Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

At university, I was shown the Schrodinger Equation, and how to solve it, including in the $1/r$ potential, modelling the hydrogen atom.

And it was then asserted that the differences between the eigenvalues of the operator were the permitted frequencies of emitted and absorbed photons.

This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation?

After all, there's no particular reason for an electron to be in an eigenstate.

What would make people think it was anything more than a (very suggestive) coincidence?

share|cite|improve this question

migrated from Apr 6 '14 at 17:40

This question came from our site for people studying math at any level and professionals in related fields.

A basic "premise" of quantum mechanics is that a system can only be in discrete physical states, which requires a specific amount of energy (within some "uncertainty") to change from one state to another. [This is the interpretation of observing spectral lines, rather than all possible frequencies of light.] This was the challenge in formulating the theory, since "classical" mechanics assumes quantities related to physical systems can change continuously, so any amount of energy can change a system from one state to another. – RecklessReckoner Apr 6 '14 at 17:03
In quantum mechanics "state" often means the function $\psi$. This can be any normalizable function, not just Hamiltonian eigenfunction and it can change continuously. You explanation would work if allowed "states" would be discrete as in the old quantum theory of Bohr. – Ján Lalinský Apr 6 '14 at 21:11
up vote 7 down vote accepted

This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate.

Good question! The function $\psi$ does not need to be Hamiltonian eigenfunction. Whatever the initial $\psi$ and whatever the method used to find future $\psi(t)$, the time-dependent Schroedinger equation $$ \partial_t \psi = \frac{1}{i\hbar}\hat{H}\psi $$ implies that the atom will radiate EM waves with spectrum sharply peaked at the frequencies given by the famous formula $$ \omega_{mn} = \frac{E_m-E_n}{\hbar}, $$

where $E_m$ are eigenvalues of the Hamiltonian $\hat{H}$ of the atom.

Here is why. The radiation frequency is given by the frequency of oscillation of the expected average electric moment of the atom

$$ \boldsymbol{\mu}(t) = \int\psi^*(\mathbf r,t) q\mathbf r\psi(\mathbf r,t) d^3\mathbf r $$ The time evolution of $\psi(\mathbf r,t)$ is determined by the Hamiltonian $\hat{H}$. The most simple way to find approximate value of $\boldsymbol{\mu}(t)$ is to expand $\psi$ into eigenfunctions of $\hat{H}$ which depend on time as $e^{-i\frac{E_n t}{\hbar}}$. There will be many terms. Some are products of an eigenfunction with itself and contribution of these vanishes. Some are products of two different eigenfunctions. These latter terms depend on time as $e^{-i\frac{E_n-E_m}{\hbar}}$ and make $\boldsymbol{\mu}$ oscillate at the frequency $(E_m-E_n)/\hbar$. Schroedinger explained the Ritz combination principle this way, without any quantum jumps or discrete allowed states; $\psi$ changes continuously in time. Imperfection of this theory is that the function oscillates indefinitely and is not damped down; in other words, this theory does not account for spontaneous emission.

share|cite|improve this answer
This looks like a very good answer! Thanks. I'm going to think about it for a bit and see what pictures I can make out of it. – John Lawrence Aspden Apr 6 '14 at 22:05
Yes, I think I see, so if we find any way to measure the electric moment, eigenvalue differences will appear in the values we get. And since the electric moment of the ground state is zero, and we expect the electric moment to be somehow connected to the field, getting a photon from the atom is effectively a measurement. Thank you very much! – John Lawrence Aspden Apr 10 '14 at 20:49
I think you're still being little bit confused. That's OK, this is rarely explained well in textbooks. The point of the above is that the average dipole moment of many atoms will oscillate in a complicated way but the oscillation will consist mainly of frequencies given by the difference formula. Oscillating electric moment is connected to radiation in a well-known way (from electromagnetic theory): the electric field of the system at great distance $r$ is given schematically by $E(r,t) \approx C\frac{\ddot{p}(t)}{r}$. – Ján Lalinský Apr 10 '14 at 21:37
The frequency of oscillation of the moment $p$ translates directly into the frequency of radiation. There are no photons in this explanation. – Ján Lalinský Apr 10 '14 at 21:38
@Lalinsky good answer. You are the only one who is remotely correct on this question. The other answers are all such nonsense. – Marty Green Dec 9 '14 at 13:38

The idea here is increasingly complex depending on how deep into modern physics you want to delve, but also key to understanding quantum mechanics. So, I'll give a bit deeper explanation than it seems you've seen, but there's plenty more.

