Reading course in statistical physics

[Philip Koeck]

This is the beginning of an online reading course of the book "Statistical physics" by Reif, volume 5 in the Berkeley physics course, using PF.
We'll start with chapter 3 and loop back to the initial 2 chapters if necessary.
All questions should be specifically about what is written in the this textbook.
Depending on how long the thread gets we might start a new thread for each chapter.

So the thread is officially open for questions about chapter 3 (assuming this is allowed on PF).

 

[Philip Koeck]

A question about section 3.1: A quantum mechanical description of an ideal gas or N particles in a 3D-box requires 3N quantum numbers (plus the information that the particles are in a box of given dimensions).
If you want to describe the same system classically you need twice as many numbers, 3N position coordinates and 3N momentum components.

Is it generally true that you need more information classically, maybe twice as much?

Somewhere in this chapter it says that you only need 3N values classically as well, but I can't find that passage at the moment.

 

[vanhees71]

It's the difference between what you call a quantum and a classical state. In quantum theory a state is completely determined if a complete set of compatible observables take determined values, i.e., then it's in a simultaneous eigenstate of all the self-adjoint operators representing the observables in this compatible set. You can choose the 3N position coordinates or (!) the 3N momenta but not both, because the position coordinates and the momenta are not compatible.

In classical mechanics the state is completely specified by a point in phase space, i.e., by 3N position coordinates and 3N canonically conjugated momenta.

Nevertheless one should be aware that this is still somewhat besides the point, because in the quantum case, if you have a gas of indistinguishable particles it doesn't even make sense to specify the 3N positition coordinates of each individual particle, because you cannot follow individual particles along some determined path as within classical mechanics. That's why you end up with bosons or fermions, where the state space are spanned by the fully symmetrized (antisymmetrized) $N$-particle products of single-particle basis states. This is why the right specification of a many-particle state is rather in terms of occupation numbers of single-particle states, i.e., you define a single-particle basis (e.g., the momentum eigenvectors) and then specify the state of the many-particle system by the number of particles in each single-particle state. This is most conveniently formalized using creation and annihilation operators, obeying bosonic or fermionic commutation relations, wrt. the chosen single-particle basis.

 

[Philip Koeck]

In section 3.3 equations 17 to 19 postulate that the following 2 statements are logically equivalent.
"The system is in equilibrium" and "All accessible states have the same probability".

Just to make sure:
For a system containing two indistinguishable particles and two single particle states (called 1 and 2) with the same energy, are the possible states the following 3?
Both particles in state 1
Both in state 2
One particle in each state (without knowing which is in which)

Does each of these states have the same probability 1/3 ?

(I have also heard the opinion that the probabilities should be 1/4, 1/4 and 1/2, respectively.)

 

[vanhees71]

It depends on whether you have fermions or bosons. Let $|u_j \rangle$ with $j \in \{1,2\}$ be the single-particle energy eigenstates with energy eigenvalues $E_1=E_2=E$. For fermions the two-particle Hilbert space are spanned by the antisymmetized product states. In this case we have only 1,
$$|U \rangle=\frac{1}{\sqrt{2}} (|u_1 \rangle \otimes |u_2 \rangle - |u_2 \rangle \otimes |u_1 \rangle,$$
while for bosons it's spanned by the symmetrized product states. In this case we have 3:
$$|U_1 \rangle=|u_1 \rangle \otimes |u_1 \rangle, \quad |U_2 \rangle = |u_2 \rangle \otimes |u_2 \rangle, \quad |U_3 \rangle=\frac{1}{\sqrt{2}} (|u_1 \rangle \otimes |u_2 \rangle + |u_2 \rangle \otimes |u_1 \rangle.$$
For fermions there's only one possibility, i.e., the two particles are with certainty in the state $\hat{\rho}=|U \rangle \langle U|$.

