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Quantized energy vs. T=Ia

  1. Jun 24, 2010 #1
    I understand quantized energy, and how it pertains to electron energy levels, but I just read that rotational energy of molecules is also quantized. Here's the original quote:

    "A macroscopic object like a child's top can store an amount of energy that, in effect, varies continuously with its rotation rate. For tiny objects like molecules in the atmosphere, the energy of rotation is quantized and can take on only discrete values. Rotational energy transitions involve energy changes that correspond to the energy of photons with wavelengths shorter than about 1 cm."

    Doesn't this mean that a given molecule can be spinning with one of several distinct angular velocities, but not with any intermediate angular velocities? Does it not accelerate from one angular velocity to the next?
     
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  3. Jun 24, 2010 #2
    It does not accelerate from one angular velocity to the next, it jumps. It is not as unintuitive as it sounds.

    For example the group velocity of a collection of wave functions may jump from one velocity to the next based on the actions of each individual normalized wave function.

    Quantized angular momentum arises from boundary conditions just as electron energy levels are quantized when the electron is bound in 1 dimension. If the path of the electron wraps back upon itself, and it does when it has angular momentum, then the same argument holds of wave function continuity at all points in space which leads to quantized energy eigenvalues.
     
  4. Jun 24, 2010 #3

    K^2

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    "Jumps" is a bit misleading, but it you're right, it doesn't accelerate either.

    Say a particle with some angular momentum absorbs a photon. Photon carries [itex]\pm \hbar[/itex] of angular momentum. So when a photon is absorbed, the particle's angular momentum will change by that quantity. Rotational energy will change appropriately as well. But absorption of a photon doesn't happen instantly. That is, it sort of does, but the moment of absorption is undetermined. So the transition is a superposition of all possible "instant jumps" between the time the absorption began and the time it ended. That means, that at any specific time, the particle is in superposition of initial state and final state. If you were to measure the angular momentum, you'd still find it in one or the other state, but the probability of finding it in the new state would get higher as time goes on.

    This as close to "gradual" transition as it gets in quantum mechanics. At no time, does the particle actually have an "in-between" angular momentum or rotational energy.
     
  5. Jun 24, 2010 #4
    Ok, yea I get that. Not in the "oh that makes sense. I could have derived that" way, but in the "I see what you're describing" way. I'm familiar with the idea of super-imposed states, but it makes me uncomfortable to imagine entire molecules doing the same spooky stuff. Water, I know you! I built you out of toothpicks and marshmallows, and now you're telling me you don't have to speed up in order to go faster?

    And dare I ask about kinetic energy, too? It can't be quantized, can it?
    I can't imagine that a molecule can only have a few distinct kinetic energies.
     
    Last edited: Jun 24, 2010
  6. Jun 24, 2010 #5

    K^2

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    That sort of depends on the shape of the universe. And on context, to an extent.

    You see, particles in a periodic potential can only have quantized kinetic energies. Electrons in a metal are a classic example. If the universe is closed on itself, then it is effectively periodic, and then you can have quantized energies.

    That would actually be a very exciting thing, because it would also mean that momentum is actually quasi-momentum, and we could build reactionless drives by happily ignoring conservation of momentum.

    Unfortunately, there is no evidence of any such strange behavior so far.

    There are also scenarios a bit closer to home. Momentum, and therefore kinetic energy, of a satellite in Earth Orbit is technically quantized. Of course, the quantum of that energy is very, very small, so macroscopically, it works the same as speeding up gradually. But on quantum level, it's still speeding up or slowing down a "jump" at a time.
     
  7. Jun 25, 2010 #6
    I am not sure the idea of a particle "being" in a super position of states should be taken so literally.

    It is always measured in an eigenstate, the super position is a collection of statistical probabilities before measurement, so it can't be said that those states exist for the particle simply because of lack of information available to the observer.
     
  8. Jun 25, 2010 #7

    K^2

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    It should be taken VERY literally. An excited atom that is now decaying into its ground state is in superposition of the two states. That means the photon is in superposition of emitted and non-emitted. But photon has an actual electric field corresponding to it. That's an observable. What's the electric field of a photon in this superposition? It's 1/√(2) of the amplitude of a single photon. In other words you actually have electric field for "half" of a photon. That's a very literal superposition of two states. Any particle that would absorb that "half" photon would also end up in a superposition of states. It only has half of the energy needed for a transition, so it's not just randomly in excited or ground state. It is very literally both-at-once.

    If that wasn't the case, things like quantum encryption and quantum teleportation would be a pipe dream, rather than experimentally verified phenomena they actually are.
     
  9. Jun 25, 2010 #8
    Statistically you have an electric field for half of a photon, this does not mean that an electron that absorbs this photon retains a new state between states. It only effects the probability at that moment of time. An electron could absorb two half photons and never gain an energy level, it could absorb 10 half photons and not gain any energy.

    Superpositions are a result of linear algebra treating functions as vectors. Linear combinations of vectors are possible therefor a function of position for an object can be result in multiple positions and so on for any other functions of observables. However when ever an interaction/measurement is taken the mathematical absurdities are removed as the state must be in an eigenstate of linear independence.
     
  10. Jun 25, 2010 #9
    I like to think that I have a decent capacity for new information, but I'm finding this to be a lot to take in. As luck would have it, I just finished a book and I'm looking for a new one. I liked "Entanglement" by Amir Aczel, but I'd like something that goes a little more into detail regarding the math involved in describing quantum phenomena. Any suggestions?
     
  11. Jun 25, 2010 #10
    I enjoyed "Quantum Theory" by David Bohm. It may be outdated but it is perfectly written.
     
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