Operador (física)

Os operadores usados na mecânica quântica são coletados na tabela abaixo (veja por exemplo,[2][3]). Os vetores em negrito com circunflexos não são vetores unitários, são operadores de 3 vetores; todos os três componentes espaciais tomados em conjunto.

Operador	Componente cartesiano	Definição geral	unidade SI	Dimensão
Posição	${\displaystyle {\begin{aligned}{\hat {x}}&=x,&{\hat {y}}&=y,&{\hat {z}}&=z\end{aligned}}}$	${\displaystyle \mathbf {\hat {r}} =\mathbf {r} \,\!}$	m	[L]
Momento	Geral ${\displaystyle {\begin{aligned}{\hat {p}}_{x}&=-i\hbar {\frac {\partial }{\partial x}},&{\hat {p}}_{y}&=-i\hbar {\frac {\partial }{\partial y}},&{\hat {p}}_{z}&=-i\hbar {\frac {\partial }{\partial z}}\end{aligned}}}$	Geral ${\displaystyle \mathbf {\hat {p}} =-i\hbar \nabla \,\!}$	J s m−1 = N s	[M] [L] [T]−1
	Campo eletromagnetico ${\displaystyle {\begin{aligned}{\hat {p}}_{x}=-i\hbar {\frac {\partial }{\partial x}}-qA_{x}\\{\hat {p}}_{y}=-i\hbar {\frac {\partial }{\partial y}}-qA_{y}\\{\hat {p}}_{z}=-i\hbar {\frac {\partial }{\partial z}}-qA_{z}\end{aligned}}}$	Campo eletromagnetico (usa momento cinético; A, potencial vetorial) ${\displaystyle {\begin{aligned}\mathbf {\hat {p}} &=\mathbf {\hat {P}} -q\mathbf {A} \\&=-i\hbar \nabla -q\mathbf {A} \\\end{aligned}}\,\!}$	J s m−1 = N s	[M] [L] [T]−1
Energia cinética	Translação ${\displaystyle {\begin{aligned}{\hat {T}}_{x}&=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\\[2pt]{\hat {T}}_{y}&=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial y^{2}}}\\[2pt]{\hat {T}}_{z}&=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial z^{2}}}\\\end{aligned}}}$	${\displaystyle {\begin{aligned}{\hat {T}}&={\frac {1}{2m}}\mathbf {\hat {p}} \cdot \mathbf {\hat {p}} \\&={\frac {1}{2m}}(-i\hbar \nabla )\cdot (-i\hbar \nabla )\\&={\frac {-\hbar ^{2}}{2m}}\nabla ^{2}\end{aligned}}\,\!}$	J	[M] [L]2 [T]−2
	Campo eletromagnetico ${\displaystyle {\begin{aligned}{\hat {T}}_{x}&={\frac {1}{2m}}\left(-i\hbar {\frac {\partial }{\partial x}}-qA_{x}\right)^{2}\\{\hat {T}}_{y}&={\frac {1}{2m}}\left(-i\hbar {\frac {\partial }{\partial y}}-qA_{y}\right)^{2}\\{\hat {T}}_{z}&={\frac {1}{2m}}\left(-i\hbar {\frac {\partial }{\partial z}}-qA_{z}\right)^{2}\end{aligned}}\,\!}$	Campo eletromagnetico (A, potencial vetorial) ${\displaystyle {\begin{aligned}{\hat {T}}&={\frac {1}{2m}}\mathbf {\hat {p}} \cdot \mathbf {\hat {p}} \\&={\frac {1}{2m}}(-i\hbar \nabla -q\mathbf {A} )\cdot (-i\hbar \nabla -q\mathbf {A} )\\&={\frac {1}{2m}}(-i\hbar \nabla -q\mathbf {A} )^{2}\end{aligned}}\,\!}$	J	[M] [L]2 [T]−2
