2 States of a Systemcosmology.berkeley.edu/Classes/S2012/Physics_112/lectures/chapter_2_quantum_states_an/...If magnetic field oriented along +/- direction, the energy = -/+mB (m=magnetic

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


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

∑


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


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( )


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


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


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∫


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( )


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


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


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


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


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


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

.......


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


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


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


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


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


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

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

Documents

2 States of a Systemcosmology.berkeley.edu/Classes/S2012/Physics_112/lectures/chapter_2_quantum_states_an/...If magnetic field oriented along +/- direction, the energy = -/+mB (m=magnetic