Physics 12c, Homework 1 Solutions

Problem 1 : Oscillator Degeneracy

We want to count the ways three harmonic oscillators can be excited to 3~above the ground state, (thats three discrete excitations). Explicitly, the answeris:

(3,0,0) with 3 permutations(2,1,0) with 6 permutations(1,1,1) with 1 permutation

For a total of 10. If you prefer we can get this directly. What follows is astandard trick to do so: We have three excitation operators which we can put inthree buckets (the oscillator one, two or three). This means we symbolically have 3E symbols and 2 B symbols (the Bs are walls that divide the buckets). Thusthe number of ways to excite is the number of permutations of EBEBE which is(

53

)which is

5!

3! 2!= 10.

More generally, this method gives the answer to the question in how many wayscan M identical particles (= M units of energy in our problem), be put into Nboxes (= distributed among N oscillators). The answer is the number of different

permutations of M E symbols and N-1 B symbols which is equal to (N+M1)!M !(N1)! ,and is exactly the same result as given by the method in the book.

Problem 2 : Heads or Tails

A coin flipped N times will show m heads with a probability(Nm

) (1/2)N

Inserting N = 500 and m = 270 , we find our answer; If we use the Gaussianapproximation, the variance is N/4 , and

P (m) =e2(mN/2)

2/Npi N/2

so P (270) = 7 103 (Can also use Stirlings approx.).

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Problem 3 : Drunken Walk

The probability to random walk R to the right and L to the left is :

P (R,L) =

(R + LR

) (1/2)R+L

or putting N = R+L , the number of steps, and x = RL , the final position,

P (x,N) =

(N

x+N/2

) 2N

Thus the probability for two random walks to terminate at the same locationafter N steps is :

P =N

x=N P (x,N)2,

P =

x

(N

x+N/2

)2 4N ,

P = 4N Nm=0(Nm)2

,

P = 4N (

2NN

).

Problem 4 : Random walk in 2 and 3 dimensions.

Recall the calculation for 1 dimension. The Root Mean Square displacementfrom the origin after N steps is

DrmsN < D2N >,

where DN is the displacement from the origin after N steps. In class we showedthat < D1 >

2= l2, and

< D2N >==< D2N1 > +l

2,

so that < D2N >= Nl2. This immediately generalizes to random walk in d dimen-

sions, provided one can walk in the direction of coordinate axes only.Then we have

D2N =di=1

(D(i))2N .

where Di is the distance traveled in the i th direction. Now at every step we havea choice of the direction in which to go, and we will assume that all of them areequally probable, and of equal length l. there is 1/d probability to go in any oneof them, so:

2

< D2N >=1

d

i

= 1d =< D2N1 > +l

2

Also, < D21 >= l2. Thus, the RMS deviation from the origin after N steps in

Nl in any number of dimensions.

Problem 5 : Birthdays

a )Let the probability that two or more people out of a group of N have the same

birthday be p. Then p = 1 q, where q is probability of all the people havingdistinct birthdays.

We count the total number of possible events DN , where D = 365. The numberof cases all people have distinct birthdays is D!(DN)! , so

p = 1 D!(D N)!D

N .

b )The task is to find N such that p(N) 12 . Since:

p = 1 D!(D N)!D

N

is 0 to the first order in N/D, we need to keep these terms. then

D!

(D N)!DN ( D

D N )DN+1/2eN Eq.A

using Stirlings approx, which is valid when both D, and D N are large. Wewill have to check the second assumption a-posteriori We want this quantity to beapprox. 1/2. Taking a logarithm:

ln2 (D N + 1/2) ln(D ND

)N

(D N)[N/D 12

(N/D)2 + . . .]N= (1/2)N 2/D + . . .

Thus,N

2D ln 2 22.

It is clear that our approximation of D N being large was valid. If we now plugN = 22 into the equation A above we get 0.48 for p. Thus we actually need alarger N . Usine N = 23 gives p = 0.51. Thus 23 is the number of people for whichthe probability of two or more coincident birthdays is 1/2.

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c )Whats the probability that 100 people all have different birthdays ?Using the Stirlings approximation:

q(N = 100) ( DD N )

DNeN 3 107

Problem 6

A neutrino detector obtains a small background signal of = 1count/day.Then,after a T = 2 yrs long experiments the physicists observe a signal of 3 eventsin one hour. Is this a statistical fluctuation? Let p be the probability of observingat least 3 events in t = 1hr. Probability P that the rate exceeds two events inone hour in the duration of experiment is:

P = 1Q,where Q is probability that the rate never does exceed 2/hr:

Q = (1 p) Tt .Detection of background neutrinos at such a small rate is governed by a Poisson

distribution. Probability of obtaining k events in time t due to random fluctua-tions is given by

P (k; t, ) = (t)ket

k!Probability of obtaining at least rate k0 is

p = P (k0; t, ) =kk0

(t)ket

k!= 1

k

Similarly the probability that 9 of the 10 people would vote for him is :

P9 =

(109

)p(1 p)9

etc. The probability that our poll will indicate that Bush will win is:

P =

k10k>5

Pk

We can easily evaluate this:

P =1623424

9765625= 0.166239

Now to estimate the probability that a poll of 1000 people will indicate thatBush will win The standard deviation for a binomial distributions is:

=Npq =

Np(1 p) 15.5

For our poll to indicate that Bush will win, we would need 500+ votes for thiscandidate. Thats more than roughly 50040015.5 = 6.5 away from the expected valueof 400 votes.

We can roughly approximate the distribution of votes for Bush as a gaussianwith mean = 400 and standard deviation = 15.5 Then the probability of thepoll being in favor of Bush is 0 to much better than 1 part in 1010: +

+6.5

12pi2

e(x)2

22 4.016 1011

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