Contents

Chapter 1
Prerequisites

1.1 Prerequisites of Thermodynamics

(I) THU 29/09/2022
In thermodynamics there are extensive quantities, that grows with the system size, and intensive quantities, which does not. Conjugation between them, means tuning intensive variables to make extensive one change.

Note that we always refer to quasi-static transformations

1.1.1 The laws of thermodynamics

0th Law Equilibrium (empirical) temperature.

We call: _A : space of state of A _B : space of state of B.
The total phase space is _A ×_B = {(a,b)}, where (a,b) denotes all possible couples.

At equilibrium, not all couples are possible. At equilibrium:

so there is a constraint. This is an equivalence relation:

This condition define the empirical temperature t_A = t_B at equilibrium.

The transitive property allow us to choose anything as a measure (e.g. thermometer).

1st Law Internal energy E

There is an internal energy, which can change in different ways, but it always conserved (conservation of energy).

where δQ is heat, δ is work, N is number of particles and μ is the chemical potential: the energy needed to add/remove a particle.

d means that ∮ dE = 0 on any cycle, so the integral does not depend on the path. δ no, so δQ,δ is not necessary 0.

E.g. Classical fluids:
δ = pdV : the work derive from the compression/expansion. So the variation of the energy: dE = δQ - pdV + μdN.

2nd Law Entropy S

In a reversible process:

Putting the two laws together,

So, a system evolves towards the maximum of entropy

3rd Law For any isothermal process, ΔS0

1.1.2 Thermodynamic potentials (for reversible processes)

Let’s see the relation between these quantities:

Internal energy dE = TdS - pdV + μdN
The internal energy can be changed by changing S,V,N so E = E(S,V,N) and:

All terms in couple are conjugate variables: T ↔ S,p ↔ V,μ ↔ N

E,S,V,N are extensive variables, so if

where λ is a scaling factor.

So:

Homogeneous function of degree 1 (linear)

That requires: E(S,V,N) = TS - pV + μN dE = TdS - pdV + μdN

Usually it’s easier to work with T then S (there are no experiment where you can tune the entropy S), so thermodynamic potentials are introduced. These are other function which are more convenient to work with:

• Internal energy:
  E(S,V,N) = TS - pV + μN dE = TdS - pdV + μdN

• Helmotz free energy:
  F(T,V,N) = E - TS = -pV + μN dF = -SdT - pdV + μdN

• Entalpy:
  H(S,p,N) = E + pV = ST + μN dH = TdS + V dp + μdN

• Gibbs free energy:
  G(T,p,V ) = E - TS + pV = μN dG = -SdT + V dp + μdN

• Granpotential:
  Ω(T,V,μ) = E - TS - μN = -pV dΩ = -SdT - pdV - Ndμ

1.1.3 Thermodynamic limit

N,V →∞ with n = N∕V fixed In this limit, all the thermodynamic potential diverges, so (e.g) E has no meaning. What we can calculate is the ratio with the number of particle (e.g E∕N : internal energy for particle). Things do not change because they depend on N, except for Ω which doesn’t.

1.1.4 Equation of state - μ-n relation

From the granpotential, one can derive p = -, which, written for a particular model (e.g. classical gas) gives the equation of state of the system/model.

Also, the μ-n relation is very important: N = -

1.1.5 Variational principles

(II) MON 03/10/2022
Suppose we set up an experiment in which S,V,N are constant. Then dE = 0, which implies that the system evolves towards the minimum of energy E.
The same happen if T,V,N are constant, in which case dF = 0, so the system evolves towards the minimum of Helmotz free energy. Etc... .

So the problem becomes a problem of minimization.
We’ll see that the minimization of F and Ω are the most important.

For reversible processes, dF = -SdT + pdV + μdN, from which:

The minimum is found with the hessian matrix:

In fact, the condition to have a minimum is that all the 3 eigenvalues of the hessian matrix should be > 0.

That is the same as saying that going in every direction the ”gradient” increase

We won’t prove this, but because of the way F = F(T,V,N), it us sufficient that

The same can be done changing the variables and it also works with other different from F (e.g. E etc.)

1.2 Prerequisites of Classical Mechanics

To represent a classical system we need:

1) Coordinates: position q, momenta p.
For 1 particle. If a particle lives in IR^d ∈ IR^d ∈ IR^d.
So the phase space ₁ is {(q,p)}∈ IR^d × R^d = 2IR^d

For N particles living in IR^d, the phase space is:

2) Observable: An observable is given by a smooth real function from the phase space to a real number.

e.g. , angular momentum, kinetic energy T are all observables

3) Measure: (ideal, without errors) A measure of an observable on a state (q_ip_i) is given by the value of the function at that particular point: f(q₁,p₁)

4) Evolution: The evolution of a system is fixed by a special observable, called Hamiltonian , through Hamilton’s equation of motion:

(1.1)

In general, = (q_i,p_i,t), but we only study the case where the hamiltonian is time-independent.

The equation 1.1 are equation of the first order in t (q_i(t)p_i(t)), so the solution is uniquely fixed by the initial conditions (q_i,p_i). Because of that we have two theorems:

Conservation of volumes (even id the shape is changed). Since each point of the phase space is a state, the volume counts the number of states. In other words, giving a volume is the same as giving a subset.

Conservation of energy. If does not depend explicitly on time, the hamiltonian is constant for each curve of motion:

1.2.1 On the notion of ”state”

We will call one of these states (q_i(t),p_i(t)) a microstate at time t. Fixing an initial microstate, completely determine the trajectory in phase space, backward and forward in time.

If the initial conditions are changed a little, allowing the system to ”choose” them between a given set, we enter the field of complex systems

In general, there is a (huge) number of possible microstates corresponding to the same macroscopic set of thermodynamic variables (the macrostate).

In order to study this, we can think of having an ensemble: a large number of copies of the system, all with the same Macroscopic State but with different microscopic realizations.

1.2.2 Probability distribution

Given an ensemble, in the limit in which the number of copies becomes very large, we can construct the probability with which, at a fixed time, a given microstate {q_i(t),p_i(t)} appears, thus recovering a probability density distribution on _N:

Which is:

• Positive: ρ(p_i(t),p_i(t)) ≥ 0

• Normalized:

  

  (implicit vector used: dq_i,dp_i should be d³ _i,d³ _i)

• For a subset U ⊂ _N, the probability to find the system in the subset U is: ∫ _Uρ(q_i,p_i)dΓ

To have an a-dimensional quantity, we replace:

where h is a constant with the dimension of an action (here we don’t say anything about it, it’s just the dimension of a little state of sides dq_i,dp_i but in quantum mechanics, it will be Plank’s constant).

