Photon emission by an excited hydrogen atom

In summary, the process of photon emission by an excited atom should not be considered as an instant event, but rather as a continuous process. This can be described by evolutionary differential equations, typically in the form of ODEs or PDEs. For a hydrogen atom, the time-dependent wave function with the Coulomb potential can be written using the Shrödinger equation. In order to describe the absent photon in the beginning and present photon in the end of the process, auxiliary functions and terms must be added to the time-dependent Shrödinger equation. The creation/annihilation operators can be used in deriving these equations, but should not be included in the final equations. The electric dipole approximation can also be used to simplify the
  • #1
Ruslan_Sharipov
104
1
Typically (in popular literature) the process of photon emission by an excited atom is considered as an instant event. But actually it is quite likely that it is a continuous process. Such processes are usually described by evolutionary differential equations (ODEs or PDEs). Assume that we consider a hydrohen atom composed by 1 proton p and 1 electron e. Let [itex]\Psi(t,x_p,y_p,z_p,x_e,y_e,z_e )[/itex] be the time dependent wave function of the atom. The time dependent Shrödinger equation with the Coulomb potential can be written for this function. How many auxiliary functions should be added for describing the photon which is absent in the beginning of the process and which is present in the end of it? How many auxilioary terms should be added to the time-dependent Shrödinger equation? It would be best to see the complete system of differential equations.

The creation/annihilation operators could be used in deriving these differential equtions, but they should not enter the ultimate equations. These should be differential equations for functions, not for operators.
 
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  • #2
Ruslan_Sharipov said:
Typically (in popular literature) the process of photon emission by an excited atom is considered as an instant event. But actually it is quite likely that it is a continuous process.
:confused: No, on the contrary, the "popular" view is closer to the continuous one. We often get posts from people with little understanding of the difference between classical and quantum mechanics, asking overly mechanistic questions like, "How many cycles are there in the photon? How long does it take for it to be emitted?"

The wavefunction evolves continuously but it describes only a probability. The measurement process causes the emission to appear instantaneous. For example for a long-lived alpha emitter, the alpha particle does not gradually seep out over billions of years, it comes out all at once, but at an unpredictable time.

To calculate the evolution of the wavefunction, just add to H the electromagnetic interaction term (e/c)p·A, and make the electric dipole approximation.
 
  • #3
Dear Bill_K. Your answer is just a hint. Please, be more specific. Please, answer the following questions:

1. How many wave functions should be written?

2. How many arguments are in each wave function, e. g. time + proton coordinates + electron coordinates? Or maybe time + proton coordinates + electron coordinates + photon coordinates?

3. What is the nature of [itex]A[/itex] in your formula? If it is an operator written in terms of creation/annihilation operators, then explain how to apply it to the wave functions.

4. How many differential equations should be written?
 
  • #4
Ruslan, I'm beginning to suspect that you're coming into this cold? If so, in quantum mechanics as with any subject, it's important to start with the basics first, then tackle the more difficult problems after you have a firm grounding.

The reason I say this is your mention of the proton. If you've learned about the states of the hydrogen atom, you've seen how the discussion starts out. The first thing you do is change to center-of-mass coordinates. Thus the proton is essentially out of the picture - the description of atomic energy levels involves only the electron wavefunction.

And we always try to simplify the problem through the judicious use of approximations. For example even though we know the electron is properly described by the Dirac equation, we don't use that fact except when it's actually necessary.

In the present case, atomic transitions are usually discussed in a semi-classical framework. That is, the electron is quantum but the electromagnetic field is not. Also we make the dipole approximation, which means that the wavelength of the electromagnetic wave is large compared to the size of the atom.

With all this said, there is only one wavefunction involved - that of the electron. We write down the interaction Hamiltonian Hint between a charged particle and a plane wave. We take the matrix element of Hint between the initial and final electron states, square it, and plug it into the Fermi's Golden Rule for time-dependent perturbation theory, to get the transition rate.
 
  • #5
Bill_K said:
Ruslan, I'm beginning to suspect that you're coming into this cold? If so, in quantum mechanics as with any subject, it's important to start with the basics first, then tackle the more difficult problems after you have a firm grounding.

Dear Bill_K, I do not like to discuss the simplified statement of the problem known from every textbook. I would like to discuss the conceptual statement from which various simplifications should follow as special cases.

Bill_K said:
The first thing you do is change to center-of-mass coordinates. Thus the proton is essentially out of the picture - the description of atomic energy levels involves only the electron wavefunction.

This is true if we study stationary states of the pair. But studying the photon emission in a strict model we should consider the center-of-mass coordinates for the trio or if we deal with the center-of-mass coordinates of the pair, we should remember that its state of motion changes as soon as the photon is emitted.

Bill_K said:
And we always try to simplify the problem through the judicious use of approximations. In the present case, atomic transitions are usually discussed in a semi-classical framework.

I would like to see the exact model in the form of differential equations, from which the approximate model is derived as some limit.

Bill_K said:
Also we make the dipole approximation.

I agree with this approximation and agree to omit spin terms from the Hamiltonian. But otherwise the model should be exact in order to clarify the conceptual statement of the problem.

Bill_K said:
We take the matrix element of Hint between the initial and final electron states, square it, and plug it into the Fermi's Golden Rule for time-dependent perturbation theory, to get the transition rate.

The power of a theory not only in describing particular phenomena using simplifications specific to them, but in unifying various phenomena into a single concept. Otherwise the theory is weak or incomplete.

Perturbation theory has its limits, so obtaining non-perturbative model in the case of such a simple object as the hydrogen atom is important for understanding the concepts and limits of the Quantum Mechanics as a whole.
 

1. What is the process of photon emission by an excited hydrogen atom?

Photon emission by an excited hydrogen atom occurs when an electron in the atom transitions from a higher energy level to a lower energy level. This transition results in the release of a photon, which is a particle of electromagnetic radiation.

2. How does the energy level of an electron affect the frequency of the emitted photon?

The energy level of an electron in a hydrogen atom is directly proportional to the frequency of the emitted photon. This means that the higher the energy level of the electron, the higher the frequency of the emitted photon.

3. What determines the color of the emitted photon?

The color of the emitted photon is determined by its frequency, which in turn is determined by the energy level of the electron. Higher energy levels correspond to higher frequencies, which typically result in shorter wavelengths and higher energies. This is why higher energy photons are often associated with blue or violet light, while lower energy photons are associated with red or orange light.

4. Can a hydrogen atom emit multiple photons at once?

Yes, a hydrogen atom can emit multiple photons at once if multiple electrons undergo transitions from higher to lower energy levels simultaneously. However, this is less likely to occur and typically only happens in highly energetic environments, such as in stars or during collisions.

5. How is the emission of photons by an excited hydrogen atom used in everyday life?

The emission of photons by excited hydrogen atoms is used in everyday life in a variety of ways, such as in fluorescent and LED lighting, lasers, and medical imaging techniques like MRI. It is also essential in understanding and studying the behavior of atoms and molecules in chemistry and physics.

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