# On the nature of zero-point energy (ZPE)

The harmonic approximation gives a non-zero kinetic energy at 0 K for which Heisenberg's uncertainty principle is used as an explanation. A particle at 0 K couldn't be stationary because that would violate its wave-like nature.

Then there is the talk of zero-point potential energy. By the virtue of its existence, a particle with a finite mass would hold some energy within itself irrespective of being or not being in motion.

The first is easy to understand. The second, though, it would make sense only if the particle was in some field that exerts potential on it. Are there implications here that come from theories that a casually thermodynamics-educated person wouldn't comprehend?

Also, am I right to assume that the term zero-point energy, when used in atomistic modeling literature, is used solely to mean the first kind described above?

• This might be a better question for the physics or chemistry SE. I think it can be answered here though. Oct 31 '20 at 21:36
• Hi @TristanMaxson, I had the use of ZPE in atomistic literature in mind. I edited the question accordingly. Nov 1 '20 at 11:38

In the Born-Oppenheimer approximation nuclei are assumed to behave as classical point-like particles. However, in reality the wave-particle duality also applies to them, and so the Schrödinger equation should also be solved for the nuclei $$I$$ in the potentials $$V^I({\bf R})$$ generated by the electrons.

The harmonic approximation implies not using the full potential to solve the nuclear Schrödinger equation, but rather using a second-order Taylor approximation $$V^I({\bf R}) \approx V^I({\bf R}_0) + \nabla V^I({\bf R})_{{\bf R}={\bf R}_0} \cdot ({\bf R}-{\bf R}_0) + \frac 1 2 ({\bf R}-{\bf R}_0) \cdot \nabla\nabla V^I({\bf R})_{{\bf R}={\bf R}_0} \cdot ({\bf R}-{\bf R}_0) + \mathcal{O}(|({\bf R}-{\bf R}_0)|^3)$$

around the minimum-energy geometry at $${\bf R}={\bf R}_0$$, where $$\nabla V^I({\bf R})_{{\bf R}={\bf R}_0}={\bf 0}$$.

The zero-point energy is simply the energy of the nuclear wave function's ground state. The ground state is stationary, since it is an eigenfunction of the Hamiltonian. However, as a quantum mechanical wave function, it naturally has both kinetic and potential energy.

The concept of "motion" does not make sense within the generally accepted Copenhagen interpretation of quantum mechanics.

• +1 very comprehensive answer. However, a point on nomenclature: I would say that the Born-Oppenheimer approximation does include the nuclear equation too. I know that nowadays BO is many times used as a synonym for the "static nuclei approximation", but if we are careful then we should say that BO separates the full condensed matter Hamiltonian into two, one for electrons and one for nuclei. I then agree with your description of the "nuclear Hamiltonian". Nov 1 '20 at 17:46
• Yes @ProfM brings up a good point about various people meaning different things when they say "BO Approximation", which I'm sure Susi is very well aware of too. I prefer to use the term "clamped nucleus" approximation in some cases. Some people also mix up the terms "adiabatic" and "BO". Nov 1 '20 at 19:04
• Yes, I mean it in the "clamped nuclei" sense. Nov 2 '20 at 0:51

The Born-Oppenheimer (BO) approximation works well at high temperatures (far from 0 K), but the quantum nature of the nuclei (i.e. the zero-point energy) is an important consideration at low temperatures.

A great example of this is the difference between the Einstein and Debye models of a solid.

In the Einstein model, the nuclei are treated as identical independent 3D quantum harmonic oscillators (masses fixed to springs attached to an imaginary rigid lattice. If you use this model to predict the heat capacity of a solid, you will find it goes exponentially to zero as $$T\to0$$. This did not match experiments.

In the Debye model,, the nuclei are attached by springs to each other and undergo collective oscillations. The collective excitation of this interacting system are phonons and the predicted heat capacity decays to zero at $$C_v \propto T^3$$, which matches experiments. • But the zero-point energy is included in the Born-Oppenheimer approximation. If you solve the nuclear problem for example within the harmonic approximation (see answer by Susi Lehtola), then the zero-point energy is included. Nov 3 '20 at 18:14