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Energy time uncertainty

  1. Sep 7, 2005 #1
    I thought I do understand the energy time uncertainty principle. I know what it can do (virtual particles) but I don't really know what the expression d(E)d(t)>=h means.
    To be a little bit more specific:
    d(t) must be the time we measure something.
    Is d(E) the standard deviation of some energy measurements on a system or is it the expectation value of the energy of a particle?
  2. jcsd
  3. Sep 8, 2005 #2
    I always found the energy-time uncertainty principle a little harder to think of in an intuitive physical way in my QM course since time's not an observable. Later in an optics course though I found an example that makes a lot of sense to me. Consider an atom with an excited electron which drops to a lower energy state, emitting a photon. Classically we just say that the energy of the photon is exactly the energy difference between the two levels in question. However, quantum mechanically there is a [itex]\Delta E[/itex] associated with the energy such that [itex]\Delta E\Delta t\geq\hbar/2[/itex], where [itex]\Delta t[/itex] is the length of time the electron was excited for.
  4. Sep 8, 2005 #3
    [tex]\Delta t[/tex] can also be interpreted as a lifetime of an
    unstable/resonant particle or entity.
  5. Sep 8, 2005 #4
  6. Sep 8, 2005 #5

    This is a good definition. But why is d(t) in the HUP the time we must have to measure the energy d(E). If d(t) is only the average time to measure d(E) of a system there must be a measurement of our measurement serie where we could measure the energy d(E) in time d(t1) with d(t1)<d(t).
    You know what I want to say?
  7. Sep 8, 2005 #6
    Let me give you this version of the time/energy HUP.

    Suppose you measure the system's energy at some time t and i measure that same system at the same time t. The HUP states that the outcome of the measurement (which will be an exact energy-value that is limited only by the accuracy of our apparatus!!!) will be different for you and me. I mean, my energy value will be different than yours. That is the uncertainty.

  8. Sep 8, 2005 #7
    I know that. But for me it is not clear why we cannot measure virtual particles. If d(t1)<d(t) is the time we measure the energy E in 100000 measurments we will get the standard deviation of energy d(E1)<d(E). This doesn't sadisfy the inequality d(E)d(t)>=h. So, if d(E) is just the standard deviation of E we should be able to measure energy values under E and over E. These under E would not sadisfy the HUP. So why can't we measure virtual particles?
  9. Sep 8, 2005 #8

    No, you got it wrong. You CAN measure virtual particles (i do not see why you even would do that, but anyhow), only if you measure several identical setups, you will acquire different energy values. That is what the spread denotes.

  10. Sep 8, 2005 #9
    The site with the URL somebody gave me before says that. Or reformulated: We cannot measure virtual particles.

    And marlon, I think it was you who said in another thread something like this:
    "HUP states that we cannot measure virtual particles."
  11. Sep 9, 2005 #10
    Maybe i did, maybe i did not. Anyway, the point of this remark is the fact that you are not able to measure the energy in an accurate manner. If you look at the position/momentum-uncertainty, the story is the same. You can measure both quantities with complete accuracy (just assume our apparatus is completely accurate), only if you measure the same system again, you get a different measurement-value. That is what the uncertainty is all about. Because of this spread, the information is not exact, but this does not mean you cannot measure it. This is a common misconception in introductory QM.

  12. Sep 9, 2005 #11
    You know, the energy-time uncertainty states that as long as the virtual particle pair exists, the energy is uncertain. This does not imply you cannot make an accurate measurement of that energy (suppose we have the technology for it). The point is that if you measure the same pair with 5 different devices at the same time, you might get 5 different energyvalues. That is the HUP in essence

  13. Sep 9, 2005 #12
    But isn't this always the case for states that consist of a superposition of eigenstates? For example if you have a state

    [tex] |\Psi \rangle = a |\alpha \rangle + b |\beta \rangle [/tex]


    [tex] \hat{H} |\alpha \rangle = E_{1} |\alpha \rangle [/tex]
    [tex] \hat{H} |\beta \rangle = E_{2} |\beta \rangle [/tex]

    Then one could calculate the expectation value [tex] \Delta E[/tex].
    But where's the uncertainty in time here?
  14. Sep 9, 2005 #13
    (delta E)(delta t) > h and (delta E1)(delta t1) > h,
    so if delta t1 < delta t, then delta E1 > delta E.

