Chapter 3: Problem 36
Suppose \(\Psi(x, 0)=\frac{A}{x^{2}+a^{2}}, \quad(-\infty< x<\infty)\) for constants \(A\) and \(a\) (a) Determine \(A\), by normalizing \(\Psi(x, 0)\) (b) Find \(\langle x\rangle,\left\langle x^{2}\right\rangle\) and \(\sigma_{x}\) (at time \(t=0\) ). (c) Find the momentum space wave function \(\Phi(p, 0),\) and check that it is normalized. (d) \(\operatorname{Use} \Phi(p, 0)\) to calculate \(\langle p\rangle,\left\langle p^{2}\right\rangle,\) and \(\sigma_{p}\) (at time \(t=0\)). (e) Check the Heisenberg uncertainty principle for this state.
Short Answer
Step by step solution
Determine Normalization Constant A
Calculate Expectation Values ⟨x⟩ and ⟨x²⟩
Compute Standard Deviation σₓ
Find Momentum Space Wave Function Φ(p, 0)
Check Normalization of Φ(p, 0)
Calculate ⟨p⟩ and ⟨p²⟩
Compute Standard Deviation σₚ
Verify Heisenberg Uncertainty Principle
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wave Function Normalization
- The integral is \( \int_{-\infty}^{\infty} \left| \Psi(x, 0) \right|^2 \, dx = 1 \).
- Simplifying using calculus, we find: \( A^2 \cdot \frac{\pi}{2a^3} = 1 \), leading to \( A = \sqrt{\frac{2a^3}{\pi}} \).
Expectation Values
- We calculate \( \langle x \rangle = \int_{-\infty}^{\infty} x \left| \Psi(x, 0) \right|^2 \, dx \).
- Given the evenness of \( \Psi(x, 0) \), it's symmetric about zero, rendering \( \langle x \rangle = 0 \).
- \( \langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 \left| \Psi(x, 0) \right|^2 \, dx \), resulting in \( \langle x^2 \rangle = \frac{a^2}{2} \).
- This informs us about the spread of possible particle locations.
Momentum Space Wave Function
- Perform the transformation \( \Phi(p, 0) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar} \Psi(x, 0) \, dx \).
- For this exercise, it resolves to \( \Phi(p, 0) = \sqrt{\frac{\pi a^3}{\hbar}} \cdot e^{-a|p|/\hbar} \).
- Integrate \( \int_{-\infty}^{\infty} |\Phi(p, 0)|^2 \, dp = 1 \) to verify proper normalization.
Heisenberg Uncertainty Principle
- Standard deviations \( \sigma_x = \frac{a}{\sqrt{2}} \) and \( \sigma_p = \frac{\hbar}{\sqrt{2}a} \) were calculated.
- Multiplying these gives \( \sigma_x \sigma_p = \frac{\hbar}{2} \).