# Quantum mechanical explanation for Le Chatelier's principle?

We know from basic chemistry that a reaction reaches equilibrium as described by Le Châtelier's principle. What's the quantum mechanical explanation to this principle? How a change in temperature/concentration causes an equilibrium to shift, from the point of view of wave functions?

• Le Chatelier's principle comes from thermodynamics, so I don't see a clear derivation of it from quantum mechanical assumptions. If there is a connection, it would likely be immensely laborious to work out. Apr 30, 2015 at 1:02
• @NicolauSakerNeto, I did not ask for a derivation from QM, but rather a separate explanation of the same phenomena. Apr 30, 2015 at 2:05

The main thing that classical and quantum statistical mechanics have in common is probability. In classical stat mech the equilibrium value of an observable is given by $$\langle A \rangle = \sum_i^N p_iA_i$$ where $A_i$ is the value of $A$ in microstate $i$ and $p_i$ is the probability of microstate $i$. The expression for $p_i$ at equilibrium depends on the ensemble (microcanonical, canonical, etc). One way to define equilibrium is that the probabilities are constant $$\frac{\partial p_i(t)}{\partial t}=0$$ So if one perturbs the system so that this equation doesn't hold the $p_i$'s will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.

In quantum stat mech the equilibrium value of an observable is given by $$\langle A \rangle = \tfrac{1}{N} \sum_\alpha^N \langle \Psi^{\alpha}| \hat{A} |\Psi^{\alpha}\rangle$$ where $\Psi^{\alpha}$ is the wavefunction for microstate $\alpha$

The wavefunctions are typically expanded in an orthonormal basis ($\phi$'s) $$\langle A \rangle = \tfrac{1}{N} \sum_{k,l} \rho_{l,k}\langle \phi_l| \hat{A} |\phi_k\rangle$$ where the density matrix is can be defined by a density operator $$\rho_{l,k} = \langle \phi_k| \hat{\rho} |\phi_l\rangle$$

The equilibrium is then defined as $$\frac{\partial \hat{\rho}(t)}{\partial t}=0$$

So if one perturbs the system so that this equation doesn't hold the $\hat{\rho}$ will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $\Psi^{\alpha}$ contributions to $\hat{\rho}$ increase of decrease.

The time dependence of $\hat{\rho}$ can be calculated by the time-dependent Schrödinger equation. So Le Chatelier's principle is a heuristic way of predicting roughly how the time-dependent wavefunctions will evolve at the simulation approaches equilibrium.

To get a more physical picture one can introduce the Born-Oppenheimer approximation in which case the numerical simulation of the time-dependent Schrödinger equation becomes a classical MD simulation (with a QM energy function) from which the time dependence of the probabilities $p_i$ can be computed and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.

• what's MD simulation? Jun 12, 2015 at 19:42
• Molecular dynamics. If you introduce the Born-Oppenheimer approximation then the time-dependent Schrödinger equation reduces to Newton's equation of motion $F=ma$, where $F=-\frac{\partial E}{\partial x}$ and $E$ is the energy obtained from the time-independent Schrödinger equation $\hat{H}\Psi = E\Psi$. Jun 13, 2015 at 9:35

Typically speaking, QM involves single systems at zero Kelvin. So as @NicolauSakerNeto indicates, temperature and concentration would usually be reflected in statistical mechanics and thermodynamics.

That said, for large systems (e.g., nanoparticles, metals, materials), finite temperatures result in thermal broadening. That is, rather than occupying the absolute lowest energy state, there is some fractional occupation of higher states given by the amount of thermal energy available and the density of electronic (or rotational or vibrational) states.

The individual orbitals themselves are purely mathematical and do not change with temperature - only the occupation of them.

