# Can the W state, or any non-stabilizer state for that matter, be considered a magic state?

Some examples of magic states include: the $$|T\rangle$$ state for implementing the $$T$$ gate is $$T | + \rangle = \frac{1}{\sqrt{2}}(|0\rangle + e^{i\pi/4}|1\rangle)$$

A $$|\text{CS}\rangle$$ state for implementing the $$CS$$ gate is $$CS | + \rangle^{\otimes 2} = \frac{1}{2} (|00\rangle + |01\rangle + |10\rangle +i |11\rangle)$$

A $$|\text{CCZ}\rangle$$ state for implementing the $$CCZ$$ gate is $$CCZ | + \rangle^{\otimes 3} \\ = \frac{1}{\sqrt{8}}(|000\rangle + |001\rangle + |010\rangle + |011\rangle + |100\rangle + |101\rangle + |110\rangle - |111\rangle )$$

A $$|\text{Toffoli}\rangle$$ state for implementing the Toffoli (also known as $$CCX$$) gate is $$CCX | + \rangle^{\otimes 2} | 0 \rangle = \frac{1}{2} (|000\rangle + |100\rangle + |010\rangle + |111\rangle)$$ It is well known that a magic state is always a non-stabilizer state. And indeed this is true for the above examples.

But is it true that every non-stabilizer state, for example the $$W$$ state, is a magic state?

• Isn't that known? Did you try a search? Commented Jun 6 at 17:51
• Observation: It should be sufficient if you can answer this for $\alpha|0\rangle + \beta|1\rangle$, since given any non-stabilizer state, you can prepare such a state from it (at least so it feels). Commented Jun 6 at 18:14
• I think every single state that you listed is incorrectly defined, except the T state. Commented Jun 7 at 8:13
• @IanGershonTeixeira A bell pair where you've applied the Hadamard gate to one of the qubits. Commented Jun 8 at 23:38
• @NorbertSchuch Yes, in the sense of magic state distillation (say they can either be distilled into an H or T state). Campbell and Browne showed that one can resort to stabilizer protocols for m.s. distillation (arxiv.org/abs/0908.0838); Reichard (dx.doi.org/10.1007/s11128-005-7654-8) and Campbell and Browne (see ref above) showed tightness in certain directions (see also Fig. 1 in doi.org/10.1103/PhysRevLett.115.030501). The remaining region of possibly bound states does not include any pure non-stabilizer states. However, this is for single qubit states. Commented yesterday