# Single bit teleportation for Hadamard gate

The circuit in Figure 13.7 of Gottesman's book https://www.cs.umd.edu/class/spring2024/cmsc858G/QECCbook-2024-ch1-15.pdf shows how to take the magic state $$T | + \rangle$$ and use single bit gate teleportation to implement the $$T$$ gate on a qubit $$| \psi \rangle$$.

The state $$H | + \rangle= | 0 \rangle$$ doesn't do anything useful. Is there some other state that we can use, with controlled Paulis and computational basis measurements, to implement a Hadamard gate on $$| \psi \rangle$$ through single bit gate teleportation?

## 1 Answer

Here's a construction that does it with a one qubit state, by using a CY gate and an MX gate. This works because Hadamard is, up to Pauli gates, equivalent to a 90 degree rotation around the Y axis.

• Just some question on notation, does $|i \rangle$ denote $\tfrac{1}{\sqrt{2}}(|0 \rangle+i|1\rangle)$ ? Does $MX$ mean measure the first qubit in the $|+ \rangle, | - \rangle$ basis? Does black circle mean $1$ turns on the control while white circle means $0$ turns on the control? And the last gate is just a CNOT/ controlled X gate right? Commented Jun 19 at 0:36
• I guess here the first black circle means $1$ turns it on, the white circle means $+$ turns it on and the 2nd black circle means $-$ turns it on? Commented Jun 19 at 0:52
• @IanGershonTeixeira Yes all those guesses are correct. You can click on the image to open the circuit in Quirk. Commented Jun 19 at 9:48