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I am very confused about how to think about the Bloch Sphere. How can we relate the concept of expectation value to the Bloch sphere? If my state lies in let's say $yz$ plane how can we say that expectation value around $x$ is 0. I can understand that we take the projections and because of the equal probability of projection onto the positive and negative axis, we get a 0. My question is how can a vector living in $yz$ plane can even be projected on $x$-axis?

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    $\begingroup$ See 3rd paragraph en.wikipedia.org/wiki/Bloch_sphere In short, Bloch sphere is just a representation of states. Orthogonal states of the quantum system are located at antipodal points. Thus any two non-antipodal states have some projection one to the other. $\endgroup$ – Alexander Jun 26 '20 at 5:43
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    $\begingroup$ Related : Understanding the Bloch sphere. Especially @CR Drost's answer. $\endgroup$ – Frobenius Jun 26 '20 at 12:51
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Imagine we're considering the spin of an electron. We denote the two possible measurement outcomes in the $z$ direction $|0\rangle =\begin{pmatrix}1\\ 0\end{pmatrix} = $ spin up and $|1\rangle=\begin{pmatrix}0\\ 1\end{pmatrix}=$ spin down, but generically, the state will be a superposition of these two states: $$ |\psi\rangle = A|0\rangle + B|1\rangle $$ Since global phases aren't physically important, we can take $A$ to be real and represent the phase factor in $B$ by $e^{i \phi}$. Normalization requires $|A|^2 + |B|^2 = 1$, which we can enforce by setting $A = \cos\theta/2$ and Re $B = \sin\theta/2$. (We use $\theta/2$ instead of $\theta$ so that our formula yields a one-to-one correspondence between a choice of $\theta,\phi$ mod $2\pi$ and a state $|\psi\rangle$. I won't prove it here, but if we had used $\theta$ instead of $\theta/2$ in our formula, the correspondence would be two-to-one). This shows that any state can be written $$ |\psi\rangle = \cos\theta/2 |0\rangle + e^{i\phi}\sin\theta/2|1\rangle $$ for an appropriate choice of $\theta$ and $\phi$. We can therefore represent a state as an ordered pair $(\theta,\phi)$, which corresponds to a point on a unit sphere. This is the Bloch sphere.

Now, by the basic formula for the expectation value, we have $$ \langle S_z\rangle = \frac{\hbar}{2}\cos^2\theta/2 -\frac{\hbar}{2}\sin^2\theta/2 = \frac{\hbar}{2}\cos\theta $$ which up to a factor of $\frac{\hbar}{2}$ is just the $z$ coordinate of the point $(\theta, \phi)$ on a sphere (since in spherical coordinates on a unit sphere, $(x, y, z) = (\sin\theta \cos\phi, \sin\theta\sin\phi,\cos\theta)$ ). You can similarly show that $$ \langle S_y\rangle = \frac{\hbar}{2}\sin\theta\sin\phi,\quad \langle S_x\rangle = \frac{\hbar}{2}\sin\theta\cos\phi $$ giving the general correspondence $$ (\langle S_x\rangle, \langle S_y\rangle, \langle S_z\rangle) = \frac{\hbar}{2}\left(x_\text{Bloch}, y_\text{Bloch}, z_\text{Bloch}\right) $$ In short: we can represent a state as a point ($\theta, \phi$) on the Bloch sphere. When we represent this point in Cartesian coordinates, the $n$-coordinate of the point gives the spin expectation in the $n$ direction. (I haven't showed it here, but this is true for all directions, not just $x, y, z$.)

Now, in response to your question: A state on the Bloch sphere in the $yz$ plane will have (Cartesian) coordinates $(0, y, z)$. By the above, the state has spin expectation zero. If you're wondering how to project a vector onto the $x$ axis, you do so as you do for any vector: dot it with a unit $x$ vector. In this case, it's clear that $(0, y, z)\cdot (1, 0, 0)=0$.

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