# B Can you find a single entangled atom in a haystack.

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1. Sep 7, 2016

### BernieM

If you have a pair of particles (perhaps atoms for example) that are entangled, and one of them is mixed in amongst a whole bunch of other atoms of the same species, yet not entangled; could you differentiate in some way the entangled atom from the rest WITHOUT destroying the existing state of entanglement?

For example, press your FLEA button (Find Lost Entangled Atom) and the entangled atom gives off a blue light so it stands out from the rest of the atoms around it.

Intuitively I think not, but I can't fully explain why I don't think it can be done. My guess is that any entanglement would have to be destroyed to accomplish it.

2. Sep 7, 2016

### .Scott

Well, if the atom in the haystack is remaining entangled to the other atom, you should look to see if there is a vacuum chamber hidden in the haystack. Follow the electric cord (the one powering the cryo-pumps and the lasers) to the vacuum chamber. Your atom will be the one levitated by the lasers.

Other than that, no. For all you know, there are many entanglement relationships in that haystack.

3. Sep 7, 2016

### DrChinese

You would be hard pressed to differentiate an entangled particle from one that is not entangled under any circumstance.

The normal technique for observing entanglement between 2 particles requires you to know WHICH two particles are entangled. Second, you must know the specific type of entanglement to look for (spin entanglement can be + or - for example). Last, there is no good way to rule out - in a single case - that the results actually represent unentangled particles (which could be misinterpreted as being entangled).

4. Sep 7, 2016

### Strilanc

No, you won't be able to find it. You won't be able to reject any of the wheat without doing a measurement that could ruin the entanglement.

With clever measurements you might be able to eliminate a surprising amount of wheat, though. For example, consider the $n=2$ case. What's effectively happening is that you're trying to distinguish between the state $a = I \otimes (|00\rangle + |11\rangle)$ and the state $b = (|00\rangle + |11\rangle) \otimes I$.

You might expect to succeed at most 75% of the time (50% of the time measuring the unentangled qubit will give a disagreeing answer, and when that fails you still have a 50% change to win by random guessing). But actually the trace distance between those two states is about 0.866:

Code (Text):

import numpy as np
def td(x, y):
return 0.5*np.sum(np.abs(np.linalg.eig(x-y)[0]))

I = np.eye(2)/2.0
E = np.mat([[1,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,1]])/2.0
a = np.kron(E, I)
b = np.kron(I, E)
print(td(a,b))
#>>> 0.866025403784

Meaning the maximum success rate for any measurement attempting to distinguish between those two state is $1/2 + 0.866/2 \approx 93\%$. (These measurements are going to decohere the system.) You get the same success rate when the extra qubit is in a known state like $|0\rangle$, instead of the maximally mixed state.

5. Sep 7, 2016

### Staff: Mentor

Entangled on what property?

6. Sep 7, 2016

### BernieM

Say polarization, horizontal or vertical.