シュレーディンガー場
] {\displaystyle {\begin{aligned}L&=(\hbar c)^{2}(\partial _{\mu }{\hat {a}}\partial ^{\mu }{\hat {a}}^{\dagger }+\partial _{\mu }{\hat {b}}\partial ^{\mu }{\hat {b}}^{\dagger }+\ldots )-(mc^{2})^{2}({\hat {a}}{\hat {a}}^{\dagger }+{\hat {b}}{\hat {b}}^{\dagger }+\ldots )\\&={\frac {1}{2mc^{2}}}\left[(\hbar c)^{2}({\frac {-imc}{\hbar }}{\hat {A}}+\partial _{0}{\hat {A}})({\frac {imc}{\hbar }}{\hat {A}}^{\dagger }+\partial ^{0}{\hat {A}}^{\dagger })-(\hbar c)^{2}\partial _{x}{\hat {A}}\partial ^{x}{\hat {A}}^{\dagger }+(A\Rightarrow B)+\ldots -(mc^{2})^{2}({\hat {A}}{\hat {A}}^{\dagger }+{\hat {B}}{\hat {B}}^{\dagger }+\ldots )\right]\\&={\frac {\hbar ^{2}}{2m}}\left[{\frac {imc}{\hbar }}(\partial _{0}{\hat {A}}{\hat {A}}^{\dagger }-{\hat {A}}\partial ^{0}{\hat {A}}^{\dagger })+\partial _{\mu }{\hat {A}}\partial ^{\mu }{\hat {A}}^{\dagger }+(A\Rightarrow B)+\ldots \right]\end{aligned}}}
ここで、 e ± 2 i m c 2 t / ħ {\displaystyle e^{\pm 2imc^{2}t/\hbar }} 項は楕円で表され、非相対論的極限では消える。