1. Section 2.4: Probability Calculations

2. The Fundamental Postulate of Statistical Mechanics:
To calculate the probability of finding a system in a given state, we use
The Fundamental Postulate
of Statistical Mechanics:
“An isolated system in equilibrium is equally likely to be found in any one of it’s accessible microstates.”
There always is an uncertainty in our knowledge of the system energy ≡ δE.
Suppose that we know that the energy of the system is in the range E to E + δE.

3. The Fundamental Postulate: “An isolated system in equilibrium is
equally likely to be found in any
one of it’s accessible states.”
Suppose that we know that the energy of the system is in the range E to E + δE.
Define: Ω(E) ≡ Total number of accessible states in this range. y ≡ A macroscopic system parameter (pressure, magnetic moment, etc.).
Define: Ω(E;yk) ≡ A subset of Ω(E) for which
y ≡ yk (yk = A particular value of y)

4. By the Fundamental Postulate of Statistical Mechanics this is:
Let P(y = yk) ≡ probability that y ≡ yk.
By the Fundamental Postulate of
Statistical Mechanics this is:
Pk  P(y = yk) ≡ [Ω(E;yk)]/[Ω(E)]
Now, calculate the mean (expected) value of y:
From probability theory, this is simply:
<y> ≡ ∑kykPk  ∑kyk[Ω(E;yk)]/[Ω(E)]
Clearly, to calculate this, we need to know both Ω(E) & Ω(E;yk). This will be discussed in detail as we go through this chapter!

5. (+,+,-) (+,-,+) (-,+,+). In principle (if we know the Ω’s) this
calculation is easy.
Example: Take 3 particles of spin ½ in
an external magnetic field again.
Suppose that we know from
measurement that the total energy is
E = - μH. As we’ve seen, there are only
3 accessible states for E = - μH. These are:
(+,+,-) (+,-,+) (-,+,+).

6. The only 3 accessible states for E = - μH are:
(+,+,-) (+,-,+) (-,+,+).
Question: What is the probability of finding spin #1 in the “up” position?
To answer, we need Ω(E) for this
problem. We had:
Ω(E)  ≡ Ω(E = -μH) ≡ 3 and
Ω(E;yk) 
≡ Ω(E = -μH; spin 1 is “up”) ≡ 2

7. P(spin 1 “up”) ≡ (⅔) P(spin 1 “down”) ≡ (⅓)
For total the energy E = - μH, the 3 accessible states are:
(+,+,-) (+,-,+) (-,+,+).
Probability of finding spin #1 in “up” position?
Formally: P(E; y = yk) ≡ [Ω(E;yk)/Ω(E)] or
P(E = -μH; spin 1 is “up”) ≡
[Ω(E = -μH; spin 1 is “up”)/Ω(E = -μH)]
We just saw that Ω(E)  ≡ Ω(E = -μH) ≡ 3 &
Ω(E;yk)  ≡ Ω (E = -μH; spin 1 is “up”) ≡ 2
So, The Probability that spin #1 is “up” is:
P(spin 1 “up”) ≡ (⅔)
The Probability that spin #1 is “down” is:
P(spin 1 “down”) ≡ (⅓)

8. <μz> = (⅓)μ <μz> ≡ ∑kμzkP(μz= μzk) = (⅔)(μ) + (⅓)(-μ)
Since we have the probability, P(E; y = yk) we can use it to find answers to problems like:
Calculate the Mean (average) Magnetic Moment of Spin # 1.
<μz> ≡ ∑kμzkP(μz= μzk)
(Sum goes over k = “up” & k = “down”)
<μz> = P(#1 “up”)(μ) + P(#1 “down”)(-μ)
= (⅔)(μ) + (⅓)(-μ) so
<μz> = (⅓)μ

9. Section 2.5: Energy Dependence of the Density of States
Note!
This section is
one of the most important sections in the chapter!!