1.2.3 Liouville’s theorem

The conservation of volumes is the Liouville’s theorem.
Given a region Ω₀ in which there is some density probability ρ(q_i,p_i) ∈ Ω₀, there is a density current of particles moving out of Ω₀:

If there is a conservation of the total probability density: + ( ) = 0 That means

which is the Liouville’s theorem

We can also use the Poisson’s bracket’s to write (ρ) = {ρ,},
reminding that {f,g} = ∑ _i

Definition: A system is called stationary iff = 0. Stationarity is a necessary condition for equilibrium: = 0, which is what we want to study. From the previous condition follows that at equilibrium: {ρ,} = 0 which can happen if

1.

ρ = const

2.

ρ = ρ((q_i,p_i))

giving, respectively, the cases of the microcanonical and canonical/grancanonical distributions.

Definition: Given an observable f, we can define its:

average	⟨f⟩ρ = ∫ Nf(qi,pi)ρ(qi,pi)
standard deviation	(Δt)2 = ⟨f2⟩ ρ - (⟨f⟩ρ)2

The subscript _ρ, indicates that we need to have defined a probability density distribution, to evaluate these two.

1.2.4 Time-Independent Hamiltonians

For time-independent hamiltonians, the energy is conserved. That means that all the trajectories will be on a (2M - 1)hypersurface in the phase space (where M is the total number of degree of freedom). The whole space can thus be foliated into different sheets according to different energies like in the figure.

So we can compute every integrals integrating firstly over an hypersurface of constant energy and then over all the possible energies:

Definition: Volume (~ number of state) in phase space with energy lower than a certain value (0 ≤(q_i,p_i) ≤ E):

Definition: The density of state is the ”area” of the hypersurface S

Σ,ω aren’t property of the space, but they depends on because the surfaces depend on .

In the case of time-independent hamiltonians:

	E ≡ ∫ SEdSEf
time average	∞ ≡ lim T→∞∫ t0t0+T dtf(q i(t),pi(t))

the latter exists for almost all initial conditions and it is independent of t₀

Definition: A system is said to be ergodic over the surface S_E iff almost all points (q_i,p_i) ∈ S_E pass through a neighbourhood U ∈ S_E during the evolution. In other words, starting from any point, if you let the system evolve for a long time, every region will be explored.

Theorem: A system is ergodic iff, for almost all initial points: _∞ = _E
This tell that the time spent on a region is proportional to the area of that region.

In the following we will consider only ergodic systems.

Chapter 2
Classical Statistical Mechanics

2.1 Microcanonical Ensemble

06/10/2022
Recap:

Phase space	N = {(qi,pi) i = 1…N}
Hamiltonian	(qi,pi) : N → IR

The microcanonical ensemble represents a closed system that doesn’t exchange either energy or matter with the environment, so in which the energy E, the number of particles N and the volume V are fixed. The evolution occurs on the hypersurface S_E ⊂_N (q_i,p_i) = E

To describe the system we need the probability distribution of the microcanonical ensemble. We assume (a priori) a uniform probability:

Definition: The probability distribution of the microcanonical ensemble is:

where C is a constant we can obtain from the normalization:

(reminding that ω(E) is the area of S = E). So:

(2.1)

Working with S is difficult (e.g. it creates some problems when integrating), so we give an operative definition. We can write the volume of a phase space:

from which,

So we can see that the microcanonical probability density function (2.1) is the limit for ΔE → 0 of:

2.1.1 Microcanonical entropy S_mc

We will see that it coincide with the thermodynamic entropy.

1) We could actually define the entropy in three different ways:

1.

S_mc⁽¹⁾ = k _B log ω(E)

2.

S_mc⁽²⁾ = k _B log Γ(E)

3.

S_mc⁽³⁾ = k _B log Σ(E)

which in general are different quantities. However in the thermodynamic limit (N,V →∞, with n = N∕V const), they represent the same quantity:

So while doing calculations, we can use one of the three.

2) Entropy should be extensive, so it should be additive: S_mc^A + S _mc^B = S _mc^A∪B

Proof: To prove it, let’s call the phase space of the two systems:
₁,₂. So _AB = ₁ ×₂

If = ₁ + ₂ E = E₁ + E₂ + E_int.
Here we make an assumption: there could be interactions in the wall that divides the two systems, however, since E₁ and E₂ both scale with the volume (L³) while E _int scale like the area of interaction (L²), we neglect the interaction term, because it disappears in the thermodynamic limit.

We can foliate the phase space with surfaces each at different energy

ω(E)	= ∫ 1×2dΓδ(- E)
	=  δ(1 + 2 - E)
	= ∫ d1 ∫ d2δ(1 + 2 - E) ∫ 1=E1ds1 ∫ 2=E2dS2
	= ∫ d1 ∫ d2δ(1 + 2 - E) ω1(E1) ω2(E2)
	= ∫ 0EdE 1 ω1(E1) ω2(E2 = E - E1)

The integrand is ≥ 0 and defined in a compact interval [0,E), so it has a maximum. Let E₁^*,E ₂^* = E - E ₁^* be the value of energy for which the integrand is maximum. So:

and for ΔE small enough:

Reminding that Γ(E) ~ ω(E)Δ(E):

Γ1(E1*)Γ 2(E2*) ≤ Γ(E)	≤Γ1(E1*)Γ 2(E2*)
log Γ1 + log Γ2 ≤ log Γ	≤ log Γ1 + log Γ2 + log (×kB)
Smc1 + S mc2 ≤ S mc	≤ Smc1 + S mc2 + k B log

In the thermodynamic limit, dividing everything by N and letting N →∞, the last term approaches zero and can be neglected. So we find:

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3) In the thermodynamic limit, this entropy coincides with the thermodynamic entropy: s_mc = s_th

Proof: At equilibrium, E = E₁^* + E ₂^*, where ω ₁(E₁) ω₂(E - E₁) is maximized.

Γ(E1*)Γ(E 2*) → dE1=E1*,E2=E2*	= 0
E 1*,E2*	= 0
(dE2 → dE1)

E 1*	= E 2*
E 1*	= E 2*
E 1*	= E 2*

since this is calculated at E₁^*,E ₂^*, it is calculated at the equilibrium (N,V fixed).

The same thing happens in thermodynamic: if T₁ = T₂,

So the equilibrium in thermodynamics requires this condition:

So we find that:

(equal and not only proportional, because k_B is the constant appositely chosen to make them equal)

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4) We saw that

We can now prove the Universal Boltzmann’s formula (works in the TD-limit):

Proof: Working from the Universal Boltzmann’s formula, we can arrive to the definition of entropy

- kBmc	= -kB ∫ NdΓ(log ρmc)ρmc ρmc = δ(- E) ρmc|Sc =
	= -kB ∫ SE(- log ω(E))
	= kB log ω(E) ∫ SdSE
	= kB log ω(E) = Smc

So we have proved that, in the TD-limit:

2.1.2 One example/exercise (1.1, 1.2): Perfect gas

We consider a gas of N non-relativistic and non-interacting monoatomic particles in 3D, confined in a volume V .

