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Is it allowed to define W-state as $|W\rangle = a |001\rangle + b | 010\rangle + c |100 \rangle$, with $a^2 + b^2 + c^2 =1$?

Edit: Assuming $0<a<1,0<b<1,0<c<1$.

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    $\begingroup$ I would say no - that is a nonstandard and potentially confusing generalization of the W state. For example up to two of $a$,$b$, or $c$ could be zero. $\endgroup$ Commented Jan 25, 2022 at 22:51
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    $\begingroup$ I find this an odd question. What do you mean with "allowed"? A W state is typically defined as the state in the case $a=b=c$. You can use a different definition if you so wish, but people are probably going to be confused by it, and possibly object to the improper use of the term "W state" $\endgroup$
    – glS
    Commented Jan 26, 2022 at 13:04
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    $\begingroup$ We could call such a state with $a,b,c\in\mathbb R$ and in $[0,1]$ a "User101 state," but it would be confusing to call it a $W$ state ior even a generalized $W$ state. $\endgroup$ Commented Jan 28, 2022 at 15:27

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The $W$ state is thought to be named after Wolfgang Dur in his paper on this subject. He and his co-authors define it as: $${\displaystyle |\mathrm {W} \rangle ={\frac {1}{\sqrt {3}}}(|001\rangle +|010\rangle +|100\rangle )}$$ Given your constraint, I could for example define $a = 1,$ $b = 0,$ and $c = 0$, but this is not an entangled or $W$ state. It would just be $|001\rangle$.

For reference, in Durer's paper he gives the definition of a generalized N-qubit W state to be:

$$\left| {{W_N}} \right\rangle \equiv 1/\sqrt N |N - 1,1\rangle$$ where $|N-1,1\rangle$ denotes the totally symmetric state including $N-1$ zeros and $1$ one. For example, when $N = 4$, you have:

$$\left| {{W_4}} \right\rangle = {\textstyle{1 \over {\sqrt 4 }}}\left( {|0001\rangle + |0010\rangle + |0100\rangle + |1000\rangle } \right)$$

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    $\begingroup$ And further, defining the $W$ state to be such that $a=b=c$ immediately enables the right generalization of a $W_n$ state, with $n\gt 3$. $\endgroup$ Commented Jan 28, 2022 at 0:23
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    $\begingroup$ See also this question regarding the name. $\endgroup$ Commented Jan 28, 2022 at 1:39
  • $\begingroup$ @MarkS, thanks. But how about if $a,b,c$ are non zero? $\endgroup$
    – User101
    Commented Jan 28, 2022 at 11:51

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