Derivation of Gibbs Ensembles

Classical Hamiltonian Mechanics

We start with Hamiltons equations.

q˙=H(q,p)p
p˙=H(q,p)q

Classical Microcanonical Ensemble

We start with some probability distribution in the phase space of the Hamiltonian H(q,p).

ρ(q,p,t)=???

The probability distribution evolves by the continuity equation.

ρ(q,p,t)t=(ρv)

Here v=q˙1,q˙2,,q˙N,p˙1,p˙2,,p˙N

In equilibrium the probability distribution at every point in phase space shouldn't change.

(ρv)=0
i=12Nρ(q,p)viai=0
i=1Nρ(q,p)q˙iqi+ρ(q,p)p˙ipi=0

Simple chain rule.

i=1Nρ(q,p)q˙iqi+q˙iρ(q,p)qi+ρ(q,p)p˙ipi+p˙iρ(q,p)pi=0

Substitute in Hamiltons equations.

i=1Nρ(q,p)H(q,p)qipi+q˙iρ(q,p)qiρ(q,p)H(q,p)qipi+p˙iρ(q,p)pi=0
i=1Nq˙iρ(q,p)qi+p˙iρ(q,p)pi=0
i=1NH(q,p)piρ(q,p)qiH(q,p)qiρ(q,p)pi=0

Next, we say that the probability distribution is a function only of the total energy, so ρ(q,p,t0)=f(H(q,p)). This means that ρ(q,p,t0)ai=f(H(q,p))HH(q,p)ai

i=1Nf(H)H(H(q,p)qiH(q,p)piH(q,p)qiH(q,p)pi)=0

That assumption creates a solution to the continuity equation. Now in the microcanonical ensemble (NVE) we say the total energy is known, so the only possibility for f is a uniform distribution over the selected energy level.

ρ(q,p)=δ(H(q,p)E)Vδ(H(q,p)E)dqdp

This equation is the principle of equal a priori for the equilibrium in a closed system.

Classical Canonical Ensemble

Define the number of states accessable by a system with N degrees of freedom at energy E to be the below.

Ω(E)=1hNVδ(H(q,p)E)dqdp

Next split the degrees of freedom into the "bath" and "system". Let the degrees of freedom for the bath be rbath=qM+1,,qN,pM+1,,pN and the degrees of freedom for the system rsys=q1,,qM,,p1,,pM.

In the Canonical ensemble the combined energy of the bath and system should be constant. The combination of the two makes a microcanonical ensemble.

Htotal(rbath,rsys)=Hbath(rbath)+Hsys(rsys)+Hcoupling(rbath,rsys)

The total accessable states with an energy should be the total number combined number of states the bath and system can be in that sum to that energy.

Ωtotal(Etotal)=1hNVΩbath(EtotalHsys(rsys))drsys

And in the microcanonical ensemble all microstates (rbath,rsys) with the correct energy have an equal probability density.

ρ(rbath,rsys)=1Ωtotal(Etotal)1hNδ(Etotal(Hsys(rsys)+Hbath(rbath)))

We can calculate the probability density of just the microstate for the system.

ρ(rsys)=Vρ(rbath,rsys)drbath
=V1Ωtotal(Etotal)1hNδ(Etotal(Hsys(rsys)+Hbath(rbath)))drbath
=Ωbath(EtotalHsys(rsys))hMΩtotal(Etotal)

Take the natural log of both sides

ln(ρ(rsys))=ln(Ωbath(EtotalHsys(rsys))hMΩtotal(Etotal))
ln(ρ(rsys))=ln(Ωbath(EtotalHsys(rsys)))ln(hMΩtotal(Etotal))

Then we expand the logarithm to its Taylor series.

f(x+c)=f(x)+df(x)dxc+12d2f(x)dx2c2+16d3f(x)dx3c3+

We set f(x)=ln(Ωbath(x)), x=Etotal, and c=Hsys(rsys) we get the following expression.

ln(ρ(rsys))=ln(Ωbath(Etotal))dln(Ωbath(Etotal))dEtotalHsys(rsys)
+12d2ln(Ωbath(Etotal))dEtotal2Hsys(rsys)216d3ln(Ωbath(Etotal))dEtotal3Hsys(rsys)3+
ln(hMΩtotal(Etotal))

By definition if the bath was a microcanonical ensemble the entropy would be Sbath(Etotal)=kBln(Ωbath(Etotal))

=ln(Ωbath(Etotal))1kBdSbath(Etotal)dEHsys(rsys)+12kBd2Sbath(Etotal)dE2Hsys(rsys)2
16kBd3Sbath(Etotal)dE3Hsys(rsys)3+ln(hMΩtotal(Etotal))

And by definition 1Tbath(E)=dSbath(E)dE.

=ln(Ωbath(Etotal))1kBTbath(Etotal)Hsys(rsys)+12kBTbath(Etotal)dTbath(Etotal)dE2Hsys(rsys)2
16kBTbath(Etotal)d2Tbath(Etotal)dE2Hsys(rsys)3+ln(hMΩtotal(Etotal))

Then if we assume the heat capacity of the bath is infinite so that dTbathdE=0 we can remove the extra terms, and the temperature just becomes a constant so Tbath=Tbath(Etotal)=Tbath(Ebath).

ln(ρ(rsys))=1kBTbathHsys(rsys)+ln(Ωbath(Ebath)hMΩtotal(Etotal))

Then exponentiate both sides.

