# Find the Probability for a "+" outcome when making a Pauli-x Measurement

So, we apply Equation 3.28 (above) to our initial vector state, following the equation below, to get $$|\psi(t)\rangle$$.

What I obtained was the basically the same equation, except now we have a $$|up\rangle$$ vector on the $$\cos$$ term (as the unity matrix returns the same vector) and "-" the $$\sin$$ term with $$|down\rangle$$, as Pauli-$$X$$ will "flip" the spin state.

Now for the following part, the equation is as follows:

But I feel I need some clarification. Obviously here our vector state is different (the one calculated above), but we have this superposition in the term (South-west pointing). Just before this exercise is asked, we are given the following relationships:

For Pauli-$$X$$ and Pauli-$$Y$$ respectively. Because in the calculated vector term, we have $$|up\rangle - |down\rangle$$, does this mean we are to use the superposition in the top right $$|North-East\rangle$$, and this would be the syntax used?

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• Main Question:
• What is the $$|\psi(t)\rangle$$ term, just to verify I am calculating properly. In the equation for $$P_+$$, we are shown the result is a $$1/2$$[1+term]. This obviously makes sense, as when calculating $$P_-$$, and adding the two terms, you should result in a "$$1$$". In my $$|\psi(t)\rangle$$ I am not getting this $$1+$$ term, I have a feeling there is something to do with an Euler identity here?.

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• Hi there, it's best to use latex or mathjax here rather than linking images. Nov 11, 2021 at 14:45

We have $$U(t) = \cos(\omega t)I -i\sin(\omega t)\sigma_{x}$$ so that: $$\begin{split} |\psi(t)\rangle & = U(t)|+\rangle \\ & = (\cos(\omega t)I -i\sin(\omega t)\sigma_{x})|+\rangle \\ & = \cos(\omega t)I|+\rangle -i\sin(\omega t)\sigma_{x}|+\rangle \\ & = \cos(\omega t)|+\rangle -i\sin(\omega t)|-\rangle, \end{split}$$ since, as you already pointed out, $$\sigma_{x}|+\rangle = |-\rangle$$.
Then, the probability $$P_{+}(t)$$ of getting a '$$+$$' outcome at time $$t$$ is: $$\begin{split} P_{+}(t) &= |\langle+|\psi(t)\rangle|^{2} \\ &= |\langle+|\Big(\cos(\omega t)|+\rangle -i\sin(\omega t)|-\rangle\Big)|^{2} \\ &= |\alpha|^{2}, \end{split}$$ where $$\alpha$$ is: $$\begin{split} \alpha &= \langle+|\Big(\cos(\omega t)|+\rangle -i\sin(\omega t)|-\rangle\Big) \\ &= \cos(\omega t)\langle+|+\rangle -i\sin(\omega t)\langle+|-\rangle \\ &= \cos(\omega t) \cdot 1 -i\sin(\omega t) \cdot 0 \\ &= \cos(\omega t). \end{split}$$ Hence $$P_{+}(t) = |\cos(\omega t)|^{2} = \cos^{2}(\omega t)$$.
As a check, you could work out the probability of getting '$$-$$' outcome, which is $$P_{-}(t) = |\langle-|\psi(t)\rangle|^{2} = \sin^{2}(\omega t)$$. Hence $$P_{+}(t) + P_{-}(t) = 1$$ as it should be.