The volume of states is:

where the integral is calculated on:

So it is the volume of a 3N-dimensional sphere of radius . So:

Σ(E)	= N ∫ 0≤∑ i2≤2mE ∏ i i2
	= Ω3N(R = )
	= (R = )3N

where Γ(x) = ∫ ₀^∞dt t^x-1e^-t is the Euler’s Γ-function, which can be seen as a generalization of the factorial. In fact, one can prove (integrating by parts) that
Γ(n) = (n - 1)! Γ(x + 1) = xΓ(x) log Γ(x) ≃ x log x - x

The previous formula is composed by an angular part (the Euler’s Γ-function) and a radial part (which scale with 3N). So finally we have that:

Calculating the derivative, one can also get the density of the states and dΓ:

and we can check that in the TD-limit, the different ways of defining the entropy are the same:

where the terms can be cancelled in the limit of N →∞ (for the last one, since E ~ N, the ratio 3N∕2E is constant).

We can then find the entropy for a perfect gas (supposing distinguishable particle)

Sdis	= kB log Σ(E)
	= kB	(TD - limit)
	= kB	(Stirling)
	= kB

(III) MON 10/10/2022
This formula has a problem: it is not extensive: if N → 2N and V → 2V we expect S → 2S, but we actually have S → 2S + N log 2.

The solution is to consider indistinguishable particles. That means that the states: (q₁,p₁,q₂,p₂,…,q_n,p_n) (q₂,p₂,q₁,p₁,…,q_n,p_n)
which are different points in the phase space _N, should be counted only once. We have N! of this equivalent vectors, so Σ(E) → Σ(E)∕N!, and we get:

Sind	= kB log
	= kB
	= kB	(2.2)

which is an extensive quantity.

So for an ideal gas in 3D microcanonical we have:

dΩ	= ∏ i=1N	distinguishable particles
dΩ	= ∏ i=1N	indistinguishable particles

From the entropy, knowing that TdS = dE + pdV - μdN, we can get:

• 

  (2.3)

  (where the 3 stands for the number of dimension)

• 

• 

Substituting 2.3 in 2.2 we get:

where we have defined the thermal wavelength λ_T and the average inter-particle distance d as:

These formulas have a problem: it might happen that for low temperature T,d λ_T ,S < 0 !. So something break down when d ~ λ_T , but there will be a quantum mechanical solution. However, what we have found is true until d > λ_T , otherwise we have interaction effects.

2.2 Canonical Ensemble

System	= {(qi(1),p i(1))}, V 1 ≪ V 2, N1 ≪ N2
Environment	ε = {(qi(2),p i(2)}, V 2,N2

We have a system and an environment with a wall in between them that permits heat and energy to pass but not particles. Energy is exchanged, but the total energy E ≡ E₁ + E₂ = const. (with E₁ ≪ E), so that the universe = ∪ ε is microcanonical.

The phase space is = ₁ ×₂ and:

The whole universe is described with a microcanonical probability distribution function, and the (canonical) probability distribution describing the system only is obtained by integrating out the environment d.o.f. So:

ρc()	∝∫ ρmc( )dΩ2 δ is selecting only the surface with E2 = E - E1
	= ∫ E2=E-E1dΩ2
	=

So ρ_c⁽⁾~ ω₂(E₂ = E - E₁) and we know that S = k_B log ω, so:

log ρc ~ log ω2(E2)	~ S2(E2 = E - E1)
(E1 ≪ E)	≃ S2(E) + E 2=E(-E1) + …
	= S2(E) -

since the first term is a constant.

We can see that it depends only on the system (E₁). Defining β = 1∕k_BT:

We have eliminated the environment, so we will avoid writing the subscript ⁽¹⁾ from now on:

where he have defined the canonical partition function Z_N, which can be determined with the normalization:

Z_N = Z_N(T,V ) depends on N,T,V .

This expression is valid for distinguishable particles. For indistinguishable particles there is a 1∕N! factor. For simplicity we write:

Using the fact that we can foliate the phase space, we can write (assuming the lowest bound to be 0):

In systems with a discrete set of energies values E_j: Z_N = ∑ _je^-βϵ_jg _j

1) For more species A,B,… of particles, we can assume that there is no interaction among different species: = _A + _B. So:

ZN	= ∫ dΩAdΩB… e-β(A+B+… )
	= …
	= ZNAAZ NBB…

So if the species are distinguishable and independent one from the others (not interacting), the partition function is the product of the partition function of all the species.

2) Given an observable f(q_i,p_i), the canonical average is:

2.2.1 Thermodynamic quantities

We recover the thermodynamic potentials, by defining:

1)

2)

Proof: We can see that both Z_N (microscopic description) and F (macroscopic) depend on N,V,T. This leads us to think that it must be related, We will see that: Z_N = e^-βF F = - log Z

Differentiating both sizes :

As an exercise, since β = β = -T

While in thermodynamic we have F_th = E_th - TS S = -_V,N.
These two expressions are the same if we identify:

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A useful formula in exercises:

3) We saw the universal Boltzmann’s formula in the microcanonical. In the canonical it is the same (that’s why it’s called universal):

Proof:

- kBc	= -kB ∫ dΩρc log ρc
	= -kB ∫ dΩ(-β- log Z)
	= kB
	= kB
	= = S

2.2.2 Equipartition theorem

Theorem: Let ξ₁ ∈ [a,b] denote one of the canonical coordinates (q) or momenta (p) and ξ_j ;  (j≠1) all other variables. Suppose that the following condition holds:

Then:

Proof:

But we can write the differential d_ξ1 = ∂ξ₁ :

So we obtain:

1 = ∫ dξ1 + ∫ ξ1

And from the hypothesis, the first term is zero. So:

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We can now see that the standard equipartition theorem is a corollary of this:
If the coordinate ξ₁ appears quadratically in the Hamiltonian, then it contributes to the internal energy with an addend of k_BT∕2.

Indeed, if: = Aξ₁² + (ξ _j) :

Let’s analyze better the condition of the theorem:

• Suppose ξ₁ = q, then: = _KIN + V (q). The requirement becomes:

  

  which is not a strange condition in physics, where usually the particle is confined.

• Suppose ξ₁ = p, then: _KIN = and the requirements becomes:

  

  There are some magnetic systems which don’t satisfy this condition, but in general it is satisfied.

2.2.3 Exercise (2.1): Find the partition function of a (classical) perfect gas in 3D

ZN	= ∫ V ∏ i=1Nd3q i ∫ IR3e-β ∑ i=1N
	= N (p2 = p x2 + p y2 + p z2)
	= 3N Gaussian integral

Generalizing in d-dimension: Z_N =

From that partition function we can get:

• 

  which agrees with the equipartition theorem: = (p_x² + p _y² + p _z²):
  3 quadratic variables 3

• F	= - log ZN = -kBT (Stirling)
  	= kBTN

  

• S = (as an exercise) or:

  

  Notice that in both cases the result is the same and implies S = 0, for T < . This is of course absurd, signalling that this model is not suited to describe the low temperature limit.

