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Suppose I have an electron that is in the spin state $$\chi =A\begin{bmatrix}3i \\4\end{bmatrix} $$ If I calculate the expectation values of its spin components $S_x$ $S_y$ $S_z$, I get $$\langle S_x \rangle= 0$$ $$\langle S_y \rangle = -\hbar \dfrac{12}{25}$$ $$\langle S_z \rangle= -\hbar \dfrac{7}{50}$$ I can do the math part easily, however I am having a bit of difficulty trying to apply a physical interpretation to my answers.

My attempt at a physical interpretation

Due to the uncertainty principle, we can not having a well defined $S_x$ $S_y$ $S_z$ component simultaneously. Hence, we can only calculate the expectation value of its components. For $S_x$, the probability that the spin will lie in the $S_x$ component will be zero. For $S_y$ and $S_z$ we the spin has either $-\hbar \dfrac{12}{25}$ or $ -\hbar \dfrac{7}{50}$.

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Average values do not tell you about probabilities (which must be non-negative numbers). In particular, if your particle is describe by a state $\vert\xi\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1 \end{array}\right)$ the average $\langle S_z\rangle=0$ but the probability of having the system in the spin-up state is $1/2$, as is the probability of having the system in the spin-down state.

Average values are weighted sums of probabilities: in the case of $\vert\xi\rangle$, the average value is $$ \langle S_z\rangle = P(\uparrow)\frac{\hbar}{2}+P(\downarrow)\left(-\frac{\hbar}{2}\right) =\frac{1}{2}\frac{\hbar}{2}+ \frac{1}{2}\left(-\frac{\hbar}{2}\right)=0 $$ where $P(\uparrow)$ is the probability of having your system in the spin-up state, and $\hbar/2$ is the value of $S_z$ when the system is in the spin-up state.

Clearly here, the fact that the average value is $0$ does not in any way mean the probability of the spin being along $\hat z$ is $0$, since the system is half the time along $+\hat z$ and half the time along $-\hat z$. In fact, the only two possible values of spin are $\pm\frac{\hbar }{2}$, which excludes $\langle S_z\rangle=0$ for this example. More generally, the only possible outcomes of $S_y$ or $S_x$ are $\pm \frac{\hbar}{2}$, and never numbers such as $-\frac{12\hbar}{25}$.

In general, the probability of a system described by $\vert \chi\rangle$ to be found in a state having definite spin along - say - $\boldsymbol{\hat y}$ is given by $$ P(\uparrow)_{\hat y}= \vert\langle\chi\vert \uparrow\rangle_{\hat y}\vert^2\, ,\qquad P(\downarrow)_{\hat y}=\vert\langle\chi\vert \downarrow\rangle_{\hat y}\vert^2 $$ where $\vert\uparrow\rangle_{\hat y}$ is the eigenstate of $\sigma_y$ with eigenvalue $+\hbar/2$. In your case, we have $$ P(\uparrow)_{\hat y}=\frac{1}{50}\, ,\qquad P(\downarrow)_{\hat y}=\frac{49}{50} $$ so $$ \langle S_y\rangle = \frac{\hbar}{2}\frac{1}{50}-\frac{\hbar}{2}\frac{49}{50}=-\frac{12\hbar}{50}\, . $$

What the average values tell you is the "direction" of the spin. The unit vector $\boldsymbol{\hat n}=(\sigma_x,\sigma_y,\sigma_z)=(0,-24/25,-7/25)$ points in a definite direction, and if you were to choose your quantization axis along this direction, the system would be in a spin-up state, i.e it would be an eigenstate of $$ \boldsymbol{\hat n}\cdot \vec S= \frac{1}{25}\left( \begin{array}{cc} -7 & 24 i \\ -24 i & 7 \\ \end{array} \right)\, . \tag{1} $$ You can indeed verify that your vector $\chi$ is an eigenvector of (1) with eigenvalue $\frac{\hbar}{2}$

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Expectation values are just average values; namely if you do a measurement of the spin at the direction in question "many times" with "identical" setups, the average values of your spin measurements at that direction should be equal to the expectation value given by quantum mechanics.

That said I think your statement about $S_x$ is not so precise. When you measure $S_x$ several times, it doesn't mean that you always get $0$; it's the average that's zero but your measurement result at a particular time may not be zero.

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