-x² wave function y(x) = € 3², (−∞0 ≤ x ≤ +∞). If the wave function is not normalized, please determine the normalization constant. Standard Integrals: A. π¹/4 B. 1/¹/4 C. 1/(2¹/4) D. 1/(4 ¹/4) +∞ ·∞0 -ax² e 2 dx = 2πT α +∞ [te-o -∞ e-ax² dx = T α

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**Transcription for Educational Website:**

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**Problem Statement:**

Consider the wave function \( \Psi(x) = e^{-\frac{x^2}{32}}, \, (-\infty \leq x \leq +\infty) \). If the wave function is not normalized, please determine the normalization constant.

**Standard Integrals:**

\[
\int_{-\infty}^{+\infty} e^{-\frac{ax^2}{2}} \, dx = \sqrt{\frac{2\pi}{a}}
\]

\[
\int_{-\infty}^{+\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}
\]

**Options:**

A. \( \pi^{1/4} \)

B. \( 1/\pi^{1/4} \)

C. \( 1/(2 \pi^{1/4}) \)

D. \( 1/(4 \pi^{1/4}) \)

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**Explanation of Diagrams (if applicable):**

There are no graphs or diagrams present in the image. The provided content consists of mathematical formulas and options related to the normalization of a given wave function in quantum mechanics. The integrals and options are standard forms and potential results for the normalization constant calculation.
Transcribed Image Text:**Transcription for Educational Website:** --- **Problem Statement:** Consider the wave function \( \Psi(x) = e^{-\frac{x^2}{32}}, \, (-\infty \leq x \leq +\infty) \). If the wave function is not normalized, please determine the normalization constant. **Standard Integrals:** \[ \int_{-\infty}^{+\infty} e^{-\frac{ax^2}{2}} \, dx = \sqrt{\frac{2\pi}{a}} \] \[ \int_{-\infty}^{+\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}} \] **Options:** A. \( \pi^{1/4} \) B. \( 1/\pi^{1/4} \) C. \( 1/(2 \pi^{1/4}) \) D. \( 1/(4 \pi^{1/4}) \) --- **Explanation of Diagrams (if applicable):** There are no graphs or diagrams present in the image. The provided content consists of mathematical formulas and options related to the normalization of a given wave function in quantum mechanics. The integrals and options are standard forms and potential results for the normalization constant calculation.
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