In another post, I claimed that there was obviously an oscillating charge in a hydrogen atom when you took the superposition of a 1s and a 2p state. One of the respected members of this community (John Rennie) challenged me on this, saying:

Why do you say there is an oscillating charge distribution for a hydrogen atom in a superposition of 1s and 2p states? I don't see what is doing the oscillating.

Am I the only one that sees an oscillating charge? Or is John Rennie missing something here? I'd like to know what people think.

  • 1
    $\begingroup$ A related problem: imagine the particle in the (quantum) harmonic oscillator is charged. Do you see an oscillating charge in this case? Schrödinger has solved the apparent contradiction between the stationary eigenstates and the harmonic motion some long time ago, see here. $\endgroup$
    – Fabian
    Commented Nov 18, 2016 at 5:53
  • 2
    $\begingroup$ Can you extend your question to explain why you think there is an oscillating charge. For example are you suggesting it is because the charge density switches between the 1s shape and the 2p shape at some frequency? $\endgroup$ Commented Nov 18, 2016 at 5:56
  • 4
    $\begingroup$ @JohnRennie Marty is in the right in this instance. The oscillation of the wavefunction is pretty easy to see from the superposition - the overall shape of the superposition will depend on the relative phase of the two components, and this changes over time whenever the eigenenergies are different. $\endgroup$ Commented Nov 18, 2016 at 14:22
  • 1
    $\begingroup$ (Also - may I suggest an explicit mention of the 1s+2p superposition pure state in the question title?) $\endgroup$ Commented Nov 18, 2016 at 14:45
  • 4
    $\begingroup$ I've reverted an edit which injected meta-commentary into the question. To ask a follow-up physics question, see How to Ask. To discuss the nature of the Physics.SE community, please post to Physics Chat or to Physics Meta. $\endgroup$
    – rob
    Commented Aug 27, 2023 at 16:56

5 Answers 5


In this specific instance you are correct. If you have a hydrogen atom that is completely isolated from the environment, and which has been prepared in a pure quantum state given by a superposition of the $1s$ and $2p$ states, then yes, the charge density of the electron (defined as the electron charge times the probability density, $e|\psi(\mathbf r)|^2$) will oscillate in time.

In essence, this is because the $2p$ wavefunction has two lobes with opposite sign, so adding it to the $1s$ blob will tend to shift it towards the positive-sign lobe of the $p$ peanut. However, the relative phase of the two evolves over time, so at some point the $p$ signs will switch over, and the $1s$ blob will be pushed in the other direction.

It's worth doing this in a bit more detail. The two wavefunctions in play are $$ \psi_{100}(\mathbf r,t) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} e^{-iE_{100}t/\hbar} $$ and $$ \psi_{210}(\mathbf r, t) = \frac{1}{\sqrt{32\pi a_0^5}} \, z \, e^{-r/2a_0} e^{-iE_{210}t/\hbar}, $$ both normalized to unit norm. Here the two energies are different, with the energy difference $$\Delta E = E_{210}-E_{100} = 10.2\mathrm{\: eV}=\hbar\omega = \frac{2\pi\,\hbar }{405.3\:\mathrm{as}}$$ giving a sub-femtosecond period. This means that the superposition wavefunction has a time dependence, $$ \psi(\mathbf r,t) = \frac{\psi_{100}(\mathbf r,t) + \psi_{210}(\mathbf r,t)}{\sqrt{2}} = \frac{1}{\sqrt{2\pi a_0^3}} e^{-iE_{100}t/\hbar} \left( e^{-r/a_0} + e^{-i\omega t} \frac{z}{a_0} \frac{ e^{-r/2a_0} }{ 4\sqrt{2} } \right) , $$ and this goes directly into the oscillating density: $$ |\psi(\mathbf r,t)|^2 = \frac{1}{2\pi a_0^3} \left[ e^{-2r/a_0} + \frac{z^2}{a_0^2} \frac{ e^{-r/a_0} }{ 32 } + z \cos(\omega t) \, \frac{e^{-3r/2a_0}}{2\sqrt{2}a_0} \right] . $$

Taking a slice through the $x,z$ plane, this density looks as follows:

Mathematica source through Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["https://i.sstatic.net/KAbFl.png"]

This is indeed what a superposition state looks like, as a function of time, for an isolated hydrogen atom in a pure state.

