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I can't seem to connect these two stories. Can you please help?

I have heard that you have to turn electron by 720 degrees in order to get the same spin state. Has this been seen experimentally? How do you turn an electron?

I know that Bloch sphere is used to explain spin states. I sort of understand how it works and how matrices can be constructed. However there you turn the vector by 360 degrees in order to get the same state. Why so? (Is that in reality you have to turn twice as much?)

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You have fallen prey to a popular simplification of spinors. The statement "you have to turn electron by 720 degrees in order to get the same spin state" does not refer to an actual rotation of an actual electron.

In quantum mechanics, we describe the states of objects as elements of a Hilbert space $\mathcal{H}$. The crucial thing is that not all elements of this space represent physically different states - if we have two elements $\phi$ and $\psi$ and they are related in such a way that one can be obtained from the other by multiplying it with any complex number $c$, i.e. $\phi = c\psi$, then they are the same state.

This is analogous to two arrows with different length pointing in the same direction describing the same direction. Only the direction of the Hilbert space element has immediate physical meaning, not the length (though it is not completely irrelevant, "phases" play a role, but this is not relevant here).

Now, it turns out that there are two different ways how such elements of a Hilbert space can behave under a full rotation by $2\pi$ - they either stay the same, $\psi \overset{2\pi}{\mapsto} \psi$, or they change their sign, $\psi \overset{2\pi}{\mapsto} -\psi$. But $-1$ is just a complex number, so $\psi$ and $-\psi$ are the same state, and a rotation by $2\pi$ does not change any state at all.

Objects whose states stay the same are called bosons and have "integer spin", objects whose states change sign are called fermions and have "half-integer spin".

The Bloch sphere you refer to is not the Hilbert space of a system, but the projective Hilbert space. The projective Hilbert space is obtained by just identifying all vectors in the Hilbert space that lie on the same ray ( = have the same direction = are complex multiples of each other).

Thus, $\psi$ and $-\psi$ are the same point in a projective space, hence in particular on the Bloch sphere, and a $2\pi$ rotation does nothing on a projective space either way - as it should, since each point of the projective space is a physically distinct state.

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    $\begingroup$ Whoa. This is news to me. I'm no expert, and I await correction, but it is my belief the the rotation referred to is in real space. The "-1" is observable in neutron interference experiments which verify that a $4\pi$ rotation in real space is necessary to reproduce the interference pattern seen with no rotation. Google is not providing any good references but see Figure 5 here. (to be cont'd) $\endgroup$ – garyp Feb 27 '15 at 20:19
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    $\begingroup$ I also found this less-than-completely-reliable source which contains the phrase "This is exactly the $4\pi$ we were looking for, and it has been seen conclusively in several experiments". I believe (for the time being) that fermions are "spinorial" objects that transform like spinors under rotation in space. $\endgroup$ – garyp Feb 27 '15 at 20:21
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    $\begingroup$ @garyp: That paper is essentially testing the spinorial nature of the neutrons by a relative phase measurement - they are "rotating" one neutron, but not the other, and thus are able to detect the relative minus because they are not rotating the entire two-neutron state. If you did a true spatial rotation, you'd have to rotate every state, and that would make it undetectable again. Their "rotation" is the action of a magnetic field that acts upon the state precisely like a rotation, but it is not a spatial rotation in the sense that the entire system is rotated. (contd.) $\endgroup$ – ACuriousMind Feb 27 '15 at 20:41
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    $\begingroup$ @garyp: The problem here is probably that natural language fails us when we try to describe what a spinor actually is and what its relation to rotations is, and that we are also not consistent in what a rotation is. In my answer, I understand "rotation" as an operation on the space the system lives in - that is, rotation is always applied to everything. If you, like in that paper, understand "rotation" as any operation that triggers the application of the rotation operator to a state, then, indeed you can measure the minus you incur after doing the $2\pi$ operation. (cont'd) $\endgroup$ – ACuriousMind Feb 27 '15 at 20:45
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    $\begingroup$ @garyp: But this, then is because you can then construct superpositions of states where you apply this operation just to one state but not the other, and you can measure this - but this, then, does not correspond to an application of the rotation operator to the total (superposition) state, since part of the state remain unrotated, so it is not a spatial rotation of the system, and hence not a rotation in the strict sense. $\endgroup$ – ACuriousMind Feb 27 '15 at 20:47
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Actually you can do an actual rotation of a spin-1/2 particle and see the change. It has been done in 1975 with neutron interferometers:

https://www.nature.com/nature/journal/v294/n5841/abs/294544a0.html

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My explanation is based on Sakurai's Modern Quantum Mechanics book (Sec 3.2 on neutron interferometry). I have taken Planck's constant and $c$ (speed of light) to be 1.

The classical rotation operator about a direction $\hat{n}$ about an angle is

$D(\hat{n}, d\phi) = 1- i (\vec{J}.\hat{n}) d\phi $,

which suggests that for spins, it should be

$D(\hat{n}, d\phi) = 1- i (\vec{S}.\hat{n}) d\phi $,

which leads to the finite angle version of the rotation operator about the z-axis as

$D( \hat{z}, \phi) = \mathrm{exp} (-i S_z \phi) $.

Now, let's consider space to be filled with a magnetic field $\vec{B}$ pointing in the z-direction so that the Hamiltonian can be easily seen to be (if you remember lessons from classical electrodynamics)

$H = - \omega S_z$,

with

$\omega = e B/(m_e)$,

which means that the time evolution operator for this Hamiltonian becomes

$U(t) = \mathrm{exp} (-i H t) = \mathrm{exp} (-i S_z \omega t) $,

which can be seen to the same as $D( \hat{z}, \phi)$ with $\phi = \omega t$.

So, here is the take-home point. An electron kept in a magnetic field pointing in the z-direction keeps on getting continuously rotated about the z-direction (the direction of $\vec{B}$) as time passes (since $\phi= \omega t$). We cannot rotate the electron spin in the old standard way in which we rotate tables and chairs by moving them around.

Indeed, it has been experimentally verified that a rotation by $2 \pi~(4 \pi)$ gives the negative of (same as) the original state, using neutron interferometry. See the above-referenced book for a more rigorous theoretical explanation and the mention of this experiment.

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