# What is the physical intuition for Bloch Sphere? [duplicate]

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?

• 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. – Alexander Jun 26 '20 at 5:43
• Related : Understanding the Bloch sphere. Especially @CR Drost's answer. – Frobenius Jun 26 '20 at 12:51

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 $$(0, y, z)\cdot (1, 0, 0)=0$$.