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Science Physics Modern Physics for Scientists and Engineers

(a) Whether the given wave function ψ = A [ e i x + e − i x ] in the region − π < x < π can be normalized. To Find: The probability of the particle being between x = 0 and x = π / 8 . The given wave function is normalized. 0.1188 Given: Wave function: ψ = A [ e i x + e − i x ] The region interval x = 0 and x = π / 8 . Formula Used: Basic trigonometry equation cos x = e i x + e − i x 2 2 cos x = e i x + e − i x ..... (1) To normalize a wave function, the condition is ∫ ψ * ψ d x = 1 ..... (2) The probability of a particle is given by P = ∫ ψ * ψ d x ..... (3) Calculation: The wave function is given by ψ = A [ e i x + e − i x ] ..... (4) Applying the trigonometric relation given in equation (1), the wave function in equation (3) can be written as ψ = 2 A cos x ..... (5) Substitute expression (5) and the given interval − π < x < π in expression (2). ∫ − π π ( 2 A * cos x ) ( 2 A cos x ) d x = 1 ⇒ 4 | A | 2 ∫ − π π cos 2 x d x = 1 ⇒ 4 | A | 2 ∫ 0 π 2 cos 2 x d x = 1 ⇒ 4 | A | 2 ∫ 0 π ( 1 + cos 2 x ) d x = 1 ⇒ 4 | A | 2 [ x + sin 2 x 2 ] 0 π = 1 ⇒ 4 | A | 2 × π = 1 ⇒ | A | = 1 2 π Substitute 1 2 π for A in expression (4). ψ = 1 2 π [ e i x + e − i x ] Substitute the given interval x = 0 and x = π / 8 and 2 A cos x for ψ in expression (3). P = ∫ 0 π 8 ( 2 A * cos x ) ( 2 A cos x ) d x = 4 A 2 ∫ 0 π 8 cos 2 x d x = 4 A 2 ∫ 0 π 8 ( 1 + cos 2 x ) 2 d x = 2 A 2 ∫ 0 π 8 ( 1 + cos 2 x ) d x = 2 × ( 1 2 π ) 2 ∫ 0 π 8 ( 1 + cos 2 x ) d x = 1 2 π [ x + sin 2 x 2 ] 0 π 8 Solve the above relation. P = 1 2 π [ x + sin 2 x 2 ] 0 π 8 = 1 2 π [ π 8 + 1 2 sin π 4 ] = 1 2 π [ π 8 + 1 2 × 1 2 ] = 0.1188 Conclusion: Thus, the given wave function ψ = A [ e i x + e − i x ] is normalized in the region − π < x < π . The probability for finding the particle in the interval x = 0 and x = π / 8 is 0.1188 . (b) The probability of the particle being between x = 0 and x = π / 4 . 0.2046 Given: The region interval x = 0 and x = π / 4 . Formula Used: The probability of a particle is given by, P = ∫ ψ * ψ d x ..... (6) Calculation: Substitute the given interval x = 0 and x = π / 4 and 2 A cos x for ψ in expression (6). P = ∫ ( 2 A * cos x ) ( 2 A cos x ) d x = ∫ 0 π 4 ( 2 A * cos x ) ( 2 A cos x ) d x = 4 A 2 ∫ 0 π 4 cos 2 x d x = 4 A 2 ∫ 0 π 4 ( 1 + cos 2 x ) 2 d x = 2 A 2 ∫ 0 π 4 ( 1 + cos 2 x ) d x = 2 × ( 1 2 π ) 2 ∫ 0 π 4 ( 1 + cos 2 x ) d x = 1 2 π [ x + sin 2 x 2 ] 0 π 4 Solve the above relation. P = 1 2 π [ x + sin 2 x 2 ] 0 π 4 = 1 2 π [ π 4 + 1 2 sin π 2 ] = 1 2 π [ π 4 + 1 2 × 1 ] = 0.2046 Conclusion: The probability for finding the particle in the interval x = 0 and x = π / 4 is 0.2046 .

