# Is there an example of a situation where you need a continuous spectrum?

If you had a hydrogen atom you could say that you want to be able to ionize them. But if you then add the potential due to the earth, e.g. $$V=\frac{-Gm_eM_\oplus}{\sqrt{(x_e-x_\oplus)^2+(y_e-y_\oplus)^2+(z_e-z_\oplus)^2}}$$ where $(x_\oplus,y_\oplus,z_\oplus)$ is the center of the earth, and $(x_e,y_e,z_e)$ is the location of the electron.

Now electrons are bound unless they achieve earth escape velocity. And if that isn't good enough you can add the potential $$V=\frac{-Gm_eM_\odot}{\sqrt{(x_e-x_\odot)^2+(y_e-y_\odot)^2+(z_e-z_\odot)^2}}$$ where $(x_\odot,y_\odot,z_\odot)$ is the center of the sun.

Now electrons are bound unless they achieve solar system escape velocity. And if that isn't good enough you can add an external potential for the mass in the galaxy. And that's good unless they have enough energy to escape the galaxy.

And you can argue that that is less than an eV of ionization so we need more energy but you could always put the whole galaxy in orbit around some super cluster far away with a huge escape velocity (we could even make the escape velocity be $c$ so it works for any non relativistic particle) and it seems unlikely to affect everything over here very much when we are a nice high quantum number angular state going about that super far away supercluster.

Not that I'm not saying that we are orbiting a giant supercluster I'm saying that it doesn't seem like it has immediate experimental consequences for most situations. So it seems like it doesn't matter much whether we are bound or are unbound.

Is there is a situation where we need unbound states rather than it just being convenient? I'm not asking for a list of them, just whether there are situations where we need them.

• Hmm, does scattering count? – arivero Aug 29 '15 at 2:16
• @arivero You can take a high n, high L state about the galactic center (or about the supercluster center) and near the scattering potential it will be almost perfectly plane wave so the math comes out the same. I want something where you need it, where it matters. I want to be a me to tell students that it is required for certain problems or else I will merely have to justify it as convenient. – Timaeus Aug 29 '15 at 2:21
• Well, in scattering, and generally in particle theory, you want to be able to argue about asymptotic states not only experimentally but also to define consistently the theory, interaction, etc. Another related thing is when the interaction area has a barrier plus well with some resonances. They can be discrete states inside of the continuum spectra. Regularising the continuum spectra makes a mess because you can not see clearly the resonant state. – arivero Aug 29 '15 at 2:28
• On the other extreme, you never need to do derivatives. Just subtract and divide epsilos of the order of planck lenght. Should you tell the students never to do calculus and keeping expanding Newton polynomials (x-e)^n? Note that this includes never to solve Schroedinger equation. – arivero Aug 29 '15 at 2:37
• @arivero Previous classes usually cover using derivatives and differentiable functions to describe things in simplified ways. For instance you approximate the entropy of a range of energies as a volume of the spherical region in phase space because it's easier than counting. But actually derivatives are usually taught in a math class so I can say "using calculus ..." or "recall that the derivative is ..." But often I have teach people new math in quantum mechanics and as a physicist I'd prefer to have a physical reason if there is one. Convenience counts for something but physics first. – Timaeus Aug 29 '15 at 2:46

For example, the escape velocity of a particle from the galaxy is about 400 km/s and in most conceivable circumstances (unless you are basically on top of the event horizon of a black hole or on the surface of a neutron star), escape velocities will be far, far below relativistic speeds (here defined as $3\times10^4$ km/s). So basically, if a particle has a relativistic speed it is almost certainly unbound.

Moreover, if you were to compute the "bound" eigenstates corresponding to an electron in the potential generated by our galaxy (for example) you would find that they were essentially identical (to an absolutely excellent approximation) to the eigenstates of a free particle.

The spectrum would be essentially continuous as can be seen in the expression for the energy spacing of a particle in a box (of length $L$):

$E = \frac{\hbar^2 k^2}{2 m}$

with allowed $k$ being:

$k = \frac{n \pi}{L}$

so that the energy states are:

$E = \frac{\hbar n^2 \pi^2}{2mL^2}$

so when we take $L\rightarrow \infty$ (which we essentially do in the case of the box being the size of a galaxy), we recover the free particle case!