• 

13/10/2022

• 

  Relation between μ ↔ n at fixed T

  Or, choosing the density: μ = μ(T) (in graph). We can see that: μ(T)∞ and

  < 0

  This condition will play an important role in a Bose-Einstein gas

• Specific heat:

  
  

Notice that c_V comes out to be constant: this is in contradiction with thermodynamic identities that require cV → 0 as T → 0. Again this shows that this model is not suited to describe the low temperature limit.

2.2.4 Magnetic systems

Another way to make work is through magnetic interaction.
So: dE = δQ - δ + μdN = TdS - pdV + dE_mag

If , are the external field and the reaction of the matter to the external field respectively, we can define as the Total magnetic field: = ϵ₀ +
For historical reasons, contains in its definition the factor ϵ₀ already.

The Maxwell equations in matter becomes:

where: = ∕μ₀ -

A charged particle is described by _el = q = V And we have:

where V (q) is the elastic potential energy.

There could be some other effects, like some atoms or molecules could have a dipole moment (e.g. water molecule). This dipole can be seen by an external field, because of the interaction ⋅. However this effect is usually not very strong ( is weak), with the exception of the ferromagnetic materials.

We have different materials with different effects: diamagnetism, paramagnetism, ferromagnetism.

Paramagnetism The assumptions for paramagnetism are that:

• There are no macroscopic charge or net current flow

• As we said, is negligible with respect to magnetisation.

• We assume ρ = 0, from which the Maxwell equations become:

We can define the total magnetization (an extensive quantity) as:

And we have: = μ₀ + μ₀(q), so:

δ	= dt∫ V d3q ⋅
	= dt∫ V d3q⋅
	= dt∫ V d3q
	= dt∂V + dt∫ V d3q + the first is a boundary term
	= ∫ V d3q ⋅ = dEmag
	= μ0 ∫ V d3q ⋅ d

If we suppose an homogeneous material, is uniform, so is uniform too. We obtain:

The first term is not relevant, since it’s the same we have in vacuum.

The total energy can thus change for more reasons:

and , are a couple of conjugated variables, just like (T,S), (p,V ), (μ,N).

If we apply this to a solid (dV = 0) in the canonical ensemble (dN = 0):

• internal energy: dE = TdS + μ₀ ⋅ d E = E(S,)
  from which: T = _M = _S (not so useful)

• Helmotz free energy: F = E - TS dF = SdT + μ₀ ⋅ d

• Gibbs free energy: G = E - TS - μ₀ ⋅ dG = -SdT - μ₀ ⋅ d
  from which: = _T (more useful, since we want the internal , given the external magnetisation )

Microscopically, paramagnetism is correlated to dipole molecules: the intrinsic magnetic moments (which is due to electrons orbiting around nuclei (microscopical current) or due to spin) can interact with an external field ( ⋅) originating paramagnetism.

Diamagnetism The magnetic forces are more difficult than the electric ones:

However, we can write: → - = (minimal coupling), from which:

The canonical partition function is:

ZN[T,V,N,]	= ∫ e-β ∑ j=1N2∕2m
	= ∫ ∏ j=1Nd3q jd3 j e-β ∑ j=1N j∕2m

We removed the dependency on (Z_N = Z_N[T,V,N]), so any thermodynamic potential that we can get does not depend on . We have obtained the following

Theorem: (Bohr-van Leeuwen) In classical theory there is no diamagnetism:

In quantum mechanics, we can get diamagnetism through the Langevin Theory, due to Larmor procession

Ferromagnetism In some materials, the vector is very strong and there are interactions between different magnetic moments: _i ⋅_j.

2.2.5 Exercise (2.5): Thermodynamics of a magnetic solid

We consider a solid of N atoms/molecules (V fixed) in a canonical setting, which have an intrinsic magnetic moment μ in a external magnetic (uniform) field = Hẑ. We assume that the field is not too intense, so |μ_i| = μ does not change (the only effect the field has on μ is to make it rotate.

We assume the particle to be distinguishable (because they are fixed at their equilibrium positions) and consider the Hamiltonian: = -∑ _i=1^N _j ⋅.

Since there is no motion degree of freedom, for each particle there are only the ones due to magnetic moment: = with ||² fixed. The phase space (of a single particle) is a 2D sphere of radius || = μ.

So it’s better to use spherical coordinates:

and the volume is: dqdp = μ² sin θdθdϕ
Indeed, if we use q = ϕ, the conjugate variable to an angle is a momentum p = cos θ, so dq = dϕ, dp = sin θdθ.

We now have everything we need to write the partition function:

And we can also calculate the total magnetization: ^z = _c = ∑ _j=1^Nμ _j^zZ

c	≡∫ ρcμjz =
	= = log Z1 =

Since all particles are the same: M^z = N_c = N

2.3 Grancanonical Ensemble

(IV) MON 17/10/2022
We consider a system S = {(q_i⁽¹⁾,p _i⁽¹⁾} and an environment ε = {(q _j⁽²⁾,p _j⁽²⁾} at equilibrium (thermal (T₁ = T₂ = T), mechanical (p₁ = p₂ = p) and a chemical (μ₁ = μ₂ = μ). These two can exchange both energy and particles, but the the total energy and the total number of particles are conserved (N = N₁ + N₂ = const), so that the whole universe = ∪ ε is canonical.

The (grancanonical) partition distribution of the system only is obtained by integrating out the environment d.o.f. So:

ρgc()	= ∫ ϵρc()
	=

(the integrand term in the denominator is the total δ + ε, the total volume V = (V ₁,V ₂))

We will choose the constant = , then we will prove later that it is the correct one (the one that satisfies the normalization). So we have:

(2.4)

Proof: Let’s now see that the normalization holds:

We can multiply the first integral at the numerator by (V ₁∕V ₁)^N₁, the second by (V ₂∕V ₂)^N₂ and the denominator by (V∕V )^N, obtaining:

If V ₁,V ₂,V are finite, the integrals are different, but in the thermodynamic limit, V ₁,V ₂,V → IR^d, so the overall ratio is 1.

However, N₁,N₂ are not fixed (only N is), so we have a different integral for each value of N₁. So actually what we want to calculate is:

	∑ N1=0N ∫ ∏ i=1Ndq i(1)dp 1(1)ρ gc
=	∑ N1=0NN1 N-N1	expansion of the binomial
=	N = 1N1

So what we have proved is that the sum over all the possible number of particle integrated over all the phase space is 1:

____________________________________________________________________________________

We can re-write the (2.4):

ρgc	=
	=
	=  eβ(F[N,V,T]-F[N2,V 2,T])

F[N,V,T] - F[N2,V 2,T]	= F(N,V,T) - F(N - N1,V - V 1,T)
(Taylor expansion)	= V,T ΔN + N,T ΔV + …
	= μ1(N1) + (-p1)(V 1) + …

So we have:

The constant can be absorbed in the measure:

(2.5)

So usually we don’t write the constant .

