Action corresponding to photon emission

In summary, the conversation revolves around the calculation of the probability of photon emission in Feynman's path integral formulation. The speaker is seeking clarification on the corresponding action for this process, but it seems that there may not be a specific action as it is derived from the perturbation expansion of the classical action.
  • #1
semc
368
5
Hi I am trying to write the probability of photon emission due to transition of electron in feynman's path integral formulation. I am stuck trying to figure out the action corresponding to the photon emission. Would anyone shed some light on this? Thanks
 
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  • #2
semc said:
Hi I am trying to write the probability of photon emission due to transition of electron in feynman's path integral formulation. I am stuck trying to figure out the action corresponding to the photon emission. Would anyone shed some light on this? Thanks

If I'm not mistaken (and I could very well be, as I'm a dilettante dabbler), the corresponding 1st-order diagram would be an incoming electron with momentum p, and an outgoing electron with a momentum (p-k) and photon with momentum k, so the perturbation expansion term would have a destruction operator for the electron with momentum p, and the electron/photon creation operators with momenta (p-k)/k.

However, if you're asking about the actual action S corresponding to that situation, as in the integral over a term in the Lagrangian, I don't think there is one; the "electron emitting a photon" and all of the other possibilities come from the perturbation expansion of the (exponential of?) the normal, (semi-)classical action, where the only term giving the interaction between the electron field and the Maxwell field is just the minimal-coupling term (##j_{\mu} A^{\mu}##?); everything else derives from that.

But, again, caveat emptor.
 

1. What is the action corresponding to photon emission?

The action corresponding to photon emission is the process by which an excited atom or molecule releases energy in the form of a photon, or particle of light. This occurs when an electron transitions from a higher energy state to a lower energy state, releasing energy in the form of a photon with a specific wavelength.

2. How does the action of photon emission relate to light?

The action of photon emission is closely related to light, as photons are the fundamental particles of light. When an electron in an excited state releases energy in the form of a photon, it creates an electromagnetic wave which we perceive as light. The wavelength of the emitted photon determines the color of the light.

3. What causes an atom or molecule to emit a photon?

An atom or molecule will emit a photon when one of its electrons transitions from a higher energy state to a lower energy state. This can happen spontaneously, or it can be triggered by an external source of energy, such as heat or light. The energy level difference between the two states dictates the energy, and therefore the wavelength, of the emitted photon.

4. How is the action of photon emission used in everyday life?

The action of photon emission plays a crucial role in many everyday technologies, such as light bulbs, lasers, and solar panels. It is also used in medical imaging techniques, such as X-rays and MRI machines. Additionally, photon emission is key in the process of photosynthesis, which is essential for sustaining life on Earth.

5. Can the action of photon emission be controlled?

Yes, the action of photon emission can be controlled by manipulating the energy levels of atoms or molecules. This can be done through various methods, such as applying an electric field, using lasers, or changing the temperature. By controlling the energy levels, scientists can control the wavelength and intensity of the emitted photons, which has many practical applications in technology and research.

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