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Why are there no well-defined energy states in nature?

  1. Dec 16, 2011 #1
    Hi all,

    My Physics tutor's reasoning for no well-defined energies existing in nature for a system in time was that if a system had a well defined energy in time, such that |psi> = |E0> then, evidently, the energy cannot vary.

    And, his logic goes on to say that if an energy cannot vary, you cannot measure the energy of the system accurately as that requires changing its energy - measuring a difference.

    Has anyone a better answer? His isn't very satisfying. Surely I could measure the energy through knowing the mass and Einstein's relation, if it were at rest in my frame?

    Much appreciated!
     
  2. jcsd
  3. Dec 16, 2011 #2

    f95toli

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    The most "fundamental" reason I can think of is that is that if we associate a frequency with the energy (as would be the case in e.g. an atom, where the energy levels decay be emitting aphoton with energy h*f)), it follows that there is no such thing as an "infinity sharp" level (which I guess is what you mean by well-defined).

    The reason for this is a mathematical result from Fourier analysis that related the uncertainty in frequency with the "uncertainty in time" (or more generally the distribution of a signal and its transform), now it follows that in order for the frequency to be "well-defined" the lifetime of the state would have to be infinite; i.e. it can't interact with anything.

    Note that this wouldn't even be true without a measurements, all states interact with the vacuum, meaning they all have a finite lifetime.

    So your tutor wasn't wrong, but I guess what I wrote above is a bit more stringent.
     
  4. Dec 18, 2011 #3
    Great answer. I've just covered Fourier so I'll look into the frequency-time relationship. Cheers, much appreciated!
     
  5. Dec 18, 2011 #4

    Fredrik

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    If |E> is an energy eigenstate with energy E, then the Schrödinger equation doesn't allow it to change with time. (The state vector will change, but the state will be the same). After a time t, the state vector has changed to [itex]e^{-iHt}|E\rangle=e^{-iEt}|E\rangle[/itex], which is in the same 1-dimensional subspace as [itex]|E\rangle[/itex], and therefore represents the same state.

    However, the Schrödinger equation only applies to systems that are isolated from their environments. For this system to be measured, it must interact with something. So the Schrödinger equation should then be applied to a larger system, that has the system we considered first as a component part. The Hamiltonian of the larger system will be* of the form [itex]H\otimes 1+1\otimes H'+H''[/itex], where H is the original Hamiltonian, H' is the Hamiltonian of the other component parts when they're not interacting with our system, and H'' represents the interactions. A state of the form [itex]|E\rangle\otimes|\psi\rangle[/itex], where [itex]|E\rangle[/itex] is an eigenstate of H, is almost certainly not going to be an eigenstate of the Hamiltonian for the combined system, and that means that the state [itex]|E\rangle\otimes|\psi\rangle[/itex] can change with time. There is no reason to think that the [itex]|E\rangle[/itex] part of it will stay the same when the H'' term is non-zero.

    *) I don't know if it can always be expressed in this form. If it can't, then it's still not plausible that the [itex]|E\rangle[/itex] part of the state vector for the larger system will remain the same for any significant amount of time. Only Hamiltonians of the form [itex]H\otimes 1+1\otimes H'[/itex] will make it stay the same forever.

    So I'd say that instead of arguing that measurements indicate that there are no energy eigenstates, we should be arguing that they indicate that there are no systems that remain isolated from their environments (including measuring devices) no matter what you do to them. (If there were, they wouldn't be of any interest to us anyway. They could be considered part of another universe).
     
    Last edited: Dec 18, 2011
  6. Dec 18, 2011 #5
    Stationary states, i.e. states that do not change with time do have definite energies. Ground states of atoms are such states. Note that excited states of atoms interact with the vacuum and are thus not stationary, even though they are energy eigenstates of the Hamiltonian.

    So if you have a state which left alone does not change in time (which even does not change by vacuum fluctuations which are always around), than you have a state with definite energy. How precise you want to measure, depends on how long you wait until you measure. When you have two hydrogen atoms in a ground state, then you can wait 2 minutes, 2 days or 2 years measuring the ground state energies of the two atoms. The deviation in the two energy measurements will become increasingly negligible the longer you wait.
     
  7. Dec 20, 2011 #6
    Can you give a reference for such a derivation? The only "derivation" of [itex] \mathcal T \Delta E \geq \frac{\hbar}{2}[/itex] that I know is by using the rather vague definition of [itex] \mathcal T = \Delta a / \frac{ \mathrm d \langle a \rangle }{ \mathrm d t}[/itex] as a characteristic time of the evolution of the observable A.
     
  8. Dec 21, 2011 #7

    f95toli

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    Well, if you google "Mathematical uncertainty principle" you should be able to find more info (e.g. http://www.ams.org/samplings/feature-column/fcarc-uncertainty)

    However, this should be covered in any book on Fourier analysis (or signal processing, since it is a fundamental property of signals and therefore important in DSP), e.g. Folland's "Fourier analysis and its applications"
     
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