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A microstate is something like a specific configuration of particles in a system. However, does this consist only of the energies of the specific particles? Or are the velocities and positions also important? Also, in quantum mechanics particles can not really be distinguished from each other. How can the definition of a microstate make sense in this situation? Finally, the positions, velocities and (kinetic) energies of the particles can take any values in a continuous interval of the real line. Does that mean that there are infinitely many microstates?

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    $\begingroup$ Have you learned about quantum states yet, such as the states of a particle in a box? $\endgroup$
    – Ghoster
    Sep 19 at 7:57
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    $\begingroup$ @Ghoster, Not yet $\endgroup$
    – Riemann
    Sep 19 at 8:22
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    $\begingroup$ Monaco is the usual example. $\endgroup$
    – hobbs
    Sep 20 at 23:12

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In classical thermodynamics a microstate is a specific set of attributes or properties of the system which unambiguously allows us to determine the evolution of the system forwards or backwards in time. The collection of possible microstates of a system is called its phase space.

For a system consisting of $n$ particles moving in $3$ dimensions the microstate is defined by the $3$ dimensional positions and velocities (or momenta) of all of those particles, so each microstate is defined by $6n$ real numbers. We say the phase space of the system has $6n$ dimensions. We don't need to include the particles' accelerations as separate independent attributes because these are determined by force laws and the force on each particle is (usually) a function of the position and velocity of itself and the other particles.

If the particles are constrained then the phase space may have fewer dimensions. For example, a bead that is constrained to move on a rigid wire has a phase space of only two dimensions - one position and one velocity or momentum - despite the fact that it may be moving in three dimensional space.

Yes, positions and velocities can take any value within a (possibly infinite) interval, so there are an infinite number of possible microstates in a system's phase space.

In quantum mechanics the phase space concept still holds, but things are a somewhat more complex because the values of the system attributes such as position and momentum cannot, even in principle, be determined exactly. This Wikipedia article describes how the phase space concept is extended to quantum mechanics.

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In classical mechanics a microstate consists of a unique set of values $\{q,p\}$ for all particles in a system. That is, to know a system's microstate is to know the exact location and momenta of all particles at an instant in time. A macrostate is some probability distribution over the accessible microstates under whatever conservation restrictions.

In quantum mechanics a microstate is the pure state $|\psi\rangle$ of a system and a macrostate is a mixed state $\hat{\rho}$ which encodes the "classical uncertainty" of a thermal system.

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In classical mechanics fully describing a microstate requires knowing all the coordinates and momenta (we are talking here about generalized coordinates and momenta, not necessarily mechanical ones) - that is the coordinates and momenta for all the degrees of freedom of the system.

In quantum mechanics full description of a microstate means knowing the full set of quantum numbers describing this state.

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A microstate is a state of an underlying mechanical theory. For classical physics it is classical mechanical state, a tuple of all particles positions and velocities. For quantum mechanics it is quantum state, wave function. For equilibrium statistical physics we can ignore non-stationary states, which means in quantum case microstates are discrete though infinite. In classical case microstates are distributed continuesly, that is the reason why quantum statistical mechanics more simple then classical one in some sense.

For statistical mechanics, a specific way of describing a microstate does not play an essential role. Basically, it is only important how the microstates are distributed at energy scale.

A properly builded quantum state of multiparticle system takes into account the indistinguishability of particles. Same can be done in the classical case too: we just consider that all states differ in particles labeling only are same state.

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    $\begingroup$ Quantum statistical mechanics has no equipartition theorem strictly speaking, that seems like a significant increase in complexity in practice...I guess maybe I see what you mean that summation is simpler than integration, though. $\endgroup$
    – Ian
    Sep 20 at 19:04

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