For the bosons we have to maximize the entropy,
$$S=-k_{\text{B}} \mathrm{Tr} \hat{\rho} \ln \hat{\rho}.$$
This must be maximized under the constraint that $\mathrm{Tr} \hat{\rho}=1$. With the Lagrange-multiplier method that means that
$$\delta S + \lambda \mathrm{Tr} \delta \rho=\mathrm{Tr} \delta \hat{\rho} (-k - k \ln \hat{\rho}+\lambda)=0.$$
Since this must hold for all $\lambda$ the expression in the bracket must vanish, leading to
$$\hat{\rho}=\exp[(\lambda-1)/k]=\text{const}.$$
From $\mathrm{Tr} \hat{\rho}=1$ we get
$$\hat{\rho}=\frac{1}{3}.$$

 

[Philip Koeck]

It depends on whether you have fermions or bosons. Let $|u_j \rangle$ with $j \in \{1,2\}$ be the single-particle energy eigenstates with energy eigenvalues $E_1=E_2=E$. For fermions the two-particle Hilbert space are spanned by the antisymmetized product states. In this case we have only 1,

Yes, I was thinking of bosons. Forgot to say.

 

[Philip Koeck]

For the bosons we have to maximize the entropy,
$$S=-k_{\text{B}} \mathrm{Tr} \hat{\rho} \ln \hat{\rho}.$$
This must be maximized under the constraint that $\mathrm{Tr} \hat{\rho}=1$. With the Lagrange-multiplier method that means that
$$\delta S + \lambda \mathrm{Tr} \delta \rho=\mathrm{Tr} \delta \hat{\rho} (-k - k \ln \hat{\rho}+\lambda)=0.$$
Since this must hold for all $\lambda$ the expression in the bracket must vanish, leading to
$$\hat{\rho}=\exp[(\lambda-1)/k]=\text{const}.$$
From $\mathrm{Tr} \hat{\rho}=1$ we get
$$\hat{\rho}=\frac{1}{3}.$$

I was also wondering whether equations 17 to 19 really are a postulate, as the book states.
Is the above a proof of this "postulate"?
If one accepts that entropy is a maximum at equilibrium then the above seems to show that the probabilities of all three possible states have to be equal, at least for the simple system considered.
Is it possible to prove this "postulate" more generally or does that only work for very simple systems?

 

[vanhees71]

I've only the German translation, but I guess the Section and Equation numbering is the same. As mentioned in a footnote of the German edition you can approach statistical physics from a dynamical point of view. For classical mechanics the tool is the Hamiltonian formulation and its application to phase-space distribution functions, whose time evolution leads to Liouville's theorem, according to which the phase-space volume is an invariant under the Hamiltonian time evolution. From this it follows that the uniform distribution on the available phase space (contrained by the conservation laws, i.e., choosing a (non-rotating) center-of-mass system as an inertial frame only by energy conservation) is the equilibrium solution.

If you read further it turns out that this is equivalent to the maximum-entropy principle for equilibrium, and also this principle can also famously be derived from the underlying microscopic dynamics, leading to Boltzmann's H-theorem, but the approach of this book is more basic, restricting the considerations strictly to equilibrium, and I think it does a good job in heuristically arguing to make the maximum-entropy principle plausible.

 

[Philip Koeck]

I'm lingering on this point because in another discussion (outside PF) I heard the opinion that for this example of 2 indistinguishable bosons the probabilities of the 3 microstates (or arrangements) are 1/4, 1/4 and 1/2.
The reason given is that the arrangement with 1 particle in each of the two single particle states actually consists of 2 arrangements, which we can't distinguish since we can't distinguish the particles, but they are still somehow there in the background.
If this is true then the arrangements of indistinguishable particles have to be counted the same way as those of distinguishable ones.

(I tried to prove that the probabilities of the different microstates are equal here https://www.researchgate.net/publication/355396651_The_transition_matrix_model, but this is also only for a very simple model.)

