# Clarification: Microstate

1. Sep 12, 2006

### ghotra

I'm looking for some insight/clarification on the definition of a microstate.

Consider two coins. Here are the possible outcomes:

T T
T H
H T
H H

Now, I have assumed something: The coins are distinct.

A better description of each microstate is:

{ (coin 1, T), (coin 2, T) }
{ (coin 1, T), (coin 2, H) }
{ (coin 1, H), (coin 2, T) }
{ (coin 1, H), (coin 2, H) }

That is, the microstate isn't _just_ the face of the coin but it also includes _which_ coin had which face.

Now suppose the coins were identical. Then, we can't distinguish coin 1 from coin 2. So we might as well drop the label. The microstates are now:

{ (T, T) }
{ (T, H) }
{ (H, H) }

That is, the microstate is specified _only_ by the face of the coin.

Now, what about real life? When I toss two coins, what is the probability that I receive exactly 1 head? Everyone knows that the answer is 1/2 and not 1/3.

Why? Why is the answer 1/2 when the coins are clearly identical (in the sense that I am restricting my interest to the "face of the coin" and not "which face was on which coin")?

2. Sep 12, 2006

### Mk

Pulling from the article on Satyendra Nathan Bose at Wikipedia:

While at the University of Dhaka, Bose wrote a short article called Planck's Law and the Hypothesis of Light Quanta, describing the photoelectric effect and based on a lecture he had given on the ultraviolet catastrophe. During this lecture, in which he had intended to show his students that theory predicted results not in accordance with experimental results, Bose made an embarrassing statistical error which gave a prediction that agreed with observations, a contradiction.

What are the possibilities of flipping two coins? Two heads/Two tails/One of each. But aren't the coins distinct? Since the coins are distinct, there are two outcomes which produce a head and a tail. The probability of two heads is one-fourth, not one-third. The error was a simple mistake that would appear obviously wrong to anyone with a basic understanding of statistics, and similar to arguing that flipping two fair coins will produce two heads one-third of the time. However, it produced correct results, and Bose realized it might not be a mistake at all.
Outcome probabilities

Coin 1
Tail TH TT

Since the coins are distinct, there are two outcomes which produce a head and a tail. The probability of two heads is one-fourth.
Physics journals refused to publish Bose's paper. It was their contention that he had presented to them a simple mistake, and Bose's findings were ignored. He wrote to Albert Einstein, who immediately agreed with him and loved the idea. Physicists stopped laughing when Einstein sent Zeitschrift für Physik to accompany Bose's, which were both published in 1924. Hee hee.

3. Sep 13, 2006

### ghotra

Maybe this didn't come through clearly.

Why are the two coins distinct?

I guess I was thinking that distinguishablity was a function of my interest. If I am only interested in the face of the coin, then all coins are identical. Obviously this line of thinking is wrong. I promise, I'm not trying to be a pain. Fundamentally, what is it that makes these two coins distinct?

Is it a premise of physics that only fundamental particles are indisinguishable? Then is it the totality of the fundamental particles that make an object distinguishable? Photons and electrons are indistinguishable particles. What about protons (these are _not_ fundamental particles)?

4. Sep 13, 2006

### ghotra

...any takers on this? Surely it isn't too difficult.

5. Sep 13, 2006

### Mk

The thing is, the coins are distinct or indistinct depending on how you want to apply.

6. Sep 13, 2006

### ghotra

There is no way that this is correct. The probability of getting a single head when tossing two real coins is always 1/2 and never 1/3.

This is _interesting_. Even if I personally am unable to distinguish two coins from one another (thus, they are identical to me and to what I care about), the coins still behave as if they are distinguishable. In theory, one can have two identical, macroscopic coins. In reality, macroscopic coins are _always_ distinguishable, depsite my ability to tell them apart. There is something intrinsic going on here, and I don't understand it.

