寫出一個問題及其解決方案，展示量子諧振子的特定選擇規則
計算振動躍遷的能量 v → v + 1 {\displaystyle v\rightarrow v+1} 和 v + 6 → v + 7 {\displaystyle v+6\rightarrow v+7} 。如果它們具有相同的能隙，請說明原因。
△ E v → v + 1 {\displaystyle \bigtriangleup Ev\rightarrow v+1} = h v 0 ( ( v + 1 ) + 1 2 ) − h v 0 ( v + 1 2 ) {\displaystyle =hv_{0}{\Bigl (}(v+1)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Bigl (}v+{\tfrac {1}{2}}{\Bigr )}}
= h v 0 ( ( v + 1 ) + 1 2 − ( v + 1 2 ) ) {\displaystyle =hv_{0}{\biggl (}(v+1)+{\tfrac {1}{2}}-{\Bigl (}v+{\tfrac {1}{2}}{\Bigr )}{\biggr )}}
= h v 0 ( v + 1 + 1 2 − v − 1 2 ) {\displaystyle =hv_{0}{\biggl (}v+1+{\tfrac {1}{2}}-v-{\tfrac {1}{2}}{\biggr )}}
= h v 0 {\displaystyle =hv_{0}}
△ E v + 1 → v + 2 {\displaystyle \bigtriangleup Ev+1\rightarrow v+2} = h v 0 ( ( v + 7 ) + 1 2 ) − h v 0 ( ( v + 6 ) + 1 2 ) {\displaystyle =hv_{0}{\Bigl (}(v+7)+{\tfrac {1}{2}}{\Bigr )}-hv_{0}{\Bigl (}(v+6)+{\tfrac {1}{2}}{\Bigr )}}
= h v 0 ( ( v + 7 ) + 1 2 − ( ( v + 6 ) + 1 2 ) ) {\displaystyle =hv_{0}{\biggl (}(v+7)+{\tfrac {1}{2}}-{\Bigl (}(v+6)+{\tfrac {1}{2}}{\Bigr )}{\biggr )}}
= h v 0 ( v + 7 + 1 2 − v − 6 − 1 2 ) {\displaystyle =hv_{0}{\biggl (}v+7+{\tfrac {1}{2}}-v-6-{\tfrac {1}{2}}{\biggr )}}
從 v → v + 1 {\displaystyle v\rightarrow v+1} 到 v + 6 → v + 7 {\displaystyle v+6\rightarrow v+7} 的振動躍遷的能量相同，其能量差為 ( h v 0 ) {\displaystyle (hv_{0})} 。這是因為量子諧振子有特定的選擇定則。該規則指出，分子在躍遷發生時只能向上或向下移動一個振動能級。如果分子偏離該規則 △ V ± 1 {\displaystyle \bigtriangleup V\pm 1} ，則它被認為是泛音，這些泛音不太可能發生。