	Rotação (I, momento de inércia) ${\displaystyle {\begin{aligned}{\hat {T}}_{xx}&={\frac {{\hat {J}}_{x}^{2}}{2I_{xx}}}\\{\hat {T}}_{yy}&={\frac {{\hat {J}}_{y}^{2}}{2I_{yy}}}\\{\hat {T}}_{zz}&={\frac {{\hat {J}}_{z}^{2}}{2I_{zz}}}\\\end{aligned}}\,\!}$	Rotação[4] ${\displaystyle {\hat {T}}={\frac {\mathbf {\hat {J}} \cdot \mathbf {\hat {J}} }{2I}}\,\!}$	J	[M] [L]2 [T]−2
Energia potencial	não aplicável	${\displaystyle {\hat {V}}=V\left(\mathbf {r} ,t\right)=V\,\!}$	J	[M] [L]2 [T]−2
Energia total	não aplicável	Potencial dependente do tempo: ${\displaystyle {\hat {E}}=i\hbar {\frac {\partial }{\partial t}}\,\!}$ Independente do tempo: ${\displaystyle {\hat {E}}=E\,\!}$	J	[M] [L]2 [T]−2
Hamiltoniano		${\displaystyle {\begin{aligned}{\hat {H}}&={\hat {T}}+{\hat {V}}\\&={\frac {1}{2m}}\mathbf {\hat {p}} \cdot \mathbf {\hat {p}} +V\\&={\frac {1}{2m}}{\hat {p}}^{2}+V\\\end{aligned}}\,\!}$	J	[M] [L]2 [T]−2
Operador de momento angular	${\displaystyle {\begin{aligned}{\hat {L}}_{x}&=-i\hbar \left(y{\partial \over \partial z}-z{\partial \over \partial y}\right)\\{\hat {L}}_{y}&=-i\hbar \left(z{\partial \over \partial x}-x{\partial \over \partial z}\right)\\{\hat {L}}_{z}&=-i\hbar \left(x{\partial \over \partial y}-y{\partial \over \partial x}\right)\end{aligned}}}$	${\displaystyle \mathbf {\hat {L}} =\mathbf {r} \times -i\hbar \nabla }$	J s = N s m	[M] [L]2 [T]−1
Momento angular de Spin	${\displaystyle {\begin{aligned}{\hat {S}}_{x}&={\hbar \over 2}\sigma _{x}&{\hat {S}}_{y}&={\hbar \over 2}\sigma _{y}&{\hat {S}}_{z}&={\hbar \over 2}\sigma _{z}\end{aligned}}}$ where ${\displaystyle {\begin{aligned}\sigma _{x}&={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\\\sigma _{y}&={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\\\sigma _{z}&={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}\end{aligned}}}$ são as matrizes de Pauli para partículas spin-½	${\displaystyle \mathbf {\hat {S}} ={\hbar \over 2}{\boldsymbol {\sigma }}\,\!}$ onde σ é o vetor cujas componentes são as matrizes de Pauli.	J s = N s m	[M] [L]2 [T]−1
Momento angular total	${\displaystyle {\begin{aligned}{\hat {J}}_{x}&={\hat {L}}_{x}+{\hat {S}}_{x}\\{\hat {J}}_{y}&={\hat {L}}_{y}+{\hat {S}}_{y}\\{\hat {J}}_{z}&={\hat {L}}_{z}+{\hat {S}}_{z}\end{aligned}}}$	${\displaystyle {\begin{aligned}\mathbf {\hat {J}} &=\mathbf {\hat {L}} +\mathbf {\hat {S}} \\&=-i\hbar \mathbf {r} \times \nabla +{\frac {\hbar }{2}}{\boldsymbol {\sigma }}\end{aligned}}}$	J s = N s m	[M] [L]2 [T]−1
Momento dipolar de transição (elétrico)	${\displaystyle {\begin{aligned}{\hat {d}}_{x}&=q{\hat {x}},&{\hat {d}}_{y}&=q{\hat {y}},&{\hat {d}}_{z}&=q{\hat {z}}\end{aligned}}}$	${\displaystyle \mathbf {\hat {d}} =q\mathbf {\hat {r}} }$	C m	[I] [T] [L]