Now that we have integrated out all the quantities related to the environment, we will drop the superscript ⁽¹⁾ from everywhere:

using the granpotential Ω = -pV and defining the fugacity z ≡ e^+βμ:

From the normalization (2.5) we obtain the grancanonical partition function:

so the grancanonical probability distribution becomes:

where - μN is sometimes denoted with : grancanonical hamiltonian.

Simple application of this formulas: perfect gas (exercise 3.1):

Definition: Grancanonical average
Given an observable f_N(q_i,p_i) (the subscript _N is because the expression of the observable can be different according to the number of particle), the grancanonical average of that observable is:

gc	= ∑ N=0∞∫ fN(qi,pi)
	= ∑ N=0∞zNZ N = ∑ N=0∞zNZ Nc

2.3.1 Thermodynamical quantities

• -pV = Ω = - log

• E = gc	= ∑ N=0∞
  	= -∑ NzN∫ e-β
  	= -z=const
  	= -z const

• 

• The entropy can be obtained from the universal Boltzmann formula:

  
  
  

  The last term can be re-written as:

  
  

  so:

  

  and since Ω = E - TS - μN, S_th = (E - μN - Ω) = S_gc

2.3.2 Virial expansion (van der Waals gases)

A real gas is a 3D gas where

We assume that the potential is a van der Waals potential: U(r = |_i -_j|), so there is interactions only between two particles.
Since this is a gas, we also assume that the interactions are weak, otherwise the system could become liquid or solid.

Starting from the grancanonical partition function, if z = e^βμ > 0 is small, we can expand the expression up to the second order (Virial expansion):

where Z₁ is the partition function where there is only 1 particle, Z₂ considers 2 particle, etc.. So:

Changing coordinates: _CM = (₁ + ₂) = ₁ -₂

So:

And we can obtain Ω, n and p:

To compute the log , we remind that log(1 + x) ≃ x -x² + … if x is small:

log	= + - + o
	= +
	= +  J2(β)

where J₂(β) ≡∫ _V d³r is the second Virial exponent.

From that we can obtain:

• Density:

  (2.6)

• Pressure:

  (2.7)

For a perfect gas, particles are non-interacting, so J₂ = 0. Thus:

So, since we have done an expansion in , we can make an expansion in density n. Assuming n small means considering a diluted gas.

(V) MON (ex.1) 24/10/2022
27/10/2022
From the density expansion (2.6) we can obtain:

In the case of no interaction (perfect gas), J₂ = 0, and for the diluted limit, N = ≪ 1.
Reminding that ≃ 1 + -x² + … for x ≪ 1:

Now we need to decide which solution to take. Since if we stop to the first order we should obtain z ∝ n (z = λ_T ³n), then we have to take the positive solution, so:

From the expansion of the pressure (2.7), one can obtain:

Now, if we assume that particles are spheres and the potential is the one in the Fig. 2.1, we can find the van der Waals equation of a real gas by expanding the term J₂:

J2(β)	= 4π ∫ 0∞dr r2
	= 4π ∫ 02r0 dr r2 + 4π ∫ 2r0∞dr r2
In the limit e-βU(r) ≪ 1, we can use the 1° order Taylor expansion: e-βU(r) - 1 ≃|-βU(r)|
	≃-4π ∫ 02r0 dr r2 + 4π ∫ 2r0∞dr r2
	= -8r03 + ∫ 2r0∞dr r2
	= <0 + >0

where b is the volume of 2 particles, and a is the average measure of the potential.

So we obtain: = b - and from the expression of the pressure:

and defining n = N∕V = 1∕v, we obtain the van der Waals equation of a real gas:

RECAP:

Microcanonical	Canonical	Grancanonical
ρmc = δ(- E)	ρc =	ρgc =
S = -kBSTh = kB log #states

2.4 State counting and Entropy

We have seen that the Boltzmann’s universal law give a value for the entropy which (in the thermodynamic limit) is the same of the thermodynamic entropy. We also know from the variational principle of thermodynamics that the equilibrium corresponds to maximum entropy.

In this brief discussion, we fix our attention to the canonical ensemble, but similar considerations hold for the grancanonical one.

In some case, the energy is discretized and we use E (instead of ) to indicate the energy level. Also, each energy level can be degenerate, meaning that more than one state have that energy. We indicate the degeneracy with g_i = g(E_i).

We define the Boltzmann’s weight as the probability to have energy E:

(2.8)

In some cases, we can measure the energy of a system, but we can’t look for its microstate. However this is a general formula and we can use it even when we don’t know the microstate.

Let’s look back at the ensemble description:

• We have a very large number N of copies of the system, all described by the same values of macroscopic variables (macrostate), but corresponding to different values of microscopic variables (microstate)

• the probability that a given microstate occurs is the N-infinity limit of the frequency with which it appears (objective interpretation)

• all physics can be derived from knowing such probability distribution

Is there a principle to derive the probability distribution describing equilibrium?

The probability distribution describing equilibrium is the one corresponding to maximum entropy, given the macroscopic constraints.

Remark. This set-up (Boltzmann, Gibbs) is grounded on the idea that
i) we have a clear identification of what a micro/macrostate is
ii) probability are defined a-priori quantities.

Previously, one constructed a theory based on the equations of motion, supplemented by additional hypotheses of ergodicity, metric transitivity, or equal a priori probabilities, and the identification of entropy was made only at the end, by comparison of the resulting equations with the laws of phenomenological thermodynamics. Now, however, we can take entropy as our starting concept, and the fact that a probability distribution maximizes the entropy subject to certain constraints becomes the essential fact which justifies use of that distribution for inference.
E.T Jaynes, Information theory and Statistical Mechanics, Phys. Rev. 106 (1957) 620

Inference problem

The ”objective” school of thoughts regards the probability of an event as an objective property of that event, always capable in principle of empirical measurement by observation of frequency ratios in a random experiment.
On the other hand, the ”subjective” school of thought regards probabilities as expressions of human ignorance; the probability of an event is merely formal expression of our expectation that the event will or did occur, based on whatever information is available.

The inference problem is the following:
If the only info we have is that a certain function of x has a given mean value = ∑ _j=1^Np _jf(x_j), what is the expectation value of another function g(x)? We must use the probability distribution which has a maximum entropy subject to the constraints:

which is obtained by maximizing (Lagrange multipliers) the function:

Remark: It can be easily generalized to more observables and/or higher moments of the distribution

2.4.1 Probability distribution from maximum entropy principle

We have a finite set of energy levels E_r, each with degeneracy g_r, on which we distribute a number n_r of ensembles with total energy E to distribute among N copies of the system. The number of ways to do that is W_{nr} = W_{nr}⁽¹⁾W _{nr}⁽²⁾.
where W⁽¹⁾ does not consider degeneracy so it counts how many ways we can put n _r with E_r; while W⁽²⁾ : the n _r particles can be distributed in g_r state.