    Virtual particles are in a sense measured, but only
    indirectly. They can't, by definition, be directly observed.
    The uncertainty relation and the speed of light define the
    limits of the virtual particle.

    Measure the mass, M, of a particle in a time interval, delta t.
    The uncertainty in the mass is then, delta M > h/c^2 delta t.
    Delta t can be reduced enough so that it's impossible to tell
    if there's just one mass M particle or more than one particle
    (of total mass M + delta M) in a particular region. One
    particle might become two particles, but there's no direct
    measurement that can confirm this. So, the possible
    extra particle is called a *virtual* particle.

    For a proton to fluctuate into a virtual proton the delta t
    would be around (4.3)(10^-24) seconds. The maximum
    distance that this virtual proton could travel (based on the
    limiting velocity, c) during delta t is about (1.3)(10^-14) centimeters
    -- which is about the size of the proton.

    Two particles separated by a certain distance might exchange
    a virtual particle provided that the mass of the virtual
    particle is such that it can traverse that distance in delta t.
    For a distance of several fermis (about the size of the
    average nucleus), a virtual particle with a mass that
    is a fraction of the proton can be exchanged.
    Yukawa showed that such exchanges between protons,
    and also between neutrons (which have no repulsive
    force to overcome), would result in a net attractive
    force binding nuclei together.

    The time - energy uncertainty relation led to the idea
    of particles that might 'exist' in observational 'gaps', which
    led to the idea of the strong nuclear force mediated by
    mesons which can be inferentially 'measured', but not directly

    The fundamental importance of this is that it led to
    the modern idea that 'forces' between spatially separated
    objects can be understood in terms of local interactions
    (ie., via models involving the exchange of virtual particles).
  15. Sep 9, 2005 #14
    Why not ? Can you prove this ?

    Why not ? How did we ever found out about pair production ?

    Why is the created particle virtual ? What if it becomes real ?
    How does a proton "fluctuate" into a virtual proton ?

    What velocity ? How do you know this velocity ?

    What the hell is this all about ?

    A virtual particle has definite momentum and therefore is basically everywhere. The spread on its position is infinite, so how can you speak about travelled distance ?

    What are these observational gaps ? Suppose we are able to measure a system in very short consecutive steps ? Why are you so certain there is an observational gap ?

    This is not what local means in QFT. It means that field only interact at points in space where these fields "touch" each other.

  16. Sep 9, 2005 #15
    I my example, i assumed the time uncertainty is amost zero (quasi simultaneous measurements). However if this is not the case, it does not change much. The nature of the explanation stays the same: take 2 identical systems and measure the energy at the same time, you will get an energy-uncertainty...this will happen during a time interval denoted by "delta t"

    Last edited: Sep 9, 2005
  17. Sep 10, 2005 #16
    Virtual particles are still *virtual* as far as I know. If
    somebody has actually observed them, then we can stop
    calling them virtual particles, can't we?

    Using the time - energy uncertainty relation, (delta t)(delta E) > h,
    and the relation between energy and mass, E = mc^2, then if
    a particle mass, M, is measured in delta t, then there's an
    uncertainty in the mass during this interval of,
    delta M > h/c^2 (delta t).

    Considering a measurement on a single proton of mass,
    (1.7) (10^-24) gram, then for there to be a proton and a virtual
    proton during the interval, delta t, then the mass uncertainty,
    delta M, must be at least (1.7) (10^-24) gram and the
    corresponding delta t > (6.6) (10^-27)/(9) (10^20) (1.7) (10^-24),
    or delta t > (4.3) (10^-24).

    For all we know, there might exist two particles, one real and
    one virtual, during the time, delta t. As long as the virtual
    particle that has fluctuated into or popped into existence also
    fluctuates or pops out of existence before delta t has elapsed,
    then no conservation rules are violated. That is, assuming
    that the conservation laws apply to this process, then the
    virtual particle must appear and vanish during delta t.
    Conservation is assumed, so the virtual particle is,
    by definition, not observable.

    You mean the first experimental observation to support
    the model? Or when the idea was invented? I don't know.
    I'm just learning about this stuff.

