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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.

glS
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QuestionEverything
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    What makes you think that there is an orthonormal W-state basis that spans the entire 8-dimensional Hilbert space of 3 qubits? – DaftWullie Jul 29 '20 at 08:09
  • @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. – QuestionEverything Jul 29 '20 at 13:23
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    So it does. It even states the basis states in equations 15-22. So what's the problem? – DaftWullie Jul 29 '20 at 13:29
  • Because in this link: https://en.wikipedia.org/wiki/Matrix_product_state#W_state" it states as the definition of w-state, "..the superposition of all the computational basis states of Hamming weight one.". But as you can see, the states they have constructed in that, do not have hamming weight of 1. – QuestionEverything Jul 29 '20 at 13:34
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    That's the difference between the W-state and the W-state class of states. – DaftWullie Jul 29 '20 at 13:35
  • Ah, I was not aware of this difference. Thanks! Anyways, they should still be orthogonal to each other, right. I have noticed in those states that <w1|w4> = .6667, not 0, for theta = 54.7356 degrees and phi = 45 degrees. Is that acceptable? – QuestionEverything Jul 29 '20 at 13:40
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    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. – DaftWullie Jul 29 '20 at 13:43
  • You are right, thanks a lot, @DaftWullie ! – QuestionEverything Jul 29 '20 at 13:46

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