Chapter 9: Q4P (page 345)

Suppose you don’t assume $H_{a a} = H_{b b} = 0$
(a) Find $c_{a} (t)$ and $c_{b} (t)$ in first-order perturbation theory, for the case
.show that , to first order in .
(b) There is a nicer way to handle this problem. Let
.
Show that
where
So the equations for are identical in structure to Equation 11.17 (with an extra
(c) Use the method in part (b) to obtain in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.

Short Answer

Expert verified

${|c_{a}|}^{2} = [1 - \frac{i}{h} \int_{0}^{t} H_{a a} (t') d t'] [1 - \frac{i}{h} \int_{0}^{t} H_{a a} (t') d t'] = 1 {[1 - \frac{i}{h} \int_{0}^{t} H_{a a} (t') d t']}^{2} = 1 {|c_{a}|}^{2} = [1 - \frac{i}{h} \int_{0}^{t} H_{a a} (t') d t'] [1 - \frac{i}{h} \int_{0}^{t} H_{a a} (t') d t'] = 0 S o {|c_{a}|}^{2} + {|c_{b}|}^{2} = 1 (t o f i r s t o r d e r)$

(b) role="math" localid="1658488366910" $d_{a} = - \frac{i}{h} e^{\frac{i}{n} \int_{0}^{t} H (t) d t} c_{b} H_{a b} e^{i ω_{0} t} d_{b} = e^{\frac{i}{n} \int_{0}^{t} H (t) d t} (\frac{i}{h} H_{b b}) c_{b} + e^{\frac{i}{h} \int_{0}^{t} H (t) d t} c_{b}$

(c) $d_{b} = e^{\frac{i}{n} \int_{0}^{t} H (t) d t} (\frac{i}{h} H_{b b}) c_{b} + e^{\frac{i}{h} \int_{0}^{t} H (t) d t} c_{b}$

Step by step solution

(a) Finding ca(t) and cb(t)

$c_{a} = - \frac{i}{h} [c_{a} H_{a a} + c_{b} H_{a b} e^{i ω_{0} t}] \Rightarrow c_{b} = - \frac{i}{h} [c_{a} H_{a a} + c_{b} H_{a b} e^{i ω_{0} t}]\}$

$c_{b} = - \frac{i}{h} [c_{a} H_{a a} + c_{b} H_{a b} e^{- i {(E_{b} - E_{b})}^{t / h}}]$

Initial conditions: $c_{a} (t) = 1, c_{b} (0) = 0$ ,

Zero order:.

First order:

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Short Answer

Step by step solution

(a) Finding ca(t) and cb(t)

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