Q6P Imagine two noninteracting parti... [FREE SOLUTION]

91影视

Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 5: Q6P (page 210)

Imagine two noninteracting particles, each of mass m, in the infinite square well. If one is in the state $蠄_{n}$ (Equation 2.28 ), and the other in state $蠄_{1} (l 鈮� n)$ , calculate localid="1658214464999" $<{(x_{1} - x_{2})}^{2}>$ , assuming (a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions.

Short Answer

Expert verified

a) The value of $< {(x_{1} - x_{2})}^{2} >$ assuming that they are distinguishable particles is $a^{2} [\frac{1}{6} - \frac{1}{2 蟺^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})]$ .

b) The value of $< {(x_{1} - x_{2})}^{2} >$ assuming that they are identical bosons is $a^{2} [\frac{1}{6} - \frac{1}{2 蟺^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})] - \frac{128 a^{2} m^{2} n^{2}}{蟺^{4} {(m^{2} - n^{2})}^{4}}$

c) The value of $< {(x_{1} - x_{2})}^{2} >$ assuming that they are identical fermions is $a^{2} [\frac{1}{6} - \frac{1}{2 蟺^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})] - \frac{128 a^{2} m^{2} n^{2}}{蟺^{4} {(m^{2} - n^{2})}^{4}}$ .

Step by step solution

Definition of identical bosons and identical fermions

According to Carroll, particles exist in two types: those that makeup matter, known as 'fermions,' and those that convey forces, known as 'bosons.

Bosons can be piled on top of one other, whereas fermions take up space.

(a) Determination of <(x1-x2)2> assuming that they are distinguishable particles

For distinguishable particles, use the following formula,

$<{(x_{1} - x_{2})}^{2}> = {<x^{2}>}_{a} + {<x^{2}>}_{b} - 2 {(x)}_{a} (x \tilde{n_{b}} = a^{2} [\frac{1}{3} - \frac{1}{2 {(蟿蟿苍)}^{2}}] + a^{2} [\frac{1}{3} - \frac{1}{2 {(蟿蟿苍)}^{2}}] - 2 脳 \frac{a}{2} 脳 \frac{a}{2} = a^{2} [\frac{2}{3} - \frac{1}{2} - \frac{1}{2 {蟿蟿}^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})] Evaluate the value of <{(x_{1} - x_{2})}^{2}> . <{(x_{1} - x_{2})}^{2}> = a^{2} [\frac{1}{6} - \frac{1}{2 {蟿蟿}^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})] Thus, the value of <{(x_{1} - x_{2})}^{2}> assuming they are distinguishable particles is a^{2} [\frac{1}{6} - \frac{1}{2 {蟿蟿}^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})] .$

(b) Determination of<x1-x22>assuming that they are the identical Bosons

For the same Bosons, the equation is as follows,

${<{(x_{1} - x_{2})}^{2}>}_{B} = {<x^{2}>}_{m} + {<x^{2}>}_{n} - 2 \overset{,}{a} x {\tilde{n}}_{n} \overset{,}{a} x {\tilde{n}}_{m} - 2 {|\overset{,}{a} x {\tilde{n}}_{n m}|}^{2} = {<{(x_{1} - x_{2})}^{2}>}_{d} - 2 {|\overset{,}{a} x {\tilde{n}}_{n m}|}^{2}$

But it is known that $\overset{,}{a} x {\tilde{n}}_{n m} = \frac{- 8 a m n}{蟿蟿^{2} {(m^{2} - n^{2})}^{2}}$ . So, that expression is as follows,

${|\overset{,}{a} x {\tilde{n}}_{n m}|}^{2} = {|\overset{,}{a} x {\tilde{n}}_{n m}|}^{2} = \overset{,}{a} x {\tilde{n}}_{n m} \overset{,}{a} x {\tilde{n}}_{n m} {<{(x_{1} - x_{2})}^{2}>}_{B} = a^{2} [\frac{1}{6} - \frac{1}{2 蟿蟿^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})] - \frac{128 a^{2} m^{2} n^{2}}{蟿蟿^{4} (m^{2} - n^{2})} Hence, the value of {<{(x_{1} - x_{2})}^{2}>}_{d} considering they are identical bosons is . a^{2} [\frac{1}{6} - \frac{1}{2 {蟿蟿}^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}})] - \frac{128 a^{2} m^{2} n^{2}}{{蟿蟿}^{4} {(m^{2} - n^{2})}^{4}}$

(c) Determination of<x1-x22>assuming that they are the identical fermions

$For Fermions that are identical, the equation is as follows, {(x_{1} - x_{2})}_{f}^{2} = x_{a}^{2} + x_{b}^{2} - 2 \overset{,}{ax} {\tilde{n}}_{a} \overset{,}{ax} {\tilde{n}}_{b} + 2 {|\overset{,}{ax} {\tilde{n}}_{ab}|}^{2} = {(x_{1} - x_{2})}_{d}^{2} + 2 {|\overset{,}{ax} {\tilde{n}}_{ab}|}^{2} = a^{2} \frac{1}{6} - \frac{1}{2 {蟿蟿}^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}}) + \frac{128 a^{2} m^{2} n^{2}}{{蟿蟿}^{4} {(m^{2} - n^{2})}^{4}} Thus, the value of {<{(x_{1} - x_{2})}^{2}>}_{d} when they are identical fermions is . = a^{2} \frac{1}{6} - \frac{1}{2 {蟿蟿}^{2}} (\frac{1}{n^{2}} + \frac{1}{m^{2}}) + \frac{128 a^{2} m^{2} n^{2}}{{蟿蟿}^{4} {(m^{2} - n^{2})}^{4}}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Definition of identical bosons and identical fermions

(a) Determination of <(x1-x2)2> assuming that they are distinguishable particles

(b) Determination of<x1-x22>assuming that they are the identical Bosons

(c) Determination of<x1-x22>assuming that they are the identical fermions

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Further Mechanics and Thermal Physics

Famous Physicists

Magnetism

Work Energy and Power

Torque and Rotational Motion

Study anywhere. Anytime. Across all devices.