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I have heard that in quantum mechanics, the term "lower-energy state" is often used as a stand-in for a concept that should really be called "higher-entropy" state. There isn't a law that says that systems decay to their lowest energy states; thermodynamics says that systems decay to their higher-entropy states.

Of the particle reactions I know of, the lower-energy state is a great stand-in for the higher-entropy state. Big particles decay into jets of little particles, excited atoms emit photons unless forbidden, and so on. Since energy is conserved, I guess they're implicitly talking about low energy density when they say low-energy.

So, are there any cases where a system (of particles) will decay in a way that raises the energy density? If not, how can the "law" of energy-lowering be derived from thermodynamics?

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  • $\begingroup$ Are you talking about how in thermodynamics we talk about a system minimizing its free energy? $\endgroup$ – Aaron Stevens Jun 28 at 13:46
  • $\begingroup$ No, I'm talking about in particle physics people talk about "decaying to a lower energy state." It's a colloquial thing and I'm wondering how true it is. $\endgroup$ – Display Name Jun 28 at 13:59
  • $\begingroup$ Sure you can! Roughly speaking, you just need to have less energy density (really, temperature) than your surroundings. A collection of atoms all in the ground state placed in outer space will spontaneously absorb energy. $\endgroup$ – knzhou Jun 28 at 18:59
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Lower energy and higher entropy

To be specific, consider the decay of an excited atom through the emission of light. When we say that an excited atom decays to a "lower-energy" state, we're referring to the energy of the atom itself. The total energy of the system is conserved (doesn't change), with the energy of the emitted light compensating for the reduced energy of the atom.

To address the question, choose a total energy $E$ and consider two classes of states:

  • Class 1: states containing only an atom with total energy $E$.

  • Class 2: states containing an atom plus propagating light with total energy $E$.

Consider any interaction that allows states in class 1 to evolve into states in class 2. Such an interaction necessarily also allows states in class 2 to evolve into states in class 1. (This is related to the fact that, in quantum theory, the operator that generates time-evolution must be hermitian.) However, regardless of the details, the system will generically still tend to end up in a state of class 2 simply because the "number" of states in class 2 is greater than the "number" of states in class 1. Entropy is a measure of the "number" of states having the given total energy $E$ (in this case), so this is how the tendency for excited atoms (or other unstable particles) to decay to lower-energy states can be deduced from thermodynamics.

A system of dipoles in an external magnetic field

The entropy $S(E)$ is an increasing function of the total energy $E$ in a "normal" system, but there are models with real-world applications in which the entropy $S(E)$ is not a monotonically increasing function of $E$. An example is a system of dipoles at fixed locations and variable orientations interacting only with a fixed external magnetic field. This system has both a single lowest-energy state and a single highest-energy state, according to whether the dipoles are either all parallel or all anti-parallel to the magnetic field. Both of these energies correspond to zero entropy, because there is only one state with each of those energies. Intermediate values of $E$ have higher entropy, with the maximum entropy corresponding to states in which the system has a net dipole moment of zero (neither net alignment nor net anti-alignment with the magnetic field). This is an example of a model in which the entropy can be increased by reducing the energy, if the initial energy is high enough.

However, in this simple dipole model, the energy of the system never changes. Energy is conserved, and the model doesn't allow for the possibility of "radiating energy away" (in the form of heat or light or anything else). In a more realistic model that does allow for those things, we are back in the situation illustrated above, because the total entropy of the whole system, including the radiation, will still be higher when the collection of dipoles has lower energy.

Formation of bound states

If an electron and proton come together to form a hydrogen atom, or if a star collapses to form a neutron star, then we have a situation where the energy density has indeed increased in one region of space, even though the total enetropy has also increased becasue of the EM and/or graviatational radiation. I'm not sure these qualify as examples, because the OP specifies decay, and because they depend on what spatial region we're considering. If we consider a region large enough to contain the original electron+proton or the original star, then the average energy density in that region still decreases after the radiation has left the region.

How to define the "number" of states

The original atom-decay example (and implicitly the other examples) referred to the "number" of states. Defining the "number" of states requires some care, but it can be done by putting the whole system in a spatial box and considering the set of states having the given energy, confined to the box. "Orthonormal" means with respect to the Hilbert-space inner product. Such a set of states has a finite-dimensional orthonormal basis, and that's one way to define the "number" of states in the preceding argument.

This is very general, because even in quantum field theory, the number of states with energy $\leq E$ confined to a box of finite volume is required to be essentially finite. This is one of the axioms of ordinary quantum field theory, often called the nuclearity property, which in turn is behind the famous Bekenstein bound on the entropy of any system with bounded energy and bounded volume.

Why this question is important

The OP's question gets to the heart of why the total energy operator (aka Hamiltonian) in quantum theory is required to have a finite lower bound. This is a prerequisite for making the preceding argument work, which in turn is a kind of stability condition.

In quantum field theory, that finite-lower-bound requirement is called the spectrum condition, one of the axioms of quantum field theory in flat spacetime. It's one of the conditions from which the Pauli exclusion principle (more generally, the spin-statistics theorem) is derived. Even more fundamentally, the spectrum condition is what allows us to define a vacuum state (the state of lowest energy), which in turn is what allows us to define "particle" in quantum field theory. (The vacuum state is the state devoid of particles, by definition.) Here's an interesting Physics SE post that emphasizes the importance of having a lowest-energy state in order to define "particle":

Do gases have phonons?

The spectrum condition makes sense in flat spacetime, which is the arena of special relativity. In a generic curved spacetime with a time-dependent geometry, there is no total energy operator in the usual sense, because the usual definition of the total energy operator starts with time-translation symmetry and uses Nöther's theorem. In a spacetime background without time-translation symmetry, the usual definition of the total energy opertor doesn't apply. This means that in a generic curved spacetime, there is no naturally-defined vacuum state that all inertial observers would agree is devoid of particles. This, in turn, is intimately related to how Hawking radiation from black holes is derived.

Altogether, the OP is asking a fundamentally important question, one that helps shape everything from the axioms of ordinary quantum field theory to questions about quantum gravity!

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