# I Is entanglement commutative?

1. Apr 5, 2017

### Kenneth Adam Miller

I've seen diagrams of quantum computer components at a high level that discusses multiplexing laser reflections over many qubits, and I have to believe that entanglement as a hardware operation has to be scaled to the many qubits by means of some operation that is applied to each of them simultaneously. That being said, if I remember correctly, there were examples of the slit experiment at a microscopic level to give readers at a introductory level an impression of what the hardware was doing. But I don't think that that was strictly what was actually at that level. Perhaps I working with a very vague understanding, but what I want to know is, if you have light that is entangled, and you strike a super cooled qubit of any kind, does that mean that that qubit is also suspended in entanglement? In other words, is entanglement commutative?

2. Apr 5, 2017

### jfizzix

If you have a pair of entangled photons A and B, and one of those photons B interacts with a qubit C, the amount of entanglement between A and the joint system BC remains the same (assuming no additional environmental interaction).

The amount of entanglement between A and B may change due to B interacting with C, but if A has no further interaction with B or C, the total entanglement between A and BC must remain constant.

3. Apr 5, 2017

### Kenneth Adam Miller

Ok, so it's as though A now shares a total entanglement with all three, but BC sort of share a subspace determined by the metrics of their interaction, is that correct?

4. Apr 6, 2017

### jfizzix

If I understand you correctly, yes.

A shares entanglement with BC, and the amount of entanglement between A and BC is the same before and after B and C interact.
What's different is how much entanglement A shares with just B, or with just C.

There's a useful concept called the monogamy of entanglement that says the amount of entanglement B shares with AC cannot be less than the sum of the entanglement between A and B, and between B and C.
$E(A:BC)\geq E(A:B) + E(A:C)$
So, as A becomes less entangled with B, A must be more entangled with C, (or at least, the maximum possible entanglement between A and C increases.

5. Apr 6, 2017

### Kenneth Adam Miller

Fascinating. Thank you.