General solution of the hydrogen atom Schrödinger equation

  • #1
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Hello everyone! I have two questions wich had bothered me for quite some time. I am sorry if they are rather trivial.

The first is about the general solution of the hydrogen atom schrödinger-equation: We learned in our quantum mechanics class that the general solution of every quantum system is a linear combination of all eigenstates (or eqivelant sets of basis functions). So a set of basis functions of the hydrogen atom are the hydrogen wave functions and the general solution of the hydrogen problem should therefore be a linear combination of this states. But every time I get in contact with atomic physics there is stated that the electrons are in the 1s, 2s, 2p,... state corresponding to the basis set of the hydrogen wave functions. But why do the electrons occupy the ‚atomic orbitals‘ and not a state, wich is a linear combination of all atomic orbitals. If an electron occupies a state wich is a linear combination, would this also mean that the energy of this electron is arbitrary (because I can add toghether atomic orbitals with different energys) and how would quantum numbers be defined in such a case (so that no two electrons have the same)?

The second question is related to the expectation value of energy: If we take the stationary schrödinger-equation
H|Φ>=E|Φ>​
we can calculate the energy E of an eigenstate |Φ> by
<Φ|H|Φ>=<Φ|E|Φ>=E​
where on the left is a expectation value <Φ|H|Φ>. Does this mean that if we measure the ground state energy of e.g. hydrogen, we do only on average get the typical ionisation energy of 13.6 eV?

I appreciate every help, thank you!
 

Answers and Replies

  • #2
anuttarasammyak
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As for the second question you can calculate squared standard deviation of energy as
[tex]<\phi|(H-E)^2|\phi>=0[/tex]
It clearly shows that the state has not in average but definite energy say -13.6 eV.
 
  • #3
PeroK
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Hello everyone! I have two questions wich had bothered me for quite some time. I am sorry if they are rather trivial.

The first is about the general solution of the hydrogen atom schrödinger-equation: We learned in our quantum mechanics class that the general solution of every quantum system is a linear combination of all eigenstates (or eqivelant sets of basis functions). So a set of basis functions of the hydrogen atom are the hydrogen wave functions and the general solution of the hydrogen problem should therefore be a linear combination of this states. But every time I get in contact with atomic physics there is stated that the electrons are in the 1s, 2s, 2p,... state corresponding to the basis set of the hydrogen wave functions. But why do the electrons occupy the ‚atomic orbitals‘ and not a state, wich is a linear combination of all atomic orbitals. If an electron occupies a state wich is a linear combination, would this also mean that the energy of this electron is arbitrary (because I can add toghether atomic orbitals with different energys) and how would quantum numbers be defined in such a case (so that no two electrons have the same)?
This is a good question. If you measure the energy of an atom, then the resulting state must be an energy eigenstate - and that means every electron being in a definite orbital. We measure the energy of an atom indirectly by looking at the absorption and emission spectrum. The energy of these photons is consistent with the atom transitioning from one eigenstate to another.

Your question is why can't an atom transition from one superposition of energy eigenstates to another superposition of energy eigenstates and, therefore, emit or absorb a photon of any energy?

I'm not sure that basic QM gives an answer to that, and you may need QFT and the quantization of the EM field to explain it fully. Part of the answer is that measuring the photon must effectively be a measurement of the atom before and after emission or absorption, and so any superposition of states is resolved probabilistically into an appropriate energy eigenstate.

Note also that basic QM requires a photon to have precisely the difference in energy levels. Which, if you think about it, is impossible. In fact, each line in the spectrum has a finite width and the probability of a photon being absorbed is at a maximum for the precise difference in energy levels and drops away very quickly. But, there is a non-zero probability of absorbing or emitting a photon of not quite the precise difference in energy levels.

Again, you need the quantization of the EM field to explain this.
 
  • #4
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This is a good question. If you measure the energy of an atom, then the resulting state must be an energy eigenstate - and that means every electron being in a definite orbital. We measure the energy of an atom indirectly by looking at the absorption and emission spectrum. The energy of these photons is consistent with the atom transitioning from one eigenstate to another.

Your question is why can't an atom transition from one superposition of energy eigenstates to another superposition of energy eigenstates and, therefore, emit or absorb a photon of any energy?

I'm not sure that basic QM gives an answer to that, and you may need QFT and the quantization of the EM field to explain it fully. Part of the answer is that measuring the photon must effectively be a measurement of the atom before and after emission or absorption, and so any superposition of states is resolved probabilistically into an appropriate energy eigenstate.

Note also that basic QM requires a photon to have precisely the difference in energy levels. Which, if you think about it, is impossible. In fact, each line in the spectrum has a finite width and the probability of a photon being absorbed is at a maximum for the precise difference in energy levels and drops away very quickly. But, there is a non-zero probability of absorbing or emitting a photon of not quite the precise difference in energy levels.

Again, you need the quantization of the EM field to explain this.
Thank you, this is helping me a lot! But I don’t understand another thing: If I measure the energy of an electron, how do I know in wich state (e.g. a superposition of eigenstates) it was before? For example: Could it be that a single electron in hydrogen is in a superposition of 1s and 2s orbitals and therefore I do sometimes measure the energy of the 1s orbital and sometimes the 2s orbital? And If it is like that, how do I know the probability of measuring a specific eigenstate or in other words the wave function of the electron beforehand (maybe by fermi-dirac statistics?).

