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#1
Jan2213, 09:31 PM

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Is it possible to excite a photon? Or bring it to a higher electronvolt?



#2
Jan2213, 10:26 PM

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In the case of massive particles, what would it mean to "excite" one?
I mean in detail  not just to give it more energy. Then try to see how that could relate to photons. 


#3
Jan2313, 02:45 AM

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The words "excite", or "bring to higher level" suggests the existence of levels in the first place, that is, you need a quantized degree of freedom. One way to create this situation for photons is to place them inside a cavity where only certain fequencies/energy levels are allowed. In this case, photons can occupy higher or lower levels yes, and you can talk about excitation.



#4
Jan2313, 05:57 AM

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Exciting a photon
@Zargon: interesting ... so what happens to the photons at the cavity walls? Isn't there a chance of being absorbed by the wall? Or were you thinking of some other way to restrict the allowed frequencies?
Lets say you have a photon in some welldefined quantum state in such a cavity. How would you excite it to the next state? Wouldn't you have to annihilate it and introduce a new photon? 


#5
Jan2313, 09:58 AM

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if you add a magnetic field to the well would the photon get excited and move to a higher energy state ?



#6
Jan2313, 10:07 AM

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#7
Jan2313, 10:12 AM

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Now you may say, what about when a light ray hits a surface and refracts or changes direction, doesn't its energy change? Or what happens when it hits a prism? The answer to that is that the abovementioned effects are due to the incident photons of white light hitting an object which absorbs, destroys, and recreates or reradiates a new photon at a different (or perhaps the same) energy.



#8
Jan2313, 12:15 PM

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Well, you could introduce a blueshift by running towards the source of the light...that's about all I can think up.



#9
Jan2313, 12:30 PM

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#10
Jan2313, 12:42 PM

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I mean, it seems as though gravity can alter the direction and energy of a photon, so maybe I was wrong with my earlier statements.



#11
Jan2313, 01:37 PM

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gravity alters spacetime the photon is still moving in a straight line with the same energy level just that straight line is spacetime curved.



#12
Jan2313, 01:52 PM

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Yeah but...if you shine a flashlight down a well the energy of the light increases the closer you get to the center of the Earth. 


#14
Jan2313, 02:34 PM

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"..if you shine a flashlight down a well the energy of the light increases the closer you get to the center of the Earth....
yes! some details below.... classical analogy: change the fixed points of a vibrating string and it has a different resonant frequency.... another way: drop a photon into a gravity well: http://www.physicsforums.com/showthr...=612091&page=3 1: A hydrogen atom is lowered into a deep gravity well. Then a photon of visible light is dropped onto the atom, which becomes ionized, although visible light does not normally ionize hydrogen. That happened because the field that keeps the atom together weakened as the atom was lowered. PeterDonis: No, it happened because the photon was blueshifted as it dropped into the gravity well. A visible light photon emitted locally, at the same altitude as the atom, won't ionize it, so the field of the atom is not "weakened" at all according to local measurements. The difference that lowering the atom gently means that it is at rest deeper inside the well, so it "sees" the blueshift of the photon. To see why that's important, consider an alternate experiment where you let both a hydrogen atom and a visible light photon freefall into the gravity well, in such a way that they meet up somewhere much deeper into the well than where you released them (you time the release of the atom and the photon from your much higher altitude to ensure this). Will the photon ionize the atom? No, because the atom is not at rest in the field; it is falling inward at a high speed, so there is a large Doppler redshift when it absorbs the photon that cancels the gravitational blueshift. 


#15
Jan2313, 04:31 PM

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I guess I could be wrong  see post #4 for questions arising from the concept, and posts #6&7 for clarification.[1] iirc the Feynman Diagrams sum to zero. Further reading. http://van.physics.illinois.edu/qa/listing.php?id=2348 http://www1.quantum.leeds.ac.uk/~almut/section3.pdf http://www.phys.ksu.edu/personal/wys.../quantumEM.pdf  [1] If I have a charged particle confined to a potential well, then change the width of the well by some means, then changing the width does not automatically change the energy eigenstate of the particle does it? Wouldn't the situation be more like making the energy of the particle uncertain  (represented as a superposition of eigenstates of the new potential) requiring a measurement of some kind to establish it? 


#16
Jan2313, 05:44 PM

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#17
Jan2313, 06:42 PM

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I think you guys are making this too hard  the best way to excite a photon is to show it a really sexy electron !



#18
Jan2313, 09:07 PM

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Wikipedia is not that great a reference  and that quote is does not actually contradict what I've been trying to say. You also have not shown how this model takes into account the other comments and questions I have referenced. Have you done the math? [1]
The particleinabox model can be solved analytically fersure  but it is not a good model of actual physical systems. It is especially problematic for light, since you have to figure out what the box is made out of that would confine a single photon without it annihilating at the walls. When it comes to energy eigenstate transitions, you still have to figure how that would come about. i.e. what would be the physical process that changes the width of such a strange box? Not everything describable in math is physically possible. You can confine a photon gas in a box though. This uses a model where photons are constantly being annihilated and created. In this case, you can raise the mean and total energy of the system by changing the width of the container. But what is it that happens to individual photons? You could be imagining a single photon bouncing between ideal, perfectly reflecting, walls [2]. In which case, the photon is being annihilated at each wall, and then a new one is created. (Though there is some philosophical hairsplitting over this point.) It is possible to arrange for the photon thus created to be a higher frequency than the one annihilated. I would assert that this process does not well fit the concept of "exciting a photon": it kinda means that it is the same photon that has more energy like an excited electroninabox is the same electron. For a single particle in a box, when you make the box smaller, the energy eigenstates raise in value, and so does the expectation value of a measurement of energy of the system. The particle, however, is not in a single energy eigenstate until a measurement of energy has been made. You can figure out the odds by expanding the initial eigenstate wavefunction in terms of eigenstates of the final potential. So the process would involve two steps  making the box smaller, and then measuring the energy level. For a perfectly reflecting box of one photon (as discussed) how would you (or the system) conduct that measurement without annihilating the photon?  [1] see this example for what happens when you change the width of a confining potential. The author has the potential increasing in width, and finesses the system so there is an eigenstate in the final system with the same energy as the ground state of the initial system. As an exercise, do the problem the other way around  making the box smaller. [2] You realize that reflection, at the photon level, is described using creation and annihilation operators right? The law of reflection is only obeyed on average and all that? (In fact D Simanek has a pmm puzzle using the idea of a photon bouncing between perfect reflectors.) [edit] @phinds +1 that! I have been resisting the pun from the start :) 


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