Using Equation (11.21), c) Complete the square: c(¹) (∞) = (t) == - i 2 ()* T ia "_H(t)ewor' dt' = + iwot' == e-(WOT/2) ²₁ -Le-u² dy= T The probability of a transition, then, is S ia hT√√T 2 2 - ( ² - ₁ WOT) 07) ²³ - (207) ². e-(t'/7)² eiwot' dt'\. ia t√√² Let t→∞ and t'/T = u: =e-(wOT/2)² √T Pa-sb ≈ [c¹" (0)² = (#)²³ exp[-(407)²] ia ħ e-(WOT) ²/4 e .
Using Equation (11.21), c) Complete the square: c(¹) (∞) = (t) == - i 2 ()* T ia "_H(t)ewor' dt' = + iwot' == e-(WOT/2) ²₁ -Le-u² dy= T The probability of a transition, then, is S ia hT√√T 2 2 - ( ² - ₁ WOT) 07) ²³ - (207) ². e-(t'/7)² eiwot' dt'\. ia t√√² Let t→∞ and t'/T = u: =e-(wOT/2)² √T Pa-sb ≈ [c¹" (0)² = (#)²³ exp[-(407)²] ia ħ e-(WOT) ²/4 e .
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Could you please write the process of getting the expression in green from the expression in red?
![Problem 11.8
(a) Using Equation (11.21),
c(¹) (t)
Complete the square:
√¹)(x) = -
=
t
- 7²
ħ
2
·()*
-∞
Ha (t')eiwot' dt'
+ iwot'
ia
hT√√T
The probability of a transition, then, is
- (WOT/2)² T
(=
28
=
.WOT
2
du
ia
ħT √√T
ia
ħ√√π
∞
2
2
- (WOT) ². Let t→ ∞ and t'/T = u:
e
e-(t'/7)² eiwot' dt'\.
- (WOT/2) ²
√T
WOT
207)2].
2
Part ≈ [c{¹) (∞)² = (^)² exp[-(₁7)²2]
Pa+b ~|c(¹)
ia -(WOT) ²/4.
ħ](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F27881e66-694c-4814-b35a-fba75a4bbb38%2F8bbede9f-7ef1-476a-a7d4-25d3eee525d0%2F3t9bja_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 11.8
(a) Using Equation (11.21),
c(¹) (t)
Complete the square:
√¹)(x) = -
=
t
- 7²
ħ
2
·()*
-∞
Ha (t')eiwot' dt'
+ iwot'
ia
hT√√T
The probability of a transition, then, is
- (WOT/2)² T
(=
28
=
.WOT
2
du
ia
ħT √√T
ia
ħ√√π
∞
2
2
- (WOT) ². Let t→ ∞ and t'/T = u:
e
e-(t'/7)² eiwot' dt'\.
- (WOT/2) ²
√T
WOT
207)2].
2
Part ≈ [c{¹) (∞)² = (^)² exp[-(₁7)²2]
Pa+b ~|c(¹)
ia -(WOT) ²/4.
ħ
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