On the other hand, a word of warning: the above statement simply states: "this is what the (square modulus of the) wavefunction looks like in this situation." Quantum mechanics strictly restricts itself to providing this quantity with physical meaning if you actually perform a high-resolution position measurements at different times, and compare the resulting probability distributions. (Alternatively, as done below, you might find some other interesting observable to probe this wavefunction, but the message is the same: you don't really get to talk about physical stuff until and unless you perform a projective measurement.)

This means that, even with the wavefunction above, quantum mechanics does not go as far as saying that "there is oscillating charge" in this situation. In fact, that is a counterfactual statement, since it implies knowledge of the position of the electron in the same atom at different times without a (state-destroying) measurement. Any such claims, tempting as they are, are strictly outside of the formal machinery and interpretations of quantum mechanics.

Also, and for clarity, this superposition state, like any hydrogen state with support in $n>1$ states, will eventually decay down to the ground state by emitting a photon. However, the lifetime of the $2p$ state is on the order of $1.5\:\mathrm{ns}$, so there's room for some four million oscillations of the superposition state before it really starts decaying.

A lot of atomic physics was forged in a time when a nanosecond was essentially instantaneous, and this informed a lot of our attitudes towards atomic superposition states. However, current technology makes subpicosecond resolution available with a modest effort, and femtosecond resolution (and better) is by now routine for many groups. The coherent dynamics of electrons in superposition states has been the name of the game for some time now.

It's also important to make an additional caveat: this is not the state that you will get if you initialize the atom in the excited $2p$ state and wait for it to decay until half of the population is in the ground state. In a full quantum mechanical treatment, you also need to consider the quantum mechanics of the radiation field, which you usually initialize in the vacuum, $|0⟩$, but that means that after half the population has decayed, the state of the system is $$ |\Psi⟩= \frac{|1s⟩|\psi⟩+|2p⟩|0⟩}{\sqrt{2}}, $$ where $|\psi⟩$ is a state of the radiation field with a single photon in it, and which is therefore orthogonal to the EM vacuum $|0⟩$. What that means is that the atom and the radiation field are entangled, and that neither can be considered to even have a pure quantum state on its own. Instead, the state of the atom is fully described (for all experiments that do not involve looking at the radiation that's already been emitted) by the reduced density matrix obtained by tracing out the radiation field, $$ \rho_\mathrm{atom} = \operatorname{Tr}_\mathrm{EM}\mathopen{}\left(|\Psi⟩⟨\Psi|\right)\mathclose{} =\frac{|1s⟩⟨1s|+|2p⟩⟨2p|}{2}, $$ and this does not show any oscillations in the charge density.

Foundational issues about interpretations aside, it's important to note that this is indeed a real, physical oscillation (of the wavefunction, at least), and that equivalent oscillations have indeed been observed experimentally.

Doing it for this hydrogen superposition is very challenging, because the period is blazingly fast, and it's currently just out of reach for the methods we have at the moment. (That's likely to change over the next five to ten years, though: we broke the attosecond precision barrier just last week.)

The landmark experiment in this regard, therefore, used a slightly slower superposition, with a tighter energy spacing. In particular, they used two different fine-structure states within the valence shell of the Kr+ ion, i.e. the states $4p_{3/2}^{-1}$ and $4p_{1/2}^{-1}$, which have the same $n$ and $L$, but with different spin-orbit alignments, giving different total angular momenta, and which are separated by $$\Delta E=0.67\:\mathrm{eV}=2\pi\hbar/6.17\:\mathrm{fs}.$$ That experiment is reported in

Real-time observation of valence electron motion. E. Goulielmakis et al. Nature 466, 739 (2010).

