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B.SadouletPhysics 112 (S2012) 02 States-Entropy 1
2 States of a System
Mostly chap 1 and 2 of Kittel &Kroemer
2.1 States / Configurations2.2 Probabilities of States
• Fundamental assumptions• Entropy
2.3 Counting States2.4 Entropy of an ideal gas
B.SadouletPhysics 112 (S2012) 02 States-Entropy 2
System States, Configurations
Microscopic: each degree of freedom
StatisticalMechanics
Moments
Thermodynamics
MacroscopicVariables
State ConfigurationState = Quantum State
Well defined, uniqueDiscrete ≠ “classical thermodynamics” where entropy was depending on
resolution ΔE
Configuration = Macroscopic specification of systemMacroscopic Variables Extensive: U,S,F,H,V,N
Intensive: T,P,µ
Not unique: depends of level of details neededVariables+ constraints=> not independent
Classical qi , pi =∂L
∂ ˙ q iQuantum state #
f (qi , pi )
< qi > , < pi >
U = Uii=1
n
∑T =
23kB
1N
Uii=1
n
∑
B.SadouletPhysics 112 (S2012) 02 States-Entropy 3
Quantum Mechanics in 1 transparencyFundamental postulates
=> A finite system has discrete eigenvalues
State of one particle is characterized by a wave function
Probability distribution =ψ x( ) 2 with ψ ψ ≡ ψ x( )∫ ψ x( )dx = 1
Physical quantity ↔ hermitian operator.
In general, not fixed outcome! Expected value of O = ψ O ψ ≡ ψ x( )∫ Oψ x( )dxEigenstate ≡ state with a fixed outcome e.g., O |ψ >= o |ψ > where o is a number.
i ∂∂t
=E
− i ∂∂x j
=pj
⇔ "State" of well defined momentum
ϕ(x,t) = 12π2
⎛⎝⎜
⎞⎠⎟
3/2
e− i Et− p⋅ x
⎛⎝⎜
⎞⎠⎟
E =p2
2M => −
2
2M⎛⎝⎜
⎞⎠⎟∇2Ψ = i ∂
∂tΨ Eigenvalues of energy ε: − 2
2M⎛⎝⎜
⎞⎠⎟∇2Ψ = εΨ
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Graphic Example
1 dimensional case, infinitely deep square well
4
ψ = Asin nxπ xL
⎛⎝⎜
⎞⎠⎟= A
exp i nxπ xL
⎛⎝⎜
⎞⎠⎟ − exp −i nxπ x
L⎛⎝⎜
⎞⎠⎟
2i ψ 2 dx = 1⇒ A =
2L0
L
∫
Solution of the Schrödinger equation − 2
2M⎛⎝⎜
⎞⎠⎟∇2Ψ ≡ −
2
2M⎛⎝⎜
⎞⎠⎟∂2ψ∂x2 = εΨ
with the proper boundary conditions ⇒ ε = 2
2M⎛⎝⎜
⎞⎠⎟nxπL
⎛⎝⎜
⎞⎠⎟
2
− i ∂∂x
ψ =pψ ⇒ superposition of 2 States of well defined momentum p = ±
nxπL ε = p2
2M
ψ
L
ψ 2
2 4 6 8 10
�0.4
�0.2
0.0
0.2
0.4
0 2 4 6 8 100.00
0.01
0.02
0.03
0.04
0.05
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Ψ (x, y,z) = Asin nxπxL
⎛ ⎝ ⎜ ⎞
⎠ ⎟ sin
nyπyL
⎛ ⎝ ⎜
⎞ ⎠ ⎟ sin nzπz
L⎛ ⎝ ⎜ ⎞
⎠ ⎟
nx ,ny ,nz integers > 0
5
Quantum StatesPrototypes:
• Atomic levels cf Fig 1.1 Kittel and KrömerHave to take into account multiplicity if we consider degenerate states (i.e. have the same
energy)
• Spin s in general 2s+1 states (exception photons s=1 but 2 states) Spin 1/2 s =±1/2 If magnetic field oriented along +/- direction, the energy = -/+mB
(m=magnetic moment)Note: Idea of an isolated system of spins is somewhat strange <= transition between higher
and lower energy states. Isolated = electromagnetic emission reabsorbed or reflected back
• Ideal gas of particlesFree particle propagating in space: orbital ψ(x) ( motion part of the wave function)Single particle in a box : Cubic side L cf. K&K p.72
Ideal (≈ non interacting): orbitals not distorted by presence of N particles. Wave function = product of single wave functions. Can bounce against each other provided interactions are short!