It's understood that a photon acts both as a particle and a wave. As a particle it has an amount of energy associated with it, and as a wave it has a wavelength and frequency. These two values are directly related; you can know one from the other.

A good first thought experiment is to consider a particle in a hypothetical one-dimensional box. It can only bounce back and forth along one direction and in a finite distance. It will settle into any one of a number of quantized states thay have a wavelength that "fit," as I'm guessing you understand from your studies.

Extend that idea to an electron, then, which is confined to "orbit" the atom. It is three dimensional and the forces involved are not infinite potential barriers, but the idea of the particle's wave settling into a frequency that "fits" still holds.

Now, when an atom absorbs or emits a photon, the energy is absorbed into or emitted by one of the quantized electrons, causing it to gain or lose energy equal to that of the photon. Since the electron can only have discrete amounts of energy, we can calculate the energy of the photons emitted!

share|cite|improve this answer
As I understand it, the particle in the box doesn't settle into an eigenstate, it just continues obeying the schrodinger equation, and so it stays in whatever mix of states it starts in, modulo constant factors. – John Lawrence Aspden Apr 6 '14 at 17:51
Well, yes, but the particle achieves a local minima in potential energy by settling into a wave that has a discrete number of half wavelengths within the boundaries of the box. – user140858 Apr 6 '14 at 17:56
I don't understand this. The particle in a box setup has the potential either 0 or infinity, and the particle is never in the infinity bit. – John Lawrence Aspden Apr 6 '14 at 18:00
Yes, that's true for the potential caused by the box. What I mean is that the particle interacts with itself, and that a wavelength that doesn't fit neatly actually interferes with itself. In other words, it creates a higher potential. In doing so, the particle adjusts itself to fit the box, if that makes any sense. – user140858 Apr 6 '14 at 18:17
user1140858, your idea of "particle adjusting to the box" is interesting but it is not a standard part of the theory. Such behaviour is in contradiction to the Schroedinger equation; once the function $\psi$ is a superposition of many Hamiltonian eigenfunctions, it will remain such. Another theory (equation?) is needed to explain such adjusting. Spontaneous emission of light is very closely related. – Ján Lalinský Apr 6 '14 at 21:15

This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation?

Your puzzlement arises because you are putting the cart in-front of the horse. The cart is the theoretical model of quantum mechanics and the horse is the data. As your question is migrated from math.SE one can understand this orientation, which is dominant also here.

The whole theoretical package of Quantum Mechanics did not arrive by a seemingly holy inspiration ( as some physical theories having to do with apples is said to have been), but was a slow accumulation of observations that forced physicists to think outside of the box of the mathematics used in classical mechanics and thermodynamics.

It started with the table of elements, the photoelectric effect, the black body radiation, the spectral lines in atomic spectra. All these could not be squeezed within the classical models. Bohr tried with his model.

The photoelectric effect forced thinking into light as particles, (once more , as Newton had proposed particles), the photons.

Then it was known and expected in classical electromagnetism that an accelerating electron would lose energy in the form of radiation into light,( so photons come into any radiation). This would be a continuous spectrum. Classical mechanics and classical electromagnetism could not produce the spectral lines, because by the classical equations the electron should fall on the nucleus emitting a continuous spectrum in the field of the protons, not the distinct spectra lines which were observed . So Bohr postulated that the electron was staying into orbits with specific energy and could only lose energy in photons ( the classical expectation) in quantized steps. This explained the phenomena mathematically by fitting series to the spectral lines, but was not satisfactory because it gave no framework for the other observations listed above, of forced states , quantized states for energy changes in the atomic micro framework.

After all, there's no particular reason for an electron to be in an eigenstate.

I explained the particular reason, if it were not in a stable orbit there would not be spectral lines to be observed and we would not have atoms, and be here discussing this in the physical form we have.

What would make people think it was anything more than a (very suggestive) coincidence?