10. A goal of the following discussion
We’ve just seen that the probability that parameter y has a value yk is:
P(y = yk) ≡ [Ω(E;yk)/Ω(E)]
Ω(E) ≡ # accessible states with energy
between E & E + δE.
So, to do probability calculations, we clearly need to know the E dependence of Ω (E).
A goal of the following discussion
is to find an approximation to the
E dependence of Ω(E).

11. GOAL: Find an Approximation
P(y = yk) ≡ [Ω(E;yk)/Ω(E)]
GOAL: Find an Approximation
for the E dependence of Ω(E).
No rigorous math! Physical arguments & “hand waving.”
Consider a Macroscopic System
with f degrees of freedom. f is huge! f ~ 1024.
Energy is in range E to E + δE. Of course, Ω(E)  δE. Extrapolating earlier discussion of the 1d oscillator:
Ω(E)  Phase space volume
in this energy range.
Define Ω(E) ≡ ω(E)δE. ω(E) ≡ Density of states

12. Goal: Estimate the E dependence of Ω(E) (or ω(E)).
So, ~ how many accessible states are
there for a macroscopic system
(f ~ 1024) at energy E?
Not interested in exact results. We want an order of magnitude estimate!! The result is abstract, but very significant!! Let
Φ(E) ≡ Total # of quantum states
for all energies E´ ≤ E.

13. φ(ε) ≡ total # of quantum states for
Φ(E) ≡ Total # of quantum states
for all energies E´ ≤ E.
Consider 1 “typical” degree of freedom.
ε ≡ energy associated with that degree of freedom. Let
φ(ε) ≡ total # of quantum states for
this degree of freedom.
Generally, φ(ε) increases with increasing ε. So, we can write:
φ(ε)  εα (α ~ 1) (1)

14. Now, use (1), (2), & (3) together:
φ(ε)  ε (1)
For the system, replace energy E by an “average energy” for a system of f degrees of freedom:
ε  (E/f) (2)
f degrees of freedom &  φ(ε) states
associated with each.
Total # of states associated with f degrees of freedom ≡ product of # associated with each:
Φ(E)  [φ(ε)]f (3)
Now, use (1), (2), & (3) together:

15. Φ(E)  [φ(ε)]f (3) A = constant
So, we have:
φ(ε)  ε (1), and ε  (E/f) (2)
Φ(E)  [φ(ε)]f (3)
Using (1), (2), & (3) together:
 The total # of states for all energies E´ ≤ E is roughly: Φ(E)  [φ(ε)]f
So Φ(E)  [E/f]f or Φ(E)  Ef or
So finally: Φ(E) ≡ AEf (4)
A = constant

16.  Ω(E)  (Φ/E)δE  [(Ef)/E]δE
Ω(E) ≡ # accessible states with
energy between E & E + δE
So, we can write:
Ω(E) ≡ Φ(E + δE) - Φ(E); δE <<< E
 Expand Φ in a Taylor’s series & keep only the lowest order term:
 Ω(E)  (Φ/E)δE  [(Ef)/E]δE
Ω(E)  Ef-1δE  Ef δE (f >> 1)

17.  Ω(E) is an incomprehensibly
 Ω(E)  Ef δE (f ~ 1024)
 Ω(E) is an incomprehensibly
Rapidly Increasing
Function of E!!

18. Briefly look at some numbers with powers of 10 in the exponent to get a “feel” for the “bigness” of Ω(E)  Ef :

19. This is a 1 with 18 zeros after it!
Briefly look at some numbers with powers of 10 in the exponent to get a “feel” for the “bigness” of Ω(E)  Ef :
1. Approximate age of the Universe
in seconds  “only”  1018 s!
This is a 1 with 18 zeros after it!