n=1
n=2
εnx ,ny ,nz =
2
2M⎛
⎝ ⎜
⎞
⎠ ⎟
πL
⎛ ⎝ ⎜ ⎞
⎠ ⎟ 2nx2 + ny2 + nz2( )
B.SadouletPhysics 112 (S2012) 02 States-Entropy 6
Fundamental Postulate
Probabilistic description of state of a systemBecause of microscopic processes, the system experiences slight
fluctuations: described by probability of being in state i.
Equilibrium = No net Flux =>Stationary No evolution of probability distribution with time
Isolated = closed : No energy/particle exchange with outside
volume constraint OK
An isolated system in equilibrium is equally likely to be in any of its accessible states Notes:
• Kittel does not specify "in equilibrium" does not matter in case where we are close to equilibrium
• "Accessible" e.g. Conservation of energy +Some states may not be accessible within time scale of experiment
B.SadouletPhysics 112 (S2012) 02 States-Entropy
ConsequencesProbability of a configuration (isolated, in equilibrium)
If we call g the number of states for a given macroscopic specification of the configuration, and gt the total number of states accessible to the system
This allow us to compute probabilities of configurations!
This computation method is conventionally called a “microcanonical” computation
Prob(configuration)=Number of states in configuration
Total number of states= ggt
B.SadouletPhysics 112 (S2012) 02 States-Entropy 8
RemarksProbabilistic description of state of a system
Two possible definitions:• Observations of different identical systems at a given time
"ensemble" of systems j• Observations at different times, system wanders over its
accessible states
Ergodicity: under very general conditions, we obtain identical results
The fundamental postulate can be demonstrated in Quantum Mechanics
Celebrated H theorem
< y >ensembles=< y >time≡T→∞lim 1
Ty(t)dto
T∫
B.SadouletPhysics 112 (S2012) 02 States-Entropy 9
EntropyEntropy Definition
≠ Kittel (ours is more general)
For an isolated system in equilibrium: identical to Kittel
where gt is the total number of accessible quantum states in the configuration
Notes: • In classical thermodynamics, definition usually used
where kB is the Boltzmann constant (uses implicitly 3rd law of thermodynamics S=0 at T=0)
Entropy requires the use of quantum mechanics: it involves the Planck constant
σ = − psstates s∑ log ps = −H
σ = log gt( )
S = kBσ =dQT∫ ⇔σ =
dQkBT∫ =
dQτ∫ with τ = kBT kB = 1.38 ⋅10−23J / K
H="Negentropy" = "Information" of Shannon
ps = 1 + equiprobable ⇒
s∑ ps =
1gt
⇒σ = log gt( )
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Theorem: An isolated system evolves towards a configuration where all the states are equiprobable and the entropy is maximum
Based on the fact that probability of transition per unit time from r to s is equal to probability of transition per unit time from s to r
Suppose we have initially Nr particles in r and Ns<<Nr in r, the number of particles going from r to s in time Δt is much bigger than from s to r.
Therefore Nr decreases and Ns increases, till they are in average equal. The probabilities of s and r are then equal!
