Today’s Outline - November 30, 2017phys.iit.edu/~segre/phys405/17F/lecture_27.pdfToday’s Outline - November 30, 2017 Final exam information Review problems from: Chapter 4 Chapter

Today’s Outline - November 30, 2017

• Final exam information

• Review problems from:

• Chapter 4• Chapter 5

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Final Exam:Tuesday, December 5, 2017

14:00 – 16:00, 212 Stuart Building

C. Segre (IIT) PHYS 405 - Fall 2017 November 30, 2017 1 / 16



















• Review problems from:• Chapter 4

• Chapter 5







• Review problems from:• Chapter 4• Chapter 5



















Problem 4.7

(a) From the definition of the Neumann functions

nl(x) ≡ −(−x)l(

1

x

d

dx

)l cos x

x

construct n1(x) and n2(x).

(b) Expand the sines and cosines to obtain approximate formulas for n1(x)and n2(x), valid when x � 1. Confirm that they blow up at the origin.

(a) Start with n1

n1(x) = −(−x)1

x

d

dx

(cos x

x

)= −cos x

x2− sin x

x


Problem 4.7


nl(x) ≡ −(−x)l(

1

x

d

dx

)l cos x

x



(a) Start with n1

n1(x) = −(−x)1

x

d

dx

(cos x

x

)= −cos x

x2− sin x

x


Problem 4.7


nl(x) ≡ −(−x)l(

1

x

d

dx

)l cos x

x



(a) Start with n1

n1(x) = −(−x)1

x

d

dx

(cos x

x

)

= −cos x

x2− sin x

x


Problem 4.7


nl(x) ≡ −(−x)l(

1

x

d

dx

)l cos x

x



(a) Start with n1

n1(x) = −(−x)1

x

d

dx

(cos x

x

)= −cos x

x2− sin x

x


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x

= −x2(

1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x

−x sin x − cos x

x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x

(x2 cos x − 2x sin x

x4+−x3 sin x − 3x2 cos x

x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x

(b) as x becomes small, we can replace sin x ≈ x and cos x ≈ 1

n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]

= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)

= xd

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)

= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)

=cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3

= −(

3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1

≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2

−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x

≈ − 3

x3−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3

−→∞


Problem 4.7 (cont.)

n2(x) = −(−x)2(

1

x

d

dx

)2 cos x

x= −x2

(1

x

d

dx

)[1

x

d

dx

(cos x

x

)]= −x d

dx

(1

x


x2

)= x

d

dx

(sin x

x2+

cos x

x3

)= x



x6

)=

cos x

x− 2

sin x

x2− sin x

x2− 3

cos x

x3= −

(3

x3− 1

x

)cos x − 3

x2sin x


n1(x) ≈ − 1

x2− 1 ≈ − 1

x2−→∞

n2(x) = − 3

x3+

1

x− 3

x≈ − 3

x3−→∞


Problem 5.3

Chlorine has two naturally occurring isotopes, 35Cl and 37Cl. Show thatthe vibrational spectrum of HCl should consist of closely spaced doublets,with a splitting given by ∆ν = 7.51× 10−4ν, where ν is the frequency ofthe emitted photon.

Consider the vibrational system ofthe HCl molecule as a harmonic os-cillator with a reduced mass µ andspring constant k , resulting in anangular frequency of vibration ω

this harmonic oscillator can maketransitions between states of differ-ing quantum number ni → nf

the frequency of the photon emit-ted by these transitions is thus

µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3


Consider the vibrational system ofthe HCl molecule as a harmonic os-cillator with a reduced mass µ

andspring constant k , resulting in anangular frequency of vibration ω



µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3


Consider the vibrational system ofthe HCl molecule as a harmonic os-cillator with a reduced mass µ

andspring constant k , resulting in anangular frequency of vibration ω



µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω

= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h

=nω

2π=

n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π

=n

2π

√k

µ


Problem 5.3





µ =mHmCl

mH + mCl

ω =

√k

µ

Ep = (ni + 12)~ω − (nf + 1

2)~ω= (ni − nf )~ω= n~ω n ≡ ni − nf

ν =Ep

h=

nω

2π=

n

2π

√k

µ


Problem 5.3 (cont.)

