Let's consider the element neon. Its ground-state electron configuration is: $1s^2 2s^2 2p^6$.

What would happen if enough energy was given for one electron in the $1s$ orbital to jump to the $2s$ orbital (i.e. exactly the $\Delta E$ between $1s$ and $2s$ was supplied)?

Would the electron from the $1s$ orbital absorb the energy? There can't be more than 2 electrons in an orbital, so what would happen to the electrons in the $2s$ orbital if the $1s$ electron absorbed the energy?

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    $\begingroup$ What does it mean to "give the electron energy" here? Energy does not move around magically, there must be a mode of transfer, and that mode of transfer will determine what happens here. The question as written is underspecified. $\endgroup$
    – ACuriousMind
    Jul 7, 2021 at 0:39
  • $\begingroup$ Does the uncertainty principle even allow delivering such a tightly constrained quanta of energy? $\endgroup$
    – BCS
    Jul 16, 2021 at 15:10

6 Answers 6



Transitions to other unoccupied states are possible but extremely unlikely, more likely that the photon will not be absorbed.


The Pauli exclusion principle prevents a third electron occupying the $2s$ state. Even if there was space in the $2s$ state a $1s\to 2s$ transition is unlikely due to selection rules and a $1s\to2p$ transition is significantly more likely if there is space in the $2p$ orbital.

Other answers here have stated that the transition to other energy levels is forbidden. Now while the probability of a transition is extremely small, it is non-zero.

A quick note on notation: I will be using a bold typeface for vectors as opposed to an over arrow so that vector operators are clearer.

Quantisation of the Electromagnetic Field

Minimal coupling of the electron to the electromagnetic field using the Coulomb potential adds a perturbation of:

$$\hat H_1=\frac{e}{m_e}\hat{\boldsymbol p}\cdot\hat{\boldsymbol A}\left(\boldsymbol r,t\right)$$

to the Hamiltonian. Where $e$ and $m_e$ are the charge and mass of the electron, $\hat{\boldsymbol p}$ is the momentum operator acting on the electron and the vector potential operator has the form:

$$\hat{\boldsymbol A}\left(\boldsymbol r,t\right)=\sum_{\lambda,\boldsymbol k}\sqrt{\frac{\hbar}{2v\epsilon_0\omega\left(\boldsymbol k\right)}}\left(\hat a_\lambda\left(\boldsymbol k\right)\boldsymbol s_\lambda\left(\boldsymbol k\right)e^{i\left(\boldsymbol {k}\cdot\boldsymbol r-\omega t\right)}+\text{h.c.}\right)$$

where $\text{h.c.}$ is the Hermitian conjugate of the preceding terms, $v$ is the volume of the cavity in which the experiment is taking place; $\omega\left(\boldsymbol k\right)$ is the angular frequency of the photon mode as a function of the wavevector $\boldsymbol k$; $\lambda$ labels the two polarisations; $\boldsymbol s_\lambda\left(\boldsymbol k\right)$ is the polarisation vector of the mode; $\hat a_\lambda\left(\boldsymbol k\right)$ is the annihilation operator for the mode; and $\boldsymbol r$ is the position of the atom (assuming the wavelength is larger than the atom the uncertainty in the electron's position can be ignored).

If we have a single wavelength and polarisation then:

$$\hat H_1=\frac{e}{m_e}\sqrt{\frac{\hbar}{2v\epsilon_0\omega}}\hat{\boldsymbol {p}}\cdot\boldsymbol s\hat a e^{i\left(\boldsymbol k\cdot\boldsymbol r-\omega t\right)}+\text{h.c.}$$

Thus, let:

$$\begin{align}\hat V&=\frac{e}{m_e}\sqrt{\frac{\hbar}{2v\epsilon_0\omega}}\hat{\boldsymbol {p}}\cdot\boldsymbol s\hat a e^{i\boldsymbol k\cdot\boldsymbol r}\\\implies\hat H_1&=Ve^{-i\omega t}+\hat V^\dagger e^{i\omega t}\end{align}$$

