Imagine two noninteracting particles, each of mass m, in the infinite square well. If one is in the stateψn(Equation 2.28 ), and the other in state ψ1(ln), calculate localid="1658214464999" (x1-x2)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<(x1-x2)2>assuming that they are distinguishable particles is a216-12π21n2+1m2.

b) The value of<(x1-x2)2>assuming that they are identical bosons is a216-12π21n2+1m2-128a2m2n2π4m2-n24

c) The value of<(x1-x2)2>assuming that they are identical fermions is a216-12π21n2+1m2-128a2m2n2π4m2-n24.

Step by step solution

01

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.

02

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

For distinguishable particles, use the following formula,

x1-x22=x2a+x2b-2xa(xnb~=a213-12ττn2+a213-12ττn2-2×a2×a2=a223-12-12ττ21n2+1m2Evaluatethevalueofx1-x22.x1-x22=a216-12ττ21n2+1m2Thus,thevalueofx1-x22assumingtheyaredistinguishableparticlesisa216-12ττ21n2+1m2.

03

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

For the same Bosons, the equation is as follows,

x1-x22B=x2m+x2n-2a,xn~na,xn~m-2a,xn~nm2=x1-x22d-2a,xn~nm2

But it is known that a,xn~nm=-8amnττ2m2-n22. So, that expression is as follows,

a,xn~nm2=a,xn~nm2=a,xn~nma,xn~nmx1-x22B=a216-12ττ21n2+1m2-128a2m2n2ττ4m2-n2Hence,thevalueofx1-x22dconsideringtheyareidenticalbosonsis.a216-12ττ21n2+1m2-128a2m2n2ττ4m2-n24

04

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

ForFermionsthatareidentical,theequationisasfollows,x1-x2f2=xa2+xb2-2ax,n~aax,n~b+2ax,n~ab2=x1-x2d2+2ax,n~ab2=a216-12ττ21n2+1m2+128a2m2n2ττ4m2-n24Thus,thevalueofx1-x22dwhentheyareidenticalfermionsis.=a216-12ττ21n2+1m2+128a2m2n2ττ4m2-n24

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 Vaia!

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

(a)Use Equation5.113 to determine the energy density in the wavelength rangedλ. Hint: setρ(ω)=ρ-(λ), and solve forρ(λ)-

(b)Derive the Wien displacement law for the wavelength at which the blackbody energy density is a maximum
λmax=2.90×10-3mKT

You'll need to solve the transcendental equation(5×x)=5e-x, using a calculator (or a computer); get the numerical answer accurate to three significant digits.

In view ofProblem 5.1, we can correct for the motion of the nucleus in hydrogen by simply replacing the electron mass with the reduced mass.

(a) Find (to two significant digits) the percent error in the binding energy of hydrogen (Equation 4.77) introduced by our use of m instead of μ.

E1=-m2h2e24π'2=-13.6eV(4.77).

(b) Find the separation in wavelength between the red Balmer lines n=3n=2for hydrogen and deuterium (whose nucleus contains a neutron as well as the proton).

(c) Find the binding energy of positronium (in which the proton is replaced by a positron—positrons have the same mass as electrons, but opposite charge).

(d) Suppose you wanted to confirm the existence of muonic hydrogen, in which the electron is replaced by a muon (same charge, but 206.77 times heavier). Where (i.e. at what wavelength) would you look for the “Lyman-α” line n=2n=1?.

The ground state of dysprosium (element 66, in the 6th row of the Periodic Table)

is listed as Is5. What are the total spin, total orbital, and grand total angular

momentum quantum numbers? Suggest a likely electron configuration for

dysprosium.

Suppose you had three particles, one in stateψa(x), one in stateψb(x), and one in stateψc(x). Assuming ψa,ψb, andψc are orthonormal, construct the three-particle states (analogous to Equations 5.15,5.16, and 5.17) representing

(a) distinguishable particles,

(b) identical bosons, and

(c) identical fermions.

Keep in mind that (b) must be completely symmetric, under interchange of any pair of particles, and (c) must be completely antisymmetric, in the same sense. Comment: There's a cute trick for constructing completely antisymmetric wave functions: Form the Slater determinant, whose first row isψa(x1),ψb(x1),ψc(x1) , etc., whese second row isψa(x2),ψb(x2),ψc(x2) , etc., and so on (this device works for any number of particles).

Chlorine has two naturally occurring isotopes,CI35and CI37. Show that

the vibrational spectrum of HCIshould consist of closely spaced doublets,

with a splitting given by v=7.51×10-4v, where v is the frequency of the

emitted photon. Hint: Think of it as a harmonic oscillator, with ω=k/μ, where

μis the reduced mass (Equation 5.8 ) and k is presumably the same for both isotopes.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free