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If I compute the trace over 1 system of the GHZ state I have a matrix with no off diagonal non-zero elements. Is this sufficient evidence that the reduced 2 qubit state is unentangled?

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3 Answers 3

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Yes, a diagonal density matrix is always separable, if that's what you're asking.

The reason is that such a two-qubit state has by definition a decomposition of the form $$\rho = \sum_{i,j=0}^1 p_{i,j} (|i\rangle\!\langle i|\otimes |j\rangle\!\langle j|)$$ for some set of probabilities $p_{i,j}$. Any such state is separable, pretty much by definition.

The same holds in arbitrary dimensions.

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  • $\begingroup$ Makes sense, thank you for your answer $\endgroup$
    – P5050
    Mar 31 at 12:55
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Essentially, yes, but I suppose that depends on who is deciding what constitutes sufficient evidence. I would suggest that you're asking "is this enough" because you suspect it's true but don't actually know how to justify it any further. In which case, it's perhaps not so obvious, and that further justification is useful.

To give the answer with a couple of the critical ideas filled in:

The mixed state is $$ \rho=\frac12(|00\rangle\langle 00|+|11\rangle\langle 11|) $$ and it is diagonal. To all intents and purposes, this means that it is a classical probability distribution, and hence not entangled.

But to formalise this a bit more, what is the definition of entangled? A two-qubit state is separable if it can be written in the form $$ \rho=\sum_ip_i\sigma_i\otimes\tau_i $$ where $\sigma_i$ and $\tau_i$ are single-qubit density matrices. If it cannot be written in this form, it is entangled. From the way that I wrote down $\rho$, it should be obvious that it can be written in a separable form: $$ p_0=p_1=\frac12, \sigma_0=\tau_0=|0\rangle\langle 0|, \sigma_1=\tau_1=|1\rangle\langle 1|. $$

Alternatively, you might like to think about whether $\rho$ can be constructed via local operations and classical communication without any shared entanglement as a starting resource (e.g. Alice tosses a coin and communicates the result. If heads, both parties produce $|0\rangle$...)

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  • $\begingroup$ Thank you this makes sense. $\endgroup$
    – P5050
    Mar 31 at 12:58
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The GreenbergerHorneZeilinger (GHZ) state is given by $$\vert GHZ \rangle = \frac{1}{\sqrt{2}}\left(\vert 000 \rangle + \vert 111 \rangle\right).$$

$$\vert GHZ \rangle \langle GHZ \vert = \frac{1}{2}\left(\vert 000 \rangle \langle 000 \vert + \vert 000 \rangle \langle 111 \vert + \vert 111 \rangle \langle 000 \vert + \vert111 \rangle \langle 111\vert\right)$$

Therefore, if we make a trace with respect to one of the systems we will have the following state,

$$\rho_{23} = \text{Tr}_1(\vert GHZ\rangle\langle GHZ \vert) = \langle 0 \vert_1 \vert GHZ \rangle \langle GHZ \vert \vert 0 \rangle_1 + \langle 1 \vert_1 \vert GHZ \rangle \langle GHZ \vert\vert 1 \rangle_1 = \frac{1}{2}(\vert\langle 0 \vert 0 \rangle\vert^2 \vert 00 \rangle \langle 00 \vert + \langle0\vert0\rangle\langle1\vert0\rangle\vert00\rangle\langle11\vert+\langle0\vert1\rangle\langle0\vert0\rangle\vert11\rangle\langle00\vert+\langle0\vert1\rangle\langle1\vert0\rangle\vert11\rangle\langle11\vert+ $$ $$+\langle 1\vert0\rangle\langle0\vert1\rangle\vert00\rangle\langle00\vert+\langle1\vert0\rangle\langle1\vert1\rangle\vert00\rangle\langle11\vert+\langle1\vert1\rangle\langle0\vert1\rangle\vert11\rangle\langle00\vert+\vert\langle1\vert1\rangle\vert^2\vert11\rangle\langle11\vert)=$$ $$=\frac{1}{2}(\vert 00 \rangle \langle 00 \vert + \vert11 \rangle\langle 11 \vert) = \frac{1}{2}(\vert 0 \rangle \langle 0 \vert \otimes \vert 0 \rangle \langle 0 \vert + \vert 1 \rangle \langle 1 \vert \otimes \vert 1 \rangle \langle 1 \vert)$$ Note that state is separable therefore because we have the subsystem as a convex combination of separable states. It is therefore incoherent in the multipartite basis.

In general, we may see that finding that a bipartite state is incoherent suffices since this state will also be separable. A state $\sigma$ is incoherent if it can be written as $\sigma=\sum_i p_i \vert i \rangle \langle i \vert$ for the reference basis $\{\vert i \rangle\}_i$. Therefore, this generalizes for bipartite scenarios in the following way: defining the reference basis in the bipartite system to be $\{\vert i \rangle \otimes \vert j \rangle \}_{ij}$ where $\vert i \rangle$ are elements of the basis by Alice and $\vert j \rangle$ are elements of the basis by Bob (it is the basis because it is the chosen reference basis to define coherence with respect to) therefore the bipartite state will be incoherent if it is of the form, $$\rho^{AB} = \sum_k p_k \sigma_k^A \otimes \tau_k^B$$ and these states are always separable (as other answers pointed out). These considerations were done following the paper Ref1, but everything generalizes for multipartite states.

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