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Given the W-state $|W\rangle = |001\rangle + |010\rangle + |100\rangle$, where $|ijk \rangle $implies $|i\rangle_A \otimes |j\rangle_B \otimes |k \rangle_C$, the partial trace over first qubit turns out to be

$$Tr_A\left[|W\rangle \langle W| \right] = {}_A\langle0 |W\rangle \langle W|0\rangle_A + {}_A\langle 1 |W\rangle \langle W|1\rangle_A = |00\rangle \langle 00| + |01\rangle \langle 01| + |01\rangle \langle 10| + |10 \rangle \langle 01| + |10\rangle \langle 10|$$

What physics does this tell us?

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The key thing that it tells you is that the W-state is partially entangled. This is perhaps a little clearer to see if you trace over two qubits: $$ \text{Tr}_{AB}(|W\rangle\langle W|)=\frac23|0\rangle\langle 0|+\frac13|1\rangle\langle 1|. $$ The state is mixed, so the overall pure state is entangled, but it's not maximally mixed, so the overall state is not maximally entangled.

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  • $\begingroup$ is there a univocal way to define what "maximally entangled" means for multipartite states? One could also argue that tracing out a single qubit from $|W\rangle$ results in the maximal amount of bipartite entanglement one can get doing so on tripartite states, hence the $|W\rangle$ should also be called "maximally entangled". Unless you just mean that the $|W\rangle$ is not maximally entangled with respect to its bipartitions, which is of course true $\endgroup$
    – glS ♦
    Nov 4, 2021 at 9:27
  • $\begingroup$ @glS perhaps not but, here, I mean with respect to tripartite entanglement as opposed to bipartite entanglement. I don't think I've ever seen a $|W\rangle$ state being referred to as maximally entangled. $\endgroup$
    – DaftWullie
    Nov 4, 2021 at 10:50
  • $\begingroup$ "The state is mixed, so the overall pure state is entangled" do we always have "mixed state $=>$ entangled" The way you phrase it makes it seems so, but afaik a mixed state may be an entangled mixed state or a separable mixed state... $\endgroup$
    – user206904
    Nov 7, 2021 at 17:49
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    $\begingroup$ @user206904 If the reduced state of an overall pure state is mixed, then the overall state is entangled. If the overall state is not pure, then you cannot make that conclusion (which is, effectively, what other methods of preparing a mixed state are describing). $\endgroup$
    – DaftWullie
    Nov 8, 2021 at 7:28

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