Then using first-order time-dependent perturbation theory which holds in the limit $\frac{t}{\hbar}\left|\langle f|\hat V|i\rangle\right|\ll1$ for all $n\ge2$. We find the probability of a transition having occured if the atom is measured after a time $t$ since the electromagentic field was applied is:

$$\begin{align}P\left(t\right)=\frac{t^2}{\hbar^2}\Bigg|&\overbrace{e^{i\left(\Delta\omega-\omega\right)t/2}\operatorname{sinc}\left(\frac{1}{2}t\left(\Delta\omega-\omega\right)\right)\langle f|\hat V|i\rangle}^\text{absorption}\\+&\underbrace{e^{i\left(\Delta\omega+\omega\right)t/2}\operatorname{sinc}\left(\frac{1}{2}t\left(\Delta\omega+\omega\right)\right)\langle f|\hat V^\dagger|i\rangle}_\text{emission}\Bigg|^2\end{align}\tag{1}$$

where $\Delta E=\hbar \Delta \omega$ be the difference in the energy levels of the initial $|i\rangle$ and final $|f\rangle$ states. This is in general non-zero even when $\Delta \omega\ne\omega$. However, we can make one more approximation to aid in understanding: if the final state $|f\rangle$ has absorbed a photon then in the limit $t\Delta\omega\gg2\pi$ the $\operatorname{sinc}$ functions do not overlap and so we need only retain the absorption term:

$$P\left(t\right)=\left(\frac{\left|\langle f|\hat V|i\rangle\right|}{\hbar}\right)^2t^2\operatorname{sinc}^2\left(\frac{1}{2}t\left(\Delta\omega-\omega\right)\right)\tag{2}$$

Further approximations from here will give you Fermi's Golden rule, one of these approximations is taking the limit such that $t\operatorname{sinc}^2$ tends to a delta function and so removes the possibility for a transition when the energy of the photon is not exactly equal to the energy gap: and so this is an inappropriate approximation to make in this case.

Energy Conservation in Quantum Mechanics

While the expectation value of the energy is conserved in the evolution of a system as described by Schrödinger equation, there may be a discontinuous jump in the energy of the system when a measurement is performed. Consider a system in a superposition of energy eigenstates, when you measure the energy the state will collapse into an energy eigenstate which in general will not have the same energy as the expectation value for the energy - the energy of the system has increased or decreased!

The energy may be transferred to or from the measurement device or surroundings to compensate.

In previous edits this section also contained a discussion of the many-words type interpretation which in my naivety I included. I apologise for anyone I have mislead and for more details you can see this question:

"Conservation of energy, or lack thereof," in quantum mechanics

@Jagerber48's answer is the most relevant to this question giving additional details that will likely be of interest to any reader of this question.
@benrg's answer gives a good explanation of why energy is conserved.
@NiharKarve's comment includes a blog post which explains why the paper may be misleading.

Putting this all Together

Equation (1) shows, in general, the when an atom is illuminated by light of a single specific wavelength and polarisation, a transition is possible even if the energy of the photons is not equal to the energy gap, which would violate energy conservation (but this is allowed); however, the probability is extremely small.

Equation (2) makes a further approximation which we can now use to find an expression for the probability:

$$P\left(t\right)=\frac{e^2}{2v\epsilon_0m_e^2\hbar\omega}\left|\langle f|\hat{\boldsymbol {p}}\cdot\boldsymbol s\hat a |i\rangle\right|^2t^2\operatorname{sinc}^2\left(\frac{1}{2}t\left(\Delta\omega-\omega\right)\right)$$

