Description
1 Ensembles of spin-1/2 particles
Consider an ensemble of 16 spin-1/2 particles. There is no magnetic field present.
a) List the possible macrostates of the ensemble.
b) Determine the total number of microstates possible.
c) Determine the number of microstates that are associated with each of the macrostates.
Your answer should be a list and you can use a calculator or Excel to compute the
numbers.
d) How does the probability of attaining the microstate
↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑
compare to the probability of attaining the microstate
↑↓↑↑↑↑↓↑↑↓↓↑↑↑↑↑?
Explain your answer.
e) Determine the probability with which each macrostate could occur.
f) Which macrostate is most probable?
g) What is the probability that a macrostate will contain at least 4 spin up particles?
h) What is the probability that a macrostate will contain at most 5 spin up particles?
2 Accessible and inaccessible microstates
Consider a system of non-interacting spin-1/2 particles. The energy of a single particle with
spin up is −µB and that with spin down is µB. If the total energy of the system is fixed,
then an accessible microstate is one that gives this total energy. Suppose that the system
consists of six particles and that the total energy of the system is −2µB.
a) Describe which macrostates are possible in terms of the number of particles with spin
up.
b) List all accessible microstates.
c) Determine the probability with which on particular given particle (e.g. the leftmost) will
be in a state with spin up.
d) Determine the probability with which any two given particles will be in states with spin
up.
1
3 Ensembles of spin-1/2 particles in a magnetic field: small number example
The energy of a particle with spin up in a magnetic field with magnitude B oriented along
the positive z axis is −µB. That for a particle with spin down is µB. Consider an ensemble
of eight spin-1/2 particles in this field. Assume that the probability with which a particle is
in the spin up state is 3/4 and the probability that it is in the spin down state is 1/4. Let
N+ represent the number of particles with spin up.
a) List all macrostates for the ensemble (in terms of N+), the energy of each macrostate
and the probability with which each occurs.
b) Which is the most likely macrostate? With what probability does it occur?
c) Determine the mean for the energy, E (this is for a large collection of such ensembles).
Is there a macrostate which has an energy exactly equal to the mean energy?
d) Determine the standard deviation for the energy, σE.
e) Suppose that one is given a single copy of ensemble without knowing its actual microstate. One way to assign an energy to it is to use the mean energy. What is the
probability that the ensemble’s microstate is that whose energy is exactly the mean
energy?
It should be clear that the ensemble could be in a different macrostate. Consider the two
other macrostates with energy closest to that of the mean energy.
f) List the energies of these macrostates and determine the fractional difference between
the energies of these and the mean energy.
g) Determine the probability that the ensemble could have either the mean energy or one
of the two other energies that are closest to the mean energy.
h) Determine the probability that the ensemble could have an energy in the range E − σE
to E + σE.
i) Based on the previous results you should be able to make a statement of the form: “The
energy of the ensemble is within ??% of the mean with probability =??.” Provide such
a statement or statements for this example.
4 Ensembles of spin-1/2 particles in a magnetic field
The energy of a particle with spin up in a magnetic field with magnitude B oriented along
the positive z axis is −µB. That for a particle with spin down is µB. Consider an ensemble
of N spin-1/2 particles in this field and assume that N is even. Let N+ represent the number
of particles with spin up. The probability with which any given particle will be in the spin
up state is different to that in which it will be in the spin down state. It will emerge that
the spin up state is favored over the spin down state. Denote the probability with which any
particle is in the spin up state by (1 + ε)/2 where 0 ! ε ! 1.
a) Show that,
E = −µBNε
2
and that the standard deviation of the energy, .
σE = µB!(1 − ε2)N.
b) Show that the magnitude of the ratio of the standard deviation to the mean approaches
zero as the size of the ensemble increases.
In order to illustrate these consider numerical examples for which µB = 1/2 and ε = 1/3.
The aim will be to determine the probability with which the energy of a macrostate is within
the range of a standard deviation of the mean.
c) Show that E = −N/6 and σE = !2N/9. Show that the value of N+ that gives the
macrostate with energy exactly equal to E is N+ = 2N/3. Show that the value of N+
that gives the macrostate with energy exactly equal to E ± σE is N+ = 2N/3∓!2N/9.
In each of the following you can use a numerical tool (e.g. Excel) to do the computations.
These all concern the range of the energies E − σE to E + σE, A measure of this range is the
ratio |σE/E|.
d) Let N = 21. Determine |σE/E|. Determine the probability with which the energy of a
macrostate is in the range E − σE to E + σE.
e) Let N = 90. Determine the fractional difference between the mean energy E and the
extremes of the energies in the range E −σE to E +σE. Determine the probability with
which the energy of a macrostate is in the range E − σE to E + σE.
f) Let N = 150. Determine the fractional difference between the mean energy E and the
extremes of the energies in the range E −σE to E +σE. Determine the probability with
which the energy of a macrostate is in the range E − σE to E + σE.
g) Does the range of energies E − σE to E + σE become more precise, relative to the mean
energy, as N increases?
h) Does the probability of falling within the range energies E −σE to E +σE become larger
or smaller as N increases? Does it appear to approach a fixed value?
