# Is the plus state a magic state for the Hadamard gate?

Is the plus state $$\left|+\right>:=\frac{\left| 0\right>+\left| 1\right>}{\sqrt{2}}$$ a magic state for the Hadamard gate $$H$$? That is, given the ability to perform (controlled) Pauli operators, can I consume an ancillia prepared with state $$\left| + \right>$$ to perform the Hadamard gate on another qubit?

To me this seems like a statement which must be true: you are given superposition in the form of $$\left| +\right>$$, and now all you have to do is transfer that superposition onto another qubit in the simplest way possible. I've been trying to come up with an explicit construction for a while now though, and I can't find one.

At the end of the paper "On the Power of Reusable Magic States", the author asks if there exists a reusable magic state for $$H$$. This seems to imply that they are aware of the existence of a non-reusable magic state for $$H$$, which I assume would be $$\left| + \right>$$.

• $|+\rangle$ isn't an entangled state. Commented Aug 11, 2023 at 16:05
• @Condo Oops, you're right. I meant "superposition" not "entanglement". The Paulis send computational basis to computational basis, and $\left| + \right>$ gives you a superposition of computational basis elements. It's fixed now. Commented Aug 12, 2023 at 4:39

Start with a state $$|\psi\rangle|+\rangle$$. Measure the operator $$X_1Z_2$$ using either lattice surgery or an ancilla qubit and $$CZ,CX$$ gates. Call this result $$m_{xz}$$ Then measure the first qubit in the $$Z$$ basis, call this result $$m_z$$. The state on the remaining qubit is then
$$X^{m_{xz}}Z^{m_z}H|\psi\rangle$$ so you have applied an H gate, up to a Pauli correction.
The above circuit shows how to inject a Hadamard gate by means of a magic state $$|+\rangle$$ and an Ising gate.