# How to rotate an electron mathematically?

Im a mathematics student who just learned about the fact that if you rotate an electron by $$2 \pi$$ its spin state changes but if you turn it by $$4 \pi$$ it stays the same.

I understand all the mathematics that is usually used to illustrate this fact, i.e. I know how to calculate the fundamental group of $$\text{SO}(3) \cong \mathbb{RP}^3$$ and I know how to show that a path in $$\text{SO}(3)$$ that corresponds to a rotation by $$2 \pi$$ can't be contracted to the constant path. I also understand the basics of the representation theory of $$\text{SO}(3)$$ and its Lie-Algebra.

What I don't get, however, is how all this is connected to physics. Im willing to accept that the spin of an electron is described by a $$v \in \mathbb{C}^2$$ and that rotating the electron changes this vector by applying a $$U \in \text{SU}(2)$$ to it (or can this be derived from first principles too?). But how do I know which element in $$\text{SU}(2)$$ corresponds to a certain "physical" rotation, described by an element in $$\text{SO}(3)$$? I know that $$\text{SU}(2)$$ is a cover of $$\text{SO}(3)$$, so I presume one could lift the rotation, but now there are two options for that lift... and why would the physical rotation even be connected to the element in $$\text{SU}(2)$$ via this cover, is this just postulated or is there a general theory behind this?

This probably boils down to the following: I understand the double-cover connection between $$\text{SO}(3)$$ and $$\text{SU}(2)$$ but from which axioms/postulates of quantum mechanics does the relevance of this to physics come from.

I think my way of stating the question shows that I probably just know too little quantum mechanics, so I'd be happy if someone could explain this to me or point me to a resource.

(I wrote this answer in a hurry yesterday, I now fixed a huge number of typos and various mistakes, sorry.)

Well, there are a number of important issues in the question.

First of all, pure quantum states are described by the elements of the complex projective space of a complex Hilbert space $$\cal H$$. We can therefore work with equivalence classes of unit vectors $$[\psi]$$, $$||\psi||=1$$, $$\psi \in {\cal H}$$. I will denote by $${\cal S}_{\cal H}$$ the set of these equivalence classes of unit vectors. A unit vector $$\psi\in {\cal H}$$ is said state vector, the associated state is the equivalence class $$[\psi]:= \{e^{ia}\psi \:|\: a \in \mathbb{R}\}\in {\cal S}_{\cal H}$$. A pure state includes all possible physical information on a given quantum system.

If $$\langle \cdot|\cdot \rangle$$ denotes the inner product in $${\cal H}$$, $$p([\psi], [\psi']):=|\langle\psi|\psi' \rangle|^2$$ is called transition probability beteween the state $$[\psi]$$ and the state $$[\psi']$$.

Evidently, it is independent of the representative elements chosen in $$[\psi]$$ and $$[\psi']$$. So it defines a structure on $${\cal S}_{\cal H}$$ (it can be rewrite in terns of a suitable metric on $${\cal S}_{\cal H}$$). It is a physical quantity describing a certain type of probability.

The action of $$SO(3)$$ on pure quantum states is performed in terms of continuous projective representations which preserve the transition probabilities. In other words, it is a map $$SO(3) \ni R \mapsto S_R$$ where

• $$S_R: {\cal S}_{\cal H} \to {\cal S}_{\cal H}$$ is bijective,

• it preserves the $$SO(3)$$ group structure ($$S_I= id$$, $$S_{R}S_{R'} = S_{RR'}$$),

• it preserves transition probabilities ($$p(S_R[\psi], S_R[\psi'])= p([\psi], [\psi'])$$),

• it is "continuous" ($$SO(3) \ni R \mapsto p([\psi], S_R[\psi'])$$ is continuous for every fixed couple of states).

In view of several theorems essentially due to Weyl, Wigner and Bargmann, all these representations can be constructed out of strongly continuous unitary representations $$SU(2) \ni g \mapsto U_g \in \mathfrak{B}(\cal H)$$ of the universal covering $$SU(2)$$ of $$SO(3)$$ acting on $${\cal H}$$.

If $$\pi : SU(2) \to SO(3)$$ is the covering map and $$U$$ is a representation as above,

$$S_{\pi(g)} [\psi] := [U_g \psi]$$

defines a representation of $$SO(3)$$ in the space of the states $${\cal S}_{\cal H}$$ which respects all requirements above. Varying $$U$$, one obtains all possible such representations.

Finally, every strongly continuous unitary representation of $$SU(2)$$ can be decomposed in terms of irreducible representations. These are finite dimensional and of dimension $$2j+1$$ with $$j= 0,1/2, 1, 3/2, \ldots$$.

Physically speaking, $$j$$ is the spin of the considered particle.

