Callen problem 15.1-1, two quantum oscillators

Callen problem 15.1–1.  (part one) "A system is composed of two harmonic oscillators, each of natural frequency ω_o and each having permissible energies (n +½) ħω_o, with n being any non–negative integer. Given that the total energy of the system is E′ = , where n′ can be any positive integer, calculate the number of microstates available to the system, and the entropy of the system." (See Chapter 15 in Thermodynamics and an Intro to Thermostatistics (1985) for background reading that leads up to this problem.)

The first thing to note is these aren't coupled oscillators.  Thus this is the simplest possible example of a Planckian oscillator system with zero-point energy. The oscillators can share the total energy available to them, but otherwise don't interact. Yeh, like photons themselves. As Peter Milonni (The Quantum Vacuum, chapter 2) says, an electromagnetic field mode is a (quantum) harmonic oscillator. Also, to be more precise, "mode" should be "normal mode".

The other thing to consider is indistinguishability versus distinguishability. Callen says nothing that I see in Chapter 15 or earlier about this, so we just rather automatically accept that the energy elements are indistinguishable. This makes it easier to count the microstates because permutations of the energy elements for a particular microstate don't need to be counted.

You can write the energy constraint of the two oscillators as

(i+½) ħω_o  +  (j+½) ħω_o  = n′ ħω_o ,

which gives

(i+½)   +  (j+½)   = n′ ,

i  +  j = n′ – 1,

j = n′ – i – 1

From this expression, you can just build a table of all possible i and j values, using the fact that the lowest “non-negative integer” is 0, not 1:

i	j
0	n′ – 1
1	n′ – 2
2	n′ – 3
…	…
n′ – 3	2
n′ – 2	1
n′ – 1	0

For any given positive integer n'  this table gives all the ways the energy can be shared between the two oscillators. So, just counting the number of rows in the table gives the number of microstates. Since we have to start counting from zero, the number of microstates is 

(n′ – 1) + 1 = n′ .

The “–1” comes from having to account for zero-point energy (ZPE) in the energy constraint, and the “+1” from starting the count at zero. So there are n′ microstates. Yeah, sort of an anticlimactic result—the number of microstates is just equal to the number of "energy elements" the system has.  With only two oscillators, whatever part of the total energy isn't in one is in the other. When there are three or more oscillators, permutations of the total number of oscillators among the energy states must be counted. (But--not to be dealt with in this problem--is this equal to the number of permutations of the total number of energy elements among the oscillators? That is, we can count oscillators per energy level, or energy elements per oscillator. Is the number of microstates the same for both methods of counting?)

 

Also, in this two-oscillator case, the equality of # of microstates and # of energy elements is only true when both oscillators have a zero-point energy, as you can see if you leave out the ZPE in the energy constraint equation above.

How many microstates does a single oscillator have? Remember, the given energy of the system is not the highest energy available because this is the thermally-isolated-system case, the microcanonical case, and the given energy is the total energy. A single oscillator with constant total energy (with or without ZPE) has one microstate.

As an example for our two oscillators, let n′ = 5.  Then i + j = n′ – 1 = 4, and there are five microstates: 

i	j
0	4
1	3
2	2
3	1
4	0

Solution to part one:  Since there are n′ microstates, the entropy is

S₁ =k log (n′) = k log (E′/ħω_o).

where log means natural log. I’ll come back later to this part of the problem to examine what kind of physical system this two–oscillator model might represent.

(What if the problem involved N oscillators without ZPE, a la Dr. Planckenstein in 1900?  I’ll look at that case below.)

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Now for part two of the problem:

"A second system is composed of two harmonic oscillators, each of natural frequency 2ω_o.  The total energy of this system is E″ = n″ħω_o, where n″ is an even integer.  How many microstates are available to this system, and what is its entropy?"

