1

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 $0<a<1,0<b<1,0<c<1$.

User101
  • 598
  • 2
  • 9
  • 3
    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. – Mark Spinelli Jan 25 '22 at 22:51
  • 2
    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 Jan 26 '22 at 13:04
  • 1
    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. – Mark Spinelli Jan 28 '22 at 15:27

1 Answers1

3

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:

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

LeWoody
  • 844
  • 1
  • 8
  • 14