Since a matter wave is associated with a particle in quantum mechanics, does the wave spins? I mean, can we visualize the spinning of wave or is it possible that the wave spins?

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    $\begingroup$ For a classical wave that has spin, read about circularly polarized light. $\endgroup$
    – rob
    Commented Apr 22, 2014 at 1:35

4 Answers 4


Just some links..

The electron is comprised of two spherical scalar waves, one inward and one outward. A phase shift of the inward wave occurs in the wave-center region near where $\tau=0$, and spin appears as a required rotation of the inward wave in order to become the outward wave. This requirement is a property of 3D space termed spherical rotation. To transform the inward wave to an outward wave and obtain the out-wave with proper phase relations requires phase shifts of the in-wave at the center. These phase shifts produce a spin value of in the entire wave structure. Because spin is the result of required wave phase shifts, a property of 3D space, spin has the same value for all charged particles independent of other particle properties.


It is shown how the spin of the electron and other charged particles arises out of the quantum wave structure of matter. Spin is a result of spherical rotation in quantum space of the inward (advanced) spherical quantum wave of an electron at the electron center in order to become the outward (retarded) wave. Wave rotation is required to maintain proper phase relations of the wave amplitudes. The spherical rotation, a unique property of 3-D space, can be described using SU(2) group theory algebra.

The Physical Origin of Electron Spin - using quantum wave particle structure

Mathematics of Electron Spin

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    $\begingroup$ Please note that it is a PROBABILITY wave that pertains to the quantum mechanical description of particles and systems of particles. The wavefunction has the wave properties described and its square gives the probability of finding the particle at (x,y,z) or measure it at specific spin. It is NOT a matter/energy wave. $\endgroup$
    – anna v
    Commented Apr 21, 2014 at 17:26
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    $\begingroup$ Beware that the first link is from Milo Wolff, a crackpot and it's likely to be wrong. $\endgroup$
    – jinawee
    Commented Jul 8, 2016 at 8:09

At the level of particles, most people don't think of "spin" in terms of actual movement, but as something else entirely. Spin is often approached as a mathematical property only. This is because some things do not make a ton of sense if you think of electrons spinning about some axis.

If you interpret spin as an electron (or other particle) actually spinning about an axis, some things don't make a whole lot of sense, like spin 1/2 particles vs. particles with other half-spins. I suppose you could think of a particle with spin as a rotating wave, moving like circularly polarized light, but I'm not sure how helpful it is to think like that.

See the Meaning of Spin. It may clarify some things.


You can think of spin abstractly as in PipperChip's answer. A system's fundamental physics is not changed if we choose to rotate the co-ordinate system we use to describe Nature with. What this means is that, for an isolated system, you can describe its evolution by the minimisation of a Lagrangian which is unchanged if you transform co-ordinates by rotation. Noether's theorem then tells us there are three conserved quantities for the system: one for each of the "generators" of rotations - think of it as one for each vector component of the rotation axis's direction. These three components are what we call the vector components of angular momentum. So the spin of a lone electron described by the Pauli or Dirac equation in the electron's centre of mass frame or of a lone photon described by Maxwell's equations is simply the abstract quantity conserved by dint of "Nature not caring about how we rotate our description".

For a photon (whose physics is more wonted to me than that of an electron), you can visualise spin nicely through the Riemann-Silberstein decomposition of Maxwell's equations. The evolution of the quantum state of a lone photon is described precisely by Maxwell's equations: accordingly you can think of $\vec{E}$ and $\vec{B}$ field vectors spinning as they do for polarised light. One must be careful to understand, in the one photon case, that the $\vec{E}$ and $\vec{B}$ field vectors are not literal electric and magnetic fields, but components of the photon's quantum state. I say more about this in this answer here.


The only way I can physically imagine an EM rotating wave is like a passing photon whose path is bent by a black hole and brought into rotation.

If an EM light wave is rotating, then the planar wave front must be going faster at greater radii. That would required a modification to the postulate of constant speed of light. At lower radii it would be going slower than the speed of light and at greater radii it would be going faster than the speed of light. At a specific radius, the Rotating Wave would be going at the known speed of light which is related to its rest mass.

The electric field is perpendicular to the magnetic field and both are perpendicular to the wave's direction of motion. Thus, the wave fronts of a rotating wave must incline and effectively shorten the wavelength of the rotating wave (length contraction). It will take longer to rotate one cycle as it travels a longer path of rotation plus forward motion (time dilation).

Furthermore, interaction of rotating waves would dissipate energy, slow down time, and lead to curvature and expansion of space. I am currently working on reconciling RWA Rotating Wave Approximation with Rotating Wave of Electron.

For video: https://youtube.com/watch?v=cc0rY8kk831 or detailed math and graphics see my article http://jnsnet.info/journals/jns/Vol_5_No_1_June_2017/1.pdf

  • $\begingroup$ Hi Bill, welcome to physics.SE. It is important to note that, while it is perfectly OK to cite yourself, you have to be upfront about it. Ideally, you would use a phrase like "see my article [...]" rather than just point to it. Also, please note that the youtube link doesn't work for some reason. Thanks! $\endgroup$ Commented Dec 9, 2018 at 23:51

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