# What's the Lagrangian for the Pauli equation?

The Pauli Hamiltonian is $$H= \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi$$

For the 1-component Schrodinger equation in absence of applied field we have the Lagrangian $$\mathcal{L} = i\hbar\psi^*\dot \psi - \frac{\hbar^2}{2m}\Big( \boldsymbol{\nabla}\psi^*\cdot\boldsymbol{\nabla}\psi \Big)- \psi^*V\psi$$ What is the 2-component analog of this which corresponds to the Pauli Hamiltonian?

• This is a good question. First thing I would do is to write the whole Pauli Hamiltonian using Cartesian tensor components. Then try to fit the sigma matrices into the interaction term particle-em field written in noncovariant fashion (i.e. again with Cartesian tensor notation). – DanielC Aug 29 '17 at 13:18

For the 1-component (classical) Schroedinger field in absence of applied external EM fields, the Lagrangian density is actually $$\mathcal{L} = i\hbar\psi^*\dot \psi - \frac{1}{2m}\Big( \psi^*\boldsymbol{p}^2\psi \Big)- \psi^*V\psi ~~;$$ (I have integrated your expression, corrected, by parts and supplanted ${\boldsymbol p}=-i\hbar {\boldsymbol \nabla}$.)
The usual degenerate canonical momentum for ψ is but $\Pi_\psi=i\hbar \psi^*$, which eliminates the time derivative term in the Legendre transform to the hamiltonian, since $\psi^* \Pi_\psi-i\hbar \psi^* \partial_t\psi=0$.
The ensuing field Hamiltonian density is then also trivial, $${\cal H}= \psi^* \Big( \frac{1}{2m} \boldsymbol{p}^2 +V\Big) \psi,$$ netting you the standard Schroedinger hamiltonian.
To generalize to Pauli's, promote the wave-fields/functions to two-spinors, insert the requisite vector potentials (in the non-quantized classical potential part), and you have $$\mathcal{L} = i\hbar\psi^\dagger\cdot \dot \psi - \psi^\dagger \cdot \Big( \frac{1}{2m}(\boldsymbol{\sigma}\cdot(\mathbf{p} - q \mathbf{A}))^2 + q \phi \Big) \psi.$$ Recall the Pauli matrices are Hermitian. The action is the space & time integral of this.