And if I may ask another question: If I measure the electrons exact location, it also has to be in a position eigenstates (if the momentum is not measured at all). But isn’t this somehow contradicting to the uncertainty relation? If the electron’s position is measured in very brief timesteps, we would have the position in relation to time and could calculate the momentum (wich would lead to a classical behaviour wich is not stable because of dipol radiation)? This is somehow reminding me of the quantum zeno effect.
 
  • #5
PeroK
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Thank you, this is helping me a lot! But I don’t understand another thing: If I measure the energy of an electron, how do I know in wich state (e.g. a superposition of eigenstates) it was before? For example: Could it be that a single electron in hydrogen is in a superposition of 1s and 2s orbitals and therefore I do sometimes measure the energy of the 1s orbital and sometimes the 2s orbital? And If it is like that, how do I know the probability of measuring a specific eigenstate or in other words the wave function of the electron beforehand (maybe by fermi-dirac statistics?).

And if I may ask another question: If I measure the electrons exact location, it also has to be in a position eigenstates (if the momentum is not measured at all). But isn’t this somehow contradicting to the uncertainty relation? If the electron’s position is measured in very brief timesteps, we would have the position in relation to time and could calculate the momentum (wich would lead to a classical behaviour wich is not stable because of dipol radiation)? This is somehow reminding me of the quantum zeno effect.
First, if you measure the position of an electron in an atom it tends to ionise the atom. There are several pages on this online.

Genuinely, therefore, knowing the position of an electron and the energy state of an atom are incompatible.

Note that generally almost all measurenents of atomic phenomena are inferred from secondary measurenents - like the wavelength of an emitted photon. Or, in the case of Stern-gerlach, you don't directly measure the spin on an electron at all but infer it from a measurenent of where the electron collides with a screen.

On the other point. Say you have a process to create atoms in a certain state. If, after a time, the atom always emits a single electron with energy ##E_2 -E_1##, where these are the first excited state and ground state energies, then you infer that you are creating atoms in the first excited state.

If instead the atoms emit photons of different wavelengths with certain probabilities, then you might infer that you are creating atoms in a superposition of energy levels.

But, always, the photons emitted are consistent with a transitions from specific energy levels.

Basic QM gives you a big clue as to why that is but it's not the whole story.
 
  • #6
hilbert2
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Note also that basic QM requires a photon to have precisely the difference in energy levels. Which, if you think about it, is impossible. In fact, each line in the spectrum has a finite width and the probability of a photon being absorbed is at a maximum for the precise difference in energy levels and drops away very quickly. But, there is a non-zero probability of absorbing or emitting a photon of not quite the precise difference in energy levels.

Even the relativistic Doppler effect from atoms moving with different velocities in the gas phase tends to broaden the spectral lines so that they aren't exactly delta function like spikes. Interactions between different atoms in a non-ideal gas of nonzero density also do something to the spectral lines. As far as I can remember, putting an EM frequency slightly different from the exact transition frequency to the Rabi oscillation model of atom-field interaction will periodically excite and de-excite the atoms. Only the exact resonance frequency will make the fraction of excited atoms increase monotonically towards 100%.
 
  • #7
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Even the relativistic Doppler effect from atoms moving with different velocities in the gas phase tends to broaden the spectral lines

You sure the relativistic version of the Doppler effect is needed for that?
 
  • #8
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And if I may ask another question: If I measure the electrons exact location, it also has to be in a position eigenstates (if the momentum is not measured at all). But isn’t this somehow contradicting to the uncertainty relation? If the electron’s position is measured in very brief timesteps, we would have the position in relation to time and could calculate the momentum (wich would lead to a classical behaviour wich is not stable because of dipol radiation)? This is somehow reminding me of the quantum zeno effect.

It seems this is a more general question and it is not limited to the Hydrogen atom.

Each position measurement disturb the state of a quantum particle. By the time you made the second position measurement, the velocity you are trying to define has nothing to do with any velocity that could has been ascribed to the particle prior the first position measurement.
 
  • #9
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It seems this is a more general question and it is not limited to the Hydrogen atom.

Each position measurement disturb the state of a quantum particle. By the time you made the second position measurement, the velocity you are trying to define has nothing to do with any velocity that could has been ascribed to the particle prior the first position measurement.
Thank you! But why is it that ‚the velocity you are trying to define has nothing to do with any velocity that could has been ascribed to the particle prior the first position measurement‘? Is that only a problem arising due to measurement ‚inperfection‘? This would be unsatisfying.
 
  • #10
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Thank you! But why is it that ‚the velocity you are trying to define has nothing to do with any velocity that could has been ascribed to the particle prior the first position measurement‘? Is that only a problem arising due to measurement ‚inperfection‘? This would be unsatisfying.

For once, a quantum particle does not have a defined velocity/momentum/position/spin or whatever prior a measurement. See the bell theorem.


But addressing the specific question of velocity, consider the case of a free particle. Classically, the free particle has constant velocity, and if you measure the position and velocity at a given time you can predict where the particle will be in the future.

This is not the case for a free particle in QM. If you do several position measurements you will see that the particle is not moving in a straight line. And the velocity you are calculating from the position measurements is not a constant, but change after each measurement.

Why? because with each measurement you are changing the state of the particle.
 

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