# How do we define qubit measurements in a plane?

When does $$\vec{a} \cdot \vec{\sigma}$$ define a measurement in x-y, y-z, and x-z planes?

• Did you mean projective measurement, i.e., $|\vec{a}|=1$? If so I think the answer is trivial. If not, can you give some references or a more detailed description of your problem? Jan 15 at 10:32

$$\vec{a}\cdot\vec{\sigma}$$ defines an operator $$a_xX+a_YY+a_ZZ$$ where $$X,Y,Z$$ are the Pauli matrices. So, for example of $$\vec{a}=(0,a_Y,a_Z)$$ then it has no component in the X plane, and we say it defines a measurement in the $$Y-Z$$ plane.
• Then a measurement in $x-y$ plane would correspond to $\vec{a}=(a_X, a_Y,0) = (\sin\theta \cos\phi, \sin\theta \sin\phi, 0)$. But this depends on both $\theta$ and $\phi$, while as we know the measurements in $x-y$ plane should only depend on $\phi$. Jan 14 at 16:32
• But remember that the vector has length 1. So the $\sin\theta$ disappears in normalisation (or, you're say that you set $\cos\theta=0$, so $\sin\theta=\pm 1$). Equally, just because a vector has been parametrised in a certain way doesn't mean that's a good parametrisation for the context. For example, the y-z plane, you're best to just parametrise it as $(0,\cos\gamma,\sin\gamma)$, automatically capturing the length 1 property. Jan 17 at 7:09