How does an $\sqrt{X}$ error propagates through an $T$ gate?

Question

I wonder whether there is a compact way of describing a noise $\sqrt{X}$ propagating through a $T$ gate. Possibly in terms of Pauli and/or Clifford operators.

Answer 1

If I understand correctly, you're saying that imagine your computation was supposed to have started in the state $|\psi\rangle$ and then had $T$ applied to it. But, instead, you actually started with $T\sqrt{X}|\psi\rangle$. You want to write this as some $VT|\psi\rangle$.

If this is the case, then it must be that $$ VT=T\sqrt{X}. $$ In other words, $$ V=T\sqrt{X}T^\dagger. $$ Note that $$ T\sqrt{X}T^\dagger=T\frac{I+iX}{\sqrt{2}}T^\dagger=\frac{1}{\sqrt{2}}I+\frac{1}{\sqrt{2}}iTXT^\dagger=\frac{1}{\sqrt{2}}I+\frac{1}{\sqrt{2}}iSXe^{-i\pi/4}. $$ You can work out what this looks like as a matrix, but it's not particularly nice. It's something like $$ \sqrt{\frac{X+Y}{\sqrt2}} $$