Show that the downward-sloping turning point as ψ ( x ) = { D ′ p ( x ) exp [ − 1 ℏ ∫ x x 1 | p ( x ′ ) | d x ′ ] , x < x 1 2 D ′ p ( x ) sin [ 1 ℏ ∫ x 1 x | p ( x ′ ) | d x ′ + π 4 ] , x > x 1 } .

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Answer 1

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Chapter 9.3, Problem 9.10P

To determine

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

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II. PROBLEM 2. LIÉNARD-WIECHERT POTENTIALS. Problem 10.15A particle of charge q moves in a circle of radius a at constant angular velocity w. (Assume that the circle lies in the xy plane, centered at the origin, and at time t = 0 the charge is at (a, 0), on the positive x axis.) Find the Liénard-Wiechert potentials for points on the z axis. It would be useful to parameterize the circle as r(t) = a (cos (wt) + sin (wt)ÿ), (1) 2 and then directly substitute the values into the Eqs. (10.46) and (10.47), i. e. V (r, t) = A(r,t) =V(r,t), (2) 4TE0 2c - 4 v where |r – w (t,)| 2 =r - w (t,), t, =t - =t - (3) and the velocity of the particle is also taken at the retarded time.

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This is the solution to problem 7.36 . Could you explain the yellow marked place ? How we got <0|y|1> and <x^2>0 ?

Answer 2

Question

Chapter 9.3, Problem 9.10P

To determine

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

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Answer 3

Question

Chapter 9.3, Problem 9.10P

To determine

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

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Answer 4

Question

Chapter 9.3, Problem 9.10P

To determine

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

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Answer 5

Chapter 9.3, Problem 9.10P

To determine

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

Answer 6

Chapter 9.3, Problem 9.10P

To determine

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

Answer 7

Chapter 9.3, Problem 9.10P

To determine

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 9.1 - Prob. 9.1P Ch. 9.1 - Prob. 9.2P Ch. 9.2 - Prob. 9.3P Ch. 9.2 - Prob. 9.4P Ch. 9.2 - Prob. 9.5P Ch. 9.3 - Prob. 9.6P Ch. 9.3 - Prob. 9.7P Ch. 9.3 - Prob. 9.8P Ch. 9.3 - Prob. 9.9P Ch. 9.3 - Prob. 9.10P

Show that the downward-sloping turning point as ψ ( x ) = { D ′ p ( x ) exp [ − 1 ℏ ∫ x x 1 | p ( x ′ ) | d x ′ ] , x < x 1 2 D ′ p ( x ) sin [ 1 ℏ ∫ x 1 x | p ( x ′ ) | d x ′ + π 4 ] , x > x 1 } .

Solution Summary: The author analyzes the downward-sloping turning point as psi '(x)=leftcDprime

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

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Chapter 9 Solutions

Show that the downward-sloping turning point as ψ ( x ) = { D ′ p ( x ) exp [ − 1 ℏ ∫ x x 1 | p ( x ′ ) | d x ′ ] , x &lt; x 1 2 D ′ p ( x ) sin [ 1 ℏ ∫ x 1 x | p ( x ′ ) | d x ′ + π 4 ] , x &gt; x 1 } .

Solution Summary: The author analyzes the downward-sloping turning point as psi '(x)=leftcDprime

Show that the downward-sloping turning point as ψ(x)={D′p(x)exp[−1ℏ∫xx1|p(x′)|dx′], x<x12D′p(x)sin[1ℏ∫x1x|p(x′)|dx′+π4], x>x1}.

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Chapter 9 Solutions

Show that the downward-sloping turning point as ψ ( x ) = { D ′ p ( x ) exp [ − 1 ℏ ∫ x x 1 | p ( x ′ ) | d x ′ ] , x < x 1 2 D ′ p ( x ) sin [ 1 ℏ ∫ x 1 x | p ( x ′ ) | d x ′ + π 4 ] , x > x 1 } .