 

[vanhees71]

No, these are not two different arrangements. The product states $|u_1 \rangle \otimes |u_2 \rangle$ and $|u_2 \rangle \otimes |u_1 \rangle$ are not even in the 2-boson Hilbert space. It's only the one symmetric combination $(|u_1 \rangle \otimes |u_2 \rangle - |u_2 \rangle \otimes |u_1 \rangle)/\sqrt{2}$. All the three states have the same energy $E_1+E_2=2E$ in this case and thus the equilibrium distribution is a probability of 1/3 to be in each state. It's the maximum-entropy state $\hat{\rho}=\hat{1}/3$.

If the particles are distinguishable you have 4 states and the equilibrium distribution is $\hat{\rho}=\hat{1}/4$.

The general property the H-theorem follows from by coarse graining is the unitarity of time evolution (or in the usual derivation of the Boltzmann equation) the corresponding unitary for the S-matrix.

 

[Philip Koeck]

I'm surprised about a result in section 4.3, equation (30).
Why is it not 1/2 k T per degree of freedom?

 

[vanhees71]

I interpret it as giving an "order-of-magnitude estimate". Of course you are right: Any phase-space-degree of freedom which occurs quadratically in the Hamiltonian provides an energy $k T/2$ (in the classical limit).

 

[Philip Koeck]

I'm looking at section 4.1 now and I've also glanced at section 4.7, where the ideal gas is discussed.

My impression from 4.1 was that when the systems A and A' are combined to A* the particles in the two systems are still kept apart by some heat-conductive barrier.
On the other hand in 4.7 a single molecule is treated as A and the rest of the gas as A'.

Is A* to be understood as a combination of A and A' where the barrier between the two has been removed?

 

[Philip Koeck]

I'm looking at section 4.1 now and I've also glanced at section 4.7, where the ideal gas is discussed.

My impression from 4.1 was that when the systems A and A' are combined to A* the particles in the two systems are still kept apart by some heat-conductive barrier.
On the other hand in 4.7 a single molecule is treated as A and the rest of the gas as A'.

Is A* to be understood as a combination of A and A' where the barrier between the two has been removed?

Maybe I should start from the other end. I'm trying to understand how the canonical approach is applied to the ideal gas in section 4.7.
Should one think of the system of interest A as one molecule that is free to fly around in the whole volume available to A* or is this molecule confined to a much smaller volume covering the immediate vicinity of the molecule before A and A' are mentally joined to form A*?

 

[Philip Koeck]

Maybe I should start from the other end. I'm trying to understand how the canonical approach is applied to the ideal gas in section 4.7.
Should one think of the system of interest A as one molecule that is free to fly around in the whole volume available to A* or is this molecule confined to a much smaller volume covering the immediate vicinity of the molecule before A and A' are mentally joined to form A*?

I'll give this another shot:

The text states that when A and A' are combined to A* the number of microstates multiplies.
If I think of A as a single atom in a small volume then the number of energy levels available to this atom is quite small since the total energy is fixed and the energy levels are widely spaced due to the limited space.

Now when A and A' are joined the atom from A suddenly has much more space so the energy levels available to A increase a lot in number.
For me that would mean that the number of microstates should be bigger (by quite a large factor) than the product of the microstates of A and those of A'.

If system A is a single atom in the total volume V* to start with then I can accept that the number of microstates simply multiply.
Is that how one should picture this example?

I'm just trying to make sense of this idea of a single atom in a gas being the system and all the other atoms the heat bath.

 

[vanhees71]

If you have two systems $A$ and $A'$ these have energy states $|E,\alpha \rangle$ and $E',\beta' \rangle$, where $\alpha$ and $\beta$ are labels which list the different eigenvectors for each eigenvalue $E$ and $E'$ of the Hamiltonians $\hat{H}$ and $\hat{H}'$.

Then the combined system is described in the direct product of the two Hilbert spaces $\mathcal{H} \times \mathcal{H}'$. The degeneracy of the energy eigenstates, which is what is to be counted as the "available microstates" becomes of course much larger in the combined system.

