# How does exciting an electron's surrounding electromagnetic field cause 'electron excitation'?

In more meaningful words than the ones above, how does adding energy to the EM field cause the electron to to change orbitals or oscillate in a different pattern.

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Are you asking for a description of what happens when light causes an electron in an atom to change energy levels? – John Rennie Aug 6 '14 at 5:08
For the hapless future grad student reading this, here's a reasonably complete demonstration of how Rabi oscillations happen when you apply an oscillating force to an electron: physics.stackexchange.com/q/138765 – DanielSank Oct 8 '14 at 3:18

how does adding energy to the EM field cause the electron to to change orbitals or oscillate in a different pattern.

Here one is using two frameworks, the classical and the quantum mechanical. The classical electromagnetic field is composed by an enormous number of photons each with energy=h*nu, nu the frequency of the electromagnetic field.

The electron is an elementary particle, a quantum mechanical entity. It interacts with photons via a well prescribed mathematically calculational mechanism , the Feynman diagrams.

The electron may be free, and then it can scatter with a photon and take part of its energy or have even more complicated interactions, but since you are framing your question in EM fields , this is another story.

Electrons bound in free atoms can be kicked up to higher energy levels by a photon with the correct energy. Electrons in a solid are also bound but complicated potentials exist which may allow many energy levels common to the solid to the point of seeming continuous.

An EM field with the appropriate energy may move the electrons around in allowed orbitals for the solid which may have different manifestations in the solid than it had before being exposed to the said EM field, as oscillations because of different collective orbitals

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In the semi-classical approach you treat the electromagnetic field of the incoming light as a (sinusoidal, with angular frequency $\omega$) perturbation $H^{\prime}$ in the Hamiltonian of the atom. The wavefunction of the atom can be expressed in terms of a linear combination of its eigenstates, each with a multiplying coefficient. If we start on the ground state, then $c_1 = 1$ and all the other coefficients are zero.

You then apply time-dependent perturbation theory to calculate the time dependence of the $c$ coefficients and the transition probabilities between the eigenstates.

The probability per unit time of a transition is $$p_{fi} = \frac{4\pi^2}{\hbar^2 T} | H^{\prime}_{fi} (\omega_{fi}) |^2,$$ where $\omega_{fi} = (E_f-E_i)/\hbar$, and $$H^{\prime}_{fi} = \frac{1}{2\pi} \int^{T}_{0} \int \phi^{*}_{f} H^{\prime} \exp(i\omega t)\phi_i\ dt\ d^3x,$$ where $\phi_f$, $\phi_i$ are the final and initial eigenfunctions and $T$ is the time over which the perturbation is applied.

What one finds is that the probability of a transition being made is maximised when $\omega = \omega_{fi}$.

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