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I recently came across a claim that the Schrödinger equation only describes spin-1/2 particles.

Is this true?

I realize that the question may be ill-posed as some would consider the general Schrödinger equation to be $$ i\hbar \frac{\partial}{\partial t} |\psi\rangle = H|\psi\rangle $$

which is tautological if an arbitrary Hamiltonian is allowed. I assume what was meant by "Schrödinger equation" is essentially a quantized version of the Newtonian relation $T = \frac{p^2}{2m}$.

I know the Dirac equation can be reduced to such a form if we take the nonrelativistic limit and assume there is no magnetic field, and besides the Schrödinger equation does predict the gross line spectrum of hydrogen. So the claim that the Schrödinger equation (approximately) describes spin-1/2 particles would seem to be justified.

Now, I often see it claimed that the Schrödinger equation is essentially the non-relativistic limit of the Klein--Gordon equation, which Schrödinger published instead of the Klein--Gordon equation because the latter predicted the hydrogen atom fine structure incorrectly. So is it not the case that the Schrödinger equation is also valid as a non-relativistic approximation for the behaviour of spin-0 particles?

For spin 1, obviously the Schrödinger equation cannot describe the photon, which is always relativistic. However it seems incorrect to me that the Schrödinger equation should not be able to describe, say, a deuteron.

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  • $\begingroup$ Came across where? $\endgroup$ – Qmechanic Aug 5 '14 at 22:24
  • $\begingroup$ @Qmechanic On Quora, but I don't want to link the post because it'll look like I'm calling out the author. $\endgroup$ – Brian Bi Aug 5 '14 at 22:28
  • $\begingroup$ Comment to the question (v1): In textbooks on non-relativistic QM, the hydrogen atom is usually first solved via the Schr. eq. with the assumption/making-a-face that the electron is spinless. $\endgroup$ – Qmechanic Aug 5 '14 at 22:37
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    $\begingroup$ Related question: Where is spin in the Schroedinger equation of an electron in the hydrogen atom? $\endgroup$ – BMS Aug 5 '14 at 23:06
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    $\begingroup$ This might be referring to an idea published by Hestnes. "Consistency with the Dirac theory is shown to imply that the Schroedinger equation describes not a spinless particle as universally assumed, but a particle in a spin eigenstate." $\endgroup$ – garyp May 27 '16 at 19:15
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As per Rob's suggestion, I decided to make this an answer.

(Addendum: I've been meditating on this very topic for some time, and have been directed to some interesting literature referenced on Streater's webpage. As per Rococo's comment, I've updated my answer, but kept the old version for posterity.)

The Answer

To talk about "spin-1/2 particles", we need the spin-statistics theorem.

The spin-statistics theorem doesn't hold for non-relativistic QM. Or more precisely, the "naive" spin-statistics theorem doesn't hold, and if we try to contrive one...it's nothing like what we'd expect.

In brief, the reason is: spin-statistics theorem depends critically on microlocality (i.e., the commutator of spacelike separated conjugate operators vanishes identically). This property holds relativistically, but not non-relativistically. (The other proofs similarly do not hold, since Lorentz invariance fails.)

But there is a trick around this. Just take a relativistic spin-1/2 particle, then take the nonrelativistic limit.

But does the Schrodinger Equation Hold?

Now, the question originally asked is: does the Schrodinger equation hold for spin-1/2 particles? To talk about spin-1/2 particles, we really are working with representations of the Lorentz group. A nonrelativistic spin-1/2 particle is obtained by taking the nonrelativistic limit (i.e., the $c\to\infty$ limit) of the Dirac equation, which is the Pauli equation:

$$\left[ \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi \right] |\psi\rangle = i \hbar \frac{\partial}{\partial t} |\psi\rangle$$

where $\boldsymbol{\sigma}$ are the Pauli matrices, $\mathbf{A}$ an external vector potential, $\phi$ an external electric potential, and $q$ the electric charge of the particle. But observe when we "turn off" electromagnetism (setting $\mathbf{A}=\phi=q=0$) we recover

$$\left[ \frac{1}{2m}(\boldsymbol{\sigma}\cdot\mathbf{p})^2 \right] |\psi\rangle = i \hbar \frac{\partial}{\partial t} |\psi\rangle$$

Uh, I leave it as an exercise for the reader to show the left hand side of this equation is $(1/2m) \mathbf{p}^{2}\otimes\boldsymbol{1}_{2}$ where $\boldsymbol{1}_{2}$ is the 2-by-2 identity matrix.

