# A variation of the Bell experiment

Tags:
1. Nov 18, 2014

### dsoodak

1/Sqrt(2)(|00>+|11>)

and (after moving the second qbit a significant distance away) apply the interferometer transformation
|0> -> 0.5(|0>+|1>)
|1> -> 0.5(|0>-|1>)

to the first qbit, we get
0.5/Sqrt(2)((|0>+|1>)|0>+(|0>-|1>)|1>)
=0.5/Sqrt(2)(|00>+|10>+|01>-|11>)
which gives equal probability of the first qbit ending up in |0> or |1>

Lets now start again with the same spatially separated Bell state but first apply the transformation

|0> -> 0.5(|0>+|1>)
|1> -> 0.5(|0>+|1>)
to the second qbit:

0.5/Sqrt(2)(|0>(|0>+|1>)+|1>(|0>+|1>))
=0.5/Sqrt(2)(|00>+|01>+|10>+|11>)

then apply the original (interferometer) transformation to the first qbit:
0.25/Sqrt(2)((|0>+|1>)|0>+(|0>+|1>)|1>+(|0>-|1>)|0>+(|0>-|1>)|1>)
=0.25/Sqrt(2)(|00>+|10>+|01>+|11>+|00>-|10>+|01>-|11>)
=0.5/Sqrt(2)(|00>+|01>)

Now, the first qbit is in state |0> with 100% (as opposed to 50%) probability as a result of what was done to the second one.

So...can anyone tell me if I made any false assumptions or stupid math mistakes here?

Dustin Soodak

2. Nov 18, 2014

### atyy

This transformation isn't reversible, so I think you have to measure the state and collapse the wave function before doing the transformation. After collapsing the wave function, the state should be a mixed state, not a pure state, ie. with 50% chance the state is |0>(|0>+|1>), and with 50% chance the state is |1>(|0>+|1>).

3. Nov 19, 2014

### dsoodak

The way I've always seen it described in this sort of experiment, something counts as a measurement if the information about the state leaks out.
I originally thought this transformation could be done (using photons as qbits) with an interferometer that has one path length 1/4 wavelength longer than the other. However, I suppose you could then get info about the photon's path information from the difference in travel time.

4. Nov 19, 2014