Because there are two isotopes, there will be two slightly differentoscillation frequencies for the molecule, resulting in small differences in theemitted photon frequencies, separated by an amount dν

∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣

=

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣

=1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ

=1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ

=1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ

=1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl

= −(

1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ

=1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)

=µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ

=1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν

≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν

= 7.51× 10−4ν


Problem 5.3 (cont.)


∆ν =

∣∣∣∣∣ ddµ(

n

2π

√k

µ

)∆µ

∣∣∣∣∣ =

∣∣∣∣ n2π√k(− 1

2µ3/2

)∆µ

∣∣∣∣=

1

2

n

2π

√k

µ

∆µ

µ=

1

2ν

∆µ

µ=

1

2νµ∆mCl

m2Cl

∆µ =d

mCl

(1

mH+

1

mCl

)−1∆mCl = −

(1

mH+

1

mCl

)−2(−∆mCl

m2Cl

)=

µ2

m2Cl

∆mCl

∆ν =1

2

µ∆mCl

m2Cl

ν ≈ 1

2

(0.973 · 2

362

)ν = 7.51× 10−4ν


Problem 5.20

Suppose we use delta function wells, instead of spikes. Analyze this case,constructing the analog to Figure 5.6. This requires no new calculation,for the positive energy solutions (except that β is now negative; useβ = −1.5 for the graph), but you do need to work out the negative energysolutions (let κ ≡

√−2mE/~ and z ≡ −κa, for E < 0). How many states

are there in the first allowed band?

This is identical to the Dirac Combproblem of last semester exceptthat the potential is composed ofnegative-going delta functions andso α→ −α

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .

This problem has two different solutions, one with E > 0 and the otherwith E < 0. The E > 0 solution is identical to the problem solved lastsemester and we will discuss it first.


Problem 5.20





x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .



Problem 5.20





x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .



Problem 5.20





x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .



Problem 5.20 - E > 0 solution

Solve the Dirac comb potential us-ing Bloch’s theorem and periodicboundary conditions by solving overa single unit

the potential can be written as asum of delta functions , where αis the “strength” of the potentialsand N is the number of repeatingperiods

in the region 0 < x < a we definek =√

2mE/~ and have

letting z ≡ ka and β = −mαa/~2we get the solution

cos(Ka) = cos(z) + βsin(z)

z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .

ψ(x) = ψ(x + Na)

= e iNKaψ(x)

K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ

the only difference is the sign of βand this justs shifts the energies ofthe bands a bit






2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .

ψ(x) = ψ(x + Na)

= e iNKaψ(x)

K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .

ψ(x) = ψ(x + Na) = e iNKaψ(x)

K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ





the potential can be written as asum of delta functions

, where αis the “strength” of the potentialsand N is the number of repeatingperiods


2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ





the potential can be written as asum of delta functions

, where αis the “strength” of the potentialsand N is the number of repeatingperiods


2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ







2mE/~ and have



z

x

-3a 0-2a-4a -a 2a 4a3aa

. . .. . .


K =2πn

Na, (n = 0,±1,±2, . . . )

V (x) = −αN−1∑j=0

δ(x − ja)

d2ψ

dx2= −k2ψ





z

The original solution with β = 1.5has bands where the right hand sideis between −1 and +1 and so hasvalues concident with the left side

when we set β = −1.5, the bandsshift to slightly lower values of zand, more importantly, there is nogap at z = 0

this emphasizes the fact that theremust be solutions for z < 0 andthus E < 0

-1

0

+1

0 π 2π 3π 4π

f(z)

z




z




-1

0

+1

0 π 2π 3π 4π

f(z)

z




z




-1

0

+1

0 π 2π 3π 4π

f(z)

z




z




-1

0

+1

0 π 2π 3π 4π

f(z)

z


Problem 5.20 - E < 0 solution

In the region 0 < x < a we defineκ =√−2mE/~ and have

d2ψ

dx2= κ2ψ

ψ(x)= A sinh(κx) + B cosh(κx), (0 < x < a)

