Transitions in QM

Considering transitions between stationary states, is it always the case that if electromagnetic radiation that produces an oscillating electric field has angular frequency which lies close to or at the atomic resonance frequency that the electric field puts the atom into a superposition of different states and induces an oscillating electric dipole moment? Does this always happen before an electron moves to a different state?

Also, how is stimulated transition rate usually defined? Would I be correct in stating that this is the probability that an electron would enter a transition state (superposition of different states) under the influence of electromagnetic radiation rather than jumping to another stationary state?

These question are motivated by the following attachment: • It would be helpful if you could specify the source of your "attachment". – freecharly Feb 24 '17 at 16:26
• @freecharly Okay, it is chapter 2 of Atomic Physics by Foot page 29. Chapter Hydrogen Atom. – user100411 Feb 24 '17 at 16:36
• Thank you. I will try to see more of the text relating to this excerpt. – freecharly Feb 24 '17 at 18:18

$\newcommand{\ket}{\left| #1\right\rangle}$ $\newcommand{\bra}{\left\langle #1\right|}$ Yes an interaction with an electromagnetic radiation can make the electrons to change of energy level. So you can see the electron as being in a superposition of states (energy levels), the interaction with the radiation leads to a non-zero transition probability. An electron can always drop to a lower energy level even though if it doesn't interact with any field (this is the spontaneous emission principle), but it cannot go to a higher energy level without interaction.

Here I will give some motivations for the derivation of the transition probability from the first excited state to the ground state in a non-relativistic approach. We consider a transition between an initial state $\ket{i,1}$ that we define to be the first excited state, and a final state $\ket{f,0}$ which is here the ground state for simplifications. Without loss of generality we can consider the oscillating electric field to be along the x-direction, so that $E(t) = |E_0|cos(\omega t) \hat{e_x}$. For one charge, the interaction Hamiltonian is given by $\hat{H_I} = -e\hat{x}|E_0|cos(\omega t)$ where $\hat{x}$ stands for the position operator. This form of the interaction Hamiltonian is derived using a Taylor expansion to zeroth order of the electric dipole and it is so called the electric dipole approximation. We note $\hat{H_0}$ the Hamiltonian for the atom without interaction and we let its eigenvalues $E_n^{(0)}$ to be given by $\hat{H_0}\ket{E_n^{(0)}} = E_n^{(0)}\ket{E_n^{(0)}}$.

Sorry, this is quite a lot of mathematical mess just to define some notations but it doesn't contain any physical interest. Here we go! In time dependent perturbation theory to first order, one gets that the transition probability from $\ket{i,1}$ to $\ket{f, 0}$ at time $t$ is given by $$P_{i->f}(t) = \Bigl|\frac{i}{\hbar}\int_{0}^{t}dt'e^{i(E_f^{(0)} - E_i^{(0)})t'/\hbar}\bra{E_f^{(0)}}\hat{H_I}\ket{E_i^{(0)}}\Bigr|^2$$ and after some algebra in our case one finds that $$P_{i->f}(t) = \frac{e^2|E_0|^2}{2m\hbar\omega_0}\frac{1}{(\omega - \omega_0)^2}sin^2(\frac{\omega - \omega_0}{2}t)$$ where $\omega_0$ is the frequency defined as $\omega_0 = (E_f^{(0)} - E_i^{(0)})/\hbar$ and can be interpreted as the angular frequency of a harmonic oscillator (since the electromagnetic field in absence of charge can be seen as a collection of independent harmonic oscillators).

To give some meaning to this result I add a figure from the very good book "A Modern Approach to Quantum Mechanics" by Townsend where $\eta$ corresponds to $\omega - \omega_0$. As you can see, the transition probability is centred on the value $\omega = \omega_0$ with a narrow peak depending on time. The transition from one state to another can be neglected except when we are close to the resonance condition $\omega \rightarrow \omega_0$.

So to answer your questions, when the frequency of the induced electric field gets close to the resonance frequency the transition probability has non negligible values. Then you can see the electron as being in a superposition of states since the states have non-zero probability values. We could have done a slightly more complicated derivation for a transition between any energy levels. In this case, if the electron goes to a higher energy state this is known as absorption (because the energy needed for the transition comes from an absorbed photon), and when it goes to a lower energy state this is the stimulated emission. Thus the stimulated transition rate is usually defined as the probability of transition by unit time when the atom interacts with an electromagnetic field. However an electron can jump to a lower energy state even without any interaction with external field. This is called the spontaneous emission and it can be derived in a pretty similar way.

