Meaning of average pressure in statistical mechanics

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SUMMARY

The discussion focuses on the derivation of average pressure in statistical mechanics as presented by Kittel and Kroemer. They establish that the pressure of a statistical state, denoted as ##p_s##, is defined by the relationship ##p_s=-\frac{d\epsilon_s}{dV}##, where ##\epsilon_s## represents the energy of the system. The average pressure, ##p##, is then expressed as ##p=-\bigg(\frac{\partial E}{\partial V }\bigg)_{\displaystyle\sigma}##, linking it to the average energy of the system. The conversation emphasizes the distinction between classical and quantum states, clarifying that in statistical physics, states are defined by probabilities rather than fixed positions and momenta.

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LightPhoton
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he fact that we are defining it in terms of such precise and singular conditions (isentropic compression of one particular state at a time) makes me have trouble understanding its physical meaning specifically, on how the actual pressure measured is related to average pressure, given the specific condition to measure ##p_s##.
Kittel and Kroemer derive the pressure of a statistical state in the following way:

They assume a volume compression of a system such that the quantum state of the system is maintained at all times; thus, the entropy ##(\sigma)## is constant in the process of compression. Now let the energy of the system with state ##s## be ##\epsilon_s##, then

$$\epsilon_s(V-\Delta V)=\epsilon_s(V)-\frac{d\epsilon_s}{dV}\Delta V+...$$

Since the work done on the system is equal to the change in energy of the system,
$$U(V-\Delta V)-U(V)=\Delta U= -\frac{d\epsilon_s}{dV}\Delta V$$

In terms of pressure, ##p_s## on the state we can write ##\Delta U = p_s\Delta V##, thus getting

$$p_s=-\frac{d\epsilon_s}{dV}$$

Defining ##p=\langle p_s\rangle ##

$$p=-\bigg(\frac{\partial E}{\partial V }\bigg)_{\displaystyle\sigma}$$

where ##E## is the average energy of the system.

Now, it makes sense to me to talk about average energy, since it's just the sum of all possible energies divided by the number of them. The same would have been the case with pressure, but the fact that we are defining it in terms of such precise and singular conditions (isentropic compression of one particular state at a time) makes me have trouble understanding its physical meaning specifically, on how the actual pressure measured is related to ##p## given the specific condition to measure ##p_s##.
 
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LightPhoton said:
Now, it makes sense to me to talk about average energy, since it's just the sum of all possible energies divided by the number of them.
No it's not. The average energy is the sum of all energies multiplied by probabilities of those energies. Those probabilities are defined by the state of the system. In the quantum canonical ensemble this state is the mixed quantum state describing a thermal equilibrium at temperature ##T##. In this sense, average energy and average pressure involve the same kind of average, which depends on the state.

Perhaps your confusion stems from thinking of a state as a classical state in which all positions and momenta are well defined. But here, in the context of statistical physics, the state is not like that. The state only determines probabilities of positions and momenta. In the quantum case, the state cannot even be defined by a single wave function. The thermal state is a mixed quantum state, in which even the wave function is uncertain.
 

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