# Energy quantization

Why does a free particle give rise to a continuous spectrum of energy eigenvalues while a particle in a bound potential gives rise to a discrete spectrum?

• because there are no finite boundary conditions to make the wavefunction periodic
– user83548
Commented Jun 5, 2016 at 22:32
• the free particle has a plane wave solution also. How would you say this Commented Jun 6, 2016 at 0:09
• the plane wave solution is not periodic in position for a free particle
– user83548
Commented Jun 6, 2016 at 0:12
• But time independent Schroedinger's equation suggests a sinusoidal or exp(ikx) kind of solutions which are periodic in nature and in position space Commented Jun 6, 2016 at 0:14
• Possible duplicates: physics.stackexchange.com/q/39208/2451 , physics.stackexchange.com/q/65636/2451 and links therein. Commented Jun 6, 2016 at 4:00

Why a free particle gives rise to Continuous Spectrum energy eigenvalues where particle in the Bound States provide the quantization?

The real answer is , because that is what has been observed experimentally.

Because of the observation of the photoelectric effect, the spectrum of light from specific elements, ( and black body radiation) it became evident that classical models of electromagnetism were inadequate in modelling mathematically the data.

The final successful quantum mechanical model of the spectrum of the hydrogen atom, the solution of the wave equation with the electric potential of 1/r led to the whole quantum mechanical framework, where the solutions of the equations complex conjugated squared gave the probability distribution of finding a quantum mechanical particle at (x,y,z,t). The solutions with a potential gave the Balmer and Lyman series of hydrogen, i.e. quantized states modeling the data. The solution without a potential, a plane wave solution gave a continuous spectrum, modelling free particles.

• Actually I must disagree on that. Given the mathematical structure of QM that result is a theorem.
– lcv
Commented Jan 18, 2019 at 12:24
• @Icy Sure, that is why I call it a "real" reason" The theory of quantum mechanics was formed because the specific quantizations were observed. If ii is a theorem in the mathematics of the model, that does not explain the existence of quntized energy, it fits the observations, and the model is validated. Commented Jan 18, 2019 at 12:54

Actually, it depends on the theory you use. In nature, however, all bound states, except for ground states, have natural width, so, strictly speaking, they belong to continuous spectrum. As for theories, the same is true for quantum electrodynamics.

Ramark This is a good question. I am not sure if the OP is still interested in it, but maybe it will be helpful to somebody else.

I interpret the question as being: Why the energy levels of bound states are quantized (e.g., discrete) while e.g., for a free particle, the energy levels form a continuum (and the 'states' are delocalized)?

The short answer is: given the mathematical structure of quantum mechanics (as we know it) this is a theorem.

Let me explain.

The description of a (pure) quantum system, say a particle, is given by a wavefunction $$\psi$$. Born's rule asserts that the probability density of finding the particle at position $$x$$ is given by

$$|\psi(x)|^2.$$

However the particle must be somewhere, so the integral of all such probability densities must be 1. I.e.,

$$\int_\Omega |\psi(x)|^2 dx =1 \ \ \ (1)$$

where $$\Omega$$ is the configuration space of the system, essentially the allowed set of coordinates of the particle. For example $$\Omega =\mathbb{R}^3$$ for a particle in real space or $$\Omega =\mathbb{R}$$ for a one-dimensional harmonic oscillator.

The requirement (1) implies that the allowed space of functions that we are considering is $$L^2(\Omega)$$ which is essentially the space of square-integrable functions over $$\Omega$$. To tell the truth in $$L^2(\Omega)$$ we must identify functions which differ on sets of measure zero but this is not important here.

Now let us come to the question. The system is described by a Hamiltonian $$H$$ (an operator acting on $$L^2(\Omega)$$). What in physics we call a bound state corresponds mathematically to an eigenstate of $$H$$, i.e., there is an energy $$E_n$$ and a state $$\phi_n$$ that satisfy

$$H\phi_n = E_n \phi_n. \ \ \ (2)$$

By our requirement, $$\phi_n$$ must live in the space $$L^2(\Omega)$$, that is its square modulus must be integrable. Take for example $$\Omega=\mathbb{R}$$ this means that at long distances $$\phi_n(x)$$ must decay sufficiently rapidly (faster than $$|x|^{-1/2}$$). For example an exponentially decaying $$\phi_n$$ would do the job. We say that the state is bound (or localized) to a certain region of space, the region for which $$|\phi(x)|$$ is appreciably different from zero.

Now consider a free particle

$$H = p^2 = -\partial_x^2$$

(I use units in which $$\hbar=1$$). We would like to find a function to satisfy Eq. (2). Obviously

$$\phi(x) = e^{i k x} \ \ (3)$$

does the job and the eigenvalue would be $$E=k^2$$. However we have a problem when say $$\Omega=\mathbb{R}$$ (or more generally when $$\Omega$$ is not compact). The function (3) clearly does not belong to $$L^2(\mathbb{R})$$ as its square modulus integral diverges. In the physics jargon we say that the 'state' is delocalized. Note that there is some amount of imprecision here as actually there is no state (no function) in $$L^2(\mathbb{R})$$ with energy $$E=k^2$$. In this case the point $$E=k^2$$ belongs to the continuum spectrum.

To summarize:

1) Bound, localized states, correspond to eigenvalue of the Hamiltonian.
Typically the set of eigenvalues is discrete.

2) The rest of the spectrum forms the continuum spectrum (there are finer characterization of the spectrum which I won't go into). In this case there is no function in the space with the given value of the spectrum. Such energies are accessed in scattering experiment.

For completeness I give here the definition of spectrum of an operator. The spectrum $$\sigma(A)$$ of an operator $$A$$, is the set of points $$\lambda$$ in $$\mathbb{C}$$ for which the operator $$(\lambda - A)^{-1}$$ is unbounded. However we start entering some heavier math here.

My favorite answer is that this is due to uncertainty principle. First of all, the energy expectation values of both a free particle and a confined particle can both take any values, i.e., they are not quantized. However for a confined particle, for the majority of states, a state of energy expectation $$E$$ is a superposition of a few energy eigenstates $$E_1, E_2, ...$$. The closer $$E_{i}$$ is to $$E$$, the higher weight the $$i$$-th state contributes. So the energy of such a generic energy-$$E$$ state is very uncertain. To the contrary, the energy of a typical free particle is quite certain — for any value of $$E$$ there is an energy eigenstate to it.

Now the uncertainty principle comes into play. A confined system has less uncertainty in $$x$$, so more uncertainly in $$p$$, and thus more uncertainty in $$E$$ (as a large proportion of which is made up by $$p^2/(2m)$$.)

You can make this argument more quantitative by investigating the energy spectrum of a harmonic oscillator.

Bound state electrons may not give discrete spectrum. For example, the high temperature black body emission is a continuous spectrum, please see the Planck formula.

Given a temperature T, the spectrum is continuous, the wave length distributes from infinite to a minimum value less than visible light.

Taking any non-metal object as an example, say a glass in room temperature, all its electrons are in bound state, it radiates infrared and radio with continuous spectrum.

In the simplest of terms (somehow addressed in some of the answers), because a particle, as a wave, has infinite modes in free space. In a bound potential, there are only certain modes in which the wave "fits" inside the given potential.