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No matter = no photon?

  1. Jan 10, 2006 #1
    this is something that bothers me:
    a professor explained today how he can preserve spin-down in a magnetic field in the +z direction even though its unstable (the energy is for spin-down in a +z field is [tex]u=\mu B[/tex] and the energy for spin-up is [tex]u=-\mu B[/tex].

    he says that if we isoltate an atom in spin-down (excited state) from the enviorement it wont be able to radiate (flip to the ground state spin-up) because there wont be anything that would absorb the photon...

    its wierd... so if i got it right, the atom only goes to ground-state if a photon was detected, so the probability it would emmit a photon in proportional to the density of absorbers around it...

    can anyone explain this?
    and id be happy to see an equation with that relation (probability vs. density)
     
  2. jcsd
  3. Jan 10, 2006 #2

    ZapperZ

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    There is a distinct possibility that you have misinterpreted this. This is especially true in the meaning of "isolating an atom from the environment".

    The probability of transition of an atom from a high to low energy state depends on 2 things: (i) the probability of going from the initial state to the final state, typically described via the Fermi Golden rule, and (ii) the probability that the emitted energy can be "taken up", either as radiation to the vacuum or via absorption by other entities.

    The confusion here comes from what is meant by "isolating an atom from the environment". What your prof. probably meant that it is decoupled even from the vacuum surrounding it, meaning it can't even radiate such energy. In such a case, just because there is a finite probability of going from the initial to the final state, since the emitted energy has nowhere to go, such a transition is prohibited.

    This is an illustration of several phenomena. In tunneling, the probability of tunneling is governed not only on the tunneling matrix element, which dictates the probability that a particle can transverse through a potential barrier, but also the probability that there is an available state at the same energy (for elastic tunneling) for the particle to end up in. In pulsed nuclear magnetic resonance (which I think is closer to the example you were given), after a 90 degree spin flip, the rate at which the magnetization grows not only depends on the probability of going from the high spin state to the low spin state, but also the ability of the surrounding lattice to take up the energy given off. So this give a good measure of the nature of the surrounding material, because there is a difference in the rate of energy absorption for different types of material.

    Zz.
     
  4. Jan 10, 2006 #3
    i see, and how can one isolate an atom from the vacuum (zero point energy i guess)?

    he said they put an atom in vacuum on an atom chip that generates the desired magnetic field which traps the atom when its in the spin-down state.
    is there any special machanism he didnt tell me about that decouples it from the vacuum itself?
     
  5. Jan 10, 2006 #4

    ZapperZ

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    In that case, the generate energy is in the form of a magnetic field, so the "receiver" in this case is not just the vacuum field but rather the magnetization. So you are transfering Delta(E) in the form of a magnetic field energy that is absorbed by another spin or angular momentum state. This is definitely what is going on an NMR.

    Zz.
     
  6. Jan 10, 2006 #5
    ok, but how can you prohibit the transition of energy and keep your atom excited forever? (thats what he does, he has an atom trapped there, and it wont escape because it cant flip its spin)

    i.e. how far does other atoms have to be so that no angular momentum could be transfered to them?
    is there a way to isolate the atom so that even if an atom is close to it there will be no way for the momentum to be transfered?
    what about the vacuum? it still can absorb angular momentum can't it (circular polarized photon)? - how do you prohibit the vacuum from absorbing it?
     
    Last edited: Jan 10, 2006
  7. Jan 10, 2006 #6

    ZapperZ

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    It's conservation of energy! A "vacuum" cannot absorb energy because that would be energy lost into nowhere.

    I am guessing that the atom is maintained inside a static magnetic field, but not knowing what the setup looks like or what is involved, that is just a pure guess. Maybe this is a good time to actually ASK your prof. the questions you just asked here. After all, he/she was PAID to do just that.

    Zz.
     
  8. Jan 10, 2006 #7
    ok, you confused me, you said
    "The probability of transition of an atom from a high to low energy state depends on 2 things: (i) the probability of going from the initial state to the final state, typically described via the Fermi Golden rule, and (ii) the probability that the emitted energy can be "taken up", either as radiation to the vacuum or via absorption by other entities.

    The confusion here comes from what is meant by "isolating an atom from the environment". What your prof. probably meant that it is decoupled even from the vacuum surrounding it, meaning it can't even radiate such energy. In such a case, just because there is a finite probability of going from the initial to the final state, since the emitted energy has nowhere to go, such a transition is prohibited."


    anyway, how can the atom "know" if there is something to absorb its photon?
    cant it just release a photon to the vacuum like the sun does? (or does the sun emmit energy only when there is a body to recieve it too?)
    if so, when our universe will get big enough all the spontaneous processes would be slower....




    and as for asking him... well, actually he isnt paid to teach me... he just came and told us about his research... and theyre all too busy there to answer my questions... (atleast thats the impression i got when i came there and asked if i could do my project there)
     
  9. Jan 10, 2006 #8

    ZapperZ

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    In an atomic dipole transition, a photon is emitted into vacuum, i.e. this is [tex]h\nu[/tex]. In the magnetic transition case, this is not a "photon" as in "light". It is a "photon" in the sense of QED field. So when you flip the spin from antiparallel to parallel (for a + magnetic moment), a "receiver" must also be able to take up the "magnetic energy" by changing its magnetic state. A vacuum can't do that.

    I strongly suggest you ask this person. It is irresponsible to simply throw out something like this and expect people to not have any ability to ask.

    Zz.
     
  10. Jan 10, 2006 #9
    i thought the only difference between dipole transition and magnetic moment transition is only the polarization of the photon (i.e. linear vs. circular), i mean we do measure photons with energy equal to [tex]\hbar \omega_L[/tex] where [tex]\omega_L=\frac{\Delta E}{\hbar}[/tex] thats the larmour freq...

    so theres a difference between the things i measure from NMR and the things i measure from excited phosphor? (other then polarization?)

    and is there a difference between the spectral lines that rise due to radius shift (the lines predicted by the primitive hydrogen model) and the ones that rise due to angular moment change (the lines predicted by the fine structure model)?

    seems odd...
     
    Last edited: Jan 10, 2006
  11. Jan 11, 2006 #10
    ok, its just an idea... but i think that around the atom there is an electric potential so that only certain energies of photon can be emmited, so if the energy difference between the up-spin and down-spin isnt one of these mods, the atom wont flip.

    so if im right, the "thing" that has to be absorbed from this transition is no different then the one that has to be absorbed from a dipole transition, and i want to call this "thing" a photon.... this means i can have an excited atom in a potential box, and he wont be able to go back to ground state.

    hmmm... i wonder if its possible to capture an atom and inflate it - i mean excite it to energy with big "radius" and it wont be able to relax... id have a gigantic atom =) not that i can do anything with it, because any photon would then blow away the electron from the atom...
     
    Last edited: Jan 11, 2006
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