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Although this question deals with the construction of a W-state, I was looking for a general way to find all the orthogonal W-states, given a number of qubits. For example, for three qubits, the first W-state I find is:

$$ W_3^1 = \frac{1}{\sqrt{3}}(|001\rangle + |010\rangle + |100\rangle). $$ An orthogonal state to this state would be (from this paper, page 4): $$ W_3^2 = \frac{1}{\sqrt{3}}(|001\rangle - |010\rangle + |111\rangle). $$ But I am not sure whether this state qualifies as a W-state or not. I want to find all the other 6 orthogonal basis states like this. Also, I would like to be able to generate such orthogonal basis states in any dimension. Of course, I can use Gram–Schmidt process to find a set of orthogonal vectors. But I'm not sure whether they would be W-states or not. What is the proper way to generate such W-basis states given a number of qubits? TIA.

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    $\begingroup$ What makes you think that there is an orthonormal W-state basis that spans the entire 8-dimensional Hilbert space of 3 qubits? $\endgroup$
    – DaftWullie
    Jul 29 '20 at 8:09
  • $\begingroup$ @DaftWullie , you are right. I have no idea about whether such states exist or not. It's just that the paper I've cited has mentioned that there is an orthogonal w-basis for 3 qubits. $\endgroup$ Jul 29 '20 at 13:23
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    $\begingroup$ So it does. It even states the basis states in equations 15-22. So what's the problem? $\endgroup$
    – DaftWullie
    Jul 29 '20 at 13:29
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    $\begingroup$ That's the difference between the W-state and the W-state class of states. $\endgroup$
    – DaftWullie
    Jul 29 '20 at 13:35
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    $\begingroup$ That depends what you want. Technically, a basis does not have to comprise orthogonal states. But we get so used to using orthonormal bases that it's easy to make errors when calculating with other bases. $\endgroup$
    – DaftWullie
    Jul 29 '20 at 13:43

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