Quantization vs. continuous energy levels

Question

I still don't get what it means for atomic energy levels to be continuous or quantitized (incontinuous). Clearing this up will really help me. Also, can anyone tell me why energy levels in solids are continuous while in gases they are quantitized? I get the part about the energy bands being more clustered due to the close proximity that the atoms in solids are, while in gases they are farther apart, but I don't get how this affects the absorption/released ray spectrums and how even THIS affects whether it is continuous or quantitized.

Answer 1

The quantization of energy levels appears both in quantum and classical mechanics, and it is not a consequence of the Schrödinger equation. It is a consequence of confinement. In fact, anytime that a wave equation (any quantum equation for the wavefunction, or a classical equation for a classical field, e.g., EM field) has periodic boundary conditions in some spatial variables, the system exhibits quantized energy levels.

As noticed in the question and in other answers, energy levels in quantum systems are not always quantized. On the other hand, also classical systems exhibits quantization of the energy levels. For example, consider the allowed frequencies of a string with fixed length (confinement), as in a guitar or violin. In such a string, the allowed "energy states" corresponds to frequencies (harmonics) which are multiples of a fundamental frequency (first harmonic).

In the quantum realm, energy levels are quantized if the wavefunction is confined in a finite space, e.g., in an atomic orbital or in a quantum well. In a solid, energy levels are also quantized, but the difference $\Delta$ between levels decreases as the system size increases. Therefore in the thermodynamical limit (large system sizes), these quantized energy levels become a continuum of states, since $\Delta\rightarrow0$.

As an example, let us consider a plane wave $$\psi(r)\propto e^{\imath k r},$$ which describes the wavefunction of a free particle (or the propagation of a sinusoidal wave of a classical field). The wavefunction has a continuous of energy levels $\omega\propto k^2$. However, if one confines the wavefunction in the segment $[0,L]$ one has that $\psi(0)=\psi(L)$ which gives $e^{\imath k L}=1$, and therefore the only wavenumber $k$ allowed are $k=2\pi n/L$. Hence, the energy levels of the confined particle are $$ \omega\propto \frac{n^2}{L^2}.$$ The gap $\Delta_n$ between energy levels goes to zero for $L\rightarrow\infty$. Therefore, if the particle is confined ($L<\infty$) the energy spectrum is quantized (finite and discrete energy levels, $\Delta_n$ is finite). If the particle is not confined ($L\rightarrow \infty$) the spectrum is continuous ($\Delta_n\rightarrow 0$). In real solids, $L$ is typically huge with respect to the typical sizes of the ion lattice, and therefore one is in the limit $L\rightarrow\infty$.

Answer 2

Technically, both solids and gasses have quantized energy levels. The difference is that molecules of a gas interact with other molecules very weakly, so the energy levels observed in emission or absorption of a collection of gas molecules are almost exactly the same as the energy levels that would be observed if you had a single gas molecule in isolation. In solids, the atoms or molecules interact very strongly, so in order to predict the absorption or emission spectrum, you have to consider the entire system at the same time, which produces energy levels that are extremely close to each other (which, for all intents and purposes, becomes indistinguishable from continuous when you have on the order of 10^23 atoms or molecules). If you considered the energy levels of an entire system of gas molecules (including the kinetic energies of each molecule), you would also get a very continuous spectrum, but there's no easy way to access this spectrum by absorption or emission of light.

Answer 3

In quantum mechanics the equation of motion is the Schrödinger equation $$ i\hbar\,\frac{\partial}{\partial t}\,|\psi\rangle = H|\psi\rangle $$ where the (self-adjoint) operator $H$, the Hamiltonian, determines its evolution. The energy levels are, by definition, the eigenvalues of such operator in its domain of definition $\mathcal{D}_H$. Spectral theory characterises what they look like according to the explicit form of $H$ and its domain; also, boundary and normalisation conditions for the state $|\psi\rangle$ may play a role. Therefore the answer to the question is: you have to explicitly calculate the eigenvalues of the Hamiltonian, which is in general a pretty hard problem to solve and no insight can be given a priori.

There are examples of physical systems presenting both behaviours: the standard example of a discrete spectrum are the energy levels of the hydrogen atom on its normalisable states. On the other hand, scattering problems often involve continuous spectra and non-renormalisable (at least in $L^2(\mathbb{R}^3)$) solutions.

Answer 4

Quantization is an experimental fact that forced physicists to consider theories that could explain the data. This happened in the beginning of the twentieth century.

1) black body radiation could only be explained by assuming that the radiation came in quanta, i.e. not in a continuous spectrum.)

2) The photoelectric effect showed that light behaved as a collection of particles, and they were called photons

3) Atoms instead of continuous spectra had discrete spectra

Hydrogen emission spectrum.

Instead of a continuous emission spectrum hydrogen gives these distinct lines. All atoms give spectra that characterize them.

Thus quantization at the atomic and particle level is an experimental fact.

First physicists tried to explain the spectrum of hydrogen with a planetary type model, the Borh model, using classical electrodynamics. The problem faced was twofold,

1) The orbits were unstable, the smallest disturbance would send the electron falling onto the nucleus radiating continually. This was solved by postulating fixed quantized orbits.

2) even though the spectrum of the hydrogen atom could be fitted, the generalization to the other atoms with more electrons could not work successfully.

With the Schrodinger equation and the postulates accompanying it Quantum Mechanics as a theory came into being and is now considered the underlying framework of all nature.

Physics theory is now in a position to fit spectra of all types of atoms, and also with certain approximation of all types of matter, from solid to liquid to gas.

The 1/2kx^2 is the potential in this simple example of a quantized system.

