# Outcomes of quantum measurements

I'm pretty new to quantum computing, and I'm wondering how I can compute the outcome of a projective measurement of a spin along the +Z axis followed by a projective measurement along the -Z axis. I think that when we make the +Z projective measurement. I know that +Z axis corresponds to $$(1, 0)^{T}$$ and -Z axis corresponds to the state $$(0, 1)^{T}$$.

I think that the answer will be $$100\%$$ probability of spin down and $$0\%$$ probability of spin up, but I'm not entirely sure why. Can someone please explain to me? I have some vague understanding that it has to do with collapsing states.

How about if we do three projective measurements: +Z followed by +X followed by -Z?

I'm not completely sure how to approach the second problem, and I'd really appreciate any help.

• There's an important piece of information missing: what is the initial spin state before the first measurement? – probably_someone Sep 21 '20 at 4:07

Let $$\Pi_{+z}=\vert+\rangle\langle +\vert$$ and $$\Pi_{-z} =\vert -\rangle\langle -\vert$$. It seems for any state $$\vert\psi\rangle$$ you want to compute \begin{align} \Pi_{-z}\Pi_{+z}\vert\psi\rangle \end{align} but this contains one piece of the form $$\langle -\vert +\rangle=0$$ so if $$\Pi_{-z}$$ immediately follows $$\Pi_{+z}$$ the result is $$0$$ so $$\vert -\rangle$$ has probability $$0$$. Alternatively, $$\Pi_{+z}\vert\psi\rangle$$ will project to $$\vert +\rangle$$ so another $$\Pi_{+z}$$ would give you $$\vert +\rangle$$ 100% of the time since the outcome of the first $$\Pi_{+z}$$ is $$\vert +\rangle$$.
You can repeat the same game with $$\Pi_{+x}$$ to get $$\Pi_-\Pi_{+x}\Pi_+\vert\psi\rangle$$ and just compute overlaps of the type $$\langle +x\vert \pm z\rangle$$.
• Ok. I see why negative z direction has probability $0$ now. I'm still trying to get the second follow-up. I'm having trouble with three measurements. – user146017 Sep 21 '20 at 5:02