• The fact that they look just like the free particle case is exact why I was asking why we need to teach people continuous spectra as a whole special case and extra thing to learn. If you have a supercluster and everything is orbiting it at a fast speed and high quantum number wouldn't everything look pretty much the same as if we had free particles with a continuous spectra? I apologize if I wasn't clear I thought I said all these things in my question. A Newtonian potential that requires an escape velocity of c would fails for relativistic reasons long before it failed for other reasons. – Timaeus Aug 29 '15 at 1:53
• Basically, plane waves end up being super useful for scattering problems and plane waves are the eigenstates of a free particle. Although it is not a perfect approximation, it is order of a magnitude below all of the other approximations of non relativistic quantum mechanics. If you didn't use plane waves for scattering calculations, lots of problems would become intractable. – Graham Reid Aug 29 '15 at 2:14
• Is so different than having an eigenstate of angular momentum about the galactic center (or supercluster center)? Near the scattering potential it's just going to look like a plane wave anyway, I see no actual difference between using the bound state and then inside an integral you note it looms like a plane wave in the region of interest. – Timaeus Aug 29 '15 at 2:16
• Sure, but if you have two ways of doing things that give identical answers in all concevable situations and one is 100x easier... :P Also I was never taught it as a separate thing. We stated with a particle in a box and showed that the free particle was infinite length generalization. If someone insists its a completely separate case they haven't thought it through. Its just a limiting case. – Graham Reid Aug 29 '15 at 2:20

I am puzzled at your leap: quantum bound state of electron to hydrogen, to the earth potential classical bound state. Bound classically and bound quantum mechanically are two different frameworks.

The electron is bound quantum mechanically to the hydrogen atom and does not "see" the gravitational coupling quantum mechanically due to the very small value of the gravitational coupling constant with respect to electromagnetic one that enters in the quantum calculation (proton electron interaction). Similar argument as that space expansion does not expand atoms.

The classical gravitational attraction of an electron to the mass of the earth will not give a quantum mechanical orbital (bound state with the center of the earth), again because of the coupling constants. If in vacuum the electron could have a classical orbit unless it had escape velocity.

• I don't know why you say that. There is nothing wrong with $V(x_e,y_e,z_e)=\frac{-e^2/4\pi\epsilon_0}{\sqrt{(x_e-x_p)^2+(x_e-x_p)^2+(x_e-x_p)^2}}+\frac{-Gm_eM_\oplus}{\sqrt{(x_e-x_\oplus)^2+(y_e-y_\oplus)^2+(z_e-z_\oplus)^2}}$ where $(x_\oplus,y_\oplus,z_\oplus)$ is the center of the earth, and $(x_e,y_e,z_e)$ is the location of the electron and $(x_p,y_p,z_p)$ is the location of the proton. You can leave $(x_p,y_p,z_p)$ and $(x_\oplus,y_\oplus,z_\oplus)$ fixed as external potentials. – Timaeus Aug 29 '15 at 5:39
• You could also throw in $V(x_p,y_p,z_p)=\frac{-Gm_pM_\oplus}{\sqrt{(x_e-x_\oplus)^2+(y_e-y_\oplus)^2+(z_e-z_\oplus)^2}}$ and let the configuration space be six dimensional (but then you need to label the potential in the previous comment as defined on a six space in the obvious way). But I agree replacing $e^2/4\pi\epsilon_0$ with $Gm_em_p+e^2/4\pi\epsilon_0$ can be unnecessary based on the values of the constants. – Timaeus Aug 29 '15 at 5:39
• Just that the corrections are insignificant within the width of the quantum energy levels due to the smallness of the gravitational constant. the electron is bound to the proton, not the center of the earth – anna v Aug 29 '15 at 5:41
• The center of mass of the atom is bound to the earth and the relative separation from the proton to the electron satisfies the hydrogen Schrödinger equation (with a reduced mass). But the point is that the energy is for the state, which is the configuration of the whole system. – Timaeus Aug 29 '15 at 5:44

One usually goes to the continuum because of its nice mathematical properties; lattice QFT is a hint of how hard a quantum theory becomes if we break the symmetries. For scattering theory, you usually want to be able to apply Lorentz invariance, or its classical counterpart, Galilean, and this already implies you are working with the continuum spectra of momentum. And remember that you need Lorentz group to classify spin.

Besides symmetries, another motivation to teach the continuum specta is because there are states with the same energy that some of the states of the continuum but very different properties. Search google for "states insde the continuum spectrum" and a lot of content in book, both in theoretical and applied physics, will surface. Randomly from the search, let me cite the abstract of this article:

Quantum mechanics predicts that certain stationary potentials can sustain bound states with an energy buried in the continuous spectrum of scattered states, the so-called bound states in the continuum (BIC). Originally regarded as mathematical curiosities, BIC have found an increasing interest in recent years, particularly in quantum and classical transport of matter and optical waves in mesoscopic and photonic systems

Of course you can still have the cake if you regularize the continuum by a very slowly decreasing potential as suggested (box is definitely bad idea, because it breaks also some discrete states of 1/r potentials) but instead of working with two different kinds of discrete sets, it is more clear to leave the continuum to be continuum and not some obscure subset of the rational numbers.

Last consideration is that we want to keep regularisation and cut-offs as a last-resource weapon. Divergences in advanced models can force us to use them, and calling for such tools before the need arises can be confusing to the student.