# Quantum Jumps are Really Smooth Slides

I gave this answer on Physics Stack Exchange to the following question:

 Does an electron move from one excitation state to another, or jump? I’m wondering, when an electron changes state, does it move from one state to another over some (very small) time period? Or does it change from one state to another in no time? If the former, what does it mean for it to be in-between states (for however short a period of time)? If the latter, how does it teleport? (Does this question make sense?)

and the answer, slightly edited, is as follows:

Leaving aside the quantum measurement problem (i.e. whether or not there is a “collapse” of quantum state to an eigenstate of an observable on measurement) and talking wholly about quantum state between “measurements” and its unitary evolution, I’d say that the transition is definitely a smooth shifting from one “eigenstate” to another, so that the electron’s wavefunction is of the form $\alpha_1(t)\, e^{-i\,\omega_0\,t}\,\psi_0(x) + \alpha_1(t)\,e^{-i\,\omega_1\,t}\, \psi_1(x)$ where $|\alpha_0|^2 + |\alpha_1|^2 = 1$, $\alpha_1(0) = 1, \alpha_0(0) = 0$, $\alpha_1(t) \to 0, \alpha_0(t) \to1$ as $t\to\infty$ and $\psi_1, \psi_0$ are the would be “jumped between” “eigenstates” (here I’m thinking of a downward transition from a raised state $\psi_1$ to a ground state $\psi_0$).

In the following I shall stick to the question of an electron as belonging to an atomic or molecular system, rather than bare electron – electromagnetic field interaction as in QED. This typifies the kind of system that your question makes sense for i.e. where the electron must have discrete, bound states.

So am using “eigenstates” in quotes because the atom (or molecule – I’ll call them all atoms for our purposes) is coupled to the electromagnetic field. So “eigenstate” means, for example, “eigenstate as calculated by the “bare” Dirac equation for an electron in an atomic system sundered from the rest of the Universe. It is no longer an eigenstate of the whole, coupled system (atom together with electromagnetic field), which is why the transition happens in the first place.

Lionel’s answer here gives you a thorough description of how light is absorbed by way of the chapter “Semiclassical theory of light-matter interactions” downloaded through his link to the “Photonics 1” section of the Faculty of Physics, Ludwig Maximilian University, Munchen download section. Here the Fermi Golden Rule is derived for spontaneous absorption rates as well as the time varying coefficients $\alpha_j(t)$ that show you how the transition, although fantastically swift, is nonetheless smooth.

A complementary process, spontaneous emission of a photon from an electron in an excited state also lets you understand this smoothness as well as why the process is one-way. You can look up Wigner-Weisskopf theory for this transition:

V.Weisskopf and E.Wigner, “Berechnung der natürlichen Linienbreite auf Grund der Diracschen Lichttheorie”, Z. Phys. 63, 54 (1930)

or you can retell this tale through my own simplification presented in:

R. Vance and F. Ladouceur, “One-photon electrodynamics in optical fiber-fluorophore systems: III. One-polariton propagation in fluorophore-driven fibers,” J. Opt. Soc. Am. B  24, 1369-1382 (2007).

The Weisskopf – Wigner paper unfortunately is in German, which is a shame (for us English speakers) because it is the very best and clearest exposition I know of (as with nigh anything Wigner had a hand in). You can try section  6.3 in chapter 6 of Scully and Zubairy, “Quantum Optics” but this doesn’t do it for me: maybe it’ll work for you.

So, for now, here is my own summary of what I wrote up in JOSA-B.

Let’s think of $\hat{a}_1^\dagger$ is thought of as the creation operator that raises the atom in question from its ground state to its first raised state and $\hat{a}_\pm^\dagger(\omega)$ the corresponding operator for a photon in a one dimensional quantized EM field at frequency $\omega$ and in right (+) or left (-) circular polarisation, the Hamiltonian has the form:

$$\hat{H} = \hbar\left(\omega_1 \hat{a}_1^\dagger \hat{a}_1 + \int_0^\infty \omega\,\left(\hat{a}_+^\dagger(\omega) \hat{a}_+(\omega)+ \hat{a}_-^\dagger(\omega) \hat{a}_-(\omega)\right)\,\mathrm{d}\omega +\\ \int_0^\infty \left(\kappa_+(\omega)^*\, \hat{a}_1^\dagger \hat{a}_+(\omega) + \kappa_+(\omega) \hat{a}_+^\dagger(\omega)\,\hat{a}_1\right)\,\mathrm{d}\omega + \int_0^\infty \left(\kappa_-(\omega)^*\, \hat{a}_1^\dagger \hat{a}_-(\omega) + \kappa_-(\omega) \hat{a}_-^\dagger(\omega)\,\hat{a}_1\right)\,\mathrm{d}\omega + const\right)\quad\quad\quad(1)$$

