B Are photons entangled with a component of the atom that emitted them?

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Photons emitted by an atom can be entangled with components of the atom, but manipulating the atom after photon emission does not affect the photon’s wavelength. Entanglement is broken when either the photon is measured or the atom is interacted with first, leading to a definitive measurement of the photon’s properties. The relationship between the atom and the emitted photon is not causal in the way initially assumed. Understanding quantum theory is crucial to grasping these concepts accurately. Ultimately, the emitted photon's characteristics remain unchanged once measured, regardless of subsequent actions on the atom.
Matthew-Champion
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If an atom were made to release a Photon, then a number of the components of the atoms nucleus were theoretically extremely quickly removed. would the previously emitted photon change wave length?
If an atom were made to release a Photon, then a number of the components of the atoms nucleus were theoretically extremely quickly removed. would the previously emitted photon change wave length?
 
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@Matthew-Champion The answer to the question in the title is yes. Generally the particles coming out of an interaction are entangled in some way at least until one of them interacts with something else (and note that a measurement is always a case of “interacts with something else”).

However, that “yes” answer does not imply what you’re asking in the body because entanglement doesn’t work the way you’re thinking. Changing one part of an entangled system (in this case, manipulating the atom after the emission) has no causal effect on the other parts of the system (in this case, the emitted photon).

There are two possibilities here. One is that we measure the wavelength (energy, frequency, they’re all related) of the photon first. This interaction breaks the entanglement so nothing we do to the atom later changes the photon energy from the value we’ve measured. The other possibility is that we interact with the atom first, breaking the entanglement, and then we measure the wavelength of the photon. Either way, we get one measurement of the photon and its wavelength is what we measure.
 
:welcome:

Your question may be the result of misunderstanding quantum theory. Although, I see @Nugatory has made some sense of it.
 
Nugatory said:
@Matthew-Champion The answer to the question in the title is yes. Generally the particles coming out of an interaction are entangled in some way at least until one of them interacts with something else (and note that a measurement is always a case of “interacts with something else”).

However, that “yes” answer does not imply what you’re asking in the body because entanglement doesn’t work the way you’re thinking. Changing one part of an entangled system (in this case, manipulating the atom after the emission) has no causal effect on the other parts of the system (in this case, the emitted photon).

There are two possibilities here. One is that we measure the wavelength (energy, frequency, they’re all related) of the photon first. This interaction breaks the entanglement so nothing we do to the atom later changes the photon energy from the value we’ve measured. The other possibility is that we interact with the atom first, breaking the entanglement, and then we measure the wavelength of the photon. Either way, we get one measurement of the photon and its wavelength is what we measure.
Thank you for your reply. yes I see where I went wrong. I got all exited before thinking properly. oh well thank you for the concise answer.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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