ψ(x)= e−iKa [A sinhκ(x + a) + B coshκ(x + a)], (−a < x < 0)

apply continuity of the wavefunction at x = 0 and, because of thediscontinuity in the derivative

rearranging the conti-nuity equation, substi-tuting for A in thed-erivative equation, andcancelling B

B

= e−iKa [A sinh(κa) + B cosh(κa)]

κA

− e−iKaκ [A cosh(κa) + B sinh(κa)] = −B 2mα

~2

A sinh(κa) = B[e iKa − cosh(κa)

][e iKa − cosh(κa)]

sinh(κa)

[κ

− e−iKaκ cosh(κa)

]

− e−iKaκ sinh(κa) = −2mα

~2




d2ψ

dx2= κ2ψ





B


κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ





B


κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ





B


κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ



apply continuity of the wavefunction at x = 0

and, because of thediscontinuity in the derivative


B


κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ






B


κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ






B = e−iKa [A sinh(κa) + B cosh(κa)]

κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ






κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ






κA


~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ






κA− e−iKaκ [A cosh(κa) + B sinh(κa)]

= −B 2mα

~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ






κA− e−iKaκ [A cosh(κa) + B sinh(κa)] = −B 2mα

~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ




rearranging the conti-nuity equation

, substi-tuting for A in thed-erivative equation, andcancelling B



~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ




rearranging the conti-nuity equation

, substi-tuting for A in thed-erivative equation, andcancelling B



~2


]

[e iKa − cosh(κa)]

sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ







~2


]


sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ







~2



sinh(κa)

[κ


]


~2




d2ψ

dx2= κ2ψ







~2



sinh(κa)

[κ− e−iKaκ cosh(κa)

]


~2




d2ψ

dx2= κ2ψ







~2



sinh(κa)


]− e−iKaκ sinh(κa)

= −2mα

~2




d2ψ

dx2= κ2ψ







~2



sinh(κa)


]− e−iKaκ sinh(κa) = −2mα

~2




sinh(κa)



~2

− 2mα

~2κsinh(κa) = [e iKa − cosh(κa)][1− e−iKa cosh(κa)]− e−iKa sinh2(κa)

= e iKa

− cosh(κa)− cosh(κa) + e−iKa cosh2(κa)− e−iKa sinh2(κa)

= e iKa

− 2 cosh(κa) + e−iKa

= 2 cos(Ka)

− 2 cosh(κa)

cos(Ka) = cosh(κa)− mα

~2κsinh(κa)

letting z ≡ κa and β = −mαa/~2we have a solution for E < 0 andthus z < 0

cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα


= e iKa


= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα


= e iKa


= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα


= e iKa


= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα


= e iKa − cosh(κa)

− cosh(κa) + e−iKa cosh2(κa)− e−iKa sinh2(κa)

= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα


= e iKa − cosh(κa)− cosh(κa)

+ e−iKa cosh2(κa)− e−iKa sinh2(κa)

= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα


= e iKa − cosh(κa)− cosh(κa) + e−iKa cosh2(κa)

− e−iKa sinh2(κa)

= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα


= e iKa − cosh(κa)− cosh(κa) + e−iKa cosh2(κa)− e−iKa sinh2(κa)

= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα



= e iKa


= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα



= e iKa − 2 cosh(κa)

+ e−iKa

= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα



= e iKa − 2 cosh(κa) + e−iKa

= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα




= 2 cos(Ka)

− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα




= 2 cos(Ka)− 2 cosh(κa)


~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα






~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα






~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα






~2κsinh(κa)


cos(Ka) =

cosh(z) + βsinh(z)

z




sinh(κa)



~2

− 2mα






~2κsinh(κa)


cos(Ka) = cosh(z)

+ βsinh(z)

z




sinh(κa)



