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To determine if the given functions \(\psi(x)\) are suitable wave functions, we need to check if they satisfy certain criteria, which include:

1. **Normalization:** A wave function \(\psi(x)\) must be normalizable, meaning the integral of the square of the absolute value of the function over the domain must converge to a finite value:
   \[
   \int |\psi(x)|^2 dx < \infty
   \]
   This ensures the probability interpretation holds.
   
2. **Continuity:** The wave function should be continuous and finite over the domain of interest.

Let's analyze each case.

### (a) \(\psi(x) = A e^{-\alpha x^2}, \alpha > 0\)
- **Domain**: \(-\infty < x < \infty\).
- **Behavior**: As \(x \to \pm \infty\), \(\psi(x) \to 0\) exponentially.
- **Normalization**: Since the function is Gaussian, it is normalizable over \(-\infty < x < \infty\).
- **Conclusion**: This is a suitable wave function.

### (b) \(\psi(x) = A e^{-\alpha x}, \alpha > 0\)
- **Domain**: \(0 < x < \infty\).
- **Behavior**: As \(x \to \infty\), \(\psi(x) \to 0\) exponentially. However, as \(x \to 0\), the function diverges.
- **Normalization**: This function diverges at \(x = 0\), so it cannot be normalized over \(0 < x < \infty\).
- **Conclusion**: This is **not** a suitable wave function.

### (c) \(\psi(x) = \frac{A}{x}\)
- **Domain**: \(0 < x < \infty\).
- **Behavior**: As \(x \to \infty\), \(\psi(x) \to 0\), but as \(x \to 0\), \(\psi(x) \to \infty\).
- **Normalization**: This function diverges at \(x = 0\), and hence cannot be normalized.
- **Conclusion**: This is **not** a suitable wave function.

### (d) \(\psi(x) = \frac{A}{\sqrt{|x|}}\)
- **Domain**: \(-\infty < x < \infty\).
- **Behavior**: As \(x \to \infty\), \(\psi(x) \to 0\), but as \(x \to 0\), \(\psi(x)\) diverges.
- **Normalization**: This function diverges at \(x = 0\), so it cannot be normalized over the whole real line.
- **Conclusion**: This is **not** a suitable wave function.

### (e) \(\psi(x) = \frac{1}{x^{1/4}} e^{-\sqrt{x}}\)
- **Domain**: \(0 < x < \infty\).
- **Behavior**: As \(x \to \infty\), \(\psi(x) \to 0\) exponentially. As \(x \to 0\), \(\psi(x)\) diverges due to the \(x^{-1/4}\) term.
- **Normalization**: Since it diverges as \(x \to 0\), it cannot be normalized over the domain \(0 < x < \infty\).
- **Conclusion**: This is **not** a suitable wave function.

### Summary:
- (a) is a suitable wave function.
- (b), (c), (d), and (e) are not suitable wave functions.

Would you like further details or have any questions about these results?

Here are five related questions to deepen understanding:
1. What is the significance of normalization in quantum mechanics?
2. How does the behavior of a wave function at infinity affect its normalizability?
3. Why is continuity an important criterion for wave functions?
4. Can a wave function with singularities at certain points be used in quantum mechanics?
5. What are the physical interpretations of different types of wave function behavior at the boundaries?

**Tip:** When analyzing wave functions, always check their behavior at both the limits and any points where they could potentially diverge (e.g., near zero).

Are the following functions suitable wave functions in the range −∞ < x < ∞ or 0 < x < ∞?

Explore the suitability of various wave functions like ψ(x) = Ae^{-αx^2} and ψ(x) = A/x in quantum mechanics. Learn about the importance of normalization and continuity in determining the validity of a wave function.

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