They prepared the superposition by removing one of the $4p$ electrons of Kr using tunnel ionization, with a strong ~2-cycle pulse in the IR, which is plenty hard to get right. The crucial step, of course, is the measurement, which is a second ionization step, using a single, very short ($<150\:\mathrm{as}$) UV burst of light.

Here the superposition you're probing is slightly more complicated than the hydrogen wavefunction the OP asks about, but the essentials remain the same. Basically, the electron is in a superposition of an $l=1,m=0$ state, and an $l=1,m=1$ state, with an oscillation between them induced by the difference in energy given by the spin-orbit coupling.

This means that the shape of the ion's charge density is changing with time, and this will directly impact how easy it is for the UV pulse to ionize it again to form Kr2+. What you end up measuring is absorbance: if the UV ionizes the system, then it's absorbed more strongly.

The absorbtion data therefore shows a clear oscillation as a function of the delay between the two pulses:

The pictures below show a good indication of how the electron cloud moves over time. (That's actually the hole density w.r.t. the charge density of the neutral Kr atom, but it's all the same, really.) However, it's important to note that the pictures are obviously only theoretical reconstructions.

Anyways, there you have it: charge densities (defined as $e|\psi(\mathbf r)|^2$) do oscillate over time, for isolated atoms in pure superposition states.

Finally, the standard caveats apply: the oscillations caused in quantum mechanics by superpositions are only valid for pure, isolated states. If your system is entangled with the environment (or, as noted above, with the radiation it's already emitted), then this will degrade (and typically kill) any oscillations of local observables. If the overall state of the world is in some meaningful superposition of energy eigenstates, then that state will indeed evolve in time. However, for heavily entangled states, like thermal states or anything strongly coupled to the environment, any local observables will typically be stationary, because each half of an entangled state doesn't even have a proper state to call its own.


Yes, the electric charge density (or more precisely, the electron's spatial probability density $p(x)$) would indeed oscillate with time, with frequency $10.2\text{ eV}/\hbar$. Energy eigenstates are stationary in time; since the state you propose isn't an energy eigenstate, it isn't stationary in time.

I don't understand anna v's answer on several levels. First of all, I don't see what the fine structure has to do with anything, because the $1s$ and $2p$ states have different principal quantum numbers, so they have a fairly large ($10.2\text{ eV} \approx 118000\text{ K}$) energy difference even in the completely non-relativistic limit. Second of all, I don't understand her claim that superpositions of different energy levels aren't allowed - if this were true, then nothing would ever change with time!

I think that what anna v's getting at is that if you include relativistic corrections from QED - i.e. you treat the electromagnetic field as quantum-mechanical - then the usual non-relativistic electron eigenstates are no longer exact eigenstates of the full relativistic Hamiltonian, and so electrons can undergo spontaneous emission or absorption of photons and change energy levels. I'm not sure what the time scales for this process are. But if you ignore relativistic QED effects (which I think is what Marty Green had in mind) then the electric charge distribution will indeed oscillate indefinitely.

  • 1
    $\begingroup$ "her claim that superpositions of different energy levels aren't allowed - if this were true, then nothing would ever change with time!" Probabilities are nomalized to 1 for each energy level, otherwise there would be no stability. Things change in time by incoming and outgoing packets of energy momentum and angular momentum; photons at the hydrogen level, (but not only in general). $\endgroup$
    – anna v
    Commented Nov 18, 2016 at 7:58
  • $\begingroup$ You cannot have a hydrogen atom at a mix of an 1s and 2p state. It would emit a photon and go to the 1s state. $\endgroup$
    – anna v
    Commented Nov 18, 2016 at 8:01
  • 8
    $\begingroup$ @annav Indeed, it would emit a photon and go to the $1s$ state - on a nanosecond timescale, which leaves room for about four million oscillations at the 0.4 fs period of the superposition. $\endgroup$ Commented Nov 18, 2016 at 14:21
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Nov 19, 2016 at 15:50

