copying unitary that can copy states ## |0\rangle ## and ## |1\rangle ## to an ancillary qubit and he can then measure this qubit to distinguish them, explicitly the unitary is [tex] U = |0\rangle\langle 0 | \otimes |0\rangle\langle 0 | + | 1\rangle\langle 1 | \otimes | 1\rangle\langle 0 | [/tex]

The unitary acts as, ## U|00\rangle = | 00\rangle## and ##U|10\rangle = | 11\rangle ## thus a measurement of the ancillary gives the state of the qubit in Bob's possession as long as it is either ##|0\rangle## or ##|1\rangle##.

Now suppose that Bob keeps applying ##U## and measuring the ancilla. The reduced density matrix on Bob's side is, ## \rho_b = \frac{1}{2} \left( |0\rangle\langle 0 | + | 1 \rangle\langle 1| \right) ##, the maximally mixed state.

So Bob will measure 0 half the time and 1 half the time. Hence he can determine that

##P(0) = 1/2## and ## P(1) = 1/2.##

Suppose now that Alice makes a measurement on her qubit giving either 0 or 1. The collapsed state

##| \phi^{\prime} \rangle ## is now ## | 00 \rangle ## or ## | 11 \rangle ## making Bob's state ## | 0 \rangle ## or ## | 1 \rangle ##.

Now as Bob keeps measuring, he will determine either,

##P(0) = 1 ## and ## P(1) = 0 ## or ##P(0) = 0## and ##P(1) = 1 ##

Hence asymptotically he can determine that Alice has made the measurement. So the information that a measurement has been made can be operationally determined by Bob even if he is light years away.

The scheme uses two things

1) An entangled Bell pair ## | \phi \rangle = | 0 0 \rangle + | 1 1 \rangle## ( neglecting normalization )

2) The fact that orthogonal states can be distinguished.

To begin with ## |\phi\rangle ## is given to Alice and Bob each having a qubit.

Bob has in his positions a copying unitary that can copy states ## |0\rangle ## and ## |1\rangle ## to an ancillary qubit and he can then measure this qubit to distinguish them, explicitly the unitary is [tex] U = |0\rangle\langle 0 | \otimes |0\rangle\langle 0 | + | 1\rangle\langle 1 | \otimes | 1\rangle\langle 0 | [/tex]

The unitary acts as, ## U|00\rangle = | 00\rangle## and ##U|10\rangle = | 11\rangle ## thus a measurement of the ancillary gives the state of the qubit in Bob's possession as long as it is either ##|0\rangle## or ##|1\rangle##.

Now suppose that Bob keeps applying ##U## and measuring the ancilla. The reduced density matrix on Bob's side is, ## \rho_b = \frac{1}{2} \left( |0\rangle\langle 0 | + | 1 \rangle\langle 1| \right) ##, the maximally mixed state.

So Bob will measure 0 half the time and 1 half the time. Hence he can determine that

##P(0) = 1/2## and ## P(1) = 1/2.##

Suppose now that Alice makes a measurement on her qubit giving either 0 or 1. The collapsed state

##| \phi^{\prime} \rangle ## is now ## | 00 \rangle ## or ## | 11 \rangle ## making Bob's state ## | 0 \rangle ## or ## | 1 \rangle ##.

Now as Bob keeps measuring, he will determine either,

##P(0) = 1 ## and ## P(1) = 0 ## or ##P(0) = 0## and ##P(1) = 1 ##

Hence asymptotically he can determine that Alice has made the measurement. So the information that a measurement has been made can be operationally determined by Bob even if he is light years away.

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