20. 1. Approximate age of the Universe This is a 1 with 18 zeros after it!
Briefly look at some numbers with powers of 10 in the exponent to get a “feel” for the “bigness” of Ω(E)  Ef :
1. Approximate age of the Universe
in seconds = “only”  1018 s!
This is a 1 with 18 zeros after it!
2. Universe volume divided by the
volume of a grain of sand = “only”  1090!
This is a 1 with 90 zeros after it!

21. Ω(E)  Ef δE, Consider f ~ 1024:
By what factor does Ω (E) change when E changes by only 1%? Let
r ≡ [Ω (E E)/Ω (E)] = (1.01)f
Evaluate r using logarithms:
log10(r) = 1024log10(1.01)   1021
So, r  10x, where x  4.32  1021, or
r ~ 1 with 4.32  1021 zeros after it!

22. r ≡ [Ω (E + 10-6E)/Ω (E)] = (1.0 +10-6)f
Ω(E)  Ef δE, Consider f ~ 1024:
Consider the same problem again, but let E increase by only 10-6E. By what factor does Ω (E) change? Let
r ≡ [Ω (E E)/Ω (E)] = ( )f
Evaluate r using logarithms:
log10(r) = 1024log10( )  4.34  1017,
So r  10y, where y = 4.34  1017, or
r ~ 1 with 4.34  1017 zeros after it!

23. C = 10u, u = 1024 Are these similar numbers??
Finally, consider 2 seemingly similar numbers:
C = 10u, u = 1024
and D = 10v, v = 2  1024
Are these similar numbers??

24. Their ratio is: (D/C) = 10u
Finally, consider 2 seemingly similar numbers:
C = 10u, u = 1024
and D = 10v, v = 2  1024
Are these similar numbers?? NO!!!
Their ratio is: (D/C) = 10u

25. C = 10u, u = 1024 u = 1 with 1024 zeros after it!
Finally, consider 2 seemingly similar numbers:
C = 10u, u = 1024
and D = 10v, v = 2  1024
Are these similar numbers?? NO!!!
Their ratio is: (D/C) = 10u
 That is, D is 10u times (u = 1024!)
larger than C
u = 1 with 1024 zeros after it!

26. The bottom line is that: Ω(E) is

27. An Almost Incomprehensibly
The bottom line is that: Ω(E) is
An Almost Incomprehensibly

28. An Almost Incomprehensibly Enormously Rapidly Varying
The bottom line is that: Ω(E) is
An Almost Incomprehensibly
Enormously Rapidly Varying

29. An Almost Incomprehensibly Enormously Rapidly Varying
The bottom line is that: Ω(E) is
An Almost Incomprehensibly
Enormously Rapidly Varying
Function of E (!!)

30. Simple Special Case: Ideal, Monatomic Gas
To make this general discussion clearer, as a simple example, lets consider a classical ideal, monatomic gas, with N identical molecules confined to volume V.
Lets calculate Ω(E) exactly for this case.
Ideal Gas  NO INTERACTION between molecules.
~ Valid approximation for real gases
in the low density limit.

31. E  (2m)-1∑(i = 1,N)(pi)2, pi = 3d momentum of particle i.
Classical ideal, monatomic gas, N identical molecules in volume V. Calculate Ω(E) for this case.
Ideal Gas  No Interaction between molecules.
In this case, the total energy E of the gas is the sum of
the kinetic energies of the N molecules, each of mass m:
E  (2m)-1∑(i = 1,N)(pi)2,
pi = 3d momentum of particle i.

32. E  (2m)-1∑(i = 1,N)(pi)2, pi = 3d momentum of particle i.
Classical ideal, monatomic gas, N identical molecules in volume V. Calculate Ω(E) for this case.
Ideal Gas  No Interaction between molecules.
In this case, the total energy E of the gas is the sum of
the kinetic energies of the N molecules, each of mass m:
E  (2m)-1∑(i = 1,N)(pi)2,
pi = 3d momentum of particle i.
Ω(E) ≡ # of accessible states in the interval E to E + δE
Ω(E) ≡ # of cells in phase space between E & E + δE.
Ω(E)  volume of phase space between E & E + δE.
In what follows, recall the 1d oscillator where
Ω(E) = area between 2 ellipses in an abstract
p-q plane (phase space)!