10
H theorem: an intuitive look
Γ rs = Γ sr
NrΓ rsΔt >> NsΓ srΔt
r s
NrΓ rsΔt
NsΓ srΔt
B.SadouletPhysics 112 (S2012) 02 States-Entropy 11
H theorem (Optional)Modern Version: see Reif Appendix A12
Consider an isolated system, and quantum states s. The probabilities of occupancy are such that
Quantum mechanics: Probabilities per unit time of transition between states r and s are symmetric
Starting from the detailed balance argumentConsider now We have
or equally well
Add the two quantities
Equality holds if : => at equilibrium quantum states have equal probabilities!
pspss∑ = 1
Γrs = Γsr ⇐∝ sW r2
dprdt
= psΓsrs∑ − prΓrs
s∑ = Γsr ps − pr( )
s∑
H ≡ ps log ps( )s∑ = −σ
dHdt
=dprdtr
∑ log pr( ) +1( ) = Γsr ps − pr( )s∑
r∑ log pr( ) +1( )
dHdt
= Γrs pr − ps( )r∑
s∑ log ps( ) +1( ) = − Γsr ps − pr( )
s∑
r∑ log ps( ) +1( )
dHdt
= −12
Γsr ps − pr( )s∑
r∑ log ps( ) − log pr( )( )
If ps > pr , log pr( ) > log ps( ) and vice versa + Γsr > 0 ⇒dHdt
≤ 0 dσdt
≥ 0
ps = pr
Opt
iona
l
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Remarks
12
Γ rs = Γ sr
Probability of transition per unit time state r→ state s = Prob of transition state s → state r
This proof is in ReifRelated, as Boltzmann already knew, to the symmetry of transition
probabilities
Quantum Mechanics -> Statistical mechanics if states loose their coherence (generally true)
=>Quantum mechanics is compatible with statistical mechanics!
Simulation (5 States ): http://cosmology.berkeley.edu/Classes/S2009/Phys112/Htheorem/Htheorem.html Diffusion in “space” state
B.SadouletPhysics 112 (S2012) 02 States-Entropy 13
Counting States: Discrete StatesPreliminaries
Number of permutations between N objects = factorial
Stirling approximation
Independent spins 1/2N spins, each of them has two states (up, down)If we define a configuration by the number of spins up, the number of
states in the configuration
We could have instead chosen to label the configuration by the difference (proportional to total total spin): suppose N is even
N!= N(N −1)(N − 2)...2.1 = Γ (N +1)
logN!N→∞≈ N logN − N +
12log 2πN( )
Total spin s = nup − ndown( ) 1
2⇒ nup = N / 2 + s ndown = N / 2 − s
g(s) = N !N2+ s⎛
⎝⎜⎞⎠⎟ !
N2− s⎛
⎝⎜⎞⎠⎟ !
→N→∞
2N 12πN / 4
exp −12
s2
N / 4⎛⎝⎜
⎞⎠⎟
Gaussian!
(from Sterling approximation)
g(nup ,ndown ) =N N −1( ).. N − nup +1( )
nup !We do not care about order!
number of ways to choose nup objects
=N !
nup !ndown ! with ndown = N − nup
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Density of spatial states per unit phase spacePhase space element for a single particle in 3 dimensions: Theorem: the density of spatial states (orbitals) per unit phase space for a
single particle in 3 dimensions is 1/h3 = density of quantum states for a spinless particleProof: Not in book! Consider a particle in a box. Its spatial wave function is
What is the x component of the momentum? It is not well defined!
Intuitively the particle is going back and forth.This is a direct consequence of the Heisenberg Uncertainty PrincipleConstraining the particle in space, you cannot have a well defined momentum!
