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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

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Most popular questions from this chapter

(a) Construct the completely anti symmetric wave function ψ(xA,xB,xC)for three identical fermions, one in the state ψ5, one in the state ψ7,and one in the state ψ17

(b)Construct the completely symmetric wave function ψ(xA,xB,xC)for three identical bosons (i) if all are in state ψ11(ii) if two are in state ψ19and another one is role="math" localid="1658224351718" ψ1c) one in the state ψ5, one in the state ψ7,and one in the stateψ17

Imagine two non interacting particles, each of mass , in the one dimensional harmonic oscillator potential (Equation 2.43). If one is in the ground state, and the other is in the first excited state, calculate (x1-x2)2assuming
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The density of copper is8.96g/cm3,and its atomic weight is63.5g/mole

(a) Calculate the Fermi energy for copper (Equation 5.43). Assume d = 1, and give your answer in electron volts.

EF=ħ22m3ρπ22/3 (5.43).

(b) What is the corresponding electron velocity? Hint: SetEF=1/2mv2Is it safe to assume the electrons in copper are nonrelativistic?

(c) At what temperature would the characteristic thermal energyrole="math" localid="1656065555994" (kBT,wherekBkBis the Boltzmann constant and T is the Kelvin temperature) equal the Fermi energy, for copper? Comment: This is called the Fermi temperature,TF

. As long as the actual temperature is substantially below the Fermi temperature, the material can be regarded as “cold,” with most of the electrons in the lowest accessible state. Since the melting point of copper is 1356 K, solid copper is always cold.

(d) Calculate the degeneracy pressure (Equation 5.46) of copper, in the electron gas model.

P=23EtotV=23ħ2kF510π2m=3π22/3ħ25mρ5/3

a) Hund’s first rule says that, consistent with the Pauli principle, the state with the highest total spin (S) will have the lowest energy. What would this predict in the case of the excited states of helium?

(b) Hund’s second rule says that, for a given spin, the state with the highest total orbital angular momentum (L) , consistent with overall antisymmetrization, will have the lowest energy. Why doesn’t carbon haveL=2? Note that the “top of the ladder”(ML=L)is symmetric.

(c) Hund’s third rule says that if a subshell(n,l)is no more than half filled,
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(d) Use Hund’s rules, together with the fact that a symmetric spin state must go with an antisymmetric position state (and vice versa) to resolve the carbon and nitrogen ambiguities in Problem 5.12(b). Hint: Always go to the “top of the ladder” to figure out the symmetry of a state.

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