As $|i\rangle\equiv|i\rangle_e|f\rangle_{EM}$ and $|f\rangle_e|f\rangle_{EM}$ where subscript $e$ is the electrons states and subscript $EM$ are the states of the electromagnetic field. Without detail $_e\langle f|\hat{\boldsymbol {p}}\cdot\boldsymbol s|i\rangle_e\equiv\boldsymbol d_{fi}\cdot\boldsymbol s$ where $\left\{\boldsymbol d_{fi}\right\}$ are the dipole matrix elements and are zero for transitions between certain orbitals independent of the energy supplied (for more details see selection rules). Finally, $_{EM}\langle f|\hat a|i\rangle_{EM}=\sqrt{N}$ if the state $|i\rangle_{EM}$ is the state for $N$ photons of the given wavelength and polarisation - but other states such as coherent states are also posible.

$$\implies P\left(t\right)=\frac{e^2N}{2v\epsilon_0m_e^2\hbar\omega}\left|\boldsymbol d_{fi}\cdot\boldsymbol s\right|^2t^2\operatorname{sinc}^2\left(\frac{1}{2}t\left(\Delta\omega-\omega\right)\right)\tag{3}$$

which holds in the limit:

$$t\ll\frac{\hbar}{\left|\,_e\langle f|\left(\hat{\boldsymbol {p}}\cdot\boldsymbol s\right)|i\rangle_e\right|}\sim 10^{-25}\text{s}$$

As the limit $t\Delta\omega\gg2\pi$ is not needed when the state $|i\rangle_{EM}$ is the state for $N$ photons of the given wavelength and polarisation because the creation operator causes the emission term to vanish anyway. However, the time is of the order of $10^{-25}\text{s}$ give or take a few orders of magnitude for Neon (obtained using the only data I could find for reduced matrix elements for dipole transitions), which is not a practical time scale to measure on.

Finally, considering your given case, given selection rules, the most likely case if the $1s$ electron did absorb a photon is a transition to the $3p$ state (as $2p$ is occupied and $3s$ is forbidden to first order by selection rules). Substituting values into equation (3) gives an order of magnitude estimate for the probability of transitioning from $1s$ to $3p$ in Neon of $10^{-12}\%\text{ per }\left(\text{photon }m^{-3}\right)$ for $t=10^{-25}\text{s}$ which is the point the approximation of first order perturbation breaks down.

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    $\begingroup$ Extremely comprehensive answer!! Thank you! $\endgroup$
    – jdbiochem
    Jul 6, 2021 at 21:40
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    $\begingroup$ This is actually the correct answer to the question. Choose this one not mine! $\endgroup$
    – SuperCiocia
    Jul 6, 2021 at 23:22
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    $\begingroup$ Using the measurement scenario as an example, does not mean that the system itself is not energy-conserving.As you have stated <E> is always constant meaning energy is conserved.Of course for any system that is not isolated (like making a measurement in this case) coherence is shared with the environment-just like energy loss to friction in classical systems. But for isolated systems, energy is always conserved.IMO, your answer,though technically correct, can be very misleading to those beginning study of physics, since you seem to emphasize that energy is always lost in quantum systems. $\endgroup$
    – joseph h
    Jul 7, 2021 at 0:30
  • $\begingroup$ @josephh I have now also added your suggestion about transfer to or from the measurement device. $\endgroup$
    – Chris Long
    Jul 7, 2021 at 7:57
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    $\begingroup$ I am not sure who adding a comment will notify, but in the chance it notifies anyone I have potentially mislead about energy conservation, I am very sorry. For more details please follow the link in the edit. I have also deleted any misleading comments I left arguing the point of what I have now discovered to be a misleading paper. $\endgroup$
    – Chris Long
    Jul 8, 2021 at 7:53

There are some good answers already, but I wanted to emphasize one more important point: There is actually no such a thing individual orbital energies in a multi-electron atom. So it does not even make sense to talk about the energy difference "$\Delta E$ between $1s$ and $2s$." The energy is the eigenvalue of the Hamiltonian for the entire system of interacting electrons.

The total energy includes electron-electron repulsion. Beyond the modifications to the potential felt by each individual electron, there are also more subtle effect, such as exchange energies, caused by the interplay of the electron-electron repulsion with the Pauli Exclusion Principle (although the exchange energies are not so important when the electron shells are filled). In general, these energy terms cannot be assigned to individual electrons—although approximations (like the Hartree approximation) that do assign each individual electron an "energy" can be extremely accurate under the right circumstances.

