# Spin 1 particle, vector and photon

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.

• 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. Feb 5 at 23:21
• Isn’t it true for any rotation axis? Why do you think it means a particular axis? Feb 5 at 23:36
• When you quote a source, you should always specify that source. Feb 5 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 6 at 1:37
• $U$ 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) Feb 6 at 9:18

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:

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

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