If $$j$$ is semi-integer, then $$U_{2\pi}$$ is $$I$$ up to a sign. But it does not matter, since we are interested in $$S_{2\pi}=id$$ on $${\cal S}_{\cal H}$$.

In my view, the claim that rotations of $$2\pi$$ can be experimentally detected is quite a physical misunderstanding.

The experiments that are claimed to do it work as follows.

One considers a pure state, represented by a vector $$\Psi$$, which is made of a superposition of two vectors $$\Psi = \psi_1+\psi_2$$. The system is a particle with spin $$j=1/2$$ here.

In the physical space — these vectors are functions $$\psi: \mathbb{R}^3 \to \mathbb{C}^2$$$$\psi_1$$ and $$\psi_2$$ have disjoint supports in the physical space, so that in particular $$\langle \psi_1|\psi_2\rangle =0$$.

Next the system interacts with some physical device which is spatially localized in the support of $$\psi_2$$ only. If the device is switched on, the net effect is to pass from $$\psi_2$$ to $$-\psi_2$$. As the spin is $$j=1/2$$, this operation corresponds to a "rotation of $$2\pi$$". Physically speaking, the procedure is in fact performed in terms of the action of a magnetic field on a system with a magnetic momentum. If the system were classical, it would be a rotation of the magnetic moment in the physical space of $$2\pi$$.

If the device is switched off, the vector $$\psi_2$$ remains unchanged.

After the operation, the overall vector passes from $$\Psi$$ to either $$\psi_1 -\psi_2$$ or $$\psi_1 + \psi_2$$. It is evident that the corresponding states are different!

The point is that the device does not act on the whole system. So, it does not rotate of $$2\pi$$ the whole particle, but just "a part".

What it is true, however is that, if the system were classical we would not observe this change of state.

That is because a classical particle cannot simultaneously stay in two disjoint regions: the supports of $$\psi_1$$ and $$\psi_2$$.

A quantum particle (in a sense) can do it, instead!

• To read more on why the universal cover appears when considering the mathematics of representations in quantum mechanics, I would recommend this excellent answer Commented Apr 4 at 7:51
• Thank you, this all makes a lot of sense and clears many things up! Do you maybe know a book that explains projective representations and how they are relevant in quantum mechanics further? Commented Apr 4 at 18:36
• There are discussions on these subjects in my books Spectral theory and QM (Springer 2017) and Fundamental mathematical structures of Quantum Theory (Springer 2019). Commented Apr 5 at 4:28

This experiment has actually been done with neutrons. The first publication is Werner et al., PRL 35, 1053 (1975).

The experiment was done with a neutron interferometer. A low-intensity beam of neutrons passes through three thin parallel sheets carved from a hand-sized chunk of perfect silicon crystal. At each of these sheets, part of the neutron beam goes straight through, but part is refracted. The geometry and the neutron wavelengths are carefully chosen so that there is a diamond-shaped path through the interferometer where the neutrons can take either leg of the diamond and end up at the same place on the final crystal vane, with the intensities of the two paths being roughly equal. Neutrons exiting the interferometer here could have reached it via either leg of the diamond, and so therefore they behave quantum-mechanically as if they have taken both legs, and interfere with each other.

(Note that the intensity in a neutron interferometer is very low. The interferometers might be ten centimeters long, and the "slow" neutrons might be traveling at a kilometer per second, so for realistic rates in the neutron detectors the mean distance between neutrons is much, much longer than the physical size of the interferometer. I don't have detection rates handy, but most neutron interferometer experiments never have two neutrons in the interferometer at the same time in their entire dataset. Each neutron takes both paths.)

You measure the interference by putting a thin neutron-transparent piece in one leg of the interferometer, to induce a phase shift, and then adjust the angle of the phase-shifter so that its thickness (and the associated phase shift) changes. In the other leg of the interferometer, you put the sample whose neutron transmission properties you want to study.

For the spin-rotation experiment, the neutrons were polarized so that they all entered the interferometer with their spins parallel to some overall magnetic field. If a neutron travels through a magnetic field which varies "slowly enough" over the neutron's path, the spin transport is "adiabatic," so that the neutron's spin will remain parallel to the magnetic field as the direction of the magnetic field changes. But if the magnetic field changes "suddenly enough," such as sending the beam through a sheet of electrical current, the neutron's spin can remain pointing in the same physical direction but reverse relative to the magnetic field.

So, in the spin-rotation experiment, the "sample" leg of the interferometer sent the neutrons through a region of slowly-changing magnetic field. There could be no magnetic field. Or the magnetic field could twist the neutron's spin through one rotation; or through two rotations. The result was that the single rotation changed the interference pattern as if the sign of the wavefunction had reversed in the two legs of the interferometer, but that two rotations put the sign back the way it had been.

My PhD advisor was at the conference where this result was announced. He told me that everyone in the room agreed with what the experiment had measured, but had a different interpretation of what exactly it meant.