5 Einstein solid statistics: small numbers
Consider an Einstein solid consisting of five particles (these are labeled A, B, C, D and E).
a) Determine the multiplicities for the four lowest energy states.
b) Suppose that there are three energy units. Determine the probability with which one
particle will have all three energy units.
c) Suppose that there are three energy units. Determine the probability with which particle
1 has two energy units.
3
6 Einstein solid statistics: large numbers
Consider two Einstein solids, labeled A and B. Solid A has 10 particles and solid B has 10
particles. The exercise will consider various energy units. Calculations can be done using the
OSP program called Einstein solid: temperature. Suppose that initially the energy number for
A is qA = 9 and, for B, qB = 3. In the following do not forget the !ω and N/2 terms in teh
energy.
a) Determine the multiplicities ΩA and ΩB. Determine the total number of microstates for
the composite system. Determine the energy per particle for each solid.
b) Now suppose that the systems can interact, exchange energy and eventually reach equilibrium. Use the program to determine and describe the most probable state for each
solid. Determine the the energy per particle for each solid.
c) In what direction did the energy flow as the systems approached equilibrium?
d) Determine the probability that the energy flows from system A to system B. Determine
the probability that the energy flows from system B to system A. How many times is it
more likely that energy flows from A to B rather than from B to A?
Now suppose that solid A initially has 20 particles and qA = 20 while solid B has 5 particles
and qB = 10.
e) Determine the equilibrium state after the solids have interacted. Determine the energy
per particle for each solid.
f) Did the energy flow from the solid with higher total energy to that with lower total
energy?
4
2020 HW14 Q1
a) Macrostates can be labeled by the number of spin up porticles,Nt.Ther
N+ =0,1,2,3,4,-, 15, 16
b) Each particde can be in ane of two stales. Thes Rare ave
xx… = 216 = 65536
Stales
c) The number of miciostates associaled with macrostate N+ is
N!
N+ (N-N+)
N+ nunber prab
이 10.000015
16 0.00024
2 120 0.00183
3 560 0.00854
4 1820 0.02777
5 4368 0.06665
6 8008 0.12219
7 11440 0.17456
8 12870 0.19638
१ 11440 0.17456
1o 8008 0,12219
11 4368 0.06665
12 1820 0.02777
13 560 000854
14 120 0.00183
15 16 0.00024
16 0.000015
2020 HW 14 QI
d) each microstate is equally likely as each spin ‘½ particle has
an equal probability of being in 4 us b
These ove equally likely.
N+ T
e) P(N)= (.)G]N (G)NN
(2)
See table in ()
fJ N+= 8
g) This is P(N+= 4) + p(N+=s) +… + P(N+=16)
From the table this is 0.98937
h) Tuis is P(N+=0) + p(N+= 1)+.. +P(N +=5) = 0.10506
2020 HW 14 Q2
2017
a) E= n+ (M8) + 1- (4s) Where 1+= number up
n-= number down
= -MB(N+-n-)
Then n- = N-n+ =D E=-MB (2n+-N)
We need E=-2MB =-1MB (2n+-N)
=D 2 = 2n+-N
2+N =D V+ =
2
Hee N=6 =4 n+ = 4. The states ave
SITUTL 1T.TLT
6 There avre (2)- =15 4.21
Stales
TTATT
个个
个个个
个个
↓个
2020 HW 14 Q2
Look at lettmost spin particle There ave 10 ways it can be up.
Prob =
102
15 3
d). Look at leftmost two particles. There ae. 6 ways they can both be
ud
6=2 Prob =
15 5
2020 HW 1493
a) E= N+ (-MB) + N-(M8)
MB(N–N+) N=N++N- =0 N-= N-N+
=MB(N-2N+)
N+ E Prob allover 65536
0 8µB
t
6μB 24/- 8
2 4MB
/252 28
3 2MB 1512/. 56
4
0 56701 70
5 -2MB 13608/ 56
=
Prob (N) (Nx) p.N p-N-
(N-
= G)NEN- (N) 3N+
= (4)8 (N)3N
L/83N+
N+ 28 65536 6 -4 MB 20412/
8 7 -6µB 17496/
8 -8μ8 6561/
20412- Prob = =0.311l
65536 b) N+=6 N=2
c) E= MB(N-2N+)
=μB(8-2N+)
=MB (8-12)
E=-4MB
d) O= E² -E2
For binomal Nt=P.N
(MB(N-2N))
2
-(MB (N-2N))
