Unbounded Hamiltonian leading to finite ground state

In summary, the hydrogen atom's Hamiltonian is bounded from below, unlike its classical counterpart. This is due to the Kato-Rellich Theorem and the existence of a minimum value for the kinetic and potential energy. Various books, such as "The Schrödinger Equation" by Berezin and Shubin and "Quantum Theory for Mathematicians" by Hall, discuss the mathematical treatment of this topic.
  • #1
geoduck
258
2
If a Hamiltonian is unbounded from below, say the hydrogen atom where the Hamiltonian is -∞ at r=0, is there a way to tell if the ground state is bounded (e.g. hydrogen is -13.6 eV and not -∞ eV)?

It seems if the potential is 1/r^2 or less, then the energy will be finite as:

[tex]\int d^3 r (1/r^2) P(r) = \int r^2 dr (1/r^2) P(r)=1[/tex]

where P(r) is the probability density.

Or is this too naive and you have to work out P(r) which can help the integral converge or make it diverge?

There are two things at play it seems, the potential and kinetic energies, and Heisenberg's uncertainty principle which keeps the hydrogen atom from falling into the nucleus.

Is there a way to minimize H(p,r)=p^2/2m+V(r) with respect to r and p, using the constraint ΔrΔp<h/2, and seeing if the solution is finite?
 
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  • #2
I think I would look to the Virial theorem in quantum mechanics.

For an interparticle potential [itex]V(r) = a r^{n}[/itex], and the system in an eigenstate of the Hamiltonian, the Virial theorem states that

[itex]\langle 2 T \rangle = n \langle V \rangle[/itex], where the brackets denote expectation values. Expressing the Virial theorem in terms of the total energy [itex]E[/itex],

[itex]\langle E \rangle = \frac{n+2}{n}\langle T\rangle = \frac{n+2}{2}\langle V\rangle[/itex].

Since [itex]n[/itex] is a finite value, we at least know that if the expectation value of the potential energy is finite, the total energy must be finite as well.
 
  • #3
Your question can be the subject of a book and it was Franz Rellich and later Tosio Kato who first gave the answer. First of all, the H-atom's Hamiltonian is bounded from below or semibounded according to the standard definition from functional analysis (you can simplify and discard the COM dynamics). Then one shows the existence of a minum value of the matrix elements for the KE+KP for the generic atomic Hamiltonian (again discarding COM dynamics) which is of course a finite sum of H-atom/ion-like Hamiltonians.

There are a myriad of books on the mathematical treatment, you can first give it a try with Berezin and Shubin's "The Schrödinger Equation".
 
  • #4
Being bounded below is an important difference between the classical and quantum Hamiltonians for hydorgen!

Following on from dextercioby's post, a good, recent book that treats all this is "Quantum Theory for Mathematicians" by Hall,

https://www.amazon.com/dp/146147115X/?tag=pfamazon01-20

With respect to the hydrogen Hamiltonian ##H##, Hall writes
It follows from Theorem 9.38 that ##H## is self-adjoint on ##\mathrm{Dom} \left( H \right)## and ##H## is bounded below. ... Note that the classical Hamiltonian ##H\left( x, \mathbf{p} \right)## for a hydrogen atom is not bounded below.

The proof of 9.38 uses the Kato-Rellich Theorem, also proved in Hall.
 
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What is an unbounded Hamiltonian?

An unbounded Hamiltonian is a mathematical operator used in quantum mechanics to describe the total energy of a system. It takes into account the kinetic and potential energy of all the particles in the system.

What does it mean for an unbounded Hamiltonian to lead to a finite ground state?

This means that the system described by the unbounded Hamiltonian has a lowest energy state (ground state) that is finite and well-defined. In other words, the system does not have infinite or undefined energy in its lowest state.

Why is the concept of an unbounded Hamiltonian leading to a finite ground state important?

This concept is important in quantum mechanics because it helps us understand the behavior and stability of physical systems. A system with a finite ground state is more likely to exist and have observable properties, as opposed to a system with an infinite or undefined ground state.

What are some examples of systems with unbounded Hamiltonians leading to finite ground states?

Examples include a particle in a finite potential well, a harmonic oscillator, and a hydrogen atom. In these systems, the unbounded Hamiltonian leads to a finite ground state energy, which corresponds to the lowest energy level of the system.

Can an unbounded Hamiltonian ever lead to an infinite ground state?

No, an unbounded Hamiltonian cannot lead to an infinite ground state. This is because the Hamiltonian is an operator that describes the total energy of a system, and an infinite energy state is physically impossible.

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