The electrons in the potential well of the atoms/molecules/solids are always in quantized levels, but the closer to the 0 of the potential well the energies are, the denser their position and experimentally one cannot distinguish them from a continuum. In solids there are collective potentials and the QM solutions may give a density of energy levels such that bands are formed where the electrons are shared by all the atoms/molecules of the solid, as in metals and semiconductors.

Wherever there exist potentials the electrons with energy below the potential (potentials go from 0 to negative values of binding energy) will be in quantized energy levels in principle, though the density of the levels together with the Heisenberg uncertainty principle may in effect display a continuum. For electrons that are free, (above zero in energy with respect to the potential), there is no quantization .

The difference between the proximity in solids and the distances in gases lies in that in gases the atoms/molecules are free, whereas in solids and liquids there exist collective binding potentials which generate more energy levels than the atomic/molecular ones.

Answer 5

If the energy levels are continuous (within a given interval of energies), then a particle (or system) can in principle have any energy in that interval. If they are quantized into say $E_1,E_2...$, then a particle (or system) can have only one of those energies, and not anything in between them.

Answer 6

Bound states have quantized energies, while unbound states have continuous energies. This can be understood by thinking of, for example, the 1D infinite square well. You can think semiclassically of the particle "bouncing back and forth" between the walls of the square well potential. At most wavelengths, the reflected particle will interfere with itself and after many trips back and forth, the wave function will average out to zero everywhere, meaning there is no solution. It's only at certain particular wavelengths (or equivalently, frequencies) where the interference is constructive that the average wave function is non-zero - and these particular frequencies make up the discrete bound-state energy spectrum.

For unbound states, there's no "bouncing back and forth", and so the self-interference doesn't eliminate any frequencies, resulting in a continuous spectrum. As Travis points out in his answer, both a gases and solids have discrete energy levels (if you neglect the kinetic energy of the gas), since both are bound states. However the solid has such a high density of states that it can be regarded as practically continuous.

Answer 7

Consider a Hamiltonian which is translationally invariant. For example,

$H = \frac{ \hat{p}^2}{2m} = - \frac{1}{2m} \frac{ \partial^2}{\partial x^2}$.

There are other options (any Hamiltonian which does not contain the operator $\hat{x}$ for example). Such a Hamiltonian forms quantum states which have definite momentum, i.e. they are eigenstates of the operator $\hat{p} = -i \frac{ \partial}{\partial x}$. We may say that the quantum states of the Hamiltonian represent free motion in space.

A crucial property of these states is that they are infinite in extent, i.e. they are not normalizable. They are delocalized. At the same time, the energy spectrum is continuous.

Now consider a Hamiltonian which consists of a translationally invariant part plus a potential term $V(x)$ which vanishes at spatial infinity. For such a Hamiltonian, there will always exist solutions which are free far away from the potential. These states are called scattering states: they have a continuous spectrum of energies. These scattering states are not normalizable.

However, the same Hamiltonian may also form bound states. A bound state decays exponentially outside of the potential region and is normalizable. Bound states can only exist at discrete energies - the reason is that they are standing waves, and standing waves must satisfy an interference condition which discretizes the spectrum. Very roughly speaking, a standing wave can only be formed when an integer number of wavelengths fits inside the orbit of the particle. But of course the real story is much more complex. There is a theorem that says in one dimension, as long as the potential is varying slowly compared to the wavelength of the particle, you have bound states at energies where the classical action

$\oint{pdx} = 2\pi(n + \frac{1}{2})$.

(This is called Bohr-Sommerfeld quantization). The integer $n$ corresponds to the number of nodes in the wavefunction.

Because the bound states decay exponentially, the energy must be below the continuum - this means that if your potential goes to zero at infinity, the energy is negative. To see this, apply the Hamilton operator $(- \frac{1}{2m} \frac{ \partial^2}{\partial x^2} + V(x) )\psi = E \psi$ and note that in the region where $V(x) \rightarrow 0$, you can disregard the second term in the Hamiltonian. Acting on any real exponential with the second derivative just gives you the same function multiplied by a positive factor, so the overal sign is negative. For energies $E < 0$, you may have discrete bound states if your potential supports them. For energies $E > 0$, you have a continuum of scattering states.

Now what about a solid? A solid is made up of a fairly complicated potential $V(x)$ which vanishes outside of the solid. So your wavefunction is confined inside a box which is the size of your solid. If your wavefunction is exponentially decaying outside the solid, then it corresponds to quantized energies $E_n$. Let's calculate the energy spectrum of a 1D particle in a box, say with size 1cm. You will get a spectrum

$E_n = \frac{ k_n^2}{2m}$ where $k_n = \frac{ \pi n}{L}$.

The difference between two energy levels $E_n$ and $E_{n+1}$ is then $E_{n+1}- E_n = \frac{ \pi^2}{2mL^2} ( (n+1)^2 - n^2) = \frac{\pi^2}{2mL^2} (2n+1)$. If you then impose a periodic potential, e.g. there is an ionic lattice with ions spaced at distance $a$ apart, the maximum value of $n$ will be

$N= \frac{L}{a}$.

Note that for a real solid, $a$ is typically a few angstrom, which is $10^{-8}$cm. Thus $N \approx 10^8$, which is a huge number. The spacing between energy levels is at most $\frac{ \pi^2}{2mL^2} ( \frac{ 2L}{a}) = \frac{ \pi^2}{mLa}$. At the same time the maximum energy is $E_{max} = \frac{k_N^2}{2m} = \frac{\pi^2}{2m} \frac{L^2}{a^2}$. The quantum states form a discrete set of points with energy between $0$ and $E_{max}$. $E_{max}$ is called the bandwith. The ratio of the splitting to the bandwidth is

$ \frac{2a}{L} \approx 10^{-8} $

Even though there is a discrete spectrum, the distance in energy between states is so small that it basically forms a continuum.