where $\kappa_\pm(\omega)$ is the coupling strength between the excited atom and the free photon electromagnetic modes. The ground state energy for the EM modes is represented by the constant I do not name here. For now, think of this as a coupling to a cavity wherein there is only one electromagnetic mode for each frequency $\omega$. Now I just write this down as a general linear coupled model and suavely make the assertion that the $\kappa_\pm(\omega)$ can be calculated in principle from quantum electrodynamics and thus haughtily give the impression that I know how to do such a thing as a triviality (I don’t fully!). With only one photon in the system (i.e. initially in the excited atom and being spontaneously emitted into the field) and given that the above Hamiltonian conserves photon number (adds a photon whenever one is taken from somewhere else), we can reduce the whole system state to the probability amplitude $\psi_1(t)$ of the emitter atom’s being excited together with the continuous functions $\psi_\pm(\omega)$ which are the probability amplitudes to find the photon in the mode with frequency $\omega$ and in left and right hand circular polarization, so we don’t end up with horrendous complexity explosion wrought by tensor products of electron and photon quantum states:

$$\begin{array}{lcl} i\,\mathrm{d}_t\, \psi_1(t) &=& \omega_1 \,\psi_1(t) + \int_0^\infty \left(\kappa_+(\omega)^* \,\psi_+(\omega, t)+\kappa_-(\omega)^* \,\psi_-(\omega, t)\right)\, \mathrm{d}\omega\\ i\,\partial_t \,\psi_\pm(\omega, t) &=& \omega\, \psi_\pm(\omega, t) + \kappa_\pm(\omega) \,\psi_1(t)\end{array}\quad\quad\quad(2)$$

You can intuitively see that this equation applies for any number of modes in a quantisation volume, not only a one-mode cavity, because we can absorb the appropriate “degeneracy” co-efficients into the co-efficients $\kappa$ (see my JOSA-B paper if you want to see the details of how this works for a full EM field, but I can assure you it’s not exactly riveting stuff!). Now, I show how to solve this system of equations in the section “The Shape of the Spectrum without a Cavity” in this answer. The outcome is:

$$\begin{array}{lcl} \psi_1(t) &\approx& \exp\left({-i\,\omega_1\,t-\frac{t}{2\tau}}\right)\\ \tau &\approx& \left.{\frac{1}{2\,\pi\,\left(\kappa_+(\omega)^2+\kappa_-(\omega)^2\right)}}\right|_{\omega=\omega_1-\omega_0}\\ \psi_+(\omega) = \psi_-(\omega) &\approx& \sqrt{\frac{\tau}{\pi}} e^{-i\,\omega\,t}\left(1-e^{-\frac{t}{2 \tau }}\right) \frac{1}{2\tau(\omega -\omega_0)+i}\\ \psi_0(t) &=& e^{-i\,\omega_0\,t+i\theta_0}\,\left(1-e^{-\frac{t}{2 \tau }}\right) \end{array}\quad\quad\quad(3)$$

and thus we get the exponential, memoryless decay of a spontaneously emitting atom and the implied, Lorentzian lineshape. The last relationship in (3) is the inferred probability amplitude that the atom is in its first raised state and so the electron’s state is the following, smoothly varying superposition of ground $\psi_{0,electron}(\vec{r})$ and raised $\psi_{1,electron}(\vec{r})$ “eigenstates”:

$$\psi_{electron}(t,x) = \exp\left({-i\,\omega_1\,t-\frac{t}{2\tau}}\right) \psi_{1,electron}(\vec{r}) + e^{-i\,\omega_0\,t+i\theta_0}\,\left(1-e^{-\frac{t}{2 \tau }}\right) \psi_{0,electron}(\vec{r})\qquad(4)$$

Here $\theta_0$ is an undetermined phase factor. Note that the linewidth depends only on the strength of coupling $\kappa_\pm(\omega)$ in the neighbourhood of the uncoupled transition frequency $\omega_1-\omega_0$ defined by the atom’s transition energy level difference. It does not depend on the shape of the coupling  $\kappa_\pm(\omega)$ as long as this latter is broadband. What’s going on intuitively? The atom is coupled to all modes roughly equally. However, it cannot emit into all equally, because if it couples to a frequency away from $\omega_1-\omega_0$, destructive interference hinders the process. So only frequencies near $\omega_1-\omega_0$ are excited. The behavior of Eq. (4) implies a Lorentzian lineshape in the frequency domain, thus we can understand the mechanisms behind the most common spontaneous emission lineshape.

The thermodynamic considerations in Lionel’s answer can be readily understood here. Here the raised state is coupled to a continuum of modes. The beginning state, namely with the excitation confined to the atom is a low entropy state (low uncertainty of where the excitation is), and it deforms smoothly and inexorably to the high entropy state wherein the excitation is in a quantum superposition spread over a huge set of electromagnetic field modes.