Take as the most simple example a gas of non-interacting particles. Then the momentum eigenvectors are the energy eigenvectors, and the degeneracy is a spherical shell in momentum space. For $N$ particles this is a $3N-1$-dimensional hypersurface in momentum space, i.e., for the combined system it's $3(N_A+N_B)-1$-dimensional, i.e., you have much more available states per particle.

 

[Philip Koeck]

If you have two systems $A$ and $A'$ these have energy states $|E,\alpha \rangle$ and $E',\beta' \rangle$, where $\alpha$ and $\beta$ are labels which list the different eigenvectors for each eigenvalue $E$ and $E'$ of the Hamiltonians $\hat{H}$ and $\hat{H}'$.

Then the combined system is described in the direct product of the two Hilbert spaces $\mathcal{H} \times \mathcal{H}'$. The degeneracy of the energy eigenstates, which is what is to be counted as the "available microstates" becomes of course much larger in the combined system.

Take as the most simple example a gas of non-interacting particles. Then the momentum eigenvectors are the energy eigenvectors, and the degeneracy is a spherical shell in momentum space. For $N$ particles this is a $3N-1$-dimensional hypersurface in momentum space, i.e., for the combined system it's $3(N_A+N_B)-1$-dimensional, i.e., you have much more available states per particle.

In section 4.1 equation (4) states that the number of microstates in the combined system, Ω* is the product of Ω and Ω', but the way you describe it I get the impression that Ω* should be much larger than this product.
Am I misunderstanding something?
Equation (4) is central to the whole argument in chapter 4, I believe.

 

[vanhees71]

It's a tensor product. If you put your particles in a (large) "quantization volume" (e.g., a cube of length $L$, imposing periodic boundary conditions for the wave function, for convenience) you have the discrete momentum eigenvalues and true momentum eigenvectors. Then the available states for given energy (micorcanonical ensemble) fufill the properties named in the book.

 

[Philip Koeck]

It's a tensor product. If you put your particles in a (large) "quantization volume" (e.g., a cube of length $L$, imposing periodic boundary conditions for the wave function, for convenience) you have the discrete momentum eigenvalues and true momentum eigenvectors. Then the available states for given energy (micorcanonical ensemble) fufill the properties named in the book.

Would you say that A consists of one atom in the volume V*?
That would mean that when A and A' are combined to A* the volume doesn't actually change.
This question only applies to how Reif discusses the ideal gas in section 4.7.

 

[vanhees71]

Do you mean Sect. 4.1? There he uses the microcanonical ensemble, and I don't see a problem with the argument there. You have two systems A and A' which are then "thermally coupled" to a total system $\mathrm{A}^*$, which means the systems can exchange energy. Then the total energy of the closed system $\mathrm{A}^*$, $E^*=E+E'=\text{const}$. Then the number of available states is indeed related by
$$\Omega^*(E^*)=\Omega(E) \Omega'(E')=\Omega(E) \Omega(E^*-E).$$
This factorization assumption of course relies on a classical picture, i.e., for any state of A, A' can be in any of its states and vice versa, i.e.

 

[Philip Koeck]

Do you mean Sect. 4.1? There he uses the microcanonical ensemble, and I don't see a problem with the argument there. You have two systems A and A' which are then "thermally coupled" to a total system $\mathrm{A}^*$, which means the systems can exchange energy. Then the total energy of the closed system $\mathrm{A}^*$, $E^*=E+E'=\text{const}$. Then the number of available states is indeed related by
$$\Omega^*(E^*)=\Omega(E) \Omega'(E')=\Omega(E) \Omega(E^*-E).$$
This factorization assumption of course relies on a classical picture, i.e., for any state of A, A' can be in any of its states and vice versa, i.e.

I'm trying to understand how the ideas of section 4.1 are applied to the ideal gas in section 4.7.