14
Counting States: Particlesd3x ⋅d3p
�
Ψ(x,y,z) = Asin nxπxL
⎛ ⎝ ⎜
⎞ ⎠ ⎟ sin
nyπyL
⎛
⎝ ⎜
⎞
⎠ ⎟ sin nzπz
L⎛ ⎝ ⎜
⎞ ⎠ ⎟ with nx ,ny ,nz integers > 0
�
nxπL
and - nxπL
Applying p = −i∇ one sees this is the superposition of 2 momenta
sin nxπxL
⎛⎝⎜
⎞⎠⎟=einxπxL − e
− inxπxL
2i
B.SadouletPhysics 112 (S2012) 02 States-Entropy
ProofLet us work first in one dimension: Consider the x direction,
For all states, the position coordinate ranges over Let us start with nx =1. The x momentum spans
The x component of the phase space volume spanned by the nx =1 state is therefore
Increasing nx by 1 will add one more state and increase the momentum space spanned by the phase space volume by an additional
and so on. Therefore the number of states per unit phase space is h.This is a totally general and exact result: the phase space density is 1/h for each of
the dimensions!In 3 dimensions the phase space volume occupied by each of the states
=> the phase space density of states is 1/h3
15
�
Δpx = πL
positive px
− − πL
negative px
= 2 πL
ΔxΔpx = L × 2 π
L= h
�
2 πL
�
Lpx
xΔx = L
ΔxΔpx = L × 2 π
L= h
nx = 1,ny = 1,nz = 1nx = 2,ny = 1,nz = 1nx = 1,ny = 2,nz = 1nx = 1,ny = 1,nz = 3
is h3
.......
B.SadouletPhysics 112 (S2012) 02 States-Entropy 16
Counting States: Particles
Proof à la KittelNot explicitly in book but many such types of calculation through outConsider again a particle in a cubic box, and compute number of states
between E and E+dE = number of integers nx,ny,nz such that
In the nx,ny,nz space, each state correspond to a volume of unityFor nx,ny,nz large enough, the number of states is given by the volume of
one quadrant of spherical shell of radius
Volume of phase space element such that the energy is between E and E+dE is
E ≤
2
2M⎛
⎝ ⎜
⎞
⎠ ⎟
πL
⎛ ⎝ ⎜ ⎞
⎠ ⎟ 2nx2 + ny2 + nz2( ) < E + dE
n = nx
2 + ny2 + nz
2 =L 2ME
π and thickness dn =
L 2ME
2πdE
L3 4πp2dp with E =p2
2Mis π 2L3 2M( )3/2 EdE
⇒# States
Phase space volume=
18π33 =
1h3
# of states =
184πn2dn = 2π
L3 2M( )3/2 E8π33
dEquadrant
nz
nxny
nContinuousapproximation
Hom
ewor
k
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Now consider N particles in weak interactions
Calculation of number of states as a function of UN particlesWeak interactions => # states =
with the constraint that total energy is U
Space integral
Momentum integral: we need to conserve energy1 particle in 1 dimension: only 1 momentum state with right energyin 2 dimensions circumference of circle
in 3 dimensions area of 2-sphere
N particles with 3 momentum components area of (3N-1)-sphere <= 3 xN dimensions in total
17
Ideal Gas
g =d3xid
3pih3∫i∏
pij2
2Mj=1
3
∑part i
N
∑ =U where M is the mass
d3xi∫i∏ = VN
p12 + p2
2 = 2MU ⇒ δ p12 + p2
2 − 2MU⎛⎝⎜
⎞⎠⎟d2pi = 2π 2MU∫
p12 + p2
2 + p32 = 2MU ⇒ δ p1
2 + p22 + p3
2 − 2MU⎛⎝⎜
⎞⎠⎟d3pi = 4π 2MU( )2∫
δ pij2
j=1
3∑
part i
N∑ − 2MU⎛
⎝⎜⎞⎠⎟
d 3pipart∏ ∝U
3N −12
for large N g ∝V NU3N2
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Ideal Gas lawWhy is it important?
