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Two electrons, or an electron and a proton, interact with each other because of the Coulomb potential, which can also be seen in the Schrödinger equation (which is the equation that describes the motion for elementary particles).

But how do an electron and a photon interact? Why can a photon kick up an electron to a higher atomic orbital, or why is a photon emitted when an electron jumps down to a lower orbital? A photon do not have any charge, so the interaction cannot be caused by the Coulomb potential. So what is it that causes the interaction, and how does that affect the Schrödinger equation? (Or well, the relativistic version of the Schrödinger equation, since we're dealing with photons.)

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  • $\begingroup$ Related: physics.stackexchange.com/q/95211 $\endgroup$ – John Rennie Jan 28 '14 at 19:57
  • $\begingroup$ @JohnRennie. Aren't photons associated with electromagnetic field? So that we can say coulomb potential also plays role. $\endgroup$ – Immortal Player Jan 28 '14 at 20:04
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    $\begingroup$ Charged particles interact with electromagnetic radiation because it is associated with an oscillating electric field. This exerts a force on the charged particle and makes it oscillate, and an oscillating charged particle in turn acts as a source of electromagnetic radiation. See Thomson scattering for a basic description of the classical interaction. $\endgroup$ – John Rennie Jan 28 '14 at 20:12
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    $\begingroup$ Extending John's comment the fully quantum description models the electromagnetic field as a collection of oscillators, the excitations of which are called "photons". Following this through you develop quantum field theory. In other words, the answer is the first chapter of a QFT book. Mandl & Shaw is very readable, but Ryder is more complete. $\endgroup$ – dmckee Jan 28 '14 at 20:17
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The details depend on exactly what level of theory you want to work at, but the basic principle is that electrons feel forces (and accumulate energies from them) whenever there are electromagnetic fields. A photon is essentially a quantum of excitation in the electromagnetic field, so when one is present there will be fields and these will affect the electron.

Anything after that depends on how you want to work: you can do a QFT type of treatment, or you can also quantize the field more naively and then heuristically give the interaction energies in analogy with the classical theory, or you can also work in a semiclassical treatment where the electron is quantized but the EM field is some fixed classical wave.

It's important to note, though, that even though the Coulomb fields of charged particles to induce electromagnetic fields, not all electromagnetic fields can (or at least should) be understood as the result of Coulomb interactions. Indeed, electromagnetic waves are fundamentally different to the Coulomb attraction fields; they are what is known as "null fields". Most importantly, trying to ascribe an origin to the field is misguided: fields create forces at different points in space based on what the value of the field is there, regardless of how that field was actually generated.


One nice way to put the problem is if you have an atom (single-electron for simplicity) interacting with photons at some wavelength below X-rays. The electric field of a given mode of the radiation can be written, classically, in the form $$ \mathbf{E}(\mathbf{r},t)=\mathbf{E}_0a(t)e^{i\mathbf{k}\cdot\mathbf{r}}+\text{c.c.}, $$ where $a(t)$ is the interesting dynamical variable, whose real and imaginary parts follow simple harmonic motion (and will therefore vary as $a(t)=e^{-i\omega t}$ classically) and $\text{c.c.}$ denotes the complex conjugate. The corresponding quantized version of this quantity is given by $$ \hat{\mathbf{E}}(\mathbf{r},t)=\mathbf{E}_0\, \hat{a} \, e^{i\mathbf{k} \cdot \mathbf{r}} +\text{h.c.}, $$ where $\text{h.c.}$ denotes the hermitian conjugate, and now $\hat{\mathbf{E}}$ is a position-dependent operator (or operator-valued function of position, however you want to see it). The 'core' dynamical variable $a(t)$ has been replaced by the operator $\hat a$, which is the annihilation operator of the corresponding quantum harmonic oscillator.

