Why does a photon emit discrete frequencies of light?

In summary, the emission of discrete frequencies of light by an atom is a result of solving the Schrodinger wave function, which yields discrete energy levels and the corresponding emission spectra. This is due to the fact that measurements collapse the wave function into a specific energy level, and the probability of transitions between levels is described by superposition. The discrete energy levels can be understood in terms of the wave nature of particles and the quantization of action, which leads to resonances and interference patterns within the atom. While there is no known necessity for nature to behave this way, it is an empirical fact that has been observed consistently.
  • #1
enwa
6
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Why does an atom emit discrete frequencies of light?

Solving the Schrodinger wave function for the hydrogen atom (that is a single particle representing an electron bound by a spherical potential) we find that it has discrete energy levels. Plotting every possible value of [itex]f[/itex] in [itex]E'-E = \hbar f[/itex] where [itex]E'[/itex] and [itex]E[/itex] are the different energy levels of eigenstates we recover the emission spectra.

What I don't understand is why we only see these discrete energy levels. According to the superposition principle, the wave function could be in a superposition with expected energy [itex]\tfrac{1}{2}(E'+E)[/itex] but differences from these levels don't show up on the emission spectra. The measurement postulate seems relevant, "measurement" collapses a wave function into an eigenstate, this seems to be happening before and after the photon emission, can anyone explain why?
 
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  • #2
I don't think there is an answer as to WHY electrons can only move between orbitals and energies in values of h, just that they do. Plank discovered this constant without having any answer to why it was this way and that particular value, only that it was.
 
  • #3
The reason is that a photon in a superposition of (|E1>+|E2>)/sqrt(2) is not the same as a photon with energy (E1+E2)/2.
 
  • #4
If you're looking for a physical rationale one of the clearest is in David Bohm's "Quantum Theory" chapter 6 or thereabouts (don't have the book here)
 
  • #5
To elaborate on LearningDG's remark, while superpositions are allowed and inevitable, when photon emissions are measured the wavefunction is collapsed according to the Born principle, which picks out a particular discrete pair of energy levels.

All the superposition tells you is the probability of particular transitions. When you observe a transition (photon emission) it will almost always correspond to the energy difference of two levels. Why so? I'm not sure it is any more than a fundamental empirical fact---this is how nature is, so it is a postulate of QM. That's why I have to say "almost always", simply because no one has ever observed otherwise, but there is no known necessity for nature to be this way.

You could ask the further question "why then are the energy levels discrete?": the answer is again deeply one of the empirical facts about nature, but we understand it in terms of quantum theory as a result of the wave nature of particle states, i.e., since the electrons are wave-like they exist in stationary orbitals only by virtue of being held in 3D standing wave formation. If they were not standing wave solutions they would not be stationary, they'd instead be time-dependent. Hence the mathematics for time independent solutions dictates spherical harmonics (standing waves in 3D), these are labeled by integer solutions, hence discrete energies. Most students understand this first from the simpler mathematical analysis of an idealized quantum particle in a box (infinite square well potential).

If you were to relax the requirement of stationary (i.e., time independent) states then presumably the wave function solutions would no longer necessarily correspond to spherical harmonics, and the energy levels could be all over the place. This might be set up, say for example, if you bombard an atom with an intense laser, or impose some other kind of drastic nonlinear dynamical perturbation of the atomic electrons. Good luck to you if you want to calculate the energy levels and transition amplitudes in such situations!
 
  • #6
I think one can give a kind of reason as to "why this is." It has to do with the importance of interference and resonance in wave mechanics, and the presence of a quantized action. An atom emitting light is usually only perturbed a very tiny amount, such that the perturbation wouldn't amount to squat over a single cycle of the atom's periodic oscillation (periodicities in the wave function are pretty much the main crux of orbital quantum mechanics). So to get the atom to emit light, one has to "tickle" it over and over, in synch with the natural frequencies that can yield constructive interference within the atom. That's called a resonance, it's pretty much why a guitar string emits notes rather than white noise. Indeed, your question is quite intimately related to "why do musical instruments emit sound at special frequencies." It comes from the combination of having a resonant frequency in the system, and obeying the interference-regulated rules of waves. Now, it should be noted that guitar strings can be any length, so can make any note, but on the atomic scale we encounter the quantization of the action-- the scale is small enough to probe the minimum scale on which action can appear. That's what creates "guitar strings" in atoms that cannot be just any length.
 

FAQ: Why does a photon emit discrete frequencies of light?

1. Why do photons emit only discrete frequencies of light?

Photons, as the fundamental unit of light, have a specific energy level depending on their frequency. This energy level is determined by the wavelength of the light, and can only take on certain discrete values. Therefore, photons can only emit light at specific frequencies.

2. What causes photons to emit light at specific frequencies?

The emission of light at specific frequencies is a result of the quantized nature of energy in atoms. When an electron in an atom absorbs energy, it jumps to a higher energy level. When it returns to its original energy level, it releases the excess energy in the form of a photon with a specific frequency.

3. How does the emission of discrete frequencies of light relate to the wave-particle duality of light?

The wave-particle duality of light states that light can behave as both a wave and a particle. In the case of photons emitting light at specific frequencies, this behavior can be explained by the particle nature of light. Each photon carries a specific amount of energy, and this energy is emitted at a specific frequency when the electron returns to its ground state.

4. Can photons emit light at any frequency?

No, photons can only emit light at specific frequencies corresponding to their energy levels. This is due to the quantized nature of energy in atoms, as mentioned earlier. Therefore, photons cannot emit light at any arbitrary frequency.

5. How do different sources emit light at different frequencies if photons can only emit at specific frequencies?

The specific frequencies of light emitted by different sources are a result of the energy levels in their atoms. Each atom has a unique set of energy levels, and thus can only emit light at specific frequencies. This is why different sources emit light at different frequencies, as their atoms have different energy levels and emission patterns.

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