• I'm not sure how the answer addresses the question. May 1, 2015 at 3:54
• The point is that quantum mechanics doesn't have much to say about temperature or concentration. QM is used to describe individual molecules or nanoparticles or proteins or whatever. It's hard to describe "concentration" of one system. Concentration is a bulk response usually treated with statistical mechanics. Stat mech does take energy levels from QM, but temperature won't change the QM properties, only the way the electrons fill in the orbitals. May 1, 2015 at 13:47
• Your question is a bit strange. Le Chatelier's principle makes sense in a thermodynamics or stat mech context. It's like you're asking me to explain how gravity affects QM. May 1, 2015 at 13:48
• Quantum gravity takes care of that :) May 1, 2015 at 14:09
• Quantum chemistry uses an 'isolated system' approximation for convenience. Le Chatelier's principle is meaningless in this approximation -- it only is sensible when the system is large enough that direct quantum computations would be overwhelmingly impractical. (Aaaand, just now read down to the first paragraph of Ian's answer...) Jul 28, 2015 at 18:29

Quantum mechanics tells us the possible stationary states of any system with a fixed energy. It would be extremely computationally difficult to use quantum mechanics to find the stationary state of a system as large as a solution but in theory this is possible.

Statistical mechanics tells us the probability that the system is in any one of the possible stationary states for some finite temperature. In particular, stat. mech. tells us which of the quantum mechanical stationary states are most likely to be occupied at equilibrium.

Le Chatelier's principle tells us how to find the new equilibrium when a system is displaced from equilibrium. In other words, Le Chatelier's principle tells us which distribution of quantum mechanical stationary states the system returns to after the system is disturbed in some way. Quantum mechanics tells us the possible states of the system, but it can tell us nothing about how those states are occupied. Only statistical mechanics can tell us how those states are occupied. Therefore, quantum mechanics alone cannot explain Le Chatelier's principle; in fact, Le Chatelier's principle holds regardless of the specific interactions between molecules and does not depend on any quantum effects at all. Le Chatelier's principle would hold even if we modeled the interactions between molecules completely classically with a Lennard-Jones or Morse potential.

There is no one because the Le Châtelier Braun principle is outside the scope of quantum mechanics.

The 'principle' formulated by Le Châtelier and Braun in year 1888 summarizes the response of an initially stable system to a perturbation that moves the system away from the equilibrium state. This is irreversibility, because the response of the system is always accompanied by a production of entropy $\Delta_i S >0$. In fact, the thermodynamic stability condition is

$$\Delta_i S = -\frac{C_V (\delta T)^2}{2T^2} -\frac{1}{T \kappa_T}\frac{(\delta V)^2}{2V} - \sum_{ij} \left(\frac{\partial}{\partial N_j} \frac{\mu_i}{T}\right)\frac{\delta N_i\delta N_j}{2} < 0$$

Effectively if perturbing the system reduces entropy, this implies that the original state was one of maximum entropy and the system will react to the perturbation by spontaneously recovering the equilibrium.

Now, it is a well-known theorem that quantum mechanics conserves entropy, and cannot explain irreversibility. To provide a solid foundation to the Le Châtelier Braun principle and similar statements about irreversible evolutions one needs to extend quantum mechanics to irreversible phenomena. This is a difficult task. There are several attempts. You can find the Brussels-Austin school work in [1]. A list of references to the MIT school work can be found in [2].

To provide a quantum explanation to the Le Châtelier Braun principle, one starts with an irreversible quantum equation of motion (for instance that from Quantum Thermodynamics). This equation has two terms: one linear and reversible (that reduces to Schrödinger form in the pure state limit) and other nonlinear and irreversible

$$\frac{d\hat{\rho}}{dt} = \left(\frac{d\hat{\rho}}{dt}\right)_{rev} +\left(\frac{d\hat{\rho}}{dt}\right)_{irr}$$

The irreversible component produces entropy $\hat{s}=\hat{s}(\hat{\rho})$ when there is dissipation (like it happens in the response of a stable system to perturbations)

$$\frac{d\hat{s}}{dt} = \frac{d\hat{s}}{d\hat{\rho}} \left(\frac{d\hat{\rho}}{dt}\right)_{irr} >0$$

and the inverse of this inequality can be used to formulate a microscopic analog of the classic principle.

REFERENCES

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.60.861

http://onlinelibrary.wiley.com/book/10.1002/0471619574