~2

− 2mα






~2κsinh(κa)


cos(Ka) = cosh(z) + βsinh(z)

z



Thus for β = −1.5 the complete solution for all z is given by

cos(Ka) =

{cosh(z) + β sinh(z)/z , z < 0

cos(z) + β sin(z)/z , z > 0

-1

0

+1

-π 0 π 2π 3π 4π

f(z)

z

and the bands still exist,are continuous, and ex-tend to negative z

because Ka = 2π/Na,with n = 0, 1, 2 . . .N − 1,each of the grey boxes hasN discrete states and thusthere must be exactly Nstates in each band




cos(Ka) =



-1

0

+1

-π 0 π 2π 3π 4π

f(z)

z






cos(Ka) =



-1

0

+1

-π 0 π 2π 3π 4π

f(z)

z






cos(Ka) =



-1

0

+1

-π 0 π 2π 3π 4π

f(z)

z






cos(Ka) =



-1

0

+1

-π 0 π 2π 3π 4π

f(z)

z




Problem 5.22

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

(b) Construct the completely symmetric wave function ψ(xA, xB , xC ) forthree identical bosons, (i) if all three are in state ψ11, (ii) if two are instate ψ1 and one in state ψ19, and (iii) if one is in state ψ5, one instate ψ7 and one in state ψ17

(a) Start with the Slater determinant

ψ =det

∣∣∣∣∣∣ψ5(xA) ψ7(xA) ψ17(xA)ψ5(xB) ψ7(xB) ψ17(xB)ψ5(xC ) ψ7(xC ) ψ17(xC )

∣∣∣∣∣∣=

√16

[

ψ5(xA)ψ7(xB)ψ17(xC )− ψ5(xA)ψ7(xC )ψ17(xB)

− ψ7(xA)ψ5(xB)ψ17(xC ) + ψ7(xA)ψ17(xB)ψ5(xC )

+ ψ17(xA)ψ5(xB)ψ7(xC )− ψ17(xA)ψ7(xB)ψ5(xC )

]


Problem 5.22




ψ =det


∣∣∣∣∣∣=

√16

[




]


Problem 5.22




ψ =det


∣∣∣∣∣∣

=

√16

[




]


Problem 5.22




ψ =det


∣∣∣∣∣∣=

√16

[




]C. Segre (IIT) PHYS 405 - Fall 2017 November 30, 2017 12 / 16

Problem 5.22




ψ =det


∣∣∣∣∣∣=

√16

[ψ5(xA)ψ7(xB)ψ17(xC )− ψ5(xA)ψ7(xC )ψ17(xB)




Problem 5.22




ψ =det


∣∣∣∣∣∣=

√16





Problem 5.22




ψ =det


∣∣∣∣∣∣=

√16



+ ψ17(xA)ψ5(xB)ψ7(xC )− ψ17(xA)ψ7(xB)ψ5(xC )]


Problem 5.22




ψ =det


∣∣∣∣∣∣=√

16



+ ψ17(xA)ψ5(xB)ψ7(xC )− ψ17(xA)ψ7(xB)ψ5(xC )]


Problem 5.22 (cont.)

(b)

ψi = ψ11(xA)ψ11(xB)ψ11(xC )

ψii =

√13

[

ψ1(xA)ψ1(xB)ψ19(xC ) + ψ1(xA)ψ19(xB)ψ1(xC )

+ ψ19(xA)ψ1(xB)ψ1(xC )

]ψiii =

√16

[

ψ5(xA)ψ7(xB)ψ17(xC ) + ψ5(xA)ψ7(xC )ψ17(xB)

+ ψ7(xA)ψ5(xB)ψ17(xC ) + ψ7(xA)ψ17(xB)ψ5(xC )


]



(b)


ψii =

√13

[



]ψiii =

√16

[




]



(b)


ψii =

√13

[



]

ψiii =

√16

[




]



(b)


ψii =

√13

[ψ1(xA)ψ1(xB)ψ19(xC )



]

ψiii =

√16

[




]



(b)


ψii =

√13

[ψ1(xA)ψ1(xB)ψ19(xC ) + ψ1(xA)ψ19(xB)ψ1(xC )


]

ψiii =

√16

[




]



(b)


