2
$\begingroup$

The X gate in the $|+\rangle$, $|-\rangle$ basis becomes the Z gate and vice versa.

What is the Pauli Y gate as a matrix transformation in the $|+\rangle$, $|-\rangle$ basis?

$\endgroup$
1
  • $\begingroup$ What do you mean by X gate become Z gate? Does it mean that Z in Hadamard basis transform basis states between each other as X gate in computational basis? If so, please note that Y works similarly in circular basis. $\endgroup$ May 15 at 7:18

1 Answer 1

2
$\begingroup$

We have,

$Y=i|+\rangle\langle -| -i|-\rangle\langle +|$

You can calculate it easily using the fact that,

$|0\rangle = \frac{1}{\sqrt 2}(|+\rangle + |-\rangle)$,

$|1\rangle = \frac{1}{\sqrt 2}(|+\rangle - |-\rangle)$, and

$Y = i|1\rangle\langle 0| -i|0\rangle\langle 1|$

So, it has the same matrix up to a sign change.


Alternative solution is to use what you already mentioned in your question:

The X gate in the $|+\rangle$, $|-\rangle$ basis becomes the Z gate and vice versa

with the fact, $Y=iXZ$, and $XZ = -ZX$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.