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In the statement:

"The rotation of the spin of a photon through 360 degrees brings it back to its originial state, and hence it is a spin 1 particle, in other words its spin is a vector"

What axis of rotation is being referred to?

If it refers to the direction of motion of the photon, then surely, since the spin of a photon is either in or opposite to its direction of motion, the spin (R or L circularly polarized), does not change for any $ \theta $; only the instantaneous direction of the photon's electric (and magnetic) field vector changes with such a rotation.

And furthermore since the spin of a photon is either in or opposite to its direction of motion, this would be nothing more than 'rotating' the photon's spin about its own axis, which would hardly be called a 'rotation' of the spin vector at all.

Qmechanic
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pete
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  • You are mistaking rotation by helicity flip. Helicity is that component of the spin $\vec{J}$ in the direction of motion; a rotation around an axis with orientation given by the vector $\vec{n}$ is the action representated by the operator $U = \exp{ i\theta \vec{n} \vec{J} }$. This operator acts over the state (or wavefunction) and can be associated to the rotation of the reference frame, i.e., to a Lorentz transformation. – Vicky Feb 05 '23 at 23:21
  • Isn’t it true for any rotation axis? Why do you think it means a particular axis? – Ghoster Feb 05 '23 at 23:36
  • When you quote a source, you should always specify that source. – Ghoster Feb 05 '23 at 23:41
  • @Vicky $ \exp{ i\theta \vec{n} \vec{J} } $ is just a helix around an axis, n⃗ . J⃗ ,normal to the imaginary axis, with a phase shift of θ. So $ \vec{n} $ can be any direction in space, and it will repeat after rotation of 2 Pi right? – pete Feb 06 '23 at 01:37
  • It’s not a helix, and $\hat n$ is in 3D space, which doesn’t have an imaginary axis. You should try to calculate $\exp{(i\theta\hat n\cdot\vec J)}$ to see what it is. – Ghoster Feb 06 '23 at 01:44
  • @Ghoster I didn't mean that either $ \vec{n} $ or 3D space had an imaginary axis and $ \exp{ i\theta \vec{n} \vec{J} } $ is a complex function and as such the space in which it resides does have an imaginary axis, i and $ \hat n\cdot\vec J $ = k so woops! the 'axis' will be $ \theta $ – pete Feb 06 '23 at 02:34
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    $U$ is not a function but a group transformation that can be represented as a matrix if you expand the exponential and take into account the matrix representation of $\vec{J} = (J_x, J_y, J_z)$. Each $J_i$ is a matrix. The complex number $i$ doesn't refer to any "imaginary" axis, it is there because $SU(D)$ (the group of rotations in a $D$-dimensional space) is unitary, this is, $U^\dagger = U^{-1}$. I have the feeling you have a lack of knowledge... (1/2) – Vicky Feb 06 '23 at 09:18
  • ... in group theory and that's why you're so confused. I recommend the course http://www.th.physik.uni-bonn.de/nilles/people/luedeling/grouptheory/, by profs. Förste and Lüdeling. (2/2) – Vicky Feb 06 '23 at 09:19
  • Extra comment: there could be people that would (rightfully) complain by my statement about $SU(D)$ being the group of rotation while, extrictly speaking, the group of rotations is $SO(D)$. As far as I know, $SU(D)$ contains $SO(D)$, there are isomorphism between them and if you want to be open to the possibility of spin, you need $SU(D)$. So call it abuse of language, if you'd like. Vid. for example https://en.wikipedia.org/wiki/Spin_(physics)#Rotations – Vicky Feb 06 '23 at 09:25
  • @Vicky Yes vicky, I confess I am completely ignorant of group theory and matrix mechanics etc. From what I remember, the latter is a more compact way of handling loads of simultaneous equations. I only know that $ J_x, J_y $ and $ J_z $ don't commute, so 'rotating' $ \vec{J} $ will probably transform $ J_x, J_y $ and $ J_z $ differently than if $ \vec{J} $ were a classical vector in 3D. But to give me a little bit of a preview on what this woule look like in one specific instance, can you answer this example for me: https://physics.stackexchange.com/questions/748828/photon-rotation – pete Feb 06 '23 at 17:27
  • $J$ does not rotate, what rotates is the state or the reference frame. You cannot address any problem or question related to spin without mastering the lectures I posted several comments above. Study group theory and matrix algebra via the link I posted before and then you'll undertand what rotates and how – Vicky Feb 07 '23 at 01:11

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A positive helicity photon with $\vec k = k\hat z$ has spin vector given by ($\hbar =1$):

$$ \vec s = - (\hat x + i\hat y) = \sqrt 2 Y_1^1(\theta, \phi)$$

It should be easy to show that is an eigenstate of rotations about the $z$-axis with eigenvalue (m=1):

$$ \lambda(\phi) = e^{im\phi} = e^{i\phi} $$

so that:

$$ \lambda(2\pi) = e^{2\pi i} = +1 $$

If I pick a transverse axis and (alias) rotate 90 degrees, I get:

$$ \vec s = -(\hat x' + \hat z') = \sqrt 2 Y_1^1(\theta', \phi') - Y_1^0(\theta', \phi') - \sqrt 2 Y_1^{-1}(\theta', \phi') $$

A 180 degree rotation gives:

$$ \vec s = -\hat x'' + i\hat y'' = -\sqrt 2 Y_1^{-1}(\theta'', \phi'')$$

I chose an alias rotation to change the coordinates and leave the spin, momentum unchanged, so in the new coordinate system, that is still a +1 helicity photon, as the momentum becomes:

$$ R(90^{\circ})\hat z = -\hat y' $$

$$ R(180^{\circ})\hat z = -\hat z'' $$

What makes this odd is that I used cartesian vectors for the momentum, and implicit spherical vector for the spin. Spherical vectors are complex combinations of cartesian vectors that look exactly like spherical harmonics in cartesian coordinates. They are useful because the basis vectors are eigenvectors of z rotations with eigenvalues $\exp{im\phi}$ for $m \in (-1, 0, 1) $.

In all cases, a 360 degree rotation is returns the state to the the original state.

It's also odd, because in the natural basis:

$$ Y_1^0(\theta, \phi) = \hat z $$

is not available to the photon spin. $m=0$ is transverse polarization.

Note an $m=2$ spin-2 particle would have a spin dyad:

$$\vec s\vec s= [-\sqrt 3(\hat x + i\hat y)][-\sqrt 3(\hat x + i\hat y)]$$

$$ = 3\left[\begin{array}{ccc} 1 & i & 0 \\ i & -1 & 0 \\ 0 & 0 & 0 \end{array}\right] = 3Y_2^2(\theta, \phi) $$

You can verify that is an eigentensor of z-rotations with eigenvalue

$$ \lambda(\phi) = e^{2i\phi}$$

For fun, a 90 degree rotation would be:

$$ = 3\left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 0 & 0 \\ i & 0 & -1 \end{array}\right] $$

$$ \frac 3 2 Y_2^2 - 3iY_2^1-\sqrt{\frac{27} 2}Y_2^0 +3iY_2^{-1} +\frac 3 2 Y_2^{-2} $$

and a 180 degree rotation would return to the original state.

JEB
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