ρ(rsys)=eHsys(rsys)kBTbath(hMΩtotal(Etotal)Ωbath(Ebath))

And the denominator is just a normalizing factor so the final expression is below.

ρ(rsys)=eHsys(rsys)kBTbathVeHsys(rsys)kBTbathdrsys

Quantum Microcanonical Ensemble

We assume a time evolution of a system of the Schrodinger equation for some Hamiltonian operator.

Ψ(x,t)t=iH^Ψ(x,t)

To measure an observable when the stat is known is

A=Ψ|A^|Ψ

but if there is a probability distrubition over some states then it is below.

=A=ipiΨi|A^|Ψi

If we have an orthanormal basis n so that k|nknk|=I we can insert it into the equation.

=ipiΨi|A^(k|nknk|)|Ψi
=kipiΨi|A^|nknk|Ψi
=kipink|ΨiΨi|A^|nk
=knk|(ipi|ΨiΨi|A^)|nk
=Tr(ipi|ΨiΨi|A^)

If we define an operator

ρ=ipi|ΨiΨi|

then the expected value of any observable is

A=Tr(ρA^)

Next we find the time evolution of ρ.

dρdt=ipid(|ΨiΨi|)dt=ipi(d|ΨidtΨi|+|ΨidΨi|dt)

Then substitute the Schrodinger equation

=ipi(iH^|ΨiΨi|+|ΨiΨi|iH^)
=iipi(H^|ΨiΨi||ΨiΨi|H^)
=i(H^ρρH^)

So it must be that for equilibrium they commute

H^ρ=ρH^

Similar to the classical case, if we have the pi be only a function of the energy for each eigenvector of the Hamiltonian, the system will be in equilibrium. Since observables are Hermetian, an orthogonal eigenbases must exist. Each operator will be defined in the eigenbasis.

H^=aEa|nana|
ρ=ap(Ea)|nana|

We can show they commute

H^ρ=aEa|nana|bp(Eb)|nbnb|
=abEap(Eb)|nana||nbnb|
=abp(Eb)|nana|Ea|nbnb|
=ap(Ea)|nana|bEb|nbnb|
ρH^

So the quantum microcanonical ensemble must have a density matrix of the form

ρ=ap(Ea)|nana|

where the n's are eigenvectors of the Hamiltonian. Since the energy is known, it must be a uniform distribution over all eigenstates with that energy

ρ=1Ωtotal(E)a=1Ωtotal(E)|na,Ena,E|

Quantum Cannonical Ensemble

We assume the total Hamiltonian is the form

H^=H^sys0^bath+0^sysH^bath+H^coupling

And we assume the coupling energy is none. From the microcannonical ensemble the total density matrix is

ρtotal=1Ωtotal(Etotal)a=1Ωtotal(Etotal)|na,Etotalna,Etotal|

And the density matrix of just the system is the partial trace

ρsys=bs1s2|ns1ns2|(ns1nb|ρtotal|ns2nb)

Where nb,ns are the eigenbasis of Hbath,Hsys. Measuing a system microstate gives.

ψ|ρsys|ψ=Trbath(ρtotal)=1Ωtotal(Etotal)bs1s2ψ|ns1ns2|ψ(ns1nb|ρtotal|ns2nb)
=1Ωtotal(Etotal)bψnb|ρtotal|ψnb

And the values of ρtotal are only the eigenvectors of eigenvalue Etotal, so it would just be the ones where the bath has the complement energy.

=Ωbath(EtotalEsys)Ωtotal(Etotal)

Take the ln of both sides

ln(ψ|ρsys|ψ)=ln(Ωbath(Etotalψ|Hsys|ψ)Ωtotal(Etotal))=ln(Ωbath(Etotalψ|Hsys|ψ))ln(Ωtotal(Etotal))

Then Taylor series again

=ln(Ωbath(Etotal))dln(Ωbath(Etotal))dE(ψ|Hsys|ψ)+d2ln(Ωbath(Etotal))dE2(ψ|Hsys|ψ)2+ln(Ωtotal(Etotal))

We are assuming that the number of states is dense enough that the derivative exists. Substitute the definition of entropy

=ln(Ωbath(Etotal))1kBdSbath(Etotal)dE(ψ|Hsys|ψ)+d2Sbath(Etotal)dE2(ψ|Hsys|ψ)2+ln(Ωtotal(Etotal))

Then the definition of temperature

=ln(Ωbath(Etotal))1kBT(Etotal)(ψ|Hsys|ψ)+d2Tbath1(Etotal)dE2(ψ|Hsys|ψ)2+ln(Ωtotal(Etotal))

Assume heat capacity is infinte

=ln(Ωbath(Etotal))1kBT(ψ|Hsys|ψ)ln(Ωtotal(Etotal))

then exponentiate both sides to get

ψ|ρsys|ψ=e1kBT(ψ|Hsys|ψ)Ωbath(Etotal)Ωtotal(Etotal)

and the extra term is just a normalizing constant so

ψ|ρsys|ψ=1Tr(Hsys)e1kBT(ψ|Hsys|ψ)

And since ψ is any eigenstate it is just the operator exponential.

ρsys=1Tr(Hsys)e1kBTHsys