Nuclear Fusion Physics and Technology/Atomic Physics summary

Definition: Linear Harmonic Oscillator Hamiltonian

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

Definition: Anihilation Operator

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

Definition: Creation Operator

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

Theorem: a, a+ compound expressions

Lets assume Linear Harmonic Oscillator Hamiltonian. Then

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

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

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

Theorem: H,a,a+ commutators

Lets assume Linear Harmonic Oscillator Hamiltonian. Then

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

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

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

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

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