Does electron have a temperature in the atom?

Question

Please correct me if I am wrong. Electron is fundamental and is zero dimensional (probably made up of strings). Electron is bound to the nucleus. Electron in an atom has velocity and position which are statistically distributed. Electron mainly interacts electromagnetically. Temperature is the measure of statistical kinetic energy of particles.

In the nucleus of an atom, proton and neutron have temperature because they are a bag bound by color forces , valence quarks and a sea of quarks and anti-quarks. Quarks interact electromagnetically. Saying that protons and neutrons have temperature means that there is a randomness in the kinetic energy of the constituents of proton and neutrons. Electrons are bound to the constituents of protons and neutrons through electromagnetic force. Therefore if there is randomness in the kinetic energy of the constituents of protons and neutrons then the electrons must be experiencing wobblyness in its orbit, that is , it must be experiencing some kind of little randomness in its motion around nucleus. And therefore the electron must have a little temperature in the atom.

My question is : Does the electron have a temperature in the atom?

Answer 1

From the way the question was formulated, I deduce that some clarification on some basic concepts is needed.

• If we speak of electrons in an atom, we can safely ignore physics at much larger energy scales (like the nuclear scale, not to mention string scale!). At the atomic scale, nuclei can be treated as an almost point-like charge and current distribution.

• the statement

Electron in an atom has velocity and position, which are statistically distributed

is not correct. As a quantum particle, an electron does not have velocity and position. Either we measure velocity (or, better, momentum) or position. There is no way of defining both at the same time.

• Consistently with the previous observation, we can speak of a statistical distribution of position or momentum encoded in the properties of the quantum state. Such a distribution is connected to the modulus squared of the wavefunction and does not depend on temperature.

• Also, the statement

Temperature is the measure of statistical kinetic energy of particles.

is not correct for quantum electrons. The connection between temperature and kinetic energy is true only in the classical regime. In a piece of metal, due to the Pauli principle, electrons may have quite a large kinetic energy (that divided by Boltzmann's constant could correspond even hundreds of thousands of kelvin degrees while being at room temperature from the thermodynamic point of view).

• In the case of an isolated atom, electrons remain in their ground state, which means that they behave as if they were at 0 K. Such a statement remains true, independently of the applicability of the concept of temperature to not more than one hundred particles.

Answer 2

I have little to say about QCD, but I will try to analyse this question from thermodynamics perspective. And I think I have a decent answer.

Save Time: In essence, protons and neutrons exhibit temperature according to Quantum Chromodynamics (QCD) because they are composed of quarks. As composite systems, akin to a balloon with air molecules, they possess a temperature, albeit small. On the contrary, electrons, being elementary particles, lack a meaningful temperature. The prospect of electrons being composed of more fundamental particles could render the concept of temperature applicable to them. L Detailed version:

Any thermodynamic property makes sense for a (i) system (ii) in state of equilibrium.

Temperature generally isn't defined on small systems.There's a couple meanings of temperature. One is that it describes the probability of the system being in states of different energy. If a system has a temperature, the probability of being in a state depends on the energy E of the state proportional to e^(-E/kT) where k is Boltzmann's constant and T is the absolute temperature. What if the probabilities don't follow that particular pattern? Then the system doesn't have a definite temperature. Usually an individual atom doesn't have a definite temperature.

A more basic definition of temperature is that it's the derivative of energy U with respect to entropy S: T=dU/dS, in thermal equilibrium. Usually this definition doesn't even make sense at atomic level.

Now protons and neutrons do have temperature as per QCD! Since they are made up of quarks. So protons and neutrons are not elementary particles they are system made up of quarks, just like a balloon is a system made up of air molecules inside it, hence they should have temperature however small it may be. But I don't think same is true for electrons. Electrons are elementary particles. They are not a system , hence temperature of electrons is meaningless. Ah! But if it is found that electrons are system made up of more fundamental particles then yeah, it will make sense.