The energy constraint is now

(i+½) 2ħω_o  +  (j+½)2ħω_o  = n″ħω_o ,

(i+½)  +  (j +½)   = n″/2 ,

i  +  j = n″/2 – 1,

j = n″/2 – i – 1

Why does Callen give us this second part of the problem?  What does it tell us that’s not told in the first part of the problem?  This is the same relation between i and j as in part one, but with n″/2 instead of n′.  And n″/2 still gives us consecutive positive integers as possible values, since we are told n″ is an even integer. The table of available microstates is

 

i	j
0	n″/2 – 1
1	n″/2 – 2
2	n″/2 – 3
…	…
n″/2 – 3	2
n″/2 – 2	1
n″/2 – 1	0

 

There are (n″/2 – 1) + 1  =  n″/2 microstates.  The entropy is

S₂ = k log (n″/2).

This part of the problem definitely has some Planck-original-quantum-paper similarity to it, although it's entirely superficial: Planck used different numbers of "primes" on his variables (E, E′, E″, etc., for energy, for instance)  to indicate different frequencies  ω, ω', ω'', etc. And that's all we have here in part two, a different frequency.  But, the big thing with Planck's theory was and is that the energy of a quantum mechanical oscillator is determined by its single, perfectly monochromatic frequency (amazing!) and by integer multiples of that frequency.  I think this is the point of Callen's problem.  Is the number of microstates going to change just because we have a higher frequency, and thus a higher energy?  What's really different about this and part one? 

Consider this:  

E″ = n″ħω_o =  (n″/2) 2ħω_o, 

where the consecutive energy levels for n″/2 are 1, 2, 3, ... , since n″ is even. But this is just like counting the number of energy levels for part one.  What we have is n″/2  = n′,  and that means 

E″ =  (n″/2) 2ħω_o=  n′ 2ħω_o

Now the energy constraint is

(i+½) 2ħω_o  +  (j+½)2ħω_o  = n′ 2ħω_o ,

(i+½)  +  (j +½)   = n′  ,

i  +  j = n′ – 1,

j =n′ – i – 1

This is the same relation between i and j as in part one. The number of microstates is the same, n′ , so the entropy is the same.

Now for part three of the problem:  What is the entropy of the system composed of the two preceding subsystems if they are separated and enclosed by a totally restrictive wall? 

Since the subsystems are noninteracting in the most restrictive way possible--they can't exchange energy or particles (oscillators) with each other--their entropies just add together to give

S₁ + S₂  =  k log (n′) + k log(n″/2)

=  k log (E′/ħω_o) + k log(E″/2ħω_o)

=  k log (E′ E″/2ħ²ω_o²  ).

This is the answer Callen gives in the text. But based on my answer to part two above, this is equivalent to 

S₁ + S₂  =  k log (n′) + k log(n″/2)

 = k log (n′) + k log(n′)

= 2k log(n′)

=2k log(E′/ħω_o).

Now that the problem is solved,  I'll just relate it to the general problem of N oscillators by using Planck's (and before him, Boltzmann's) combinatorial formula for the total number of permutations of a system's "complexions" . This is the total number of microstates, and is the formula Planck used to calculate the entropy of his discrete-energy-level oscillators, with P being the number of energy elements and N being the number of oscillators:

(P+N–1)! / P!(N–1)!  

In part one of this problem we have N = 2 oscillators and P = n′ energy elements, giving the number of complexions as

(n′  + 2 – 1)!/ n'!1!  =  (n′ + 1)! / n′!  = n′ + 1.

We don't have the 1/2 + 1/2 ZPE terms here. With the zero-point energy included, the minus one as shown above in part one of the problem would cancel the plus one here, and we'd again have n′ microstates.

Now let N=3 oscillators and keep P=n′, no ZPE: 

(n′  + 3 – 1)!/ n'!2!  =  (n′ + 2)! / 2n′!  = (n′ + 2)(n′ + 1)/2.

Now let N=4 oscillators:

(n′  + 4 – 1)!/ n'!3!  =  (n′ + 3)! / 6n′!  = (n′ + 3)(n′ + 2)(n′ + 1)/6.  

(To be continued in a later post using Planck’s example with ZPE from his Theory of Heat Radiation, 2^nd edition, 1912.)

Planck included or postulated or uncovered the zero-point energy in a paper published in 1911, and included this derivation in his 1912 book The Theory of Heat Radiation (the 2nd edition, which is very different from the 1st edition of 1906, with ZPE being one of the big differences). Einstein and Stern in their 1913 paper discussed some consequences of ZPE, although its physical meaning caused confusion theoretically until the 1920s when quantum non-commutativity of certain operators representing physical quantities was discovered. 

