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Does observing a particle add energy to it?


Baby Astronaut

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I do not know if it adds energy, but I know that observing a particles will change some pairs of the physical properties of those particles. This is the Uncertainty Principle, this is how it is described on wikipedia:

 

"In quantum mechanics, the Heisenberg uncertainty principle states that certain pairs of physical properties, like position and momentum, cannot both be known to arbitrary precision. That is, the more precisely one property is known, the less precisely the other can be known. This statement has been interpreted in two different ways. According to Heisenberg its meaning is that it is impossible to determine simultaneously both the position and velocity of an electron or any other particle with any great degree of accuracy or certainty. According to others (for instance Ballentine[1]) this is not a statement about the limitations of a researcher's ability to measure particular quantities of a system, but it is a statement about the nature of the system itself as described by the equations of quantum mechanics."

 

As far as adding energy, I would guess observing something calls for light, and most of the time when light hits something it heats it up, even if it is just minute.

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Say you observe an electron or photon to detect which slit it entered through. Does the observation add more energy to the particle than it began with?

 

Photon detection is usually destructive — no more photon. Detection of electrons can remove energy from them. It can also add, in the case of a multiplier system, like a microchannel plate.

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He means number of bosons, of which photons are one kind. In other words, it doesn't have to go anywhere, it simply no longer exists. Usually you say it is "absorbed," which basically just means that its energy has been transferred to whatever it "hit." And that's how you would detect it.

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He means number of bosons, of which photons are one kind. In other words, it doesn't have to go anywhere, it simply no longer exists.

I thought if photons were to disappear into a particle, they'd be spat out eventually. Or is that not the same photon as the absorbed one?

 

Usually you say it is "absorbed," which basically just means that its energy has been transferred to whatever it "hit." And that's how you would detect it.

Confused a bit here. For such a detection, wouldn't the photon have to bounce off whatever it "hit", and then return to the detection unit?

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What the heck is a boson number?

 

I did a search on Wikipedia and it returned: Did you mean: bacon number

 

Damn, that's awesome, but I can't make it do that. I wanted to get a screenshot.


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I thought if photons were to disappear into a particle, they'd be spat out eventually. Or is that not the same photon as the absorbed one?

 

 

Confused a bit here. For such a detection, wouldn't the photon have to bounce off whatever it "hit", and then return to the detection unit?

 

 

Sometimes a photon is spat back out. But there are a number of non-radiative ways an excited system can relax. In the case of, say, a reverse-biased semiconductor (PiN photodiode), the photon will excite an electron which will create a current which is measured. The electron loses the energy in the detection circuit, probably causing resistive heating.

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Damn, that's awesome, but I can't make it do that. I wanted to get a screenshot.

Done. First try again, I got the bacon.

 

20p26iu.jpg


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whoops, realized that I left out a quotation mark in the url of post #6 so the link doesn't work. Mind fixing it? Thanks.

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