ψii =

√13


+ ψ19(xA)ψ1(xB)ψ1(xC )]

ψiii =

√16

[




]



(b)


ψii =√

13


+ ψ19(xA)ψ1(xB)ψ1(xC )]

ψiii =

√16

[




]



(b)


ψii =√

13


+ ψ19(xA)ψ1(xB)ψ1(xC )]

ψiii =

√16

[




]



(b)


ψii =√

13


+ ψ19(xA)ψ1(xB)ψ1(xC )]

ψiii =

√16

[ψ5(xA)ψ7(xB)ψ17(xC ) + ψ5(xA)ψ7(xC )ψ17(xB)



]



(b)


ψii =√

13


+ ψ19(xA)ψ1(xB)ψ1(xC )]

ψiii =

√16




]



(b)


ψii =√

13


+ ψ19(xA)ψ1(xB)ψ1(xC )]

ψiii =

√16



+ ψ17(xA)ψ5(xB)ψ7(xC ) + ψ17(xA)ψ7(xB)ψ5(xC )]



(b)


ψii =√

13


+ ψ19(xA)ψ1(xB)ψ1(xC )]

ψiii =√

16



+ ψ17(xA)ψ5(xB)ψ7(xC ) + ψ17(xA)ψ7(xB)ψ5(xC )]


Problem 5.24

Check the following equations for the example in Section 5.4.1 where threeparticles are in a one-dimensional infinite square well withn2A + n2B + n2C = 363.

Q(N1,N2,N3, . . . ) = N!∞∏n=1

dNnn

Nn!

Q(N1,N2,N3, . . . ) =∞∏n=1

dn!

Nn!(dn − NN)!

Q(N1,N2,N3, . . . ) =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!

The three equations are for distinguishable particles, fermions and bosonsand there are 4 possible configurations, which will be treated for each case

in all cases, N = 3 and dn = 1


Problem 5.24


Q(N1,N2,N3, . . . ) = N!∞∏n=1

dNnn

Nn!

Q(N1,N2,N3, . . . ) =∞∏n=1

dn!

Nn!(dn − NN)!

Q(N1,N2,N3, . . . ) =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!




Problem 5.24


Q(N1,N2,N3, . . . ) = N!∞∏n=1

dNnn

Nn!

Q(N1,N2,N3, . . . ) =∞∏n=1

dn!

Nn!(dn − NN)!

Q(N1,N2,N3, . . . ) =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!




Problem 5.24 - distinguishable case

Q = N!∞∏n=1

dNnn

Nn!

= 6∞∏n=1

1

Nn!

since most terms have Nn = 0, which result in values of 1, only the termswith Nn 6= 1 are enumerated

N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3

Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2

Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1

Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1

Q = 6× 11! ×

11!

11! = 6



Q = N!∞∏n=1

dNnn

Nn!= 6

∞∏n=1

1

Nn!


N11 = 3 Q = 6× 13! = 1

N5 = 1; N13 = 2 Q = 6× 11! ×

12! = 3

N1 = 2; N19 = 1 Q = 6× 12! ×

11! = 3

N5 = N7 = N17 = 1 Q = 6× 11! ×

11!

11! = 6


Problem 5.24 - fermion & boson cases

Q =∞∏n=1

dn!

Nn!(dn − Nn)!

=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1

All four cases give Q = 1



Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3

Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2

Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1

Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1

Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!

= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1




Q =∞∏n=1

dn!

Nn!(dn − Nn)!=∞∏n=1

1

Nn!(1− Nn)!

N11 = 3 Q = 13! ×

1(−2)! = 0

N5 = 1; N13 = 2 Q = 11!0! ×

12!(−1)! = 0

N1 = 2; N19 = 1 Q = 12!(−1)! ×

11!0! = 0

N5 = N7 = N17 = 1 Q = 11!0! ×

11!0!

11!0! = 1

Q =∞∏n=1

(Nn + dn − 1)!

Nn!(dn − 1)!= 1

All four cases give Q = 1C. Segre (IIT) PHYS 405 - Fall 2017 November 30, 2017 16 / 16

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