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

Question

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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Answer 1

and welcome to the stack exchange!

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.