Viewing a single comment thread. View all comments

Dr-Luemmler t1_jd7e31c wrote

Reply to comment by lizardweenie in Can a single atom be determined to be in any particular phase of matter? by Zalack

An ensemble can also be one atom or one molecule. So yes, it certainly has a well defined temperature.

0

lizardweenie t1_jd92qsl wrote

As other posters have mentioned, temperature is a property of a distribution. It tells us the probability of populating an excitation of a given energy. This isn't up for debate, it's just a matter of definitions. If you want to come up with some new concept that is well defined for a single particle, that's cool, but temperature doesn't work for single particles.

0

Dr-Luemmler t1_jd9m4t4 wrote

Oh, the distribution works if there are multiple available states at a given energy level. And there are. You can even calculate it for a single particle in a box. We actually did in class...

0

lizardweenie t1_jd9xip3 wrote

I don't mean to be rude, but it really seems like you haven't learned about temperature in a rigorous way (Like you would in a statistical mechanics class). It sounds like you've at least had some sort of exposure to undergrad level quantum mechanics, which is great. But recognize that your knowledge may not apply to this, and consider taking a statistical physics class.

If you did take such a class, you would learn that beta (which is propositional to 1/T) can be defined in terms of the partition function of the system of interest, but the entire concept relies on having multiple particles, (not simply one particle that transitions from state to state).

1

Dr-Luemmler t1_jdaglj7 wrote

I dont want to be rude, but you just need different states the particle could be, which you get with the quatification of impulse for each direction and the electronic states. Having multiple undistinguishable particle, and measuring their states is just one way to calculate the partition function. Another one is to track the trajectory of a single particle. In other words, we just need different states with different probabilities. I see no reason why that would not hold for a single atom. Temperature itself is also not a relative measurement as you can also see temperature dependent radiation from only a single atom.

The temperature in thermodynamics is defined as $T = dE/dS$. As $S \approx log(\Omega)$, the amount of "accessible" states need to increase with increasing temperature to hold the first formula. As a single atom has three dofs, we fullfill it.

Sorry, I really see no reason you could be right. I have also studied a bit advanced statistical thermodynamics and wrote my BA in that field. But I can be wrong, I cant say I was excellent in that field and some years have past since then. Maybe you can give me some hints for proper literature.

1

lizardweenie t1_jdamc26 wrote

No worries, you're not being rude. As for references, this a matter of basic definitions so I'd recommend some good textbooks, depending on your background.

I'd say that Chandler's book is pretty good: (I used it at the beginning of my PhD) http://pcossgroup.xmu.edu.cn/old/users/xlu/group/courses/apc/imsm_chandler.pdf

If you're looking for a different perspective, I've heard good things about Reichl: "A Modern Course in Statistical Physics"

Fun fact about this statement: > the amount of "accessible" states need to increase with increasing temperature to hold the first formula

This need not be the case. In certain scenarios, you can actually obtain negative temperatures which are perfectly valid. https://en.wikipedia.org/wiki/Negative_temperature

2

lizardweenie t1_jdapyy8 wrote

I just thought of a reasonable thought experiment that might clarify your confusion:

Say you have a bath of non interacting hydrogen atoms (consider for a moment, only electronic excitation), and we are able to measure the state of each atom.

Say we measure this bath and find that f_0 fraction are in the ground electronic state E0, and f_1 are in the first excited state E1. We could then infer a temperature by comparing these populations to a Boltzmann distribution, which tells us the relative probability of finding an atom in a state at a given energy (for a given temperature). In this case temperature is a well defined and meaningful concept.

Now say instead that we have a single hydrogen atom, we measure its state, and we find that it's in the first excited state. What then is the temperature? If we try to infer a temperature from this, (using a Boltzmann distribution), we get -inf. Say instead we measure it, and it's in E0. In this case, our inferred temperature will be 0. So for this single atom system, any temperature that we try measure can only give two values, (0, or negative infinity). In this system, clearly temperature isn't behaving how we would like it to.

This troublesome result points to a larger problem with the question: asking "what is the probability distribution for state occupation" doesn't really work well for the example: the atom was measured and determined to be in state E1, its probability distribution is a delta function, which is an inherently non-thermal distribution.

2