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EDIT : I answer the question you asked in comments here to be able to write a long text.

You ask a very interesting question and a complete answer would probably require Quantum Field Theory where one can see an electron as a charged particle surrounded by a “cloud” of virtual particles such as photons. The interactions are then described as exchanges of virtual particles. Even without the QFT description, different approaches can be used to describe the interaction between the EM field and an atom.

In a fully quantized theory (where the EM field and the atom are treated quantum-mechanically) there is a coupling between the atomic transitions and the EM field as I tried to show it in my answer.

Using a classical theory (where both the EM field and the atom are treated classically) the interaction induces an oscillating dipole moment whose frequency determines the rates of absorption and stimulated emission. However, this oscillation still exists (although differently) in absence of external field, so that there is a coupling between the initially excited atom and the vacuum EM field. Then spontaneous emission is also allowed, as in the fully quantized description. From this point of view, one can say that if the atom is initially excited then it has an oscillating dipole moment even without any exposure to EM radiation. The atom can thus produce an EM radiation if it transits from one energy level to a lower one.

I hope it answer your questions. If you want a better understanding you can read chapters 2 and 4 of “The quantum theory of light” from R. Loudon. I wish you all the best with perturbation theory :)

• Thanks a lot for a very nice answer. I am starting to learn perturbation theory now (time independent first), but would like to continue a discussion once I have done so, if possible. In the text I am using it states that the em radiation induces an oscillating electric dipole moment on the atom, at what stage from the exposure to the em radiation does the atom have an oscillating dipole moment? And does the atom then produce em radiation itself, due to the oscillating dipole moment? – user100411 Feb 27 '17 at 10:47
• I add an edit to answer your question :) – Pigeon Mar 12 '17 at 16:10

Yes, absolutely. Everthing that matter does in its interactions with ordinary thermal light can be understood from the point of view of the light driving the matter into a state with oscillating charge density. The subsequent absorption or scattering of the incident light can be correctly calculated by using Maxwell's equations to calculate the resulting e-m field as the superposition of the incident field plus the new field generated by the oscillating charge.

The only thing wrong with your description is that there is really no need to assume that the effect must be to drive the matter into a pure eigenstate. It is much more normal for the matter to be driven into a superposition, and then re-radiate its excess energy back into the field without ever reaching the pure level of an excited state. Or even necessarily returning all the way to the ground state before being re-driven into a new superposition.

The whole process is purely continuous and it is not helpful in the slightest to try and analyze it in terms of discrete transitions. Yes, you can also do it that way, with photons and quantum jumps, and it is mathematically correct, but it totally obscures the physics of what is really happening.

I discuss all this in my blog, "Why I Hate Physics".

EDIT: Of course, it's not always an electric dipole moment. Sometimes it's an oscillating magnetic moment, as in the 21-cm line of hydrogen. And sometimes it's even a quadrupole moment. Those transitions are of course much weaker than the electric dipole transitions.

DISCLAIMER: I am a recognized crackpot whose answers are routinely and massively downvoted on this site by people who know much more than me.

• Thanks for your answer. Just a few things. So am I the correct in stating that the Bohr idea of electrons moving up and down energy levels due to absorption and emission is incorrect, but rather the electrons go from an energy eigenstate to a superposition of energy eigenstates under the influence of EM radiation? Also, would you defined transition rate? – user100411 Feb 24 '17 at 13:09
• The "transition rate" is, in the traditional interpretation, the probability per unit time of an energy jump between two eigenstates. In the continuous picture, this quotient is simply the power output of the atom as a classical antenna, as it relates to the energy difference between the eigenstates. It's exactly the same thing. – Marty Green Feb 24 '17 at 14:48
• Hah! I got one upvote among all those downvotes. – Marty Green Feb 24 '17 at 18:31
• @Marty Green - I looked into the cited book by "Atomic Physics" by C. J. Foot. It explicitly states that the transition involves a superposition of states and also gives the oscillating electric dipole moment based on the charge interpretation of the wave function just like Schrödinger did in his first treatment of the subject. See exercise 2.10 of the book fulviofrisone.com/attachments/article/403/… . – freecharly Feb 24 '17 at 18:56