References

For more thorough reviews on this matter, I can heartily refer the reader to:

  • A.S. Wightman, "The spin-statistics connection: Some pedagogical remarks in response to Neuenschwander's question" Eprint, 7 pages
  • R. E. Allen, A. R. Mondragon, "No spin-statistics connection in nonrelativistic quantum mechanics". Eprint arXiv:quant-ph/0304088, 2 pages

The (Old) Answer

The "vanilla" Schrodinger's equation (from non-relativistic QM) does not describe a spin-1/2 particle.

The plain, old Schrodinger's equation describes a non-relativistic spin-0 field.

Case Studies

If we pretend the wave function is a classical field (which happens all the time during the "second quantization" procedure), then it turns out to describe a spin-0 field. See Brian Hatfield's Quantum Field Theory of Point Particles and Strings, specifically chapter 2 --- on "Second Quantization".

But wait, there's more! If we consider other non-relativistic fields and attempt quantizing, e.g. the Newton Cartan theory of gravity, we also get spin-0 boson! For this result (specific to quantizing Newtonian gravity), see:

The Reason

Well, this should not surprise us, since the Schrodinger equation is the nonrelativistic limit to the Klein-Gordon equation. And we should recall the Klein-Gordon equation describes spin-0 bosons!

The nonrelativistic limit $c\to\infty$ should not affect the spin of the particles involved. (That's why the Pauli model is the nonrelativistic limit of the Dirac equation!)

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  • $\begingroup$ This is interesting, but it also seems at best incomplete, because people use slight generalizations of the Schroedinger or Schroedinger-Pauli equation all the time to describe atoms and composite particles with all sorts of spins. $\endgroup$ – Rococo May 27 '16 at 18:56
  • $\begingroup$ @Rococo I am starting to have second thoughts about what I've written since reading "No spin-statistics connection in nonrelativistic quantum mechanics" arXiv:quant-ph/0304088, being alerted to it by Streater's Lost Causes in Physics. $\endgroup$ – Alex Nelson May 28 '16 at 17:04
  • $\begingroup$ that aligns more with my (limited) understanding of the topic. I would suggest that you add that comment to your main answer body, since comments don't always persist. $\endgroup$ – Rococo May 28 '16 at 18:21
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    $\begingroup$ Also FYI your link did not work for me, but this one did: web.archive.org/web/20160101001254/http://www.mth.kcl.ac.uk/… $\endgroup$ – Rococo May 28 '16 at 18:21
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    $\begingroup$ @Rococo thanks for digging up that link...I swear, I will learn how to use the internet one day! ;) $\endgroup$ – Alex Nelson May 29 '16 at 14:37
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Schrödinger simply does not account for spin. For spin 1/2, you need the Pauli equation or Dirac. A spinless particle, meaning spin zero such as the Higgs boson or pions (ignoring their internal quark/gluon structure) is described by the Klein-Gordon wave equation.

I've read a claim that, to the extent that it does describe a particle with spin, Schrödinger's is for a particle stuck in an $S_z$ eigenstate. I'm not sure if this holds up to closer analysis, and don't recall where I read that, so I leave this to others to critique.

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    $\begingroup$ You certainly can use the nonrelativistic Schrödinger equation to describe the evolution of a spin-half particle; all you need to do is to specify the Hamiltonian. What's special about the Dirac equation is that the two spin states, and the antiparticle's two spin states, arise on their own. $\endgroup$ – rob Aug 6 '14 at 3:19
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    $\begingroup$ For those interested in more on this, geocalc.clas.asu.edu/pdf/Consistency.pdf argues that "Schrodinger theory is identical to the Pauli theory when the electron is in an eigenstate of the spin" $\endgroup$ – Luke Burns Nov 10 '17 at 15:52
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It is possibly a reference to the Pauli equation or Schrödinger–Pauli equation: the formulation of the Schrödinger equation for spin-½ particles, which takes into account the interaction of the particle's spin with an external electromagnetic field. It is the non-relativistic limit of the Dirac equation and can be used where particles are moving at speeds much less than the speed of light, so that relativistic effects can be neglected. (wiki , Pauli equation)

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It's true. The Schrodinger equation is a differential equation whose solutions are wavefunctions of spin-1/2 particles - and spin-1/2 particles only. The question is quite deep. The Schrodinger equation for a zero-spin particle (which models energy itself), or a spin-one particle (i.e. a photon) is similar but does not have the 1/2 factor in the 'diffusion constant'. That $1/2$ factor is the only difference. The elementary wavefunction for spin-zero and spin-one particles - when using natural units - involves mass, energy and momentum concepts that ensure E = p = m. For spin-1/2 particles, we have E = p = 2m, because we now have both linear momentum as well as angular momentum (and both use the mass factor, with the equipartition theorem explaining why energy is divided equally over both). For further detail, see this. It's somewhat speculative - but true.

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