Local operations and classical communication (LOCC) on quantum system
I'm currently busy learning about the basics of quantum information theory. Does anyone know how the measurement described in the wiki link LOCC is a measurement on the product space $\mathbb{C}^2 \otimes \mathbb{C}^n$ as stated?
entanglement quantum-information
John DoeJohn Doe
$\begingroup$ Could you add some details from the wiki page to your question? It would be better if it is self-contained, and it would help clarify what you would like to know about. $\endgroup$ – James Wootton Apr 17 '18 at 21:12
I assume the question refers to how LOCC is used to distinguish the two states $$ (|00\rangle+|11\rangle)/\sqrt{2}\qquad(|01\rangle+|10\rangle)/\sqrt{2} $$ when Alice and Bob each hold one qubit, and are separated by a great distance.
There are many different measurement operators that could achieve the same task. If Alice just held both qubits herself, she could simply implement a measurement in the Bell basis, described by measurement operators $$ M_0=\frac12(|00\rangle+|11\rangle)(\langle00|+\langle11|)\qquad M_1=\frac12(|00\rangle-|11\rangle)(\langle00|-\langle11|)\qquad M_2=\frac12(|01\rangle+|10\rangle)(\langle01|+\langle10|)\qquad M_3=\frac12(|01\rangle-|10\rangle)(\langle01|-\langle10|) $$ However, these measurements don't have a tensor product structure, and so cannot be implemented by LOCC. The results $M_0$ and $M_2$ distinguish the two states given.
As an alternative, Alice and Bob do as described at the original link; both make $Z$ basis measurements. So, their measurement operators are described by $$ M_0'=|0\rangle\langle0|\otimes |0\rangle\langle0|\qquad M_1'=|0\rangle\langle0|\otimes |1\rangle\langle1|\qquad M_2'=|1\rangle\langle1|\otimes |0\rangle\langle0|\qquad M_3'=|1\rangle\langle1|\otimes |1\rangle\langle1| $$ If they get either answers $M_0'$ or $M_3'$, they had the first state, while if they get $M_1'$ or $M_2'$ they had the second state. Note that while Alice and Bob each act locally, on their qubit (as you can tell from the tensor product structure), they only know which overall result they got by comparing their measurement results, which requires classical communication.
Now a brief comment about the dimension of the Hilbert space. I have specifically talked about two qubits, $\mathbb{C}^2\otimes\mathbb{C}^2$. The first set of measurement operators, $\{M_k\}$, cannot be described under this tensor product structure; they are operators on $\mathbb{C}^4$, while the $\{M_k'\}$ operators can be described under $\mathbb{C}^2\otimes\mathbb{C}^2$. As for why the original source was talking about $\mathbb{C}^2\otimes\mathbb{C}^n$, for that specific example it seemed to be an unnecessary complication. Yes, you can always restrict to a qubit inside a larger Hilbert space, but there's no reason not to just take $n=2$ as far as I can see.
$\begingroup$ Thanks for your response. One query, what do you mean by "these measurements don't have a tensor product structure, and so cannot be implemented by LOCC", is $M_0=\frac12(|00\rangle+|11\rangle)(\langle00|+\langle11|)$ not simply the tensor product $M_0=\frac12(|0\rangle \otimes |0\rangle+|1\rangle \otimes |1\rangle)\otimes(\langle0| \otimes \langle0|+\langle1| \otimes \langle1|)$? Hence does this not have a tensor product structure? $\endgroup$ – John Doe Apr 19 '18 at 11:12
$\begingroup$ I mean that you cannot write $M_0=A\otimes B$ where $A$ is a 1-qubit operator corresponding to something that Alice can apply on her qubit (if you like, a 2x2 matrix) and B is something he can apply on his qubit. $\endgroup$ – DaftWullie Apr 19 '18 at 11:24
$\begingroup$ Just one thing to confirm. Are your measurement operators $\{M'_n\}_{n}$ a way of combining the measurements of both Bob and Alice, where equivalently we could consider the measurements as being local operations in the sense of measurements of the form $|0\rangle \langle0| \otimes I$, $I \otimes |0\rangle\langle 0|$, $|1 \rangle \langle 1| \otimes I$ and $I \otimes |1 \rangle \langle 1 |$? $\endgroup$ – John Doe Apr 19 '18 at 11:53
$\begingroup$ Yes. What it gives one the opportunity to succinctly describe the case (although it's unnecessary in this particular case) where Bob performs different measurements based on Alice's measurement result. For example, you could say, "If Alice gets 0 answer, Bob measures in Z basis, and he measures in X basis if she gets answer 1". Then you have the measurement operators $|0><0|\otimes|0><0|,|0><0|\otimes|1><1|,|1><1|\otimes|+><+|,|1><1|\otimes|-><-|$, which you can't do in the way you wrote it. $\endgroup$ – DaftWullie Apr 19 '18 at 12:02
$\begingroup$ The above refers to the use of LOCC equivalence. What is the basic difference between LOCC equivalence and SLOCC equivalence? $\endgroup$ – John Doe Apr 19 '18 at 13:53
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