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{Particle Set}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:

      Last modified on 24 February 2012, at 23:20