The superposition of eigenstates in a hydrogen atom results in an oscillating wave function in time with a frequency corresponding to the difference of energies of the eigenstates. Schrödinger considered for a time the wave function squares as charge density which resulted in an oscillating charge distribution. As this corresponded to an electric dipole oscillation and also explained intensities and polarization of observed light emission, he assumed heuristically that this interpretation explained the origin of light emission. See E. Schrödinger "Collected Papers on Wave Mechanics ", Blackie & Son Ltd., London and Glasgow 1928

  • $\begingroup$ Odd that no one has commented on your answer so far. I am going to post a follow-up as a separate question: is it true that you get the correct intensity for the emitted light by applying Maxwell's equations to the oscillating charge density? $\endgroup$ Commented Nov 19, 2016 at 13:23
  • $\begingroup$ And here is the follow-up question: physics.stackexchange.com/questions/293577/… $\endgroup$ Commented Nov 19, 2016 at 14:44
  • $\begingroup$ Re "...considered for a time...": Are you suggesting it is not correct? $\endgroup$ Commented Nov 20, 2016 at 10:09
  • 2
    $\begingroup$ @Marty Green - After finding his famous equation, Schrödinger for some time heuristically assumed that the (square of the single) electron wave function scalar corresponded to the particle and charge being smeared out in real space. He himself realized that this meaning of the wave function could not be maintained in multi-electron wave functions because they described the electrons in higher dimensional configuration space. Eventually he accepted the electron location probability interpretation suggested by Born. $\endgroup$
    – freecharly
    Commented Nov 20, 2016 at 14:56
  • 2
    $\begingroup$ @Marty Green - I found this recent article on the charge density interpretation of Schrödinger where it is argued that it is actually correct: philsci-archive.pitt.edu/9696/1/electroncloud_v9.pdf $\endgroup$
    – freecharly
    Commented Nov 27, 2016 at 5:46

First, I respectfully disagree with @anna v's statement that there cannot be a superposition of two states with different energy (although she seems to withdraw this statement in her comment). The superposition principle rules supreme, so if each of two states is possible, any superposition of the states is possible. Stability of energy levels does not seem relevant, as only the ground level is stable anyway.

Now, let us consider some (non-normalized) superposition, say $$\psi_1(\vec{r}) \exp(iE_1 t)+\psi_2(\vec{r}) \exp(iE_2 t),$$ of two eigenstates of energy $\psi_1(\vec{r}) \exp(iE_1 t)$ and $\psi_2(\vec{r}) \exp(iE_2 t)$. The probability density and the charge density (averaged over an ensemble) for this superposition will equal (up to a constant factor) \begin{align} (\psi_1(\vec{r}) \exp(iE_1 t) +\psi_2(\vec{r}) & \exp(iE_2 t))^* (\psi_1(\vec{r}) \exp(iE_1 t)+\psi_2(\vec{r}) \exp(iE_2 t)) \\&=|\psi_1(\vec{r})|^2+|\psi_2(\vec{r})|^2+2\Re(\psi_1(\vec{r})^*\psi_2(\vec{r})\exp(i(E_2-E_1)t)). \end{align} Therefore, the charge density (averaged over an ensemble) for the superposition does have an oscillating part in (pretty much) any point.

