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I saw that the Pauli matrices really work, that's rotating a state by 180 degrees, only if you take the density matrix for example with X it only works if first we take X $|\psi\rangle$ and multiply it by itself dagger. Then you can see it rotating around the X-axis. Same with Y and Z. Also, in error correction, you use the R mapping only after you created a density matrix, but you don't use it on the vector state itself. Why is it?

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Describing the Pauli-$X$ matrix as a rotation by $\pi$ about a particular axis is specifically referring to how you would visualise the action of the gate with regards to the Bloch Sphere picture. This most naturally maps to the density matrix, which is why it's most apparent there. But you can apply it to the pure state picture as well. Note that you can parametrise any single-qubit pure state as $$ |\psi\rangle=\cos\frac{\theta}{2}|0\rangle+\sin\frac{\theta}{2}e^{i\phi}|1\rangle. $$ This has a Bloch vector of the form $$ \vec{n}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta). $$

Now, let's consider $X|\psi\rangle=e^{i\phi}\left(\cos\frac{\theta}{2}e^{-i\phi}|1\rangle+\sin\frac{\theta}{2}|0\rangle\right)=e^{i\phi}\left(\sin\left(\frac{\pi}{2}-\frac{\theta}{2}\right)e^{-i\phi}|1\rangle+\cos\left(\frac{\pi}{2}-\frac{\theta}{2}\right)|0\rangle\right)$. By the same mapping, this has a Bloch vector $$ \vec{n}'=(\sin(\pi-\theta)\cos\phi,-\sin(\pi-\theta)\sin\phi,\cos(\pi-\theta))=(\sin\theta\cos\phi,-\sin\theta\sin\phi,-\cos(\theta)). $$ This is exactly a $\pi$ rotation (or reflection) about the x axis.

In error correction, it's a completely different reason for using the density matrix - you're trying to deal with possible errors, and that means you need an ensemble of states (one for each possible error), not just a pure state.

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Most error correction schemes suppose that the transmitted vector state becomes mixed on the receiver end. For example, the transmitted state $|00\rangle$ can turn to $|11\rangle$ with probability $p^2$, to $|10\rangle$ with prob $p(1-p)$, to $|01\rangle$ with prob $(1-p)p$ and to $|00\rangle$ with prob $(1-p)^2$. So, the state in the end is a mixed state which can't be associated to a single vector state.
You still can work with probability distributions of vector states instead of density matrices (and apply whatever rotations you like just on vectors), but generally this is not that fruitful. Density matrices capture the essence of the physical state.

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  • $\begingroup$ Thanks. Strange thing is that I tried to work out the Pauli rotations. I found that they really rotate by 180 only when applied to the density matrix of a Bloch sphere vector from both sides $\endgroup$ – bilanush Mar 7 at 0:14
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    $\begingroup$ Of course. If $|\phi\rangle$ is a vector state and $U$ is some rotation then the result is $U|\phi\rangle$. But in terms of density matrices the result is $(U|\phi\rangle) \cdot (U|\phi\rangle)^\dagger = U |\phi\rangle \langle\phi| U^\dagger$. $\endgroup$ – Danylo Y Mar 7 at 9:09
  • $\begingroup$ Yeah I know. It's just that I couldn't see that x|ϕ⟩ is doing the rotation. Only on density matrix $\endgroup$ – bilanush Mar 9 at 10:11

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