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 1


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.

enter image description here

  • $\begingroup$ 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? $\endgroup$ Commented Jun 19 at 0:36
  • $\begingroup$ 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? $\endgroup$ Commented Jun 19 at 0:52
  • $\begingroup$ @IanGershonTeixeira Yes all those guesses are correct. You can click on the image to open the circuit in Quirk. $\endgroup$ Commented Jun 19 at 9:48

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