写出一个问题及其解决方案，展示量子谐振子的特定选择规则
计算振动跃迁的能量 v → v + 1 {\displaystyle v\rightarrow v+1} 和 v + 6 → v + 7 {\displaystyle v+6\rightarrow v+7} 。如果它们具有相同的能隙，请说明原因。
△ 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 )}}
从 v → v + 1 {\displaystyle v\rightarrow v+1} 到 v + 6 → v + 7 {\displaystyle v+6\rightarrow v+7} 的振动跃迁的能量相同，其能量差为 ( h v 0 ) {\displaystyle (hv_{0})} 。这是因为量子谐振子有特定的选择定则。该规则指出，分子在跃迁发生时只能向上或向下移动一个振动能级。如果分子偏离该规则 △ V ± 1 {\displaystyle \bigtriangleup V\pm 1} ，则它被认为是泛音，这些泛音不太可能发生。
