I am unable to grasp the following statements which I found in the literature.

For a closed system (no transfer of matter), heat in statistical mechanics is the energy transfer associated with a disordered, microscopic action on the system, associated with jumps in occupation numbers of the quantum energy levels of the system, without change in the values of the energy levels themselves.

Work is the energy transfer associated with an ordered, macroscopic action on the system. If this action acts very slowly, then the adiabatic theorem of quantum mechanics implies that this will not cause jumps between energy levels of the system. In this case, the internal energy of the system only changes due to a change of the system's energy levels. - Wikipedia

Here is what I understand about heat and work:

Consider a gas in a rigid container. The average internal energy E of the system is given by:

$$ E = \Sigma_{i} p_{i}E_{i} $$ $$ p_{i} = n_{i}/N $$

Here, $N$ is the total number of micro states and $n_{i}$ is the number of micro states with energy $E_{i}$.

Suppose I heat the container in a stove, the heat absorbed by the gas manifests as an increase in the random jiggling of particles of the gas. Therefore, $n_{i}$ goes up for those micro states which have a high $E_{i}$, while $E_{i}$ doesn't change for each micro state. This is what the first block quote states.

In another experiment, I don't heat the gas, but compress the container to a new volume by an infinitesimal amount. My understanding is that this would lead to an increased jiggling too. If this is the case, how does doing work on the gas change the energy level $E_{i}$ of each micro state? (According to the second block quote)

Where am I wrong?


1 Answer 1


Consider a one dimensional infinite square well for simplicity. Treating the system quantum mechanically, the micro states are the eigenstates of the hamiltonian. These eigenstates are sinusiods, where the number of half wavelengths determines the energy level. The more wavelengths fit in the box (in other words, the shorter the wavelength), the faster the wavefunction phase oscillates. This increased oscillation with increased energy can be thought of as jiggling faster. So we see we have several different microstates with different energies.

Now if you were to heat the system up, then, as you said, particles would move from the low energy states to the high energy states, and so they would be jiggling faster. I think you understood this part already.

But now what happens if you compress the box? Then you would be making the wavelength of each eigenstate decrease, and so the frequency of oscillation of each eigenstate increases. Thus decreasing the box size also causes an increase in jiggling. However, where before each particle jiggled more because it moved to a higher energy state, we now have each particle jiggling more because it stays in the same state, but this state that it stays in starts jiggling faster.

I don't think you said anything wrong in your question, so I am not really sure where you are confused, but hopefully I cleared it up.

  • $\begingroup$ Your answer illustrates the dependence of microstate energy on physical dimensions for quantum system. Can work raise the microstate energies for a classical system with near continuous energy levels. Can't work affect distribution too? Thank you for the answer, you clarified much! $\endgroup$ Commented Mar 23, 2016 at 4:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.