We can take a different example to start with:
If system A is a small lump of a solid that is brought into thermal contact with another solid A', then I completely accept that Ω* = Ω x Ω'.
In that case A doesn't acquire more space in any way due to being combined with A'.

In the case of the ideal gas the situation seems different to me.
If A is one atom in a small volume then it will have a much larger volume available after being combined with A'. To me that means that A has more available microstates afterwards and the total number of of microstates after combining should be larger than the product Ω x Ω'.

If, however, A consists of one atom in the combined volume V* then I can accept the argument, but Reif doesn't actually say one way or the other.

 

[vanhees71]

For a gas the volumes are just given "external parameters", it's just the volume of the vessel you use to contain your gas.

For the grand-canonical ensemble you can simply consider some partial volume (which is just thought, i.e., it isn't somehow materially realized) within a big volume given by your container. There both energy exchange and particle exchange is possible, i.e., in the grand-canonical ensemble only the averages of the relevant additive conserved quantities (energy, momentum, angular momentum, conserved particle number or charges) are given, and then the entropy is maximized under these constraints. The Lagrange multipliers then define temperature, center-of-momentum velocity, thermal vorticity and chemical potential(s).

For the canonical ensemble you have to consider the partial volume as somehow realized such that the particle number (or conserved charge) is fixed within this partial volume but energy can be exchanged. Then the fixed particle number (conserved charge) is a fixed external parameter but only the expectation value of the energy is fixed and you maximize entropy under this constraint.

 

[Philip Koeck]

For the canonical ensemble you have to consider the partial volume as somehow realized such that the particle number (or conserved charge) is fixed within this partial volume but energy can be exchanged. Then the fixed particle number (conserved charge) is a fixed external parameter but only the expectation value of the energy is fixed and you maximize entropy under this constraint.

I think Reif only considers the canonical ensemble in section 4.7.

So one should imagine that, for some reason, the atom that is treated as system A somehow doesn't escape its limited partial volume even after A and A' have been combined.
Is that what Reif is thinking of?

To me it's a strange idea, but I guess I can accept it.

 

[vanhees71]

If you have canonical ensemble by definition the particles in the partial volume can't leave this volume, i.e., particles cannot be exchanged between the parts of the larger system but only energy. One can think of a container made of a metal with good "heat conduction" kept at a given temperature through a much larger surrounding "heat bath".

 

[Philip Koeck]

If you have canonical ensemble by definition the particles in the partial volume can't leave this volume, i.e., particles cannot be exchanged between the parts of the larger system but only energy. One can think of a container made of a metal with good "heat conduction" kept at a given temperature through a much larger surrounding "heat bath".

Sure, but that isn't what Reif is discussing.
In 4.7 he writes that the system A is 1 atom of an ideal gas.
So the gas as a whole is A* and there is no boundary between A and A'.

Right in the beginning of the section Reif defines conditions (i) and (ii).
A bit later he writes: "By virtue of condition (ii) we can consider a particular atom as a small system in thermal contact with a heat bath (A', my addition) consisting of all the other atoms."

 

[vanhees71]

Yes, this is the justification for using a classical "point-particle picture", i.e., if the mean distance between the molecules/atoms is much larger than the de Broglie wavelength of these particles you can count them as individually identifiable single particles. This means the gas is "nondegenrate", i.e., the indistinguishability in the sense of quantum theory does not play much of a role, i.e., Bose-Einstein or Fermi-Dirac statistics is well-approximated by the classical Boltzmann statistics. At least in my German edition of the Berkeley Physcis Course Reif does not discuss quantum statistics.

 

[Philip Koeck]

I've found another online-book that might be better.
Here's the link: https://farside.ph.utexas.edu/teaching/sm1/Thermal.pdf
It was also suggested by Greg Bernhard some time ago (https://www.physicsforums.com/threa...echanics-an-intermediate-level-course.220906/). He gives a link with a pdf of an older version of the same book.