33. 2mE = ∑(i = 1,N)∑(α = x,y,z) (piα)2
Ω(E)  volume of phase space between E & E + δE.
2mE = ∑(i = 1,N)∑(α = x,y,z) (piα)2
The energy is independent of the particle positions!
piα = α component of momentum of particle i.
Ω(E)  ∫(E  E + δE) d3r1d3r2…d3rNd3p1d3p2…d3pN
A 6N dimensional volume integral!

34.  Ω(E)  VN ∫(E  E + δE) d3p1d3p2…d3pN
Ω(E)  volume of phase space between E & E + δE.
2mE = ∑(i = 1,N)∑(α = x,y,z) (piα)2
The energy is independent of the particle positions!
piα = α component of momentum of particle i.
Ω(E)  ∫(E  E + δE) d3r1d3r2…d3rNd3p1d3p2…d3pN
A 6N dimensional volume integral!
The limits E & E + δE are independent of the ri’s
 The position integrals for each ri
can be done immediately:
∫d3ri ≡ V  ∫d3r1d3r2…d3rN ≡ VN
 Ω(E)  VN ∫(E  E + δE) d3p1d3p2…d3pN

35.  Ω(E)  VN ∫(E  E + δE) d3p1d3p2…d3pN (1)
A 3N dimensional volume integral in p space
Consider the sum
2mE = ∑(i = 1,N)∑(α = x,y,z) (piα)2 (2)
(2) ≡ “sphere” in 3N dimensional momentum space.

36.  [R(E)]3N  (E)(3N/2)  Ω(E)  VN ∫(E  E + δE) d3p1d3p2…d3pN (1)
A 3N dimensional volume integral in p space
Consider the sum
2mE = ∑(i = 1,N)∑(α = x,y,z) (piα)2 (2)
(2) ≡ “sphere” in 3N dimensional momentum space.
Briefly consider the case of 1 particle only. (2) is:
2mE = (px)2 + (py)2 + (pz)2
This is a “sphere” in momentum space of “radius”
R(E) = (2mE)½
For 1 particle, the 3d “sphere volume”  [R(E)]3  (E)(3/2)
For N particles in 3N dimensional momentum space,
(2) ≡ a “sphere” of “radius” R(E) = (2mE)½
So, the 3N dimensional “sphere volume” is
 [R(E)]3N  (E)(3N/2)

37. This is shown schematically for 2 dimensions in the figure:
Lets write: Ω(E)  VNG(E)
where: G(E) ≡ ∫(E  E + δE) d3p1d3p2…d3pN (3)
G(E) ≡ Volume of “spherical shell” between E & E + δE
This is shown schematically for 2 dimensions in the figure:
G(E)  [R(E)]3N  (E)(3N/2)
 Ω(E)  VN(E)(3N/2)
Write: Ω(E) = BVN(E)(3N/2)
B = constant, Ω(E) = # of accessible states for an ideal
gas in the energy interval E to E + δE

38. Ω(E) = BVN(E)(3N/2)δE In summary, for the Ideal Gas, we found:
= # accessible states of an ideal gas in the energy interval E to E + δE, B = constant

39. Ω(E) = AEf δE, A = constant
In summary, for the Ideal Gas, we found:
Ω(E) = BVN(E)(3N/2)δE
= # accessible states of an ideal gas in the energy interval E to E + δE, B = constant
In general, we found for f degrees of freedom,
Ω(E) = AEf δE, A = constant
For the ideal gas, f = 3N, so we got an E dependence of E(½)f instead of Ef, but, again, all of this was
Approximate & Order of Magnitude.
So, no worries about the difference between f & (½)f.