We first consider classical particles, so they are distinguishable if they have different energy. So:

From the maximum entropy principle, the equilibrium distribution corresponds to max entropy S = log W_{nr}
with the constraints

So, in the classical case:

A	= kB log W{nr} + α + β
	= kB + α + β
	= kB + α + β
	= kB + α + β

To maximize this we derive:

and we find the number of particle that maximizes the entropy:

And we can get the probability to get a particle in the energy level r:

So we get the same expression as in 2.8. Also, we obtain the Lagrange multiplier β =

Quantum particle instead are always indistinguishable, not only when they have the same energy. Thus there is only one way to have n₁ particles with energy E₁, n₂ particles with energy E₂, etc... So: W_{nr}⁽¹⁾ = 1 and

For bosons, we can imagine of putting the particles in a line and draw boundaries to select in which energy level they are. So we have n_r indistinguishable particles and g_r - 1 indistinguishable boundaries. In total we have n_r + g_r - 1 objects:

For fermions, we can put at most 1 particle for each ”box”, which is like saying that each box can be empty or with a ball (n_r < g_r). So it’s like selecting n_r objects out of g_r possibilities:

We will derive again this distributions in the next chapter.

Chapter 3
Quantum Statistical Mechanics

3.1 Review on Quantum Mechanics and Statistics

3.1.1 Quantum System

(VI) THU 03/11/2022
The degree of freedom of quantum particle are described in terms of a vector of the Hilbert space. With Dirac’s notation, vectors are denoted like: ∈. We can have a linear superposition: λ + μ∈.
The scalar product of two vector is called a braket.

A (pure) quantum state is a ray (an equivalence class):

So ~ λ λ = |λ|e^iα≠0

The projection operator ℙ represents uniquely a quantum state:

ℙ is a projection operator if

• ℙ is bounded

• ℙ^† = ℙ (self adjoint)

• ℙ² = ℙ (idempotent)

Proof:

____________________________________________________________________________________

This operator projects on the linear subspace generated by :

An observable is given by a self-adjoint operator: A : → A^† = A
for which the spectral theorem holds: A = λ_j:

• The eigenvalues are real: λ_j ∈ IR

• The eigenvectors (normalized) are orthogonal: ψ_m⟩ = 0 n ⁄= 0

• _n is an orthonormal (o.n) complete set, thus it’s an orthonormal basis of .

Since we’re only referring to bounded operators, the words hermitian and self-adjoint are equivalent.

So each vector of the Hilbert space can be expressed as a linear combination of the basis vectors:

Also, the sum ℙ₁ + ℙ₂ is the projection over the span (linear combination) of ψ₁,ψ₂, so:

Let’s recap better the spectral theorem: if an operator A is self-adjoint, there exists a set of projection operators that diagonalize the operator:

(3.1)

with: ℙ_n^† = ℙ _n ℙ_nℙ_m = δ_nmℙ_n ∑ _nℙ_n = I ∀n ∈ IR

The Evolution of a system is fixed by a special observable, called hamiltonian H, through the Schrodinger equation:

We will consider only cases where the hamiltonian is time-independent. In this case the evolution is fixed by a unitary operator U:

Unitary means: U(t)^† = U(t)^-1 = U(-t).
That means that: ψ(t)⟩ = ⟨ψ(t = 0)|ψ(t = 0)⟩, so normalization is preserved (thus probability is conserved).

The dynamic of a quantum system is perfectly deterministic: if we know and apply the equation above, we have the evolution. The probabilistic aspect arises in the measurement.

A measure of an observable A on a state yields a set of possible outcomes {λ_n} corresponding to its eigenvalues, with probabilities p_n given by:

So: p_n = |c_n|² = ℙ _n|ψ⟩

Remark: after the measurement, the state collapses into

Also, we can note that, using the spectral decomposition of the operator A (eq. 3.1) we can write:

which is the statistical average.

This kind of measure is called projective measure, because one can also generalize the notion of measure by not starting with A decomposed using the projection operators.

3.1.2 Density matrix

Suppose we have two particles described by ₁,₂. Then _TOT = ₁ ⊗₂ and dim(₁ ⊗₂) = n⋅m. This is different from classical mechanics, where the total space is the Cartesian product between two spaces: ₁ ×₂ and dim(₁ ×₂) = n + m.

Let _n,_m be the o.n basis of ₁,₂ respectively. Then ₁ ⊗₂ is generated by = o.n basis.
That means that every object of this space can be written as a linear composition of this following object:

If we take a vector of the Hilbert space: ∈, its projection is

Theorem: If ρ_ψ = ℙ_ψ = , then ρ_ψ is:

i.

A bounded operator: ∥ρ_ψ∥≤ 1

ii.

Self-adjoint: ρ_ψ^† = ρ _ψ

iii.

Positive: ρ_ψ|α⟩≥ 0, ∀|α⟩

iv.

Unit-trace: Tr[ρ_ψ] = 1

v.

Idempotent: ρ_ψ² = ρ _ψ

Proof: Let’s prove the new ones (iii and iv):

iii.

ρ_ψ =

iv.

[M]_min = M|em⟩ {en} o.n. basis

This is:

(1)

linear Tr[M₁ + M₂] = Tr[M₁] + Tr[M₂]

(2)

cyclic Tr[M₁M₂…M_k] = Tr[M_kM₁…M_k-1]

So it’s independent on the chosen o.n. basis. If U is the matrix of basis change:

So we have a trace-class operator A such that:

We want to prove that ρ_ψ = is a trace-class operator with
Tr[ρ_ψ] = 1.
We can choose this o.n. basis: { = ,,} such that:

____________________________________________________________________________________

Let’s now prove the other way:
Theorem: If ρ is such that (i) ↔ (v) are satisfied, then exists ∈ such that ρ =

Proof: From (i) and (ii) follows that ρ is bounded and self-adjoint. So we can write it using the spectral decomposition: ρ = ∑ λ_nℙ_n and states that:

From (iii): λ_n ≥ 0
From (v):

and since λ is positive ⇐⇒ λ_n = 0, λ_n = 1 ∀n
From (iv): Tr[ρ] = ∑ _nλ_n = 1. So that means that all the λ are 0 apart from one of them which is 1.

If we suppose λ₁ = 1, λ₂ = λ₃ = = 0 ρ = λ₁ℙ₁ = ℙ₁ =

____________________________________________________________________________________

This allow us to give the following:

Definition: A pure state of a quantum system is described by ρ such that ρ = ⇐⇒ (i) ↔ (v)
ρ is called density operator (matrix).