I think this has something to do with the classical limit, but I'd like someone to comment on this. Photons and electrons are identical....when you have an ensemble of photons and electrons, somehow the ensemble becomes distinguishable. What is responsible for making something distinguishable. Is a proton distinguishable or indistinguishable---it is not a fundamental particle in the sense that it is composed of quarks, which are identical.

7. Sep 14, 2006

### marcusl

Writing down the ensemble state names or descriptors is not the whole story, you also need their multiplicities. Your example then becomes
1 { (T, T) }
2 { (T, H) }
1 { (H, H) }

and now your probabilities come out correct. These are more appropriately called macrostates (as opposed to the microstates you correctly identified in your email) and if you calculate the probabilities of finding each macrostate (they are 1/4, 1/2, 1/4) you are starting to do statistical mechanics. For the simple case of indistinguishable particles as found both in your case and in typical thermodynamics problems, all microstates in a macrostate are equally probable so the probabilities are simply found from the multiplicities. This is how Boltzmann calculated the entropy of a macrostate.

8. Sep 14, 2006

### ghotra

You have implicitly assumed the coins are distinct. If the coins are identical, then there are only 3 microstates. This is exactly what Bose-Einstein statistics is about. From http://en.wikipedia.org/wiki/Satyendra_Nath_Bose: [Broken]

"Because photons are indistinguishable from each other, one cannot treat any two photons having equal energy as being different from each other. By analogy, if the coins in the above example behaved like photons and other bosons, the probability of producing two heads would indeed be one-third. Bose's "error" is now called Bose-Einstein statistics"

So let's state this clearly:

If the coins are identical, then the probability of getting a single H is 1/3 and not 1/2. There are only 3 microstates. In this case, the macrostates for "total number of heads" has a 1-1 correspondence with the microstates.

If the coins are distinct, the the probability of getting a single H is 1/2 and not 1/3. There are 4 microstates. In this case, there are 2 microstates for the macrostate "total number of heads is 1" and there is not a 1-1 correspondence with the microstates.

My question is this: Why are two macroscopic coins always considered to be distinct, even if I cannot distinguish them? That is, it is _impossible_ to have two identical, macroscopic coins; thus, the probability of getting a single head when flipping two real, physical coins is always 1/2 and not 1/3.

Last edited by a moderator: May 2, 2017
9. Sep 17, 2006

### ghotra

Here are some thoughts. Please critque them.

When speaking informally, we write the microstate as:

(for distinct coins)
HH
HT
TH
TT

However this just shorthand as marcusl implied. In physics, we cannot specify a microstate as above.

We are treating the coins as independent. In principle, we can construct two incredibly complex wave functions describing each of the coins. If the overlap of these wavefunctions is essentially zero, then perhaps we take this to mean that the coins are distinguishable. That is, the coins are distinguishable in the sense that their wavefunctions will never put them in the same expected position/momentum.

If we take this as true, then we can just treat the coins as distinguishable and the "microstates" derive nicely as HH, HT, TH, TT.

Please comment on the above. If acceptable, it seems the moral is:

In the classical limit, the wavefunctions have no overlap and thus, we can treat any macroscopic object as distinct from any other macroscopic object. Thus, we always care about permutations in the real world, even if one is unable to distinguish the objects and even if one does not care to distinguish the objects. We must treat them as distinguishable, and this is due to quantum mechanics!

10. Sep 19, 2006

### marcusl

Identical does not necessarily mean indistinguishable. The two coins, while equal in composition, shape, mass, inertia tensor, etc., remain distinguishable. If tossed sequentially, the results are certainly distinguishable. If tossed together, their trajectories may be followed and separated by high speed camera, for example. The appropriate statistics (as typically applied to ideal gas atoms, e.g.), are those of Maxwell-Boltzmann.

Although I am not an expert in such matters, I believe your observation about non-overlapping wavefunctions is pertinent.

11. Sep 21, 2006

### Chronos

It does not matter how you define what distinguishes one coin from another. The resulting orientation of the coins will behave randomly because no two coins can occupy the same state at the same time.