I am studying on entanglement detection applying stabilizer operators. In page 4 of this paper https://arxiv.org/abs/quant-ph/0501020 ,"for the detection of $N$-qubit entanglement we have to make measurements on all qubits and have to measure a full set of generators. This is because we need that the expectation value of the witness is minimal only for the GHZ state. If the witness does not contain a full set of generators then there are at least two of the elements of the GHZ basis giving the minimum." I do not understand this argument, how can we prove the existence of at least two GHZ basis elements for a given witness which does not contain a full set of generators? Thank you.
1 Answer
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Imagine you have a generating set of stabilizers for a state $|G\rangle$ (i.e. it's a +1 eigenstate of all the generators). Now take one of them away. You will be able to find a sequence of Pauli operators, $P$, that anti-commute with the stabilizer you have removed, but commute with everything else (you can set this up as a binary programming problem, and you'll find it always has a solution). By construction, $|G\rangle$ and $P|G\rangle$ have the same expectation on all the remaining generators, and hence will return the same value for the witness.
(Put another way, $P|G\rangle$ is the state which is a +1 eigenstate of the remaining stabilizers and -1 eigenstate of the one you removed.)
Consider the example of a 4-qubit GHZ state. (For simplicity, I'm going to choose something that's locally equivalent, rather than using exactly the GHZ state). I have generators $$ X_1Z_2Z_3Z_4, Z_1X_2, Z_1X_3, Z_1X_4. $$ I'm going to write these in a 4x8 matrix: one row for each stabilizer. The first 4 columns indicate positions of Xs, the second 4 indicate positions of Zs. $$ G=\left(\begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ \end{array}\right) $$ Now imagine some other string of Pauli operators $P=(x|z)$. I want to find one the commutes with the first 3 rows and anti-commutes with the last one. We set this up as $$ G\left(\begin{array}{c} z^T \\ x^T \end{array}\right)=\left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array}\right). $$ Notice the change in order between $x$ and $z$. This is because we already know that $X$-type operators commute with $X$s in the stabilizer. They must be compared with $Z$s instead. Also note that all the algebra is being done Mod 2.
We just have to find something that solves this equation. I can immediately tell you that $P=Z_4$ will work. Now that I tell you that, you can verify the relations. But if I hadn't know, I'd have gone to Mathematica (or equivalent, but I find Mathematica particularly convenient), and calculated
G = {{1, 0, 0, 0, 0, 1, 1, 1}, {0, 1, 0, 0, 1, 0, 0, 0}, {0, 0, 1, 0,
1, 0, 0, 0}, {0, 0, 0, 1, 1, 0, 0, 0}};
LinearSolve[G, {0, 0, 0, 1}, Modulus -> 2]
which gives an answer
{0, 0, 0, 1, 0, 0, 0, 0}
exactly as I predicted.
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$\begingroup$ Thank you very much for the solution! I have two questions: 1. Why P|G⟩ is necessarily a GHZ basis element? 2. Could you please explain more about constructing of P and binary programming problem? $\endgroup$– Star21Jan 31 at 14:52
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$\begingroup$ Part 1: if you apply Pauli matrices to a GHZ state, you get something that's in the GHZ basis. (Up to a point, that's a matter of definition about how you extend the GHZ state into a basis.) $\endgroup$ Feb 1 at 7:31
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$\begingroup$ Thank you very much! As you wrote, we can find more than one solution for P, because here we have 8 variables and 4 equations. Am I right? And this argument holds for general case, similarly. $\endgroup$– Star21Feb 2 at 13:25
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$\begingroup$ Well, there might be multiple solutions, but they all work out equivalently, because they're terms like $Pg_i$ where $P$ is the Pauli string I already found, and $g_i$ is a generator. So the action on $|G\rangle$ is the same. $\endgroup$ Feb 2 at 14:06