91影视

For example (c):

$$\begin{gathered} c_N\left(t\right)=1-\frac{i}{魔}{H}_{NN}^{\text{'}}t; \\ {c}_m\left(t\right)=-2i\frac{H_{mN}^{\text{'}}}{(E_m-E_N}{e}^{i(E_m-E_N)t/2魔}\sin \left(\frac{E_m-E_N}{2魔}t\right)\left(m鈮�N\right) \end{gathered}$$

And,

$$\begin{gathered} {\left|c_N\right|}^2=1+\frac{t^2}{{魔}^2}{\left|H_{NN}^{\text{'}}\right|}^2 \\ {\left|c_m\right|}^2=4\frac{{\left|H_{mN}^{\text{'}}\right|}^2}{{\left(E_m-E_N\right)}^2}{\sin }^2\left(\frac{E_m-E_N}{2魔}t\right) \end{gathered}$$

So,

$\underset{m}{鈭�}{\left|c_m\right|}^2=1+\frac{t^2}{{魔}^2}{\left|H_{NN}^{\text{'}}\right|}^2+4\underset{m鈮�N}{鈭�}\frac{\left|H_{mN}^{\text{'}}\right|}{{\left(E_m-E_N\right)}^2}{\sin }^2\left(\frac{E_m-E_N}{2魔}t\right)$

This is plainly greater than 1! But remember: The c鈥檚 are accurate only to first order in $H^{\text{'}}$; to this order the ${\left|H^{\text{'}}\right|}^2$ 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):

$$\begin{gathered} c_N=1-\frac{i}{魔}{V}_{NN}{鈭�}_0^t\cos \left(蝇t^{\text{'}}\right)d{t}^{\text{'}}=1-\frac{i}{魔}{V}_{NN}{\overline{)\frac{\sin \left(蝇t^{\text{'}}\right)}{蝇}}}_0^t;c_N\left(t\right)=1-\frac{i}{魔蝇}{V}_{NN}\sin \left(蝇t\right) \\ {c}_m\left(t\right)=-\frac{V_{mN}}{2}\left[\frac{e^{i(E_m-E_N+魔蝇)t/魔}-1}{\left(E_m-E_N+魔蝇\right)}+\frac{e^{i\left(E_m-E_N-魔蝇\right)t/魔}}{\left(E_m-E_N-魔蝇\right)}\right]\left(m鈮�N\right) \end{gathered}$$

So, ${\left|c_N\right|}^2=1+\frac{{\left|V_{NN}\right|}^2}{{\left(魔蝇\right)}^2}{\sin }^2\left(蝇t\right);$ and in the rotating wave approximation

${\left|c_m\right|}^2=\frac{{\left|V_{mN}\right|}^2}{{\left(E_m-E_N\underline{+}魔蠔\right)}^2}{\sin }^2\left(\frac{E_m-E_N\underline{+}魔蝇}{2魔}t\right)\left(m鈮�N\right)$

Again, ostensibly$\underset{}{鈭�{\left|c_m\right|}^2>1}$, but the \extra鈥� terms are of second order in, and hence do not belong (to first order). You would do better to use$1-\underset{m鈮�N}{鈭�}{\left|c_m\right|}^2$Schematically:

$$\begin{gathered} c_m=a_{1}H+a_2{H}^2+路路路,{\left|c_m\right|}^2=a_1^2{H}^2+2a_1{a}_2{H}^3+路路路路,whereas{c}_N=1+b_{1}H+b_2{H}^2+路路路, \\ {\left|c_N\right|}^2=1+2b_{1}H+\left(2b_2+b_1^2\right)H^2+路路路 \end{gathered}$$

Thus knowing $c_m$ to first order (i.e., knowing ) gets you ${\left|c_m\right|}^2$ to second order, but knowing $c_N$ to first order (i.e.$b_1$ ) does not get you ${\left|c_N\right|}^2$ o second order (you鈥檇 also need $b_2$. It is precisely this term that would cancel the 鈥渆xtra鈥� (second-order) terms in the calculations of $\underset{}{鈭�}{\left|c_m\right|}^{2}above..$ .