    Anyway, the original poster asked why virtual particles
    can't be observed. My understanding is that it's because
    such an observation would violate the conservation of energy.
    They can, however, 'exist' because of the time - energy
    uncertainty relation ... and certain experimental effects
    that fit the model have apparently been observed.

    Virtual particles are called 'virtual' because they
    can't be observed. I don't know enough of the specifics
    of the model(s) to answer the other questions.

    The speed of light, c = (3) (10^10) centimeters/second.

    delta t > 0 and delta E > 0 -- there's no infinite spread involved.

    Wrt how the time - energy uncertainty relation pertains to
    virtual particles, and why virtual particles aren't observable,
    the simple example I used involved a single stationary

    It seemed interesting to touch on the idea of a virtual particle
    moving from one particle to another during it's virtual lifetime.
    This is one of the basic concepts involved in the strong interaction,
    isn't it?

    A virtual particle that has the same mass as a proton could
    move a distance corresponding to the diameter of a proton
    during a delta t of (4.3) (10^-24) seconds. A virtual particle
    that has a fraction of the mass of a proton could move
    even farther, eg. the diameter of a nucleus, during that
    delta t.

    There's always some time interval, delta t > 0 involved in a
    measurement. There's an uncertainty in our knowledge of
    the state of the system at any instant in that interval.
  18. Sep 11, 2005 #17
    Well define observe ? I mean if you can see effects that are directly caused by virtual particles, would you not classify those as "seeing virtual particles" ?

    Anyhow, your definition of a virtual particle is totally wrong.

    I really do not see what your point is here. Why is there a proton and a virtual proton. A proton does not just fluctuate into a virtual proton and a virtual proton will; not just pop out of the vacuum because there is no charge conservation. Please be more specific on this. What process are you talking about ?

    No, this is the whole point of virtual particles. During their existence energyconservation can be violated thanks to the Heisenberg uncertainty principle. Only between the inital and final state, you will have energy conservation. Momentum is conserved at all times.

    Yes they will always apply...
    Why ? Besides, keep in mind my previous remark on "what you call observing"

    Yeah, but delta x and delta p = 0, there is an infinite spread involved. What about that ? That was my point. You cannot chose for the HUP when it fits you, you know ?

    Are you talking about a stationary virtual particle ? This is not possible because this implies you know the position of this particle. This is not possible because of the HUP and momentum-conservation. This is something i also told you when you were talking about trajectories of virtual particles...

    Well, not necessarily, a virtual particle can transform into other virtual particles or it can become real whene enough "external" energy is available.

    and the EM-interaction, and the weak interaction and, err, gravity ??? Yes ofcourse

    Again, you cannot speak about travelling distances like this for the above reasons. You can only talk about energy uncertainties and time uncertainties...

    No there is not. Why can't i measure the same system twice at the same time ? Ofcourse this is possible...The HUP does NOT arise because of the guy who measures the system or because of the used apparatus. It arises because of the "quantum"-nature of the system we are measuring.

    This is just the same as if you measure the momentum of a particle on a certain position x (delta x = 0). If you redo the exact same measurement and measure this particle again at x, you can get a different value for the momentum. This is the spread in momentum-values predicted by the HUP. It has nothing to do with the measurement itself, or the observer or the accuracy of the apparatus. But my point is that you CAN acquire an exact value for both x and p in one single measurement. The HUP states that if you redo this experiment with same x, then you will acquire a spread in p-values. The notion of "we cannot measure p and x at the same time with absolute accuracy" is often misinterpreted by many people.

  19. Sep 11, 2005 #18
    Hello Marlon,

    I do not understand what you said:
    To be exact:
    Suppose you and me, we have a both each a system, both identical. Then if I measured the energy, I will not necessarily get the same value as you. Till here, ok.

    But what do you mean with
    Maybe you could elaborate on this by giving a physical example.
  20. Sep 11, 2005 #19
    Well, i just meant that the delta t denotes the time period in which a system's energy can be uncertain. This is not however the time we need to measure the system. That was my point. Nothing special.

  21. Sep 12, 2005 #20
    Concerning the OP, one should note that the uncertainty principle comes mathematically from valid statement on wave mechanics. Fourier analisys.

    Hearing a stationary sound for a time dt allows you to infer its frequency to within the precision dE.

    Best Regards
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