Question

For the examples inProblem 11.24(c) and (d), calculate ${\mathit{c}}_{\mathbf{m}}\left(t\right)$to first order. Check the normalization condition:

$\mathbf{\sum }_{\mathbf{m}}{\left|c_m\left(t\right)\right|}^{\mathbf{2}}\mathbf{=}\mathbf{1}$,

and comment on any discrepancy. Suppose you wanted to calculate the probability of remaining in the original state ${\mathbf{\psi }}_{\mathbf{N}}$ ; would you do better to use ${\left|c_N\left(\mathit{t}\right)\right|}^{\mathbf{2}}\mathbf{,}\mathbf{or}\mathbf{}\mathbf{1}\mathbf{-}\mathbf{\sum }_{\mathbf{m}\mathbf{\ne }\mathbf{N}}{\left|c_m\left(\mathit{t}\right)\right|}^{\mathbf{2}}$ ?

Short answer

The probability of remaining in the original state is

$c_m\left(t\right)=-\frac{V_{mN}}{2}\left[\frac{e^{i\left(E_m-E_N+ħ\omega \right)}-1}{\left(E_m-E_N+ħ\omega \right)}+\frac{e^{i\left(E_m-E_N-ħ\omega \right)t/ħ}-1}{\left(E_m-E_N-ħ\omega \right)}\right]\left(m\ne N\right)$

Step-by-step solution

Calculating  cm(t)

For example (c):

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

And,

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

So,

$\sum _m{\left|c_m\right|}^2=1+\frac{t^2}{{ħ}^2}{\left|H_{NN}^{\text{'}}\right|}^2+4\sum _{m\ne N}\frac{\left|H_{mN}^{\text{'}}\right|}{{\left(E_m-E_N\right)}^2}{\sin }^2\left(\frac{E_m-E_N}{2ħ}t\right)$

This is plainly greater than 1! But remember: The c’s 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.

Step 2: Calculating the first order

For example (d):

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

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

${\left|c_m\right|}^2=\frac{{\left|V_{mN}\right|}^2}{{\left(E_m-E_N\underline{+}ħ\varpi \right)}^2}{\sin }^2\left(\frac{E_m-E_N\underline{+}ħ\omega }{2ħ}t\right)\left(m\ne N\right)$

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

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

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’d also need $b_2$. It is precisely this term that would cancel the “extra” (second-order) terms in the calculations of $\sum _{}{\left|c_m\right|}^{2}above..$ .