However, at a conceptual level, the question is really asking about what happens if the energy supplied to the atom is exactly the $\Delta E$ between the $1s^{2}2s^{2}2p^{6}$ state and the $1s^{1}2s^{3}2p^{6}$ state—except that the latter state does not exist, making this quantity undefined. This is actually a real practical problem for experimental tests of the Pauli Exclusion Principle that look for transitions to states with overfilled electron orbitals, because we do not have any reliable method for calculating the energies of the overfilled orbital states and must rely on some rather crude approximations.

  • 3
    $\begingroup$ Excellent points! $\endgroup$
    – Chris Long
    Jul 6, 2021 at 22:49
  • $\begingroup$ Very nice. I do not envy OP's task of having to pick between this and @ChrisLong 's answer. :) $\endgroup$
    – AnoE
    Jul 7, 2021 at 8:36
  • $\begingroup$ @AnoE Nah, Chris Long's answer is definitely better. $\endgroup$
    – Buzz
    Jul 8, 2021 at 4:17

Nothing happens.

Fermi golden rule says that, to first-order approximation, the transition probability from an initial state $|i\rangle$ to $|f\rangle$ is: $$ \Gamma_{i\rightarrow f} = \frac{2\pi}{\hbar} |\langle f|H'|i\rangle|^2 \rho(E_f),$$ where the $\langle f|H'|i\rangle$ is the matrix-element and $\rho(E_f)$ is the density of states at the energy $E_f$ of final states.

If the transition is allowed by the selection rules, the matrix element is non-zero. But if the target shell is full, the density of final states $\rho(E_f)$ is zero (there are no final states available). So the transition rate is zero.

  • $\begingroup$ I believe Fermi's Golden rule applies in the limit that the perturbing potential is applied for a long time and so is just an approximation of: $\Gamma_{i\to f}=\frac{\text{d}}{\text{d}t}\int\text{d}\omega'\rho\left(\omega'\right)\left(\frac{\left|\langle\phi\left(\omega'\right)|H'|i\rangle\right|}{\hbar}\right)^2t^2\operatorname{sinc}\left(\frac{1}{2}t\left(\omega'-\omega_i-\omega\right)\right)$ where $H'$ is the perturbing potential that oscillates at frequency $\omega$ and $E_i=\hbar\omega_i$ Thus, should the probability of tranisiton not be small but finite? $\endgroup$
    – Chris Long
    Jul 6, 2021 at 17:26
  • $\begingroup$ That will probably fine tune the matrix element side of the formula. The density of states bit is not an approximation $\endgroup$
    – SuperCiocia
    Jul 6, 2021 at 18:31
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    $\begingroup$ My point was that the questioner was asking what would happen if the photon was absorbed by the $1s$ electron (which is physically possible due to the $\operatorname{sinc}$ function) just the probability of doing so is extremely small - my order of magnitude estimate is approximately $10^{-12}\%\text{ per }\left(\text{photon }m^{-3}\right)$. $\endgroup$
    – Chris Long
    Jul 6, 2021 at 21:36
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    $\begingroup$ yes you are right and your answer is the actual correct answer to the question. Mine is approximate and simplistic. $\endgroup$
    – SuperCiocia
    Jul 6, 2021 at 23:23
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    $\begingroup$ Do you mean "nothing happens" as in "no energy is even absorbed in the first place from whatever source", or "the electron now has more energy but stays in its original orbital because it can't reach a higher one that's not full". @ChrisLong's answer says "more likely that the photon will not be absorbed", assuming that the unspecified energy source is a photon. $\endgroup$ Jul 7, 2021 at 4:12

A transition from $1s^2 2s^2 2p^6 \rightarrow 1s^1 2s^3 2p^6$ is forbidden because there is no such excited state in the anti-symmetric Hilbert space of 10 photons due to Pauli exclusion principle.