(a) Whether the given wave function ψ = A [ e i x + e − i x ] in the region − π < x < π can be normalized. To Find: The probability of the particle being between x = 0 and x = π / 8 . The given wave function is normalized. 0.1188 Given: Wave function: ψ = A [ e i x + e − i x ] The region interval x = 0 and x = π / 8 . Formula Used: Basic trigonometry equation cos x = e i x + e − i x 2 2 cos x = e i x + e − i x ..... (1) To normalize a wave function, the condition is ∫ ψ * ψ d x = 1 ..... (2) The probability of a particle is given by P = ∫ ψ * ψ d x ..... (3) Calculation: The wave function is given by ψ = A [ e i x + e − i x ] ..... (4) Applying the trigonometric relation given in equation (1), the wave function in equation (3) can be written as ψ = 2 A cos x ..... (5) Substitute expression (5) and the given interval − π < x < π in expression (2). ∫ − π π ( 2 A * cos x ) ( 2 A cos x ) d x = 1 ⇒ 4 | A | 2 ∫ − π π cos 2 x d x = 1 ⇒ 4 | A | 2 ∫ 0 π 2 cos 2 x d x = 1 ⇒ 4 | A | 2 ∫ 0 π ( 1 + cos 2 x ) d x = 1 ⇒ 4 | A | 2 [ x + sin 2 x 2 ] 0 π = 1 ⇒ 4 | A | 2 × π = 1 ⇒ | A | = 1 2 π Substitute 1 2 π for A in expression (4). ψ = 1 2 π [ e i x + e − i x ] Substitute the given interval x = 0 and x = π / 8 and 2 A cos x for ψ in expression (3). P = ∫ 0 π 8 ( 2 A * cos x ) ( 2 A cos x ) d x = 4 A 2 ∫ 0 π 8 cos 2 x d x = 4 A 2 ∫ 0 π 8 ( 1 + cos 2 x ) 2 d x = 2 A 2 ∫ 0 π 8 ( 1 + cos 2 x ) d x = 2 × ( 1 2 π ) 2 ∫ 0 π 8 ( 1 + cos 2 x ) d x = 1 2 π [ x + sin 2 x 2 ] 0 π 8 Solve the above relation. P = 1 2 π [ x + sin 2 x 2 ] 0 π 8 = 1 2 π [ π 8 + 1 2 sin π 4 ] = 1 2 π [ π 8 + 1 2 × 1 2 ] = 0.1188 Conclusion: Thus, the given wave function ψ = A [ e i x + e − i x ] is normalized in the region − π < x < π . The probability for finding the particle in the interval x = 0 and x = π / 8 is 0.1188 . (b) The probability of the particle being between x = 0 and x = π / 4 . 0.2046 Given: The region interval x = 0 and x = π / 4 . Formula Used: The probability of a particle is given by, P = ∫ ψ * ψ d x ..... (6) Calculation: Substitute the given interval x = 0 and x = π / 4 and 2 A cos x for ψ in expression (6). P = ∫ ( 2 A * cos x ) ( 2 A cos x ) d x = ∫ 0 π 4 ( 2 A * cos x ) ( 2 A cos x ) d x = 4 A 2 ∫ 0 π 4 cos 2 x d x = 4 A 2 ∫ 0 π 4 ( 1 + cos 2 x ) 2 d x = 2 A 2 ∫ 0 π 4 ( 1 + cos 2 x ) d x = 2 × ( 1 2 π ) 2 ∫ 0 π 4 ( 1 + cos 2 x ) d x = 1 2 π [ x + sin 2 x 2 ] 0 π 4 Solve the above relation. P = 1 2 π [ x + sin 2 x 2 ] 0 π 4 = 1 2 π [ π 4 + 1 2 sin π 2 ] = 1 2 π [ π 4 + 1 2 × 1 ] = 0.2046 Conclusion: The probability for finding the particle in the interval x = 0 and x = π / 4 is 0.2046 .

Solution Summary: The author explains how the given wave function can be normalized. The probability of a particle is given by psi =2Amathrm

Modern Physics for Scientists and Engineers

Modern Physics for Scientists and Engineers

4th Edition

ISBN: 9781133103721

Author: Stephen T. Thornton, Andrew Rex

Publisher: Cengage Learning

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Modern Physics for Scientists and Engineers

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