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En notant :
- {\displaystyle \Psi (t,{\vec {r}})={\binom {\Psi _{+}}{\Psi _{-}}}} la fonction d'état de la particule, où {\displaystyle \Psi _{\pm }} est l'amplitude de probabilité d'observer le spin {\displaystyle \pm 1/2},
- {\displaystyle \ q} la charge de la particule, {\displaystyle \ m} sa masse,
- {\displaystyle \mathbb {A} =\left(U({\vec {r}},t),{\vec {A}}({\vec {r}},t)\right)} le quadripotentiel du champ électromagnétique ambiant, {\displaystyle {\vec {B}}=\nabla \times {\vec {A}}} le champ magnétique,
- {\displaystyle {\vec {\sigma }}=\left(\sigma _{1},\sigma _{2},\sigma _{3}\right)} le vecteur des matrices de Pauli.
L'équation de Pauli est :
{\displaystyle i\hbar {\partial \Psi ({\vec {r}},t) \over \partial t}=\left({1 \over 2m}\left({\vec {P}}+q{\vec {A}}({\vec {r}},t)\right)^{2}+qU({\vec {r}},t)-{q\hbar \over 2m}{\vec {\sigma }}.{\vec {B}}({\vec {r}},t)\right)\Psi ({\vec {r}},t)}
De l'expression précédente se déduit l'Hamiltonien de Pauli:
{\displaystyle H={1 \over 2m}\left({\vec {\sigma }}.[{\vec {P}}-q{\vec {A}}({\vec {r}},t)]\right)^{2}+qU({\vec {r}},t)}
Démonstration
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円&0\end{pmatrix}}\quad \sigma _{y}={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\quad \sigma _{z}={\begin{pmatrix}1&0\0円&-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)}