Nuclear Fusion Physics and Technology/Atomic Physics summary

Definition: Linear Harmonic Oscillator Hamiltonian

Linear Harmonic Oscillator Hamiltonian ${\displaystyle {\hat {H}}_{LHO}:\mathbb {V} \rightarrow \mathbb {V} }$ of a particle ${\displaystyle p\in {\mathfrak {ParticleSet}}}$is defined as
${\displaystyle {\hat {H}}(|n>)={\frac {{\hat {p}}^{2}(|n>)}{2m}}+{\frac {1}{2}}m\omega {\hat {x}}^{2}(|n>)}$
where operators ${\displaystyle {\hat {x}}(|n>),{\hat {p}}(|n>):\mathbb {V} \rightarrow \mathbb {V} }$

Definition: Anihilation Operator

Anihilation Operator ${\displaystyle {\hat {a}}:\mathbb {V} \rightarrow \mathbb {V} }$ is defined as
${\displaystyle {\hat {a}}(|n>)={\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}+i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}}$

Definition: Creation Operator

Creation Operator ${\displaystyle {\hat {a}}^{+}:\mathbb {V} \rightarrow \mathbb {V} }$ is defined as
${\displaystyle {\hat {a}}^{+}(|n>)={\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}-i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}}$

Theorem: a, a+ compound expressions

Lets assume Linear Harmonic Oscillator Hamiltonian. Then

${\displaystyle {\hat {a}}[{\hat {a}}^{+}(|n>)]={\frac {{\hat {H}}(|n>)}{\hbar \omega }}+{\frac {{\hat {1}}(|n>)}{2}},\qquad {\hat {a}}^{+}[{\hat {a}}(|n>)]={\frac {{\hat {H}}(|n>)}{\hbar \omega }}-{\frac {{\hat {1}}(|n>)}{2}},}$

Proof:
Directly form a, a+ definitions with respect to LHO Hamiltonian definition and [x,p] commutator

${\displaystyle a[a^{+}(|n>)]{\overset {def\,a,a^{+}}{=}}\left({\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}+i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}\right)\left({\sqrt {\frac {m\omega }{2\hbar }}}{\hat {x}}-i{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}\right)=}$
${\displaystyle ={\frac {m\omega }{2\hbar }}{\hat {x}}^{2}(|n>)-i^{2}{\frac {1}{2\hbar m\omega }}{\hat {p}}^{2}(|n>)-i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {x}}[{\hat {p}}(|n>)]+i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}{\hat {p}}[{\hat {x}}(|n>)]=}$
${\displaystyle =|{\hat {H}}={\frac {{\hat {p}}^{2}|n>}{m}}+{\frac {1}{2}}m\omega {\hat {x}}^{2}|n>|={\frac {\hat {H}}{2\hbar \omega }}-i{\sqrt {\frac {m\omega }{2\hbar }}}{\sqrt {\frac {1}{2\hbar m\omega }}}({\hat {x}}{\hat {p}}|n>-{\hat {p}}{\hat {x}}|n>)=}$
${\displaystyle ={\frac {\hat {H}}{2\hbar \omega }}-{\frac {i}{2\hbar }}[{\hat {x}},{\hat {p}}]|n>={\frac {\hat {H}}{2\hbar \omega }}-{\frac {i}{2\hbar }}(i\hbar {\hat {1}})|n>={\frac {{\hat {H}}|n>}{2\hbar \omega }}+{\frac {{\hat {1}}|n>}{2}}}$

Theorem: H,a,a+ commutators

Lets assume Linear Harmonic Oscillator Hamiltonian. Then

${\displaystyle [{\hat {a}},{\hat {a}}^{+}](|n>)={\hat {1}}(|n>),\qquad [{\hat {H}},{\hat {a}}^{+}](|n>)=\hbar \omega {\hat {a}}^{+}(|n>)\qquad [{\hat {H}},{\hat {a}}](|n>)=-\hbar \omega {\hat {a}}(|n>)}$

Proof:
The first expression may be evaluated from commutator definition directly from a, a+ compound expression theorem

${\displaystyle [{\hat {a}},{\hat {a}}^{+}]={\hat {a}}{\hat {a}}^{+}-{\hat {a}}^{+}{\hat {a}}{\overset {theorem}{=}}{\frac {{\hat {H}}|n>}{2\hbar \omega }}+{\frac {{\hat {1}}|n>}{2}}-{\frac {{\hat {H}}|n>}{2\hbar \omega }}-{\frac {{\hat {1}}|n>}{2}}={\hat {1}}|n>}$

The second and the third expressions needs ${\displaystyle {\hat {H}}={\hat {H}}({\hat {a}},{\hat {a}}^{+})}$ first, which can be obtained from a,a+ compound expression theorem again: