This problem is often paradoxically called the 1D particle in a box, yet at the end of the calculation, one attains a wave!  This shouldn't surprise us since nature is only comprised of wave quanta which we often think of as particles, but should not.  We should really be calling it "The 1D Matter Quantum in a Box" or something similar.  In any case, the matter quantum of which we are speaking is, in this example, an electron.  It doesn't have to be an electron, but that's the usual situation since otherwise there are very few physical analogs to this problem.  The best real life application of this problem is likely a pi-bond in molecules where the p-orbitals overlap.  In fact, a long chain of carbon atoms connected by alternating single and double bonds acts roughly like a very long 1D box for electrons.  The energies we are about to calculate would relate to absorption frequencies in such molecules. If it is not clear, what we mean by 1D is just that the electron wave is confined to travel back and forth in a linear fashion by the nature of the system. It is a common problem to solve because 1D makes the math manageable. Assume the box has a length L.  An electron standing matter wave inside this box is an exact analogy to the standing waves that persist on a guitar string.  The guitar string harmonics correspond to the various energy levels of the electron, where higher harmonics of higher frequencies are of higher energy, as we should expect from E=hf.  The electron wave needs to meet itself constructively after two reflections, just as the wave on the guitar string had to.  This is all accommodated by the WS quantization condition:

We relate the momentum of the particle by using an energy argument. The total energy E=K+U, is the sum of kinetic and potential energies of the quantum. The potential energy is uniformly zero in this problem - like roaming around the flat bottom of a box. Outside the box the electron cannot roam since we assume it'd require infinite energy. Because of this assumption we need not worry about the region beyond the box's boundaries. In later sections we will discuss more realistic boundaries. In any case, the kinetic energy for non-relativistic masses is , where p=mv from classical mechanics. Since potential energy is unchanging at all locations within the box where the electron will be (flat), we can infer that kinetic energy will be constant just as you would have expected from first semester mechanics. If kinetic energy is constant then the magnitude of the momentum too is unchanging. Momentum being a constant, we can remove it from the integral. The integral just gives us , or . Lastly, since E=K+U=K+0=K, the total system energy is just . Combining these two expressions leads to the energy states of our electron:

These are the well-known energies of a "particle" in a 1D box, where m is the mass of the confined electron and L is the box size. These energy levels exactly match the results of modern quantum calculations.

You will notice that the energy is proportional to the square of the integer given by the set of natural numbers. This means the energies are discrete, or quantized. When a measurement is taken, the system will always be found in one or the other of these energy states, and as such, it takes a precise amount of energy to make the electron transition from one state to another. The energy that will cause the transition comes from the absorption of energy from some other system such as a quantum of incident light. Light gets absorbed as individual quanta as we learned from the photoelectric effect, and they have energies given by E=hf. So to raise the energy level of the electron in our present system from E₁ to E₂, where , we need a photon that provides that amount of energy. Thus we can equate . This allows us to find the possible frequencies of light that this system would absorb. By using for the light, we can find the corresponding wavelengths of light that this system would absorb. The set of all of these absorbed frequencies (or wavelengths) is called the system's absorption spectrum. Correspondingly, when the electron goes down to a lower energy state, it will emit light according to the same relations. The set of emitted frequencies is the emission spectrum. While these are related (obviously), upon measurement they aren't always equal. The reason for this is that you have to have excited states present in order for them to emit light. To excite electrons into high states might take a tremendous amount of energy in the form of a very high temperature - one which the system for natural reasons may never reach. For this reason, the absorption spectrum will always tend to be more complete than the corresponding material's emission spectrum, which invariably will be missing some transition lines. In the case of a gas of molecules you could imagine the following: Shine broad spectrum light on the system (may be outside the visible spectrum). Look at what gets transmitted through the gas. You will find that the absorption spectrum gets subtracted out of the continuous spectrum such that there are missing or dimmed frequencies. The emission spectrum assumes that the gas IS the source of light. But at ambient temperatures nearly every electron is in its lowest energy state and no transitions are taking place. To help the matter, we could heat the gas. It turns out the temperature needs to be very, very high in order to even make an appreciable fraction of the electrons occupy the first excited state. This probability can be approximated using Boltzmann statistics as

,

where the numerator of the exponent contains the energy necessary to excite the state above the ground state and the denominator contains the system's temperature in kelvin and a constant called Boltzmann's constant where . It is convenient to write Boltzmann's constant in terms of electron-volts for atomic problems. In these units it is . What this probability tells us, for instance, is that if the first excited state is 2 eV above the ground state for some gas phase atoms, that even at a temperature of 1200 K, only roughly 4 parts per billion will be in the first excited state, and the rest in the ground state! This is why we never see the emission spectrum of the atmosphere unless lightning lends a whole lot of energy to excite and ionize the atoms and molecules.

What wavelength of light will excite an electron in a 1D box of width 1nm from the first excited to the second excited state? The energy of a quantum of light (photon) is . The energy states of the particle in the box as we calculated above are . The photon energy needs to be . Solving for wavelength gives 659 nm.

These waves look just like waves on a guitar string. While mathematically this shouldn't come as a surprise, physically it might. First of all, what does a wavy electron look like? What is meant by a variation in amplitude. The idea is that the electron is spread out in space. Its presence at different points in space which would correspond to both its mass and charge's locations are in the form of standing waves. Where there are nodes of the wave form there is no mass nor charge. Both of those quantities are proportional to the square of the amplitude as we've often discussed for waves. I should caution you to not think of this as an electron bouncing between walls since here it is a standing wave, which as you recall is made up of a wave going to the left with a corresponding wave also going to the right at the same time. Thus we can see that the momentum, while useful for finding the wavelength, actually is zero for the standing wave! This is completely in agreement with the full quantum treatment of the problem. If you don't remember what these standing waves look like, they are depicted below. The squared waves are pictured (as dashed lines) since they correspond to the physically measurable quantities such as charge and mass densities. Lastly, and yet very importantly we spoke about the nature of field quanta. The idea of field collapse means that when something interacts with the quantum, it will collapse and act local. Thus the common quantum interpretation of these squared waves is related to probability of finding a particle at a particular location. We will discuss more of that when we repeat this calculation using the full quantum treatment.