# Can we define W state as $a |001\rangle + b | 010\rangle + c |100 \rangle$?

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.

• 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. Commented Jan 25, 2022 at 22:51
• 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"
– glS
Commented Jan 26, 2022 at 13:04
• 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. Commented Jan 28, 2022 at 15:27

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 $$|001\rangle$$.
$$\left| {{W_N}} \right\rangle \equiv 1/\sqrt N |N - 1,1\rangle$$ where $$|N-1,1\rangle$$ denotes the totally symmetric state including $$N-1$$ zeros and $$1$$ one. For example, when $$N = 4$$, you have:
$$\left| {{W_4}} \right\rangle = {\textstyle{1 \over {\sqrt 4 }}}\left( {|0001\rangle + |0010\rangle + |0100\rangle + |1000\rangle } \right)$$
• And further, defining the $W$ state to be such that $a=b=c$ immediately enables the right generalization of a $W_n$ state, with $n\gt 3$. Commented Jan 28, 2022 at 0:23
• @MarkS, thanks. But how about if $a,b,c$ are non zero? Commented Jan 28, 2022 at 11:51