I know that your question is more mathematical than physical. But hopefully this will put you into parts of the literature where people have actually thought seriously about the problem, since there is data to compare it with.

• Thank you so much for your detailed answer, this certainly helps me understanding what physical implications this sign change has. I’ll try to look at the literature! But im a bit confused because wouldn’t this mean that the spin isn’t fully described by a projective space (as implied in the comment under my question), because there the sign information would be lost and these effects couldn’t be explained? Commented Apr 4 at 5:16
• I'm afraid your question about projective spaces is a little too deep in the mathematical weeds for me. It rings some bells, but I haven't thought about this carefully in several years.
– rob
Commented Apr 4 at 5:20
• Is the experiment you're thinking of Observation of gravitationally induced quantum interference or Observation of the phase shift of a neutron due to precession in a magnetic field? Both have a similar enough experimental setup to what you describe, include SA Werner, are pre-1990, and feel like they could leave a conference audience with agreement about measurement and disagreement about interpretation. Commented Apr 7 at 15:16
• @AdamHyland Thanks for finding that! I've edited a citation into the answer. The gravitational phase shift experiment (sometimes called the C.O.W. paper, hah) is also an interesting story, because the measurement doesn't agree with the prediction. I think the consensus on that one is that the interferometer underwent some torsion or other deformation as it was rotated out of the horizontal plane, but (later) efforts to model this torsion were unsuccessful.
– rob
Commented Apr 7 at 15:36

The simple answer is that physical rotations are continuous, so they can be described not by a point in $$SO(3)$$ or $$SU(2)$$ but by a path from the origin. Such paths are unambiguous since $$SO(3)$$ and $$SU(2)$$ are locally isomorphic.

The only situation in which you might have a rotation in $$SO(3)$$ without a disambiguating path is if you are performing a (physically meaningless) change of coordinates. In that case you can pick either representative in $$SU(2)$$; the result is the same up to a (likewise meaningless) global factor of $$-1$$.

Valter Moretti's answer suggests that this is an inherently quantum phenomenon. That happens to be true in the real world because of the spin-statistics theorem, but there is nothing in the mathematics of it that would prevent a spin-½ wave from existing in a classical world. The Dirac equation is a classical field equation, in fact, and if you did a classical Young's double-slit experiment on a classical Dirac wave with a stack of 90°-rotation plates at one of the slits, you would need eight plates to return the fringes to their original position.

The other answers in this post invoking "quantum mechanics" or "quantum states" to explain the classical spinor behavior are totally misguided.

When the great mathematician Élie Cartan first introduced the concept of spinor in 1913, it's a purely classical concept of geometry. Quantum mechanics has not been fully developed yet back in 1913. The classical spinors have NOTHING to do with quantum mechanics. The rotation property of a spinor is a purely classical spinor behavior, not like the other quantum properties of a spinor when the spinor field is quantized in QED for example.

Let's look at the different ways a classical spinor $$\psi$$ and a vector $$V$$ transform under a rotation: $$\quad \psi \rightarrow e^{iT\theta}\psi, \\ \quad V \rightarrow e^{iT\theta}Ve^{-iT\theta}= e^{2iT\theta}V,$$ where $$T$$ is a rotation generator of the Lie algebra which anti-commutes with $$V$$. Because of the double-sided nature of $$e^{iT\theta}Ve^{-iT\theta}$$, a $$\theta = 2 \pi$$ rotation of a spinor/fermion is translated to a $$2\theta = 4 \pi$$ rotation of a vector.

Therefore it's wrong to claim that "if you rotate an electron by $$2 \pi$$ its spin state changes but if you turn it by $$4 \pi$$ it stays the same". Actually, a $$2 \pi$$ single rotation is necessary to bring a spin 1/2 particle back to its initial state. The proper question is "why a $$2 \pi$$ single rotation of a spin 1/2 particle is translated to $$4 \pi$$ double rotation of a vector". As you can see, the "double rotation" is on a vector, not on a spinor!

You may want to argue that since the rotation angle is perspective (spinor vs. vector) dependent, why do we take the spinor's perspective (a $$2 \pi$$ rotation), rather than the vector's perspective(a $$4 \pi$$ rotation)? It has to do with the fact that spinor is more fundamental than vector. From physics perspective, vector is "the square of spinor": $$V_\mu = \bar{\psi}\gamma_\mu \psi$$ In other words, a vector is a composite quantity derived from bi-spinor or spinor bilinear. Therefore, a lot things about vector are doubled, including the rotation angle.

I have to emphasize again that the above $$\psi$$ is the classical spinor field, NOT a quantum wave function. The classical spinor field $$\psi$$ (without field quantization such as in QED) and its rotation properties have NOTHING to do with "quantum mechanics" or "quantum states".

See more explanations here.