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I Understanding ΔEΔt uncertainty

  1. Jul 1, 2016 #1
    Would it be meaningful to think of it as the 0 component of the 4-vectors x and p? In which case the regular uncertainties between position and momenta are the 1,2,3 components of the 4 vectors and the ΔEΔt is just the 0 component?
     
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  3. Jul 1, 2016 #2

    ShayanJ

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    No, that can't be correct. The rigorous derivation of the uncertainty relation for position and momentum considers them to be operators. But there is no time operator in QM, at least not in the standard formulation. And when you go to QFT, even position is no longer an operator, let alone time.

    Some say that the uncertainty relation between energy and time is only a myth. And to some extent its true because it can't be thought of in the same way you think about the uncertainty relation for position and momentum. But there are some situations where it is true in a different meaning. You can check "Modern Quantum Mechanics" by Sakurai for such different meanings of this relation but I don't remember which section.

    And at the end, there is the famous saying of Griffiths:
    "In general, when you hear a physicist invoke the uncertainty principle, keep a hand on your wallet."
     
  4. Jul 1, 2016 #3
    I can't speak to the QFT part of your response, because I don't know QFT.

    I know that the derivation of x and p uncertainties doesn't exactly transfer over identically to the t and E uncertainty. But the pattern is so simple and clean, it seems like there should be some sense in which meaningful.

    I see no reason not to have a time operator which would just be the projection into the t (or 0) coordinate in Minkowski space, it's even hermitian. The issue is in considering E its fourier transform.
     
  5. Jul 1, 2016 #4

    ShayanJ

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    In non-Relativistic QM, Minkowski spacetime is not used, but a three dimensional space with a Newtonian absolute time.
     
  6. Jul 1, 2016 #5
    You're correct of course, but am I to conclude then that this is just a coincidence?
     
  7. Jul 1, 2016 #6

    ShayanJ

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    The two uncertainty relations are not on the equal footing. So there is nothing for you to consider as a coincidence or not!
     
  8. Jul 1, 2016 #7
    I understand that, the coincidence I was referring to was that each component of the 4 momentum has an uncertainty relation (of sorts) with its corresponding component of 4 position. Notwithstanding their differences, notwithstanding that for three components we have quantum operators and for the fourth we don't. Notwithstanding all that, we can still write them all as

    Δχ^μ * Δp_μ >= h/2

    Might be meaningless coincidence , might be worth thinking about.
     
  9. Jul 2, 2016 #8

    Demystifier

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    This is not meaningful in standard formulation of QM, but makes sense in a non-standard one:
    https://arxiv.org/abs/0811.1905
     
  10. Jul 2, 2016 #9

    radium

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    Im one way of understanding it is to look at transition probability amplitudes, which depend on the energy difference. Frequency is proportional to this difference. So if you plot the amplitude versus frequency. You get a width proportional to 1/t. As t becomes large, the width becomes smaller and center around states where ΔE is proportional to 1/t.

    The uncertainty principle can be thought of as a property of Fourier transforms and the bandwidth theorem, which is a kind of uncertainty relation. So while t is not an operator, it is the conjugate observable to energy in the Fourier transform. This also can be used to explained momentum position uncertainty principle as they are also conjugate variables.
     
  11. Jul 3, 2016 #10

    Physics Footnotes

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    Although Schrodinger's equation can be made relativistic by using e.g. the Dirac Hamiltonian, it is not manifestly relativistic. An important consequence of this is that although you know that deep down space and time are behaving symmetrically, the formalism does not reveal that symmetry explicitly and can therefore be very misleading.

    Heisenberg's Uncertainty Principle (HUP) is a perfect example of this nefarious state of affairs because it is proved by exploiting the asymmetric character of the formalism (namely by the use of operators on a Hilbert Space of functions defined over the spatial configuration of the system).

    More specifically, although for each time ##t##, ##\psi_t(x)## belongs to a Hilbert Space, the space-time functions ##\Psi(x,t)## do not. This means that the operator theory (i.e. Hilbert Space functional analysis) available to us in the traditional form of the HUP is no longer meaningful when we introduce operators (like ##T## and ##E##) acting on the ##\Psi(x,t)## instead of the ##\psi_t(x)##.

    There may well be a Lorentz covariant uncertainty relation defined over space-time in QM, but we will not be able to establish it (at least not rigorously) in the language of orthodox Hilbert Space quantum mechanics.
     
    Last edited: Jul 3, 2016
  12. Jul 7, 2016 #11

    bhobba

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    How true that is.

    Its based on the Galilean transformations as carefully explained in Ballentine. The Galilean transformation are explicitly not relativistic.

    In classical mechanics this is explained, admittedly in Landau's usual terse style, in his very beautiful book on Mechanics:
    https://www.amazon.com/Mechanics-Third-Course-Theoretical-Physics/dp/0750628960

    However for some reason it's not usually found in textbooks - don't know why because its very important.

    To merge QM with relativity you need QFT where locality is a bit more subtle than in SR being part of the so called cluster decomposition property:
    https://www.physicsforums.com/threads/cluster-decomposition-in-qft.547574/.

    Its an interesting discussion relating that to Bells theorem but needs a new thread.

    Thanks
    Bill
     
    Last edited by a moderator: May 8, 2017
  13. Jul 11, 2016 #12
    Watching this thread I cant resist to ask the same about the spin-phase relation,something like:[tex]\Delta s\Delta\theta\geq \hbar[/tex] [tex]\Delta E\Delta t\geq \hbar[/tex][tex]\Delta x\Delta p\geq \hbar[/tex]

    There is too much beauty in these three relations, relating rotational invariance, time reversal and translational symmetry. Should they be interpreted in terms of "fluctuations" maybe? since x and p are standard operators in QM, and E (the hamiltonian) too, but what about time¿ Can time at some scale fluctuate? (Like a quantum theory of gravity would suggest)
     
  14. Jul 12, 2016 #13

    vanhees71

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    All three relations are wrong. The first one doesn't make sense, because you cannot define an angle observable so easily. This question has been very satisfactorily answered in another recent thread in this forum, boiling down to the recommendation to read the papers by Kastrup on this topic:

    http://arxiv.org/abs/quant-ph/0510234
    http://arxiv.org/abs/quant-ph/0307069

    The "natural" uncertainty relation for angular momenta are
    $$\delta l_j \delta l_k \geq \frac{\hbar}{2} |\epsilon_{jkl} l_l|.$$

    The last uncertainty relation should read
    $$\Delta x \Delta p \geq \frac{\hbar}{2},$$
    and the 2nd one has to be derived with great care in the context of the proper meaning of ##\Delta E## and ##\Delta t## since time is not an observable in quantum theory, and the general Heisenberg-Robertson uncertainty relation
    $$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}]\rangle|$$
    is not applicable.

    For a thorough discussion of this issue, see

    http://arxiv.org/abs/quant-ph/0105049v3
     
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