A wave function is A(e"* + e*) in the region -π
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![### Problem Statement: Wave Function Normalization and Probability Calculation
A wave function Ψ is given by \( \Psi = A(e^{ix} + e^{-ix}) \) in the region \(-\pi < x < \pi\) and zero elsewhere. Normalize the wave function and find the probability of the particle being
(a) between \( x = 0 \) and \( x = \frac{\pi}{8} \), and
(b) between \( x = 0 \) and \( x = \frac{\pi}{4} \).
### Solution Approach
#### Step 1: Normalization of the Wave Function
To normalize the wave function, we must impose the following condition:
\[
\int_{-\pi}^{\pi} |\Psi(x)|^2 \, dx = 1.
\]
Given the wave function \(\Psi = A(e^{ix} + e^{-ix})\), we first compute \(|\Psi|^2\):
\[
|\Psi|^2 = |A(e^{ix} + e^{-ix})|^2 = A^2 (e^{ix} + e^{-ix})(e^{-ix} + e^{ix}) = A^2 (2 + 2\cos(2x)) = 2A^2 (1 + \cos(2x)).
\]
The normalization condition then becomes:
\[
\int_{-\pi}^{\pi} 2A^2 (1 + \cos(2x)) \, dx = 1.
\]
Splitting up the integral:
\[
2A^2 \int_{-\pi}^{\pi} (1 + \cos(2x)) \, dx = 1.
\]
Note that \(\int_{-\pi}^{\pi} \cos(2x) \, dx = 0\). Thus, we have:
\[
2A^2 \int_{-\pi}^{\pi} 1 \, dx = 1 \implies 2A^2 [x]_{-\pi}^{\pi} = 1 \implies 2A^2 (2\pi) = 1 \implies A^2 = \frac{1}{4\pi} \implies A = \frac{1}{2\sqrt{\pi}}.
\]
#### Step 2: Probability Calculations
For \( x = 0 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F093e7116-3e01-49ad-9157-7a22a5caeb16%2F03cfffca-8aba-4e2d-93b5-044b8f1cfe39%2Friincdq_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement: Wave Function Normalization and Probability Calculation
A wave function Ψ is given by \( \Psi = A(e^{ix} + e^{-ix}) \) in the region \(-\pi < x < \pi\) and zero elsewhere. Normalize the wave function and find the probability of the particle being
(a) between \( x = 0 \) and \( x = \frac{\pi}{8} \), and
(b) between \( x = 0 \) and \( x = \frac{\pi}{4} \).
### Solution Approach
#### Step 1: Normalization of the Wave Function
To normalize the wave function, we must impose the following condition:
\[
\int_{-\pi}^{\pi} |\Psi(x)|^2 \, dx = 1.
\]
Given the wave function \(\Psi = A(e^{ix} + e^{-ix})\), we first compute \(|\Psi|^2\):
\[
|\Psi|^2 = |A(e^{ix} + e^{-ix})|^2 = A^2 (e^{ix} + e^{-ix})(e^{-ix} + e^{ix}) = A^2 (2 + 2\cos(2x)) = 2A^2 (1 + \cos(2x)).
\]
The normalization condition then becomes:
\[
\int_{-\pi}^{\pi} 2A^2 (1 + \cos(2x)) \, dx = 1.
\]
Splitting up the integral:
\[
2A^2 \int_{-\pi}^{\pi} (1 + \cos(2x)) \, dx = 1.
\]
Note that \(\int_{-\pi}^{\pi} \cos(2x) \, dx = 0\). Thus, we have:
\[
2A^2 \int_{-\pi}^{\pi} 1 \, dx = 1 \implies 2A^2 [x]_{-\pi}^{\pi} = 1 \implies 2A^2 (2\pi) = 1 \implies A^2 = \frac{1}{4\pi} \implies A = \frac{1}{2\sqrt{\pi}}.
\]
#### Step 2: Probability Calculations
For \( x = 0 \
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