This post relates to the two preceding posts concerning counting
the number of microstates available to two quantum harmonic oscillators.
The total energy is shared by the oscillators, but the
problem doesn’t say what the mechanism of energy transfer between oscillators
is, or what kind of energy it is (acoustic = phonons, or electromagnetic =
photons). As stated, the problem is not
physical. It’s merely a counting problem based on having two oscillators
(boxes), one with i units of energy and the other with j energy
units, and n’ total energy units (balls) to distribute in the
oscillators. Plus there’s always the ZPE of one-half-a-ball in each of the two
boxes. This ½-ball of ZPE in each box sums up to be one ball that can’t be
counted when we are asked how many microstates there are, as shown by the
energy constraint equation
i + j = n′ – 1
There’s also the undiscussed issue of the
quantum energy elements being identical and indistinguishable. Using the
example of n′ = 5, we have n′ – 1 = 4 indistinguishable balls to
be distributed, which I did using a table in my post of Callen's problem 15.1-1. Now I’m going to use a different method of counting the number
of microstates for indistinguishable balls and then use the same method to
count states for distinguishable balls.
If we had four distinguishable balls we could label them
ABCD, but in the quantum oscillator problem with n′ – 1 = 4, we
have indistinguishable balls, which we can label AAAA. We can call them letters
now instead of balls and use a divider | to show the number of ways these
letters can be distributed in the two boxes (two oscillators), starting with
zero letters in the left box and ending with zero letters in the right box:
|AAAA A|AAA
AA|AA
AAA|A AAAA|
It’s easy to see from this partitioning that there are
only five microstates, because permutations of the letters are not
distinguishable. This indistinguishability applies to energy elements in the
two quantum oscillator problems, Callen's 15.1-1 and 15.1-2.
For a (supposedly) classical physics comparison, we can
do a similar partitioning with four distinguishable letters ABCD, starting with
the five arrangements as above
|ABCD A|BCD
AB|CD
ABC|D ABCD|.
But now we must also count the distinguishable
permutations of the letters among the two boxes.
First we can visually count the number of permutations in
each of the above arrangements. Only permutations or exchanges of letters
between the boxes (across the partition) are counted. For multiple letters inside
a box, their permutations among themselves are not counted as part of the total
number of microstates. So we DO have indistinguishability at this single-box
level in classical physics.
The first and last arrangements have zero balls in
one box, so no exchanges between boxes are possible, and each of these
arrangements represents ONE microstate. The
second and fourth arrangements each have FOUR microstates, because each of the
four letters can be permuted into the single-letter box. The middle
arrangement’s permutations can be visually counted
by showing all the possible distinguishable arrangements between the two boxes.
There are SIX such arrangements:
AB|CD AC|BD AD|BC CB|AD DB|AC CD|AB
.
The total number of microstates is thus 2 +
8 + 6 = 16 when we consider the energy elements to be distinguishable.
Since 16 equals 24, and we have
two boxes and four distinguishable letters, this would seem to indicate a
general way of calculating the number of microstates, namely mn,
where n is the number of distinguishable energy elements to be
distributed in m oscillators. Nice formula! Where does it come?
Going back to four “balls” and two “boxes”,
we know we have to put all four balls into the two boxes (because each
arrangement must have the same total energy). How many ways are there to do
this? There's no preference for putting any ball in either box, so there are
two ways of placing the first ball, two ways of placing the second ball
(independent of where the first ball is placed), and so on, giving 2x2x2x2
= 24 ways of
putting the four distinguishable balls in the two boxes. This is due to the
randomness of the selection of the boxes. For each ball we can flip a coin to
determine which box it goes in.
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This section was rewritten on 24 Oct 2020. In fact, this method will work for indistinguishable balls also as long as we have them lined up, as we must if we want to write them down: AAAA. They are distinguishable by their positions, even though they aren’t labeled. We could say they are localized by the somewhat unavoidable process of writing or printing them. We’re now talking about an ordered process.
Before beginning the process, the four A’s can be rearranged and these arrangements are indistinguishable. But once we begin the process of putting the A’s in boxes, we can observe which A is put in which box: Pull out the right-most A and put it in a box, pull out the next one, the next and then the left-most A. This selection and placement process is reversible, because the information of where each A has been placed is necessarily known.
Well, actually, there are two cases to consider as the A's are being placed inside the boxes. Normally, there is no order within the box, so when more than one A is placed in a box, there's no way to know if the same A is chosen when the process is reversed. To make the process reversible, we have to violate this usual condition by, for instance, stacking the letters on top of each other, a "first in, last out" arrangement. If we do this, the process is reversible, the number of unique ways of placing the four A's can be counted. Therefore, in this ordered process the four otherwise indistinguishable A's have 24 possible arrangements just like the distinguishable letters ABCD.
This is similar to labeling the corners of a square with the letters AAAA versus labeling them with ABCD. Normally, rotations by 90 degrees of the AAAA square about its center are "changes that don't make a difference,"* and these preserve the symmetry of the square. But if you actually perform the rotations (this is the ordered process), and count the number of rotations, then the process can be reversed, because the recorded information can be used to distinguish the otherwise indistinguishable positions of the square.
Even with the corners labeled ABCD, there is no way to tell whether the rotation was 90 degrees, for instance, or (360 + 90) degrees if the rotations haven't been observed.
But for the moment I will consider selections to be random and not ordered, so that the AAAA arrangements are indistinguishable except for the five partitions
|AAAA A|AAA AA|AA AAA|A AAAA|,
as discussed at the beginning of this
now-about-to-end screed.
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Being “identical” or not identical is a
property we think of as independent of observation. Being indistinguishable or
distinguishable is more of a result of an observation, or an interaction, a
result dependent on the method of observation and more applicable to arrangements
than to individual particles.
However, the concepts of identity and
distinguishability both have limits of observation that prevent us from knowing
if we’re missing something when we make our judgement. Identity—the
property of two or more things being identical or not—belongs more to classical
mechanics, and indistinguishability—the concept of two or more things not being
distinguishable from each other—belongs more to quantum mechanics. But does
either of the concepts belong totally to classical mechanics or totally to
quantum mechanics? There’s ambiguity involved that maybe depends on the method
of measurement, sort of like the ambiguity involved in the wave-particle
duality.
*This is Sean Carroll's definition of symmetry, expressed in his early books and DVDs. More recently he has said on his website, "I think of symmetries as transformations that don't change anything physical.".