Energy partitions part one

Now it’s time for a more general description of what Boltzmann and Planck did with their different versions of discrete energy elements distributed in equilibrium among their respective systems of molecules (Boltz) or resonators (Planck). This description goes by the name “energy partitions”. I'm not sure who invented this generic terminology, but it was put to good use in 1922 by Darwin and Fowler in their paper "On the partition of energy".

The reason for the invention of energy partitioning by Boltzmann and its later use by Planck was to calculate entropy.

Energy partitions describe the number of ways in which discrete energy elements can be distributed among a collection of particles capable of absorbing and emitting energy. You may not have encountered the name “energy partitions” before, but if you’ve studied thermodynamics and kinetic theory beyond the freshman level, you’ve encountered the equipartition theorem and the partition function. Since I’m providing links to those terms, I won’t describe them further except to say they are based on the idea of energy partitions.

 

Occupancy numbers and occupation numbers

 

There are two equivalent ways energy partitions can be expressed: the number of energy elements per particle, or the number of particles per energy level. The first use of energy partitions in a way that later became standard was by Boltzmann in 1877, in his paper “On the Relationship between the Second Fundamental Theorem of the Mechanical Theory of Heat and Probability Calculations Regarding the Conditions for Thermal Equilibrium.”

In this paper, Boltzmann described how to calculate the number of different ways kinetic energy elements could be partitioned among the molecules of a gas, and he considered his “molecules” to be structureless particles possessing only translational kinetic energy. In his model, portions of kinetic energy that are integer multiples of a very small indivisible element of energy, ε, are transferred between the molecules during elastic collisions. (Check on Goodstein or d’Abro or somebody and 1/3 of molecules traveling in each x, y, z direction, no angles of collision needed in calculations.)

 

A more rigorous mathematical description of Boltzmann’s energy partitions was developed by Alexander Bach in the late 1980s. (Bach claims that Boltzmann used the idea of indistinguishable classical particles, and also introduced Bose-Einstein statistics, which was supposedly not discovered until 1924 by Bose and Einstein. I’ll discuss both of these controversial claims later.) Bach’s scheme describes energy partitions in terms of cells that contain a certain number of particles, where the cells and particles can be either energy elements or physical particles (molecules or resonators in our case).  

 

Bach uses the term “occupancy number” for the number of energy elements per physical particle, and “occupation number” for the number of physical particles per energy level. (Bach also has a category called "configurations," but that would get us into the problem of particle distinguishability, which I'm not ready to do yet.)

 

Let’s look at occupancy numbers first. In Boltzmann’s notation the total kinetic energy of his gas molecules is L = λε and the number of molecules is n.  The overall occupancy number for his gas is therefore λ energy elements (particles) distributed among the n molecules (cells).  Individual-cell occupancy numbers are needed to calculate probabilities, and those are dealt with in the following section.

It may seem that just inverting the overall occupancy number description would give the overall occupation number as n molecules (particles) distributed over λ energy levels (cells), but here we run into a little counting problem. Or maybe I should say an accounting problem. 

 

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One part of the problem is that Boltzmann introduced another constant in addition to λ to count the number of energy elements, which, remember, are necessarily kinetic energy elements. He says “we assume that the kinetic energies of each molecule form an arithmetic progression, such as the following:  0, ε, 2ε, 3ε,…, pε.  We call P the largest value of the kinetic energy, pε."

But this little p seems to me to be the same as λ, because the largest possible kinetic energy any one molecule can have is the total kinetic energy. Then all others would have zero energy, and p would equal λ.

 

(This would never happen in a realistic setting with many molecules, with “never” meaning an infinitesimal probability of something like 1/n.  But we can imagine a video-game (not much of a game, but never mind that) in which n-1 molecules are at rest at the bottom of the screen, and we send in the n^th molecule from the top of the screen with kinetic energy pε = λε.  Eventually, this total energy would be distributed evenly, in an average way, among the molecules. Equilibrium would be established, with the usual mean-square fluctuations about the equilibrium value or average value of the kinetic-energy-per-molecule.)