So what's a photon in this formalism? By the term "photon", we refer to an elementary excitation of the field, which is the state with one quantum of energy, i.e. the first Fock state, $$|1\rangle=\hat a^\dagger|0\rangle.$$ You can of course have more than one photon (e.g. a higher number state like $|7\rangle$), or superpositions of different numbers of photons (like $\frac1{\sqrt{2}}(|0\rangle+|1\rangle)$), or more complicated states such as classical-like coherent states.


OK, so that's how the photons look. (Or at least, they can be modelled like this.) Now for the interaction.

It turns out that atoms are much smaller than most of the wavelengths they interact with. That is, if $R$ is the largest distance an electron might get from the nucleus - a few Bohr or so - and $k=2p/\lambda$ is the wavenumber of the radiation, then the only way not to have $kR\ll 1$ is to be using fairly hard X-rays, in the sub-nanometre regime.

This means that as far as any electron will be concerned, the electric field is constant, and the $e^{i\mathbf{k}\cdot \mathbf{r}}$ term can be completely neglected. Taking this a bit further, you can use the classical formula for the corresponding potential energy, the dipole coupling $$ V_\text{int}=e\mathbf{r}\cdot\mathbf{E}, $$ and simply upgrade all the quantities in it to operators to get the corresponding coupling operator: $$ \hat{V}_\text{int}=e\hat{\mathbf{r}}\cdot\left(\mathbf{E}_0\hat a+\mathbf{E}_0^\ast\hat a^\dagger\right). $$

This gives you, finally, your hamiltonian. You need to combine it with whatever the atomic hamiltonian, $\hat H_0$ of the electron is (including its kinetic energy and its Coulomb interaction with the nucleus) as well as the energy contained in the field, which is given by the good old harmonic oscillator formula. That means that the total energy, and the total hamiltonian, is $$ \hat H =\hat H_0 +e\hat{\mathbf{r}}\cdot\left(\mathbf{E}_0\hat a+\mathbf{E}_0^\ast\hat a^\dagger\right) +\hbar\omega(\hat a^\dagger\hat a+\tfrac12). $$ Note that the hatted position $\hat{\mathbf{r}}$, is of course the electron's position, which is now an operator. The corresponding Schrödinger equation is $i\hbar\tfrac{d}{dt}|\Psi\rangle=\hat H|\Psi\rangle$, and you must be careful to note that the state $|\Psi\rangle$ is now the combined state of both the atom and the field, which includes such superpositions as $$ |\Psi\rangle=\frac1{\sqrt{2}}\left( |\text{ground}\rangle|1\rangle+|\text{excited}\rangle|0\rangle \right). $$


Finally, then, what does one mean, for example, by a phrase such as "the atom was excited by absorbing a photon"? Suppose you start off with your field in the one-photon state $|1\rangle$ and the atom in the ground state $|g\rangle$. This is no longer an eigenstate of the system, because the coupling does not respect it: $$ \hat V_\text{int}|g\rangle|1\rangle=e(\mathbf{E}_0\cdot \hat{\mathbf{r}})|g\rangle|0\rangle. $$ Therefore, the state of your system must evolve. What will it evolve into? To a first approximation, you can take the time-evolution operator to first order, $$\hat U(t)=e^{-i\hat H t}\approx1-i\hat H t;$$ this means, for example, that you will have transitions to any excited state $|e\rangle$ with amplitude $$ \langle e|\langle 0|\hat V_\text{int}|g\rangle|1\rangle=e\mathbf{E}_0\cdot \langle e|\hat{\mathbf{r}}|g\rangle\langle e|0\rangle, $$ which will tend to be nonzero.

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  • $\begingroup$ How come up until this point I have never seen the complex conjugate included in the equation for a classical electric field? I never saw this in undergrad physics...could you elaborate why c.c. should be there? Thanks. $\endgroup$ – physicsmajor May 9 at 5:01
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    $\begingroup$ @physicsmajor ask separately. $\endgroup$ – Emilio Pisanty May 9 at 5:58

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