(VII) MON (ex.2) 07/11/2022
10/11/2022
So we’ve seen that a pure state is defined by a ray [] or equivalently by its associated (rank-1) projector:
∋ normalized ~ e^iϕ ⇐⇒ density op. ρ _ψ = iff (i) ↔ (v)

We can also have a mixed state, for instance an electron produced in a lab which is neither spin up nor spin down.
A mixed state is defined by a statistical ensemble of pure states:

and it’s represented by means of the density operator

A mixed state satisfies (i) ↔ (iv):
(i),(ii) is trivial because it’s sum of (i) and (ii)
(iii) ρ ≥ 0 because 0 ≤ ρ_k ≤ 1
(iv) Tr[ρ] = Tr[∑ _kρ_kp_k] = ∑ _kp_kTr[ρ_k] = ∑ _kp_k = 1

However, (v) does not hold (ρ²≠ρ).
In fact: {} orthogonal ψ_k′⟩ = 0 if k≠k′ .

which is equal to ρ = ∑ _k=0^Mp _kρ_k, only if:
∃k : p_k≠0, p_k = 1 with p_k = 0 ∀k≠k

So (v) is true only if ρ = is a pure state.

So we can say that a (generic) state is described by a density operator ρ that satisfies (i) ↔ (iv) and:

Theorem: A density matrix is pure (∃, ρ = ) ⇐⇒ ρ² = ρ

Expectation value:
The expectation value of an observable A in the case of ρ_ψ = can be written as:

We can generalize this to a mixed case ρ = ∑ _k=0^Mp _kρ_k:

So in general, ∀ρ : _ρ = Tr[ρA]

An example: the Qubit
A classic bit is just a number that can be 0 or 1, while the quantum bit is a 2-level system for which:

An equivalent way to describe it is to chose an o.n basis {,}
where: = ,  =

A qubit is a generic state of this space, which is a linear superposition of the 0 and 1 state:

= α + β with |α|² + |β|² = 1

03/11/2022 (c)
We can describe the evolution on this system:

trough a unitary operator U, which in this case represents rotations on the Bloch sphere. We can also define:

1.

I = I =

2.

NOT = = X Pauli matrix → →

3.

Z = → →-

which construct a quantum gate on a single qubit.

With 2 qubits we have:

and the o.n basis of _TOT is 4 dimensional, made from:

A generic state of 2 qubits is described by:

We can also have separable or entangled states:

1.

α₁₀ = α₁₁ = 0
= α₀₀ + α₀₁ = ₁ = ₁₂
In this case, the state is called separable, because it can be separated into a multiplication of a state of particle 1 ⋅ a state of particle 2.

2.

α₀₁ = α₁₀ = 0 = α₀₀ + α₁₁
This state is not separable, so we say it’s entangled. An example of an entangled state is a Bell state: =

Suppose these are spin ↑(0) or ↓(1). Alice and Bob can measure it and the output will be unpredictably ↑ or ↓ with 50% of probability. However, if Alice measure ↑, she knows for sure that Bob will measure ↑ too.

10/11/2022 (c)
A qubit is a pure state. In fact (i)↔(v) are satisfied. In particular, (v) follows from the condition |α|² + |β|² = 1.

Note that being a pure state, it means that all the particles are in the state = α + β, then it’s the measure procedure that makes it collapse to or - ↑ or ↓.

• Pure density matrix

  

  The diagonal pieces (p_α = |α|²,p _β = |β|² represents the probability of a measure.
  The non-diagonal pieces are instead responsible for the quantum interference phenomena.

• Mixed density matrix
  A mixed state would be a state where some particles are prepared as ↑, some others as ↓ (a so-called classical mixture):

  
  
  

We can notice that if we do a measure, the probabilities of the outcome are the same as before, even if the matrices are different.

Exercise: Design an experiment which is able to determine if the system is in a pure on in a mixed state.

3.1.3 Identical particles - Permutation group

Let’s consider a system composed by N subsystems (N particles), each described by _j j = 1,…,N. The system will be described by the Hilbert space:

If the subsystems are identical: _j ≡_tot = ^⊗N
If they are also indistinguishable, the states span only a subspace of _tot, whose vector have special properties under the action of the permutation group: under the action of a permutation (i.e. swapping particles around), the state should be invariant, up to a phase. In the following we will see why and also that there are two way to achieve this result: this will lead to the definition of bosons and fermions.

We need to see the properties of a quantum system under the effect of permutation group. Let’s see the permutations of the N objects we have:

The set of all possible permutations of N elements is a group. Let’s call it ℙ_N: the permutation group on N elements. This is a group because:

• it’s closed under composition (a composition of permutation is a permutation)

• has an identity I : (1, 2,…,N) → (1, 2,…,N)

• ∀σ, ∃σ^-1(inverse) s.t. σ^-1 ∘ σ = I

We can notice that ℙ_N has a finite number of elements (= N!).

A transposition (or elementary permutation) σ_j j = 1,…,N - 1 is a swap between the j and the j + 1 elements. Then a permutation can be decomposed into transpositions. In other words, the N - 1 transpositions are the generator of the group:

Theorem: ∀σ ∈ℙ_N : σ = σ_α1σ_α2…σ_αk k finite
This decomposition is not unique and also k is not an unique value. However, all decomposition of the same element has always an even/odd number of transposition (k is always even or odd)

This allows us to divide the permutations into even and odd permutations.

Definition: sgn(σ) =

The transpositions are not all independent, but there are relations between them. In fact, they satisfy the identities:

i.

σ_iσ_j = σ_jσ_i if |i - j|≥ 2 (visual proof in figure 3.1a)

ii.

σ_iσ_i+1σ_i = σ_i+1σ_iσ_i+1 (visual proof in figure 3.1b)

iii.

(σ_i)² = I (trivial)

ℙ_N, the group generated by the N - 1 transpositions σ_j, satisfies as well the properties (i) (ii) (iii).

3.1.4 Quantum statistic

If some objects are indistinguishable, it means that a permutation among those

doesn’t affect the physical content of a wave function, which is |ψ(x₁,x₂,…,x_N)|². That means that the wave function should be the same up to a global phase:

Remark: this is a physical property, not a mathematical one.

If we then decompose a permutation into transpositions that generate it:

with: ϕ_σ = ϕ_α1 + ϕ_α2 + + ϕ_αk

Let’s now analyze a single transposition: σ_j : ψ(1,…,N)e^iϕ_j ψ(1,…,N), which we will simply write as σ_je^iϕ_j.
This must satisfy the (i),(ii),(iii) relations of a transposition and this leads to some considerations on the phase:

i

if |i - j|≥ 2, then σ_iσ_j = σ_jσ_i, so:

⇐⇒ ϕi + ϕj = ϕj + ϕi Trivially satisfied

ii

σ_iσ_i+1σ_i = σ_i+1σ_iσ_i+1, so:

So ϕ_j = ϕ ∀j and the single transposition σ_je^iϕ_j becomes simply: σ _je^iϕ

iii

(σ_j)² = I, so:

⇐⇒ 2ϕ = 2πn n ∈ℤ

So there are only 2 possibilities (since ϕ ∈ [0, 2π[):

This applies for a single transposition. For the whole permutation σ = σ_α1σ_α2…σ_αk and ϕ_σ = ϕ_α1 + ϕ_α2 + + ϕ_αk. So we have:

For bosons, the way function is completely symmetric. For fermions it is completely anti-symmetric. Remark: what we call bosons and fermions are just due to the statistic and has nothing to do with spin. Only in relativistic quantum mechanics one can prove the spin-statistic theorem.