My impression is that the main chapter (chapter 5) follows Reif very closely, so all the discussion so far is applicable.
It also treats quantum statistics very early on and it's a bit more modern and quite compact.

So far I've read chapters 3 and 4 and found them very clear. There's a nice discussion of the H-theorem for example.
Chapter 1 is intro and 2 is mathematical basis, so I just glanced through those.

Would this book be interesting?
I'm definitely looking forward to chapter 5.

 

[vanhees71]

I've seen some of Fitzpatrick's manuscripts and found all excellent.

 

[Philip Koeck]

Actually I don't get the last step of equation 4.34 on page 50 in Fitzpatrick's book (the current online-version).

Why is ∂σ/∂x = X' / τ (and also ∂σ/∂y = Y' / τ) ?

 

[vanhees71]

It's defined by
$$\mathrm{d} G=\mathrm{d} x X' + \mathrm{d} y Y'=0$$
leading to curves defined implicitly by
$$\sigma(x,y)=c=\text{const}$$
Then you have
$$\mathrm{d}_x \sigma=\partial_x \sigma + y' \partial_y \sigma=\partial_x \sigma -X'/Y' \partial_y \sigma=0$$
From this
$$Y' \partial_x \sigma = X' \partial_y \sigma,$$
and this is a function of $(x,y)$, which I can write in the Form $X' Y'/\tau$. This shows that for functions with 2 independent variables for any inexact differential there's always an integrating factor, $\tau$, such that
$$\mathrm{d} G/\tau=\mathrm{d} \sigma$$
is an exact differential.

 

[Philip Koeck]

From this
$$Y' \partial_x \sigma = X' \partial_y \sigma,$$
and this is a function of $(x,y)$, which I can write in the Form $X' Y'/\tau$. This shows that for functions with 2 independent variables for any inexact differential there's always an integrating factor, $\tau$, such that
$$\mathrm{d} G/\tau=\mathrm{d} \sigma$$
is an exact differential.

Got it. The two expressions are the same so I can replace both with a 3rd expression.
The integration factor τ, which depends on x and y in general, is used to make the replacement correct. Something like that.

It will be interesting to read in the next chapter how this is applied to dS = d^-Q / T.
(Couldn't find a crossed-out d.)
Somehow one also has to realize that this equation is only valid for a d^-Q in a reversible process.

 

[Philip Koeck]

I have a rather general question about the definition of entropy used in chapter 5:
S = k ln Ω, where Ω is the number of available microstates.
Boltzmann wrote W rather than Ω, and I believe this stood for probability (Wahrscheinlichkeit).
Obviously this is not a number between 0 and 1, so it's more like something proportional to probability.

Probability would be number of available microstates divided by total number of microstates (including those that are not available).
Now for distinguishable particles both these numbers are bigger than for indistinguishable particles, by a factor N!, where N is the number of particles, in the case of low occupancy.

Would it make sense therefore to use the following definition of entropy for distinguishable particles to make sure that this "probability" W is calculated correctly?
S = k ln (Ω / N!) for distinguishable particles at low occupancy.

 

[Philip Koeck]

I have a rather general question about the definition of entropy used in chapter 5:
S = k ln Ω, where Ω is the number of available microstates.
Boltzmann wrote W rather than Ω, and I believe this stood for probability (Wahrscheinlichkeit).
Obviously this is not a number between 0 and 1, so it's more like something proportional to probability.

Probability would be number of available microstates divided by total number of microstates (including those that are not available).
Now for distinguishable particles both these numbers are bigger than for indistinguishable particles, by a factor N!, where N is the number of particles, in the case of low occupancy.

Would it make sense therefore to use the following definition of entropy for distinguishable particles to make sure that this "probability" W is calculated correctly?
S = k ln (Ω / N!) for distinguishable particles at low occupancy.

Maybe I should move this to a new post.