The postulates of Quantum Mechanics imposed on the mathematical solution of the Schrodinger equation brought logic and a causal path to the random efforts for a theoretical framework, outside the box of classical theories. So the appropriation of the differential equation now called "Schrodinger equation" to interpret the data was not a coincidence but a great think outside the box of classical theories. By imposing the physical postulates on the interpretation of the solutions, the fortuitous fits of the Bohr model series could be understood as derived from a formal mathematical physical theory.

share|cite|improve this answer
This is an excellent answer and it most directly answers the question on the level at which it was asked. – J... Apr 7 '14 at 8:58
@Anna v A very interesting answer, thanks a lot. To the OP: if you like the sound of this answer, you may find an excellent introduction of the mathematical description of quantum mechanics from the experimental principles done in a book by A. Connes, available on his website (Non-Commutative geometry, chapter 1, note that the rest of the book is absolutely not understandable for me, but the first chapter is really nice, and gives a short and profound discussion :-) – FraSchelle Apr 9 '14 at 8:32

Conservation of energy.

If we measure the energy of an atom, we will always report an eigenvalue, because we are forcing it into an eigenstate (this is something like the quantum mechanical definition of measurement). Now suppose that we measure the energy of an atom twice, before and after it emits a photon. For conservation of energy to hold, the energy of the photon must be the difference of the two eigenvalues.

It may be that the atom is not in an eigenstate exactly when it emits the photon, but an emission with energy level not a difference of eigenvalues would produce apparent contradictions as soon as we attempted to measure the change in energy.

share|cite|improve this answer
But we don't normally measure the energy of atoms before and after they emit photons. So why would they behave as if we did? – John Lawrence Aspden Apr 6 '14 at 17:57
@JohnLawrenceAspden This is much less strange than it would be if every atom remembered if it had ever had its energy measured, and anticipated whether it would ever be measured again. The point is that if we measured, certain changes would be nonsensical from the perspective of conservation of energy. Of course, this is an approximate picture (and, as Ján Lalinský points out, the answer is approximate as well). – you-sir-33433 Apr 6 '14 at 23:26

To have emission (or absorption) of photons you must have a Hamiltonian that includes those degrees of freedom also. If your system consists of (a) the electromagnetic field and (b) a hydrogen atom, you can specify the state with (a) for each frequency, the number of photons with that frequency and (b) the state of the hydrogen atom, in your favorite way, for example $1s$ or $2p$. You could write $\vert n_\omega=1, 1s\rangle$ for a state with 1 photon of frequency $\omega$ and the atom in the state 1s.

To calculate the probability for a transition between the states $\vert i\rangle$, meaning no photons and hydrogen atom in initial state $i$, and $\vert n_\omega =1, f\rangle$ where $f$ is some final state, you need to calculate an inner product like $$P = \langle n_\omega =1, f|O|i\rangle$$ where $O$ is some operator. The probability for the transition is then something proportional to $|P|^2$. The most significant contribution comes from the electric dipole moment operator and this is a standard calculation in textbooks. The result is that $P$ is proportional to $$P\propto \frac{\sin(t(\omega + \omega_f - \omega_i)/2)}{(\omega + \omega_f - \omega_i)/2}$$ where $\omega_f, \omega_i$ are the related to the initial and final energies by $\hbar\omega_f = E_f$ and similarly for $i$, and $t$ is the elapsed time. Clearly $P$ can be non-zero even if energy isn't conserved.

However, in the limit $t \to \infty$, $|P|^2$ approaches something proportional to $t\delta(\omega + \omega_f - \omega_i)$ where the $\delta$ is a Dirac delta. This is where conservation of energy comes from. The statement that atoms can emit photons only at specific frequencies is false if taken literally, each spectral line comes with a natural width corresponding to that $P$ for finite $t$ is non-zero even away from $\Delta E = 0$.

You can find a detailed calculation of $P$ in any textbook on quantum mechanics. I learned from Townsend's A Modern Approach to Quantum Mechanics, but I think you will find this calculation Sakurai's or Griffiths's books also.

share|cite|improve this answer
This looks like a really interesting answer. I'm going to think about it for a bit. Thanks. – John Lawrence Aspden Apr 6 '14 at 19:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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