VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end:2 This problem utilizes the results from (x), (x), (p) , (x?) , (p?) for the Particle on a Line, which you calculated in Homework 2 Problem 3. a) Calculate oy and 0}. b) Calculate their product 0,0. Write your final expression in the form 0x0, = 1 - f(n). (In other words, make sure you final answer has a factor of #/2 in front. This makes grading easier.) c) Show that, for n = 1, your result from ?? is in accord with the Heisenburg uncertainty principle. d) For n > 1, will your result from (b) still be in accord with the Heisenburg uncertainty principle? Explain why or why not.
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:
2 This problem utilizes the results from (x), (x), (p) , (x?) , (p?) for the Particle on a Line, which
you calculated in Homework 2 Problem 3.
a) Calculate oy and 0}.
b) Calculate their product 0,0. Write your final expression in the form 0x0, = 1 - f(n). (In
other words, make sure you final answer has a factor of #/2 in front. This makes grading
easier.)
c) Show that, for n = 1, your result from ?? is in accord with the Heisenburg uncertainty
principle.
d) For n > 1, will your result from (b) still be in accord with the Heisenburg uncertainty
principle? Explain why or why not.
This is AI generated Response.
Asked by: Rahul
Created at: 2026-02-06 04:28:26
Asked by: Rahul
Created at: 2026-02-06 04:28:26
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Let's solve the problem step by step as you requested:
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## **Step 1: Calculate \(\sigma_x\) and \(\sigma_p\)**
Recall the definitions of standard deviations:
- \(\sigma_x = \sqrt{ \langle x^2 \rangle - \langle x \rangle^2 }\)
- \(\sigma_p = \sqrt{ \langle p^2 \rangle - \langle p \rangle^2 }\)
For a particle in a box (or on a line) in the \(n\)-th stationary state:
- \(\langle x \rangle = \text{(center of box)}\) (usually or \(L/2\), but variance does not depend on center)
- \(\langle p \rangle = \) (since the wavefunctions are sines or cosines)
Let’s use the standard results for a box of length \(L\) (assuming box from \(x=\) to \(x=L\)):
- \(\langle x \rangle = \frac{L}{2}\)
- \(\langle x^2 \rangle = \frac{L^2}{3} - \frac{L^2}{2 n^2 \pi^2}\)
- \(\langle p \rangle = \)
- \(\langle p^2 \rangle = \left( \frac{n \pi \hbar}{L} \right)^2\)
So,
\[
\sigma_x = \sqrt{ \langle x^2 \rangle - \langle x \rangle^2 }
= \sqrt{ \left( \frac{L^2}{3} - \frac{L^2}{2 n^2 \pi^2} \right) - \left(\frac{L}{2}\right)^2 }
= L \sqrt{ \frac{1}{3} - \frac{1}{2 n^2 \pi^2} - \frac{1}{4} }
= L \sqrt{ \frac{1}{12} - \frac{1}{2 n^2 \pi^2} }
\]
\[
\sigma_p = \sqrt{ \langle p^2 \rangle - \langle p \rangle^2 }
= \frac{n \pi \hbar}{L}
\]
**Explanation:**
We used the standard quantum results for a particle in a box and directly applied the variance formulas for position and momentum.
---
## **Step 2: Calculate \(\sigma_x \sigma_p\) and Write in Required Form**
\[
\sigma_x \sigma_p = \left(L \sqrt{ \frac{1}{12} - \frac{1}{2 n^2 \pi^2} }\right) \cdot \left( \frac{n \pi \hbar}{L} \right)
= n \pi \hbar \sqrt{ \frac{1}{12} - \frac{1}{2 n^2 \pi^2} }
\]
Let's rewrite this in the form \(\frac{\hbar}{2} f(n)\):
\[
\sigma_x \sigma_p = \frac{\hbar}{2} \cdot 2 n \pi \sqrt{ \frac{1}{12} - \frac{1}{2 n^2 \pi^2} }
\]
So,
\[
f(n) = 2 n \pi \sqrt{ \frac{1}{12} - \frac{1}{2 n^2 \pi^2} }
\]
**Explanation:**
We multiplied the expressions from Step 1 and factored out \(\hbar/2\) as requested.
---
## **Final Answer**
\[
\sigma_x \sigma_p = \frac{\hbar}{2} \cdot 2 n \pi \sqrt{ \frac{1}{12} - \frac{1}{2 n^2 \pi^2} }
\]
where \( f(n) = 2 n \pi \sqrt{ \frac{1}{12} - \frac{1}{2 n^2 \pi^2} } \).
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Let me know if you'd like the answers for parts (c) and (d) as well!
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