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The Pauli equation is the generalization of the Schrodinger equation to charged particles in an electromagnetic field. It is written in a very compact way with heavy use of operators, so it can be helpful to "unpack" it. Namely, given access to $\psi_+$ and $\psi_-$, the electric potential $\phi$, and the magnetic potential $A$, what equation would be used to find $\dot\psi_+$ and $\dot\psi_-$? Said equation should only include "elementary" operators (gradient, laplacian, curl, and divergence, etc) as well as linear algebra (matrix multiplication, dot-products, etc). No "composite" operators are allowed!

Breaking down the "standard" equation's left hand side into three terms yields:

$q\phi\psi$: This is just the potential energy V of the Schrodinger equation, both $\psi_+$ and $\psi_-$ are treated independently.

$(1/2m)q \hbar \sigma \cdot B\psi$: This appears to be the quantum version of a spinning magnet in a magnetic field. If the magnetic field is inhomogeneous it will separate electrons based on their spin. If the magnetic field is aligned with your spin basis the separation will be based on spin. If the magnetic field is in a different direction (homogenous or not) as your basis it will act to "swap" $\psi_+$ and $\psi_-$ with each-other.

Unpacking it we have a 2 by 2 matrix-vector multiplication: $(q \hbar/2m) \sigma \cdot B\psi = (q \hbar/2m)(\sigma_xB_x+\sigma_y B_y+\sigma_zB_z)\psi$

$(p-qA)^2 \phi, p=-i\hbar\nabla$. This is the hardest to unpack and appears to represent charged particles orbiting in a magnetic field. My recipe is to "take the gradient of $f$ (and multiply by $-i\hbar$), then subtract $qA$ times $f$". $\psi_-$ and $\psi_+$ are independent. Inputing $\psi_+$ or $\psi_-$ as our $f$ and applying this recipe twice gives us a 3x3 matrix. A contraction gives us scalars:

$(1/2m)(p-qA)^2 \phi = (-\hbar^2/2m)\nabla^2 \psi+(iq\hbar/2m)A \cdot \nabla \psi+(iq\hbar/2m)\psi \nabla \cdot A+(q^2/2m)(A \cdot A)\psi$

The first term is the "inertia" term of the standard Schrodinger equation and involves the laplacian. The second and third term appears to be analogous to the magnetic field curving the trajectory (and it may include the Aharanov-Bohm effect as well).

The last term seems wrong. Firstly, it involves squaring the charge which is not involved in any classical or quantum effect, which sounds like a gauge invariance of some sort. Suppose $A_x = x$ and $A_y = A_z = 0$. There is no magnetic field. This choice of A turns the last term into a potential well in the $x$ direction, and I don't see how this would be canceled out by any other term given it has the same sign for electrons and positrons.

Is this unpacking and analysis correct?

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  • $\begingroup$ The term you are having trouble just comes from the classical Hamiltonian for a charged particle in a magnetic field, with the $\vec{p}=-i\hbar\vec{\nabla}$ replacement to quantize it. $\endgroup$ – Buzz Nov 6 '19 at 2:01
  • $\begingroup$ @Buzz: good point as the classical Hamiltonians are often reused by the quantum equations of motion. $\endgroup$ – Kevin Kostlan Nov 6 '19 at 5:10
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    $\begingroup$ Yup, that $A^2$ term really is supposed to be there; you would get the wrong answer if it weren't there. It combines with the terms linear in $A$ to give the correct magnetic interaction. $\endgroup$ – knzhou Nov 6 '19 at 18:25
  • $\begingroup$ @knzhou: There seems to be a sign problem. The terms linear in A are opposite sign for positrons vs electrons, but the A^2 term isn't. How can it combine correctly for both particles? $\endgroup$ – Kevin Kostlan Nov 6 '19 at 21:03
  • $\begingroup$ In your example with no magnetic field, the two particles will indeed have different solutions for $\psi$, but this is fine because they are gauge equivalent — the gauge transformation of a wavefunction depends on the charge of the particle. $\endgroup$ – knzhou Nov 6 '19 at 21:19

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