# Why is it that the Pauli matrices and error correction operator act only on $|\psi\rangle\langle \psi|$ and not on state vector $|\psi\rangle$ itself?

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?

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.
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.
• 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 $(U|\phi\rangle) \cdot (U|\phi\rangle)^\dagger = U |\phi\rangle \langle\phi| U^\dagger$. – Danylo Y Mar 7 '19 at 9:09