(VIII) MON 14/11/2022
An example: System of N=2 particles in IR³
We can describe two particles in IR³ with two vectors: ₁,₂ ∈ IR³.
The Hilbert spaces respectively for a single particle and for two particles are

N=1 = L2(IR3) = {ψ( 1) square integral}
N=2 = L2(IR6) = {ψ( 1,2), square integral}

where L²(IR⁶) = L²(IR³) ∘ L²(IR³).
Permutations are described by the permutation group: ℙ₂ = {I,σ} with σ : x₁ ↔ x₂
According to our rules, the wave function should be symmetric in the bosonic case and anti-symmetric in the fermionic case. So we define two operators:

• Symmetrizer: : ψ(x₁,x₂) = ψ₊(x₁,x₂) (symmetric by construction)

• Antisymmetrizer: : ψ(x₁,x₂) = ψ_-(x₁,x₂) (antysimmetric)

It is easy to show (as an exercises), that and are projection operators. In fact ^† = ² = ^† = ² =
If we call _S and _A respectively the spaces of symmetric and anti-symmetric wave functions: : _S : _{A S},_A ⊂ = L²(IR⁶)

Also: = = 0 →_S ⊥_A
In fact:

ϕ-⟩ def. of scalar prod. =	∫ d3 1d3 2 ψ+*(x 1,x2)ϕ-(x1,x2)
=	∫ symm ⋅ antisymm = ∫ anti-symmetric function = 0

So we can write = _S ⊕_⊥A.  In fact every function can be written as the sum of a symmetric and an anti-symmetric function:

Notice that since fermions are described by an anti-symmetric wave function, they can’t occupy the same state (Pauli exclusion principle is automatically included in this construction):

Generic N > 2 particles in IR³
With N particles, we can have more transpositions. Let’s indicate with P ∈ℙ_N a permutation: : ψ(x₁,x₂,…,x_n)ψ. This just re-shuffle the order of particles.
We define:

= ∑ P∈ℙN	: ψ(x1,x2,…,xN) ∑ P ψ
= ∑ P∈ℙNsgn(P)	sgn(P) =

As before, and are orthogonal projector operators:
^† = ² = ^† = ² = . and: = = 0
Also: : _N →_S : _N →_{A S} ⊥A

with 3 or more particles there are function that ar neither symmetric nor anti-symmetric.

We can have an example of ′ in a system of N particles in IR^d. A single particle is described by ₁ = L²(IR^d) with the orthonormal basis: {u _α(x)}_α=1^∞.
N particles are described by _N = _{N times} with the o.n basis
_α1α2…αN
Notice that the order is important, because it indicates that there is particle 1 in α, particle 2 in β etc..

We aim at describing in an intrinsic way each _N^{( )}, _N^{( )}. We can define the symmetrizer and antisymmetrizer as:

: ψα1α2…αN(x1,x2,…,xN) = ψn1,n2,…,nk…(x 1,x2,…,xN)
: ψα1α2…αN(x1,x2,…,xN) = ψn1,n2,…,nk…(x 1,x2,…,xN)

Here we can see that the order is no longer important and what matters is just how many particle are in each state. n_k is called occupation number and counts just that: how many particles are in the k state. Notice that for bosons there are no constraints (n_k = 0, 1, 2,… ), while for fermions there can only be 1 particle at maximum for each state (n_k = 0, 1).
Also, the total number of particles should be constant, so

∑ _k=1^∞n _k = N

For example, with N = 3 we have: u_α(x₁)u_β(x₂)u_γ(x₃) as the o.n basis and the possible permutations are:

So we have:

=	uα(x1)uβ(x2)uγ(x3)
±	uα(x1)uβ(x3)uγ(x2)
±	uα(x2)uβ(x1)uγ(x3)
+	uα(x2)uβ(x3)uγ(x1)
+	uα(x3)uβ(x1)uγ(x2)
±	uα(x3)uβ(x2)uγ(x1)

(and similar for )

Bosons are described by taking all the signs above (all +), fermions are described by the bottom one (so + - - + + -). However, one could take another combination of plus an minus, obtaining something non-physical (∈′)

3.2 Second quantization

We can’t work with a wave function spit in N! peaces (in a gas N ~ 10²³). So, we see the second quantization, an algebraic and abstract approach that will lead to very powerful results.

The approach is different from what it is followed in QFT

We follow the right path. Also, we are not interested on time: time is fixed and independent.

3.2.1 Creation/Annihilation operators

There are two kinds of these operators: bosonic and fermionic

• Bosonic creation/annihilation operators

  We define a,a^† such that = aa^†- a^†a = I on some whose dim= ∞
  If that wasn’t the case (and dim = n) then Tr = Tr = n But also:
  Tr = Tr- TrTr- Tr = 0
  So dim = ∞.
  Notice that they are not self-adjoint operators, so they aren’t observables.

  We also define = a^†a = -a = a^†

  For instance, this is the operator used in the 1D harmonic oscillator problem:

  	=	a = a† =
  	= ℏω	= i =
  	= ℏω

  The o.n basis is given by:

  

  We won’t prove it (it’s just algebra), but

  

  is called number operator, because the eigenvalue is just the number of quanta. The base _n=0^∞ is called Fock basis and it is an o.n. basis on .
  Of course, one can chose another base, but we know very well how this base here behaves.
  Applying the creation/annihilation operators just consists in going up or down in the ladder in Fig. 3.2:

  	a =	Going down the ladder (destroying a particle)
  	a† =	Going up the ladder(creating a particle)

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  Figure 3.2: Creation/annihilation ”ladder” for bosons.

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  Also, the fact that we can apply a^† many times is another reason why the Hilbert space should have dim = ∞.

• Fermionic creation/annihilation operators

  We define c,c^† such that = cc^† + c^†c = I and c² = ² = 0.
  As before we also define the number operator
  = c^†c = -c = c^†
  

  For instance, this is the operator used in a spin 1∕2 particle system. Given the Pauli matrices:

  

  satisfying = ϵ _αβγσ^γ , one can construct:

  

  such that = I _2×2 ^† = σ^{+ 2} = ² = 0

  The Fock basis is the o.n. basis given by:
  vacuum = c^† = c^†c^† = 0 (since ² = 0)

  So we have only two possible states: = 0 = 1 ⋅
  as shown in Fig. 3.3.

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  Figure 3.3: Creation/annihilation ”ladder” for fermions.

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  As before = n is called number operator, but this time n = 0, 1