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Usually, when a wave or wave-like object or system goes through a $180^{\circ}$ twist or turn or whatever, we say it is opposite to how it was oriented before, and if it came across its former self like this it would destructively interfere.....

But, theorists claim that a spin-$\frac{1}{2}$ particle that rotates or travels through a full circle or wavelength, is now in its opposite 'phase' orientation....

If a particle is now 'opposite' or 'sign-flipped from where it was before, why not just say it has phase-shifted by $180^{\circ}$?

Doesn't the sign flip or potential for destructive interference DEFINE a change of $180^{\circ}$?

If a spin-$2$ particle changes sign after $90^{\circ}$, why not define that much 'phase-shift' as $180^{\circ}$, and say that its wavelength is $\frac{1}{4}$ of an otherwise-identical spin-$\frac{1}{2}$ particle, and that it's inherent angular momentum is $4$ times as strong?

Kurt Hikes
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    If I turn my car radio dial by one full turn, I receive a different station than before. Should I therefore define a full turn of my radio dial to be something other than 360 degrees? – WillO Nov 18 '21 at 02:46
  • Related: https://physics.stackexchange.com/q/597902/2451 – Qmechanic Jan 14 '22 at 20:24

2 Answers2

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Consider the function $\phi_s(\theta) = e^{i s \theta}$.

Note that \begin{eqnarray} \phi_{2}(0) &=& \phi_2\left(\pi\right) = \phi_2\left(180^o\right) \\ \phi_{1}(0) &=& \phi_1\left(2\pi \right) = \phi_1\left(360^o\right)\\ \phi_{1/2}(0) &=& \phi_{1/2}\left(4\pi \right) = \phi_2\left(720^o\right) \end{eqnarray} Now call s the spin (helicity would actually be a better word but let's put that aside), call $\theta$ the angle of rotation in physical space, and call $\phi$ the phase of the wavefunction. We see that

  • for spin-2, we need to rotate the particle in space by 180 degrees for the particle's wavefunction's phase to advance by 360 degrees,
  • for spin-1, we need to rotate the particle in space by 360 degrees for the particle's wavefunction's phase to advance by 360 degrees,
  • for spin-1/2, we need to rotate the particle in space by 720 degrees for the particle's wavefunction's phase to advance by 360 degrees.

Also note that since $\phi_{1/2}(2\pi)=-\phi_{1/2}(0)$, if we rotate the particle in space by 360 degrees -- which you might think would just be returning the particle to it's original state -- we actually change the sign of the particle's wavefunction. This is pretty weird! But that's how spin-1/2 particles work.

Andrew
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    It's not that weird. The same thing happens in the plate trick. https://en.wikipedia.org/wiki/Plate_trick – PM 2Ring Nov 18 '21 at 03:12
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    @PM2Ring Sure, weirdness is subjective, and spin-1/2 particles are mathematically self-consistent and have mechanical analogies and eventually you just learn to trust the formalism and not think about it. Still I think most people go through a moment when learning about spin-1/2 particles where they think "that's pretty counterintuitive." – Andrew Nov 18 '21 at 03:25
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    Agreed. Note that I didn't say that it's not weird. :) – PM 2Ring Nov 18 '21 at 03:27
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You need to differentiate between the angle of rotation and the angle of a phase shift.

Everyone agrees that a transformation that sends $x$ to $-x$ is a phase shift by 180°. That's just what phase shift means - it's the angle $\phi$ in $x\mapsto \mathrm{e}^{\mathrm{i}\phi}$. This is not the angle we're talking about when talking about rotations.

Every abstract rotation can be presented as $\mathrm{e}^{\mathrm{i}\phi\vec n\cdot \vec \sigma}$ where $\phi$ is the angle of rotation, $\vec n$ a unit vector and $\vec \sigma$ the vector of generators of rotations around the coordinate axes. Every physical object transforms now in a particular representation of the rotation algebra such that the $\vec \sigma$ corresponds to a particular vector of three matrices (with varying dimensions $2s+1$ depending on the spin $s$).

The statement "spin-1/2 particles change sign under a 360° degree rotation" means that the representation of rotations on spin-1/2 objects is such that the abstract rotation $\mathrm{e}^{2\pi i\vec n\cdot \vec \sigma}$ is mapped to the concrete value $-1 = \mathrm{e}^{\pi \mathrm{i}}$ in the representation space for that spin. Note that really $\mathrm{e}^{2\pi\mathrm{i}\vec n \cdot \vec \sigma} = 1$ should hold abstractly, so this means we're not actually looking at a proper linear representation of the rotation group, but something else that is able to map a full rotation to something other than the identity without becoming inconsistent. I discuss the technical aspects of such projective representations and why they arise in quantum mechanics in this Q&A of mine.

A perhaps more physical and less misinterpretable way to say what "flipping the sign under a full rotation" means is to think about a spin-1 and a spin-1/2 object side by side: The same abstract rotation that returns the spin-1 object to its original state (and is hence a full rotation in the generic case where the object has no rotational symmetry) will flip the sign of the state of the spin-1/2 object. Since we have refrained from talking about any angles at all in this formulation, there is no ambiguity here.

ACuriousMind
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