= (UB) 2[N²-4NN +NEJ-(UB)°[N2-4NN++4N[
=(48)2 [N2-ANN+N] MB)Y [N²-4NNT +4N²)
OE²= (UB)²ON
= MB N P+(1-p+)
3.1 OE= MB√8 2216
= MB
2020 н 14 (3
JE= 2.45 Mв
e) This is N+=6 -0 prob = 0.311
f) These are
N+= 5 =0 E=-2MB
N+=7 = E=-6MB
The diff in erergy from N+=6 is 24B
Frachinal dilf is 1)= tac deff =%2 2μB=
g) Tuis is prob (N+= 5) + prob(N+=6) +prob(N+=7) = 0.788
b) The values of E range the same as =0 U.788
as for g
i) The erergy is within 50% of – 4MB withe prch 0.788
3+1
a) P=
2020 HW 14 G4
P& = 1-P4 =
3-1
Z
Let N+ be the nunber of particles withe spin
NTher the erergy is
Now
E = -N+MB + N-MB = MB(N–N+)
= MB (N-N+-NJ)
E= MB(N-2N+)
E= MB(N-2N+)
and N+ = NP4
=0 E = MS/N – 2N )3[
= NBM [3-1-1]
=D E =-MBNE
الى
Then OE=-E2 and here
E² = M²B² (N-2N )² = M’B² (N²-4NN+4N+²)
E= M²B² (N²-4NN+ + 4NE)
2020 HW 14 Q4
For binomial distibutions
N = N P4
So
2
N² = N+²+ N PAps
= Np+N ())
= N² (13) +N()
E = N8² [N²- 4N²(1) + N°(1+3) +N(1-35]
= M²B²[N²-202-2NZE + N²)1423+32) +N)1-32)]
Then E*-E’= Mt8²[uSe +N(1-32)-NE[
=D OE = B² N (1-E3)
=D JE= MB√N(1-83)”
b) E = HBVN(1-6)
E -ABNE
V1-62
VNE 0 as N-8
-N
3
6
2N N =
a
In generad E = uB (N-2N+) = (N-2N+)
We reed E=E
=(N-2N+) = –
=4 N-2N+ =-N/3
=D 4 2N+ =D N+=N
Ther the value of N+ that gives EtOE is
d) For the probabilities
N
P/N+)= (N) Pa PN
(= +NZ
Am/% 2N 6
=
N 3
N
E
prob range of
N+
prob
21 0.617 11,8 < N+ < 16.2 0.754
90 0.298 55.5 < N+ < 64. 4
0.686
150 0.231 94.2 N+ <105.8 0.659
A
g) Becomos smaller
h) Gets smaller, maybe approaches fixed volue
22020 HW 14 Q5
a) howest erergy otates:
q=0,1,2,3
2 (N,9)= (Ng~) = (Nr= !
(N+q-1)!
q (N+q-1-q)! g! (N-D! g!
(4+q)!
Hee N=5 = R(N,9)= 4:9!
Л (N,9)
0
5
2 15
3 35
475
b) with 9=3 there ave 35 equally likely microstates. of hese the following
have one particle with all erergy wnits Thes
A B C D E
Ο 0 0 U 3
O 0 U 3 σ
0 0 3 0 C
0 3 Ο
C 0
3
0 0 0 0
c) The nicrostates cue:
5
prob = /35 =
2 10 co prob = 4/55
total 20100
a) Da= (Nadoi) =G)-G) 9
2020 HW 4 Q6
= 48620
No +98-1 13-1
= = 22О
9B 3
ЛЛАЛ = 10696400
The eregy for a system s
E= tw (9+ N/½). So thee
erugy per pauhicle is
5/10)+
For A D EA EW (+) my= 21
NA
EAE5
NA
For B EsN= hw (i+) tw =
ER EB = hw 4
NB ड
6) The most probable has qA= 6 =0 98=6
EA Then for A EAN= (+) –
10 NA 10
Similaly for B
ER =0 =hw NB 10
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0.16 row Ea Po
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2 2 0.036
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0.12 4
4 0.123
5 5 0.162
6 6 0.178
0.10 7 7 0.162
8 8 0.123
9 0.076
0.08 Π 10 10 0.036
11 11 0.012
0.06 12 12 0.002
0.04
0.02
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0 1 2 3 4 5 6 7 8 9 10 11 12
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It flowed from system A to system B
(one with most erurgy po patide to one with least erigy per pahicle)
d) Prob (A – 8)= Prcb (A eugy deceases)
=-Prob (9^=0) + Proo (G^=1)+… + Prob (94e 8)
= 0.002+ 0.012 +.. + 0.123
= 0.874
Prob (B-DA) = Prob (A enogy incraies)
こ Prcb(9=10) + Prob(qA=11) + Prob (9A=12)
= 0.050
e)
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2 0
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10 10 0
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12 12 0.001
13 13 0.001
14 14 0.003
15 15 0.005
16 16 0.009
17 17 0.015
18 18 0.023
19 19 0.034
20 20 0.049
0 5 10 15 20 25 30 21 21 0.067 E 22 22 0.086
Drag table columns to yellow (horizontal axis) or green (vertical axis) for curve fitting non-editable
Most probable 9A=25 9B= 30-25=5
erergy EA= hw /9 NAs) = tw (25410) +hw EANA W -tw
EB=thw (5+s)
EENE thw2
+) Erergy flowed from B-DA So not from higher o lower eregy
It flowed from higher energy per partice (8) to lower oigy per
Pasticle (A)