 

But Boltzmann was no dummy.  He introduced p and λ for a reason.  He wanted “the highest possible value of the kinetic energy” to be different from the constant total energy. Apparently he meant this to be interpreted as the highest possible kinetic energy of any single distribution, with p = λ being only one such p, and a highly unlikely one at that.

 

But the strangest thing is that Boltzmann later gives an example in which he allows p to go to infinity while  λ = 7 (see below), so there’s some kind of madness in his method (rather than method in his madness), but it’s too much for me to juggle at the moment, and I’ll just continue with occupancy numbers and occupation numbers now.

 

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I have to backtrack a little bit, because of the accounting problem. Boltzmann’s states the main calculational purpose of his 1877 paper this way:  “I want to establish the probability of a state distribution regardless of how it is created, or more specifically, I want to find all possible combinations of the p+1 kinetic energy values allowed to each of the n molecules and then establish how many of these combinations correspond to each state distribution.”

 

But nobody counts zero as a value when we’re talking about occupancy numbers.  There are λ energy elements. There are n molecules.  No matter how many of them have zero energy, there are still λ energy elements distributed over n molecules.  Zero as an energy element is not counted as an occupancy number. However, again, Boltz was no dummy, and his (English-translated) statement is certainly correct: each molecule when considered independently of the others, has equal access to p+1 kinetic energy values.

But this is not very significant, since nobody uses occupancy numbers in calculations of probability anyway, except me (see below).  Occupation numbers are used instead, and since occupation numbers have cells that are energy levels, the zero-energy level must be counted. 

For occupation numbers, then, we have p+1 cells, and the overall occupation number is n molecules (particles) distributed over p+1 energy levels (cells). The counting problem I mentioned above is now accounted for.

 

Occupancy versus occupation calculations for n=3 and λ=p=5

 

Before any calculations are done, the total-number-of-particles constraint and the total-energy constraint must be expressed in equations. An occupancy-number variable and an occupation-number variable are needed in order to do this.

 

Occupancy number variable:  k_j = the number of energy elements in molecule j. The sum of the k_j’s from j = 1 to j = n is the total number of energy elements λ. Let’s say there are three molecules and five energy elements. Then

 

k₁ + k₂ + k₃ = 5.

 

This equation is both a constraint on the energy (the sum = λ = 5) and a constraint on the number of molecules (the upper limit of the sum = n = 3)

 

Occupation number variable: w_i is the number of molecules in energy level i. The sum of the w_i’s from i = 0 to i = λ is the total number of molecules n. Using three molecules and five energy elements tambien, we have

w₀ + w₁ + w₂ + w₃ + w₄ + w₅ = 3

 

0·w₀ + 1·w₁ + 2·w₂ + 3·w₃ + 4·w₄ + 5·w₅ = 5.

 

Now let’s make a table.  It's helpful to show occupancy and occupation number tables together.  So, here’s my two-tables-in-one, with occupancy numbers on the left, occupation numbers on the right, using n=3 and λ=5:

        occupancy #’s                                                occupation #’s

k1	k2	k3	---	w0	w1	w2	w3	w4	w5
5	0	0		2	0	0	0	0	1
0	5	0		2	0	0	0	0	1
0	0	5		2	0	0	0	0	1
4	1	0		1	1	0	0	1	0
4	0	1		1	1	0	0	1	0
0	4	1		1	1	0	0	1	0
1	4	0		1	1	0	0	1	0
1	0	4		1	1	0	0	1	0
0	1	4
3	2	0		1	0	1	1	0	0
3	0	2		1	0	1	1	0	0
0	3	2		1	0	1	1	0	0
2	3	0		1	0	1	1	0	0
2	0	3		1	0	1	1	0	0
0	2	3		1	0	1	1	0	0
2	2	1		0	1	2	0	0	0
2	1	2		0	1	2	0	0	0
1	2	2		0	1	2	0	0	0
3	1	1		0	2	0	1	0	0
1	3	1		0	2	0	1	0	0
1	1	3		0	2	0	1	0	0

 

(to be continued with permutations, combinations, particle distinguishability (or not), and the multinomial theorem)