I suggest the following thought experiment that describes a machine which makes everybody happy.

Suppose a lottery is conducted. The winner is awarded a billion dollars plus the title of eternal Earth dictator.

The winner is determined in the following way.

• There is a thorough isolated black box inside which a pretender is placed.

• Inside the box a quantum experiment is conducted with equal probability of the two outcomes: 0 or 1 (it may be based on an atom decay)

• The pretender measures the result with an apparatus.

• The black box is connected by a quantum link (in form of an optical fiber) to an external quantum computer

• After the measurement by the pretender is done the result is automatically transmitted by the optical link to the quantum computer

• The transmitted qubit is used by quantum computer as an initial state of the quantum register instead of usual Hadamard transform before performing a Shor factorization algorithm.

• After the algorithm is performed, the result is checked whether it is correct.

• If it is correct the pretender is declared the looser. If the factorization returned a wrong result, the pretender is declared the winner.

What happens.

Any average person placed in the black box will measure the result of the atom decay and similarly to a Schroedinger cat will enter the superposition of two states: the one which saw 0 and the one which saw 1.

Similarly the measuring apparatus will enter the superposition. So the qubit sent via the optical fiber will be a superposition of two states:

$$\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$$

As this value is exactly the same what Hadamard gate returns after being applied to an empty qubit, this will allow the quantum computer to perform the factorization as usual and return the correct result, dispite that the initial step of applying the Hadamard gate has been replaced by loading the qubit obtained from the black box.

Now suppose the observer himself is placed in the black box. After the atom decay he will perform measurement which will collapse the wave function. The observer will be in either state where he saw 0 or where he saw 1. So either basis state 0 or basis state 1 is sent via the cable to the quantum computer.

Since this is different from the usual value which should be loaded in the register before performing the factorization, the factorization will return a wrong result. So the pretender will be declared the winner.

It follows that any person, when he or she is observing the lottery from outside will see all other pretenders to loose.

At the same time any participant that tries the lottery himself will find himself a winner.

This is a fascinating lottery which will make all the participation pretenders the winners in their own eyes. Each person who goes through this machine will see other people to congratulate them, to proclaim them the eternal emperor and so on.

I would like to see whether there any mistake in the reasoning.

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The mistake is that the trasmitted Qubit is entangled with the degrees of freedom of the internal pretender/observer, so that it is not coherent when it is presented to the quantum computer from the point of view of the computer. The quantum computer will fail independently of whether the observer is internal or external, because of lack of coherence in the qubit. In order to get the coherence back, the computer will have to undo the measurement by the observer, and recohere the external qubit.

But it's a nice idea anyway.

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You are the second person who tells me so and I think you are right. This is because however well the box is isolated, it will be unsealed in the future. So to make it to have effect one has to turn all internal degrees of freedom of the predender entangled with the qubit into basis states (i.e. remove any entanglement) before the box is unsealed. –  Anixx Apr 24 '12 at 12:20
@Anixx: That's one way of looking at it, but the removal of entanglement happens when you extract the entangled qubit out of the box in the first place--- your intuition is probably failing because of the inability to visualize a two-state system entangled with two completely classically different states--- that's a post-measurement outcome translated to the many-worlds picture. In this case, the ability to switch pictures to subjective Copenhagen should convince you that what you want to do is impossible. –  Ron Maimon Apr 24 '12 at 16:35
The entanglement between qubit and internal observer only becomes observable-detectable after the quantum computer has received information via a classical channel that doesn't go faster than light. If the quantum computer fails even if no classical information is able to arrive to it, then a FTL bit has been transmitted from the internal observer and the observers that confirm that the quantum computer just fails –  lurscher Dec 30 '13 at 0:54

Even if all your other arguments were correct, note that being in either one of two states with equal probability is not the same as being in a quantum state $\frac{1}{\sqrt{2}}| 0\rangle + \frac{1}{\sqrt{2}}| 1\rangle$. In particular, why is it not the orthogonal state $\frac{1}{\sqrt{2}}| 0\rangle - \frac{1}{\sqrt{2}}| 1\rangle$?

In dealing with problems like this (and as a general rule in almost any other scenario) it is important to apply the formalism carefully in each step. This usually prevents making unwanted assumptions along the way.

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In this case, Annix was trying to exploit this difference using a macroscopic entanglement, to see if he could make an outcome which depended on whether it was the superposition state 1 or 2. This fails, because for macroscopic objects there is no difference between the two superpositions, because you can't make interference to measure the relative phase. –  Ron Maimon May 9 '12 at 23:47