There is one more question, I said that Usually an individual atom doesn't have a definite temperature. But atom is a system then why don't it possess temperature, the answer is : An atom really does have a small number of moving parts. Given what our particular universe is made of, there simply aren't parts available to put a lot of different things in that space without having an enormous amount of energy available. Now in the very early universe things were so hot that many different particles were around that could fit in the space of what would now be an atom. So it makes sense to talk about the temperature on the scale of an atom at that point in the story. However, the temperature then was so high that atoms themselves couldn't form- their parts would all shake loose.

One question is prominently asked again and again " if electron is related to proton and neutrons and they have temperature then electron should have too" I think innocent mistake has been committed in reasoning, citizens make a nation but is individual citizen a nation? No!, Gas has a temperature but temperature of molecules is meaningless. Similarly atom may have a temperature but it is not necessary that fundamental constituents of atom have temperature, also: protons and neutrons are subsystems not fundamental constituents quarks are.

But I still didn't understood one thing, number of downvotes on such a nice question.

Answer 3

You cannot speak of a temperature of a single particle, Temperature is an ensemble property. The electron is - as far as we can tell - a fundamental particle unlike the proton (which is a composite particle consisting of quarks).

Answer 4

There is technically a way to incorporate the concept you are trying to talk about; I gave you an upvote to undo the downvotes you are getting on this question.

There is a reason why there is a kneejerk reaction for the others to be downvoting your question. Under the wavefunction form of quantum mechanics, there is no way to incorporate the temperature. Temperatures are captured by classical mixtures, and in the wavefunction formalism, there is no way to talk about classical mixtures, only quantum superpositions. Another way of expressing this is that wavefunctions can only talk about pure states, and temperature is expressed by mixed states. As such, we have to first tell you that the thing you are looking for is not even expressible inside the standard form.

On top of that, most physicists would be aware of the fact that the nuclei's temperature is different from the electron's, sometimes dramatically so. For example, we know that there is the phenomenon of para- and ortho- states, e.g. in hydrogen gas and in oxygen gas. The spin of nuclei is the most experimentally visible part of nuclear physics, and yet even there, it easily can take days as the nuclei lags behind the electrons at thermal equilibriation. So, even if we take your concept at face-value, it is quite important to stress to you that the coupling between nuclei and electrons are weak enough that they can exist at different temperatures for really long times. I mean, if it is so weak that it takes millions of years to equilibriate, then for all intents and purposes you should consider them as never going to equilibriate, and have to deal with their separate temperatures as a fact of life. After all, as Feynman put in his little-known statistical mechanics textbook, what we are interested is an intermediately infinite amount of time, since, given really ridiculously long times, any gas will poke holes in their containers and exit.

The correct way to study this problem is to consider the density operator formalism. There are ways to "directly" get these things via some complicated Wick rotated path integral, and because it is Euclidean, it is actually better mathematically motivated than the usual path integral in Minkowski metric. Then you can talk about the nucleus having a temperature and also the electrons having a temperature, and even have them be the same temperature. Needless to say, the result is obvious: Basically for all the temperatures we care about in everyday life, the deviation from just assuming $T=0\,$K is negligible. This more than justifies why textbook treatments ignore the complication.

If we want to work purely with wavefunctions, that is also possible. Simply compute the ground state wavefunction and the first excited state, and compute the contribution of the first excited state's contribution to the density operator. It will be so small as to be irrelevant.

Of course, none of the above considerations will work in extreme conditions. For example, if we are considering a star, especially when we want to simulate a supernova, then we have to treat everything relativistically, including both the temperature and the pressures. The excited states will no longer be as negligible as they are in everyday life.

Now that I have outlined the way to consider what it is you wanted to consider, I would like to offer yet another reason to just stick with the simpler stuff. The nucleus is necessarily a dense soup of quarks and gluons. We talk about a temperature distribution for them, no problem. However, condensed matter physics tells us that the very fact of fermions being condensed, implies that there would be an equivalent huge Fermi temperature that we can define for them. This is exactly the same thing as in the sea of electrons in a metal, they essentially are at some $10000\,$K, such that room temperature is a tiny alteration on the big picture behaviour. It will barely be perceptible. So, really, just stop worrying about this.