Équation de Pauli

On rappelle que: ${\displaystyle {\vec {\sigma }}.{\vec {B}}={\vec {\sigma }}.({\vec {\nabla }}\times {\vec {A}})=\sigma _{x}[{\vec {\nabla }}\times {\vec {A}}]_{x}+\sigma _{y}[{\vec {\nabla }}\times {\vec {A}}]_{y}+\sigma _{z}[{\vec {\nabla }}\times {\vec {A}}]_{z}}$

Avec les matrices de Pauli: ${\displaystyle \sigma _{x}={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\quad \sigma _{y}={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\quad \sigma _{z}={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}}$

Le rotationnel a pour expression: ${\displaystyle {\vec {\nabla }}\times {\vec {A}}={\begin{pmatrix}\nabla _{y}A_{z}-\nabla _{z}A_{y}\\\nabla _{z}A_{x}-\nabla _{x}A_{z}\\\nabla _{x}A_{y}-\nabla _{y}A_{x}\end{pmatrix}}}$

${\displaystyle \Rightarrow {\vec {\sigma }}.{\vec {B}}={\begin{pmatrix}({\vec {\nabla }}\times {\vec {A}})_{z}&({\vec {\nabla }}\times {\vec {A}})_{x}-i({\vec {\nabla }}\times {\vec {A}})_{y}\\({\vec {\nabla }}\times {\vec {A}})_{x}+i({\vec {\nabla }}\times {\vec {A}})_{y}&-({\vec {\nabla }}\times {\vec {A}})_{z}\end{pmatrix}}}$

${\displaystyle ={\begin{pmatrix}\nabla _{x}A_{y}-\nabla _{y}A_{x}&\nabla _{y}A_{z}-\nabla _{z}A_{y}-i(\nabla _{z}A_{x}-\nabla _{x}A_{z})\\\nabla _{y}A_{z}-\nabla _{z}A_{y}+i(\nabla _{z}A_{x}-\nabla _{x}A_{z})&-\nabla _{x}A_{y}+\nabla _{y}A_{x}\end{pmatrix}}}$

En appliquant cet opérateur au spineur ${\displaystyle [\psi ]({\vec {r}})={\begin{pmatrix}\psi _{+}({\vec {r}})\\\psi _{-}({\vec {r}})\end{pmatrix}}}$ on obtient:

♦ Pour le terme ${\displaystyle [{\vec {\sigma }}.{\vec {B}}]_{11}:}$

${\displaystyle (\nabla _{x}A_{y}-\nabla _{y}A_{x})\psi _{+}=\nabla _{x}A_{y}\psi _{+}-\nabla _{y}A_{x}\psi _{+}}$

${\displaystyle =\nabla _{x}(A_{y}\psi _{+})-A_{y}\nabla _{x}\psi _{+}-\nabla _{y}(A_{x}\psi _{+})+A_{x}\nabla _{y}\psi _{+}}$

${\displaystyle =[\nabla _{x}A_{y}-\nabla _{y}A_{x}+A_{x}\nabla _{y}-A_{y}\nabla _{x}]\psi _{+}=[[{\vec {\nabla }}\times {\vec {A}}]_{z}+[{\vec {A}}\times {\vec {\nabla }}]_{z}]\psi _{+}}$

♦ Pour le terme ${\displaystyle [{\vec {\sigma }}.{\vec {B}}]_{12}:}$

${\displaystyle \nabla _{y}A_{z}\psi _{-}-\nabla _{z}A_{y}\psi _{-}-i(\nabla _{z}A_{x}\psi _{-}-\nabla _{x}A_{z}\psi _{-})=:\nabla _{y}(A_{z}\psi _{-})-A_{z}\nabla _{y}\psi _{-}-\nabla _{z}(A_{y}\psi _{-})}$

${\displaystyle +A_{y}\nabla _{z}\psi _{-}-i(\nabla _{z}(A_{x}\psi _{-})-A_{x}\nabla _{z}\psi _{-}\nabla _{x}(A_{z}\psi _{-})+A_{z}\nabla _{x}\psi _{-})}$

${\displaystyle =[({\vec {\nabla }}\times {\vec {A}})_{x}+({\vec {A}}\times {\vec {\nabla }})_{x}]\psi _{-}-i[({\vec {\nabla }}\times {\vec {A}})_{y}+({\vec {A}}\times {\vec {\nabla }})_{y}]\psi _{-}}$

Ce qui permet de conclure: ${\displaystyle {\vec {\sigma }}.{\vec {B}}={\vec {\sigma }}({\vec {\nabla }}\times {\vec {A}})+{\vec {\sigma }}({\vec {A}}\times {\vec {\nabla }})}$

On rappellera:${\displaystyle {\vec {P}}={\frac {\hbar }{i}}{\vec {\nabla }}\Rightarrow }$ ${\displaystyle {\vec {\sigma }}.{\vec {B}}={\frac {i}{q\hbar }}.{\vec {\sigma }}[({\vec {P}}\times q{\vec {A}})+(q{\vec {A}}\times {\vec {P}})]=-{\frac {i}{q\hbar }}.{\vec {\sigma }}[({\vec {P}}-q{\vec {A}})\times ({\vec {P}}-q{\vec {A}})]}$

On rappelle la relation, avec ${\displaystyle {\vec {M}},{\vec {N}}}$ deux opérateurs quelconques et I l'opérateur unitaire:

${\displaystyle ({\vec {\sigma }}.{\vec {M}})({\vec {\sigma }}.{\vec {N}})={\vec {M}}.{\vec {N}}I+i{\vec {\sigma }}({\vec {M}}\times {\vec {N}})}$

${\displaystyle \Rightarrow i{\vec {\sigma }}[({\vec {P}}-q{\vec {A}})\times ({\vec {P}}-q{\vec {A}})]=({\vec {\sigma }}.({\vec {P}}-q{\vec {A}}))^{2}-({\vec {P}}-q{\vec {A}})^{2}I}$

${\displaystyle \Rightarrow {\vec {\sigma }}{\vec {B}}=-{\frac {1}{q\hbar }}[{\vec {\sigma }}({\vec {P}}-q{\vec {A}})]^{2}+{\frac {1}{q\hbar }}[{\vec {P}}-q{\vec {A}}]^{2}I}$

Il vient alors:

${\displaystyle H={\frac {1}{2m}}[{\vec {P}}-q{\vec {A}}]^{2}+qU-{\frac {q\hbar }{2m}}{\vec {\sigma }}.{\vec {B}}}$

${\displaystyle ={\frac {1}{2m}}[{\vec {P}}-q{\vec {A}}]^{2}+qU+{\frac {1}{2m}}[{\vec {\sigma }}({\vec {P}}-q{\vec {A}})]^{2}-{\frac {1}{2m}}[{\vec {P}}-q{\vec {A}}]^{2}I}$

${\displaystyle ={\frac {1}{2m}}[{\vec {\sigma }}({\vec {P}}-q{\vec {A}}({\vec {r}},t))]^{2}+qU({\vec {r}},t)}$