For example (c):
c_{N}(t)=1-\frac{i}{\hbar} H_{N N}^{\prime} t ; \quad c_{m}(t)=-2 i \frac{H_{m N}^{\prime}}{\left(E_{m}-E_{N}\right)} e^{i\left(E_{m}-E_{N}\right) t / 2 \hbar} \sin \left(\frac{E_{m}-E_{N}}{2 \hbar} t\right)(m \neq N) .
\left|c_{N}\right|^{2}=1+\frac{1}{\hbar^{2}}\left|H_{N N}^{\prime}\right|^{2} t^{2}, \quad\left|c_{m}\right|^{2}=4 \frac{\left|H_{m N}^{\prime}\right|^{2}}{\left(E_{m}-E_{N}\right)^{2}} \sin ^{2}\left(\frac{E_{m}-E_{N}}{2 \hbar} t\right) , so
\sum_{m}\left|c_{m}\right|^{2}=1+\frac{t^{2}}{\hbar^{2}}\left|H_{N N}^{\prime}\right|^{2}+4 \sum_{m \neq N} \frac{\left|H_{m N}^{\prime}\right|^{2}}{\left(E_{m}-E_{N}\right)^{2}} \sin ^{2}\left(\frac{E_{m}-E_{N}}{2 \hbar} t\right) .
This is plainly greater than 1! But remember: The c’s are accurate only to first order in H^{\prime} ; to this order the \left|H^{\prime}\right|^{2} terms do not belong. Only if terms of first order appeared in the sum would there be a genuine problem with normalization.
For example (d):
c_{N}=1-\frac{i}{\hbar} V_{N N} \int_{0}^{t} \cos \left(\omega t^{\prime}\right) d t^{\prime}=1-\left.\frac{i}{\hbar} V_{N N} \frac{\sin \left(\omega t^{\prime}\right)}{\omega}\right|_{0} ^{t} \Longrightarrow c_{N}(t)=1-\frac{i}{\hbar \omega} V_{N N} \sin (\omega t) .
c_{m}(t)=-\frac{V_{m N}}{2}\left[\frac{e^{i\left(E_{m}-E_{N}+\hbar \omega\right) t / \hbar}-1}{\left(E_{m}-E_{N}+\hbar \omega\right)}+\frac{e^{i\left(E_{m}-E_{N}-\hbar \omega\right) t / \hbar}-1}{\left(E_{m}-E_{N}-\hbar \omega\right)}\right](m \neq N) . So
\left|c_{N}\right|^{2}=1+\frac{\left|V_{N N}\right|^{2}}{(\hbar \omega)^{2}} \sin ^{2}(\omega t) ;
and in the rotating wave approximation
\left|c_{m}\right|^{2}=\frac{\left|V_{m N}\right|^{2}}{\left(E_{m}-E_{N} \pm \hbar \omega\right)^{2}} \sin ^{2}\left(\frac{E_{m}-E_{N} \pm \hbar \omega}{2 \hbar} t\right) \quad(m \neq N) .
Again, ostensibly \sum\left|c_{m}\right|^{2}>1 , but the \extra” terms are of second order in H^{\prime} , and hence do not belong (to first order).
You would do
better to use
1-\sum_{m \neq N}\left|c_{m}\right|^{2}
. Schematically:
c_{m}=a_{1} H+a_{2} H^{2}+\cdots, \quad \text { so } \quad\left|c_{m}\right|^{2}=a_{1}^{2} H^{2}+2 a_{1} a_{2} H^{3}+\cdots , whereas c_{N}=1+b_{1} H+b_{2} H^{2}+\cdots, \quad \text { so } \quad\left|c_{N}\right|^{2}=1+2 b_{1} H+\left(2 b_{2}+b_{1}^{2}\right) H^{2}+\cdots
. Thus knowing
c_{m}
to first order (i.e., knowing a_{1}
) gets you
\left|c_{m}\right|^{2} to second order, but knowing c_{N} to first order
(i.e.. b_{1} ) does not get vou
\left|c_{N}\right|^{2} o second order (vou’d also need b_{2} . It is preciselv this b_{2} term that would } [/latex] cancel the “extra” (second-order) terms in the calculations of \sum\left|c_{m}\right|^{2} \text { above } .