[Physics] Schrödinger-Pauli Equation Solutions

electromagnetismquantum mechanicsquantum-spinspinorswavefunction

The Schrödinger-Pauli equation is the non-relativistic limit of the Dirac equation, and therefore describes spin-1/2 particles in an external electromagnetic field. It is given by:

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

Are there any analytical solutions to this equation? I have searched online but have unfortunately been unable to find any.

Best Answer

I cannot be sure, but I suspect that you can get analytical solutions of the Pauli equation by taking a non-relativistic limit of analytical solutions of the Dirac equation. The latter can be found in many books, say Bagrov, Vladislav G. / Gitman, Dmitry, The Dirac Equation and its Solutions (http://www.degruyter.com/view/product/177851) (you can find a Google preview). One example of an analytical solution of the Pauli equation can be found in http://arxiv.org/abs/physics/9807019 .

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:

Joy Christian's "Exactly Soluble Sector of Quantum Gravity". Phys.Rev.D 56 (1997) 4844-4877, eprint arXiv:gr-qc/9701013

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!)

[Physics] How to derive the Pauli equation from the Hamiltonian of a charged particle in an electromagnetic field

The Schrödinger equation is always the time evolution equation in the Schrödinger picture and is given by $$ \mathrm{i}\hbar\partial_t\lvert\psi(t)\rangle = H\lvert \psi(t)\rangle.$$ This means that the Pauli equation is obviously not the evolution equation for the Hamiltonian $H$ you write down - because that Hamiltonian is the Hamiltonian of a spinless charged particle, while the Pauli equation describes a spin-1/2 charged particle. The correct Hamiltonian for such a particle is simply the l.h.s. of the Pauli equation, i.e. $$ H_P = \frac{1}{2m}(\vec\sigma\cdot(\vec p - q\vec A))^2 + q\phi.$$ The correct form of this Hamiltonian, like all Hamiltonians, is eventually guessed by matching it to experiment, in this case to the Stern-Gerlach experiment, as described on the Wikipedia page for the Pauli equation.