Non-diagonal components of matrices of operators (Pauli matrices as an example)

Question

I understand the derivation of the pauli matrices. But there is something else I dont understand. Lets take $\hat{s}_x$ as an example. $$ \hat{s}_x=\frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} $$ and in gerneral the matrix elements of $\hat{s}_x$ are $$ \hat{s}_x=\begin{pmatrix} \langle\uparrow|\hat{s}_x|\uparrow\rangle & \langle\uparrow|\hat{s}_x|\downarrow\rangle \\ \langle\downarrow|\hat{s}_x|\uparrow\rangle & \langle\downarrow|\hat{s}_x|\downarrow\rangle \\ \end{pmatrix} $$ and now what I dont understand is that for example $$ \langle\downarrow|\hat{s}_x|\uparrow\rangle=\frac{\hbar}{2}\neq0 $$ I'm suprised this is not euqal to 0 because (quote from feynman lectures vol. 3 chapter 8)

We have also talked about what happens when particles go through an apparatus $A$. If we start the particles out in a certain state $ϕ$, then send them through an apparatus, and afterward make a measurement to see if they are in state $χ$, the result is described by the amplitude $$ \langle\chi|A|\phi\rangle $$

And now I dont have an intuitive understanding of what the apparatus $\hat{s}_x$ does. So by looking at what the forumulas say $\hat{s}_x$ is an apparatus that switches the spin orientation with a probability of $\frac{\hbar}{2}$. But that doesnt sound that right. Is there some intuitive explanation of why this isnt =$0$? Or maybe isnt there even an intuitive understanding of the matrices of operators with the method I used (comparing the values with the bra-ket formulas of the corresponding entries)?

Answer 1

You have 4.5 questions in here, which you seem to be connecting in obscure ways. Let me parse out the correct statements. You appreciate you are working in the basis of eigenstates ("spin orientations") of another operator, $\hat s_z$. Indeed,

So by looking at what the formulas say, $\hat{s}_x$ is an apparatus that switches the spin orientation with a probability of $\frac{\hbar}{2}$.

Yes! So $$ \langle\downarrow|\hat{s}_x|\uparrow\rangle=\frac{\hbar}{2}\neq 0 $$ is the nonvanishing probability amplitude as Feynman says, that if you start with a spin up state you end up with the spin down state; while the up-up amplitude you also found to be 0. If you measure your outcome state (with, e.g., $\hat s_z$), it will be down (it will have the negative gadget eigenvalue)!

Alternatively, and equivalently, you may think of $\hat s_x$ itself as a measuring gadget. You recall the eigenstates of it are, for positive or negative eigenvalues $\pm \hbar/2$ respectively, $$ |+\rangle=\frac{1}{\sqrt 2}. ( |\uparrow\rangle +| \downarrow\rangle) ,\\ |-\rangle=\frac{1}{\sqrt 2} ( |\uparrow\rangle -| \downarrow\rangle) ~~~\leadsto \\ |\uparrow\rangle = \frac{1}{\sqrt 2} ( | +\rangle +|-\rangle) \\ |\downarrow\rangle= \frac{1}{\sqrt 2} ( | +\rangle -|-\rangle). $$ The operator $\hat s_x$ will put the $|\uparrow\rangle$ state to one of its own eigenstates as a measuring apparatus. If its dial reads $\hbar/2$, it will have ascertained it is reading the $| +\rangle $ component, which will happen with probability amplitude $$ \langle +|\hat{s}_x|\uparrow\rangle=\frac{\hbar}{2\sqrt{2}} ~. $$ If minus that, then it is picking the $| -\rangle $ component, which will happen with probability amplitude $$ \langle -|\hat{s}_x|\uparrow\rangle=-\frac{\hbar}{2\sqrt{2}}~. $$

But if you, instead, as above, you measure the states with a z-axis gadget, then you do not read (observe) the $\pm$ x-orientation amplitudes so they add (the essence of QM interference) to yield an amplitude
$$ \langle\downarrow|\hat{s}_x|\uparrow\rangle=\frac{\hbar}{2}, $$ as per above. The operator acts like a 2-slit gadget.

Each operator measures the eigenvalues of its own eigenstates.