# What does the partial trace of $|W\rangle$ states represent physically?

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?

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.
• 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
• @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. Commented Nov 4, 2021 at 10:50
• "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... Commented Nov 7, 2021 at 17:49