EDIT: (11/19/2016) In my initial answer above, I tried to avoid interpretational issues. However, as the OP accepted @freecharly's answer (and expressed interest in comments on that answer) and as @annav added in her answer that "It is very clear that the space charge distribution is not oscillating at the individual electron level, the charge sticks to the electron as the spot shows", I conclude that there may be clear interest in the interpretation, so let me add a few words.

freecharly mentioned the well-known interpretation by Schrödinger, where the squared magnitude of the wave function is the charge density. This interpretation has some weaknesses. For example, a wave packet in free space spreads steadily, which is in tension with the charge of electron being integer. In my article http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf (published in the European Physical Journal C) (at the end of section 3) I proposed another tentative interpretation: "the one-particle wave function may describe a large (infinite?) number of particles moving along the trajectories defined in the de Broglie–Bohm interpretation. The total charge, calculated as an integral of charge density over the infinite 3-volume, may still equal the charge of electron. So the individual particles can be either electrons or positrons, but together they can be regarded as one electron, as the total charge is conserved." This seems to be compatible with the notion of vacuum polarization and can provide the same charge density as in Schrödinger's interpretation (while the total charge in any volume is integer, there can be a fractional average charge density in the limit of decreasing volume), however, wave packet spreading is not as problematic.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Nov 19, 2016 at 15:52

Here are hydrogen atom energy levels


For hydrogen and other nuclei stripped to one electron, the energy depends only upon the principal quantum number n.


This fits the hydrogen spectrum unless you take a high resolution look at fine structure where the electron spin and orbital quantum numbers are involved. At even higher resolutions, there is a tiny dependence upon the orbital quantum number in the Lamb shift.

fine struct h

So your:

when you took the superposition of a 1s and a 2p state.

in a quantum mechanical frame does not make sense. There cannot be a superposition because these are two distinct energy levels. To go from the 2p to the 1s energy and angular momentum will be radiated away, and to go from the 1s to the 2p energy is needed.

Energy levels are posited as stable in the quantum mechanical frame, i.e the probability of an electron that was in a 1s state to remain a 1s state is 1, unless energy is supplied. Quantum mechanics is continually validated.

Thus there cannot be an oscillation between 1s and 2p states and energy conservation at the same time.

Edit after comment by Emilio Pisanty

Me: You cannot have a hydrogen atom at a mix of an 1s and 2p state. It would emit a photon and go to the 1s state. – anna v

Reply: @annav Indeed, it would emit a photon and go to the 1s state - on a nanosecond timescale, which leaves room for about four million oscillations at the 0.4 fs period of the superposition. - Emilio Pisanty

In conclusion I will accept that my answer holds for times larger than nanosecond scales. It seems that technology is reaching much faster times than I was aware, and superposed states can within these time limits exist.

Now whether the oscillation of a wavefunction squared ( i.e. probability), can be considered as oscillating charges is not clear. The double slit experiment one particle at a time shows that the interference pattern is a probability wave , not a mass wave. There is one spot on the screen for the one electron at a time, the spot coming from the interaction of the electron with the atoms of the screen. The accumulation shows the interference, i.e. the oscillation over space of the accumulated charge distribution. It is very clear that the space charge distribution is not oscillating at the individual electron level, the charge sticks to the electron as the spot shows.

Thus I expect that the time oscillation of the probability distribution shown in Emilio's answer cannot be interpreted as other than a probability distribution , "how probable it is that the single electron of the hydrogen atom will be found at the (x,y,z,t) point". It is another way of saying that the electron is not in an orbit, but in an orbital in the hydrogen atom.

Thus I will be very skeptical that it is the charge that is waving in the femtosecond scales. Not the least that classically oscillating charges radiate (see the answer by freecharly )

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Nov 19, 2016 at 18:09
  • 1
    $\begingroup$ Regarding whether oscillation of probability distribution can be considered as charge oscillation, consider that in the classical limit the electron's very localized wave packet will orbit the nucleus, still being a probability cloud. Its mean value will indeed obey classical equations of motion. (See my related self-answer for an animation of such a wave packet evolution). Thus for all intents and purposes, this probability cloud does represent an orbiting electron. $\endgroup$
    – Ruslan
    Commented Dec 20, 2017 at 11:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.