If the photon has energy $\Delta E$ equivalent to the energy difference between $1s$ and $2s$ then it is likely far detuned from other transitions. If we neglect all possible other excited state then in this situation absolutely nothing would happen. The atom would just pass by.

However, if we consider other states, such as $1s^1 2s^2 2p^6 3p^1$, we can now refer to the answer from @Chris Long. Even though the photon is probably very far detuned from this transition, there is still a small probability that the transition will be driven. The same statement is true for a range of other electronic energy states. These transitions are all driven off-resonantly so the probability of transition is very small.

But, in any case, the atom goes from being purely in the ground state, to being mostly in the ground state with small superposition components of various excited states. These small superposition components contribute to the total electron wavefunction slightly changing shape. The amount the shape changes decreases the further detuned we are. Normally this effect is neglected, but I raise it here because it allows us to recover our intuition that SOMETHING ought to happen when the electric field of the photon passes by the atom.

To say "nothing happens" is correct at the same level of approximation that it is correct that "a harmonic oscillator driven far off resonance undergoes no motion".

edit: As @Ruslan points out, what is more likely than a transition to a higher energy bound state such as $1s^12s^22p^6e3p^1$ is the transition to an ionized state such as $1s^12s^22p^6$ where one electron is lost to the continuum. See the image:

enter image description here

The $1s\rightarrow 2s$ photon in Hydrogen is 10 eV. If, instead of a $1s$ electron absorbing the photon, one of the $2s$ or $2p$ electrons absorbs the photon then those electrons will end up with ~20 eV of energy which exceeds the ionization threshold of 13.6 eV.

So in this case the system will evolve from the ground state into a superposition of ground state, ionized states, and (a very small component of) excited bound states. It's worth noting that while the excited bound state contribution will be small because of the large detuning, the probability of ionization may in fact be large.

  • $\begingroup$ I'd expect ionization (by driving higher-shell electrons) to be far more likely than a transition of a $1\mathrm s$ electron due to a detuned EM wave. $\endgroup$
    – Ruslan
    Jul 7, 2021 at 11:06
  • $\begingroup$ @Ruslan yes, very fair point. Adding that in. $\endgroup$
    – Jagerber48
    Jul 7, 2021 at 15:21

Due to Pauli's exclusion principle although the energy given corresponds to energy gap between $1s$ and $2s$, there is not final state as in $2s$ possible from the transition, hence that particular transition is FORBIDDEN due to Pauli's Exclusion principle and nothing happens in the context of this particular transition.


  1. Notice that I have taken the liberty to use "corresponds to" instead of "exact energy difference". Although theoretical providing an exact energy would have resulted in a transition (assuming a situation where $2s$ is empty), in practice, there are lot phenomena that broadens this energy different, so "corresponds to" is meant to accommodate that as well.

  2. Although the transition for an electron at $1s$ to $2s$ is not possible, it should be remembered that, there can be other transitions that are not forbidden, which can take place, say an electron in the outer shell getting excited to a higher state.

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    $\begingroup$ Fermi's exclusion principle? Isn't it Pauli's exclusion principle? $\endgroup$ Jul 6, 2021 at 16:49
  • $\begingroup$ @ThomasFritsch ohh lol sorry... I had "fermions" in mind. Anyways someone Joseph has corrected it. Thanks! $\endgroup$ Jul 7, 2021 at 17:16

To reinforce Ruslan's comment, and try to address the question without too much reference to full-up quantum complexity, the energy in question is the Kα energy, 848.6 eV. If, in an experiment, you illuminate neon with photons of that energy, the thing that you see is ionization. The ionization energy is only 21.6 eV. There is no spectroscopic absorption feature at the Kα energy, demonstrating that the atom in its ground state doesn't have any special tendency to be excited by that energy. The energy absorption increases above the K edge energy of 870 eV, when you have enough energy to remove a K shell electron.


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