Timeline for Can we characterise the general structure of two-qubit witness operators?
Current License: CC BY-SA 4.0
11 events
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| Oct 12, 2022 at 13:50 | comment | added | glS♦ | @Sherlock it's just a way to specify the partition we're referring to. While $\mathbb{C}^4$ and $\mathbb{C}^2\otimes\mathbb{C}^2$ are clearly the same space, when discussing things like entanglement you're always referring to some choice of bipartition for the space. So writing $\mathbb{C}^2\otimes\mathbb{C}^2$ I'm saying that we're talking about entanglement between two two-dimensional spaces (in this case the bipartition is trivial, as there's nothing to say about $\mathbb{C}\otimes\mathbb{C}^3$, but in more general cases it's not) | |
| Oct 12, 2022 at 11:56 | comment | added | Sherlock | Just want to learn something, why will we use $\mathbb{C}^2\otimes\mathbb{C}^2$ notation instead of $\mathbb{C}^4$? | |
| Aug 13, 2022 at 12:05 | history | became hot network question | |||
| Aug 13, 2022 at 10:18 | comment | added | glS♦ | @MarkS glad to hear it =). As already pointed out, mind that the standard definition of witness op is slightly different, if equivalently, to the one I wrote here (you generally ask for positive expvals on all separable states, and negative expval for some non-separable state). A product operator will never be a witness, because it will admit product eigenstates, and it'd have to be in the form $A\otimes B$ with $A,B\ge0$ to get non-neg expval on product states. I think $W$ being a permutation matrix is mostly by accident: most other such matrices won't be witnesses (nor Hermitian, really) | |
| Aug 13, 2022 at 9:52 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 13, 2022 at 9:38 | comment | added | narip | @MarkS And mind that, entanglement witness does not say witness itself is entangled. You may notice that the condition required in the problem is a separable pure state, in fact, all the separable state is a convex combination of separable pure states, so the condition is equivalently saying that, for all separable state the inner product will be positive, so once you found a negative inner product, the state must be entangled state. | |
| Aug 13, 2022 at 9:28 | comment | added | narip | @MarkS It's not entanglement witness, since by choosing $\mathbb{P} _u=|+\rangle \langle +|$ and $\mathbb{P} _v=|-\rangle \langle -|$, we have $\langle \mathbb{P} _u\otimes \mathbb{P} _v,X\otimes X\rangle =-1< 0$. | |
| Aug 13, 2022 at 9:11 | answer | added | Danylo Y | timeline score: 3 | |
| Aug 12, 2022 at 23:35 | comment | added | Mark Spinelli | I gotta say, your questions are always well written and informative - although I have almost no chance of providing any helpful observation, I feel like I learned something just by reading and understanding! Apart from SWAP $X\otimes X$ is another permutation matrix but yet is not an entanglement witness, as by definition it factors, correct? | |
| Aug 12, 2022 at 19:13 | history | edited | glS♦ | CC BY-SA 4.0 |
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| Aug 12, 2022 at 19:07 | history | asked | glS♦ | CC BY-SA 4.0 |