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

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.

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 $$\sqrt{\frac{X+Y}{\sqrt2}}$$