let us assume that the entropy we have defined is the entropy of thermodynamics
From previous slide
18
σ = − pii∑ log pi = log gt( ) = S
kbDefine τ = kbTThe Thermodynamic identity First law( ) is then dU = TdS
quantity of heattransfered to system
− pdVwork by pressure on the system
(no change of number of particles N)
≡ τdσ − pdV ⇒τ =∂U∂σ V ,N
p = −∂U∂V σ ,N
σ = log V NU 3/2N( ) + const.⇒U = Aexp 23N
σ⎛⎝⎜
⎞⎠⎟V −2 /3 (A is a constant)
τ =∂U∂σ V ,N
=23N
U ⇔U =32Nτ ≡
32Nk bT indeed looks like the temperature
p = −∂U∂V σ ,N
=23UV
⇔ pV = Nτ ≡ Nk bT indeeds looks like the pressure
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Our and Kittel’s approach
We will define in chapter 3
and show that these definitions behaves as the temperature and pressure
19
1τ=∂σ U,V ,N( )
∂Upτ=∂σ U,V ,N( )
∂U
⇒ dσ ≡1τdU +
pτdV ⇔ dU ≡ τdσ − pdV
B.SadouletPhysics 112 (S2012) 02 States-Entropy 20
Ideal Gas2 technical notes (optional)
1) Exact formula: It can be shown that the surface area of 3N-1 sphere of radius r is:
See e.g. Wolfram site: http://bbs.sachina.pku.edu.cn/Stat/Math_World/math/h/h456.htm or http://en.wikipedia.org/wiki/Hypersphere
2) If you follow carefully the dimensions, we need a factor h/L
We can compute (painfully) the number of states and then get probabilities: “Microcanonical” methods
g = hLδ pij
2
j=1
3∑
part i
N∑ − 2MU⎛
⎝⎜⎞⎠⎟
d3xid3pi
h3i∏∫ =
V N −1/3 2MU( )3N −1
2 Ω3N
h3N −1
r3N −1Ω3N where the solid angle factor is
Ω3N =2 π( )
3N2
Γ 3N2
⎛⎝⎜
⎞⎠⎟=2(π )
3N2
( 32N −1)!
with
m2
⎛⎝⎜
⎞⎠⎟!= m
2⎛⎝⎜
⎞⎠⎟m2−1⎛
⎝⎜⎞⎠⎟... 1( ) m even
m2
⎛⎝⎜
⎞⎠⎟!= m
2⎛⎝⎜
⎞⎠⎟m2−1⎛
⎝⎜⎞⎠⎟... 12
⎛⎝⎜
⎞⎠⎟π 1/2 m odd
δ functions have a dimension! δ x( )∫ dx = δ f x( )⎡⎣ ⎤⎦∫
dfdx
dx = 1
Opt
iona
l
B.SadouletPhysics 112 (S2012) 02 States-Entropy
Hom
ewor
k
21
Sackur Tetrode FormulaWe are now in position to compute the entropy
we could try
Only problem: violently disagrees with experiment!Solution: In quantum mechanics, particles are indistinguishableWe have over-counted the number of particles by N! (Gibbs)
Sackur Tetrode Monoatomic, spinless: cf Kittel chapter 6
Writing n = N
V,nQ =
2πMh2
2U3N
⎛⎝⎜
⎞⎠⎟3/2
we get σ =SkB
= N lognQn
⎛⎝⎜
⎞⎠⎟+52
⎡
⎣⎢
⎤
⎦⎥
g = 1N !
hLδ pij
2
j=1
3∑
part i
N∑ − 2MU⎛
⎝⎜⎞⎠⎟
d3xid3pi
h3i∏∫ =
V N −1/3 2MU( )3N −1
2 Ω3N
N !h3N −1
For large N we can use the Stirling approximation both for integer and half integer
log m2
⎛⎝⎜
⎞⎠⎟!≈ m
2log m
2⎛⎝⎜
⎞⎠⎟−m2
and we get
σ (U,V ,N ) = logg N→∞⎯ →⎯⎯ N log 2πMh2
2U3N
⎛⎝⎜
⎞⎠⎟
3/2⎡
⎣⎢⎢
⎤
⎦⎥⎥+ N log V
N+ 5 / 2⎛
⎝⎜⎞⎠⎟
σ = logg = logV N −1/3 2MU( )
3N −12 Ω3N
h3N −1
⎡
⎣
⎢⎢
⎤
⎦
⎥⎥