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Quantum myth 2. time-energy uncertainty

  1. Apr 22, 2008 #1
    We are discussing the Demystifier's paper "Quantum mechanics: myths and facts". http://xxx.lanl.gov/abs/quant-ph/0609163

    Myth 1 is discussed here:

    The myth 1 thread is still alive but, personally, I would like to press on. I will be offline for most of the month of June and want to get through all these, if possible.

    In QM, there is a time-energy uncertainty relation

    The topic is the claim that the common statement that time and energy are related by an uncertainty relation similar to that for position and momentum is a myth. By myths we mean widely repeated statements which, true or false, are not something we can validly assert given our current understanding.

    I have a preliminary question. I have seen the statement that the energy operator is the hamiltonian and not [tex]i\hbar \frac{\partial}{\partial t}[/tex]. That is not how I would put it, though I am quite ready to admit I know little about it. But I would have supposed that [tex]i\hbar \frac{\partial}{\partial t}[/tex] really is the energy operator [tex]E[/tex] and that the condition

    [tex]H|\psi\rangle = i\hbar \frac{\partial}{\partial t}|\psi\rangle [/tex]

    is a constraint of sorts, telling us a relation between the energy and the other variables which physical systems must satisfy.

    Could someone else elaborate on the relationship between E and H?
  2. jcsd
  3. Apr 22, 2008 #2


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    It comes from classical E&M -- see Jackson, or the author of your choice --; with a few wrinkles here and there. It very evident when you do time dependent perturbation. Check out Quantum Optics.What's to worry in practice? (You won't get it unless you deal with the E&M origins; there's a certain amount of physical reasoning involved.)

    Re the above: the Hamiltonian is the infitesimal generator of translations in time.
    Reilly Atkinson
  4. Apr 22, 2008 #3
    To add to what Reilly said, the unitary operator [tex]e^{-itH}[/tex] is the time evolution operator (I set [tex]\hbar=1[/tex]). So H is the generator and you get back the equation that you thought was a constraint. That equation works that way because of the time evolution of operators in the Heisenberg picture is given by commuting with the Hamiltonian-- [tex]\frac{dA}{dt} = [H,A][/tex] (where I assume there is no explicit time dependence). I think that it's all because we are dealing with a Lie algebra. But I dunno I don't recall that well. I thought there was a connection between that commutation relation and infinitesimal transformations represented by exponentials of the small parameter with the generator of the symmetry.

    Can someone clear this up?
  5. Apr 23, 2008 #4


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    Pellman, see again Eqs. (7) and (8) in my paper, and the discussion around them.
  6. Apr 23, 2008 #5


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    For a better understanding, it is also useful to have a classical analog. The best classical analog of the Schrodinger equation is the Hamilton-Jacobi equation. This is Eq. (16), which (for a time-independent potential) can also be written as
    [tex] H(\nabla S,x)=-\partial_t S .[/tex]
    The solution has the form
    [tex] S(x,t)=S(x)-Et ,[/tex]
    so the Hamilton-Jacobi equation can be written as
    [tex] H=E .[/tex]
    Last edited: Apr 23, 2008
  7. Apr 23, 2008 #6
    Is this not the same thing as saying [tex]E=i\hbar \frac{\partial}{\partial t}[/tex]? "The momentum is the infinitesimal generator of translations" is the same thing as saying that [tex]p=-i\hbar \nabla[/tex] , right? (The sign difference goes away if we start distinguishing covariant and contravariant vectors.)
  8. Apr 23, 2008 #7


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    Are you familiar with the Poisson brackets in classical mechanics?
  9. Apr 23, 2008 #8
    D, I am very intrigued by the points made around eqs 7 and 8. The more I think about it, the more it bugs me. You would think that dealing with this asymmetry in how time and space are related in QM would be first thing to do before attempting a relativistic theory.

    Do you know of any papers or texts which discuss this question? Searching through papers for fundamental stuff is nearly impossible. what keywords am I going to search for? quantum, probability and time? Ha!

    Have patience with me here. I am a total amateur, not even a student.

    I know,

    On the other hand,

    .. my first thought was that maybe classically and non-relativistically, instead of formulating the a problem in terms of [tex](x_j,y_j,z_j),j=1,..N[/tex] parametrized by t, one could formulate the problem in terms of [tex](t_j,y_j,z_j)[/tex] parametrized by x with [tex]E_j[/tex] being the canonical conjugate to the time for the jth particle. Then some function of the x-components of the momentum would act as the "hamiltonian" similar to the way that, using time as the parameter, the Hamiltonian is equal to the sum of individual particle energies.

    ending up with something like

    [tex] \frac{dt_j}{dx}\propto\frac{\partial H}{\partial E_j} [/tex]
    [tex] \frac{dE_j}{dx}\propto\frac{\partial H}{\partial t_j} [/tex]
    [tex] \frac{dy_j}{dx}\propto\frac{\partial H}{\partial p_{y,j}} [/tex]
    [tex] \frac{dp_{y,j}}{dx}\propto\frac{\partial H}{\partial y_j} [/tex]

    But after looking at it briefly, I can only say that if this is possible, it is not easy.

    My point would be ... is time-as-parameter a necessity, whether classically or quantum-mechanically, or is it just more convenient?

    yes, but same question.
    Last edited: Apr 23, 2008
  10. Apr 23, 2008 #9
    The points I touched on in the last post are cute but of course it wouldn't change the topic question. The point is, whatever our parameter is, will it have an uncertainty relation?
  11. Apr 23, 2008 #10

    Ken G

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    You might find this snippet by John Baez to be relevant: http://math.ucr.edu/home/baez/uncertainty.html
    It's a clever prescription for eliminating time as anything but the parameter we need to take a derivative with respect to to get the Schroedinger equation. Then you just look at whatever state you are analyzing when you ask about the time-energy uncertainty relation, find some observable A that satisfies d<A>/dt = 1 on that state of interest, and the mathematical proof Baez cites then tells you that
    (Delta A) (Delta H) >= hbar/2
    (where < > means expectation value and Delta stands for the standard deviation of the observable in the state in question)
    in a way perfectly consistent with all the formalism of quantum mechanics. In short, you don't need a universal time operator to get the uncertainty relation you seek, you only need to find an operator A that is "an observable that is functioning like the time parameter" in that particular situation.

    In terms of the myth of a time-energy uncertainty relation, what this means is that it is indeed a myth to think there is such a universal uncertainty relation, but it is not a myth to think that in any given situation, any real operator that functions like the time parameter (i.e., d<A>/dt = 1) will itself obey a timelike-energy uncertainty relation in that situation.
    Last edited: Apr 23, 2008
  12. Apr 24, 2008 #11
    Properly, the time energy "uncertainty" isn't an "uncertainty" relation at all. Properly, it is an inequality. For whatever reason, "time-energy uncertainty" has propogated into the literature and has become the term of choice, but in no way does the time-energy inequality have the same physical meaning and interpretation of say the position-momentum uncertainty. For a good discussion of this, and all things related to funadmental quantum mechanics, see "Understanding Quantum Physics: A User's Manual" by Michael A. Morrison.

    I'm trying to load the article that is referenced in this thread, but it is taking quite a while. Ergo I cannot check equations 7 and 8 and the relevent discussions on them. However, I can say with some certainty, that the temporal-spatial asymmetry with the Schrödinger Hamiltonian lead Paul Dirac et al to search for a full relativistic theory. :/
  13. Apr 25, 2008 #12
    The asymmetry in question is the view of the wave-function as probability at a specific time. That is [tex]|\Psi(x,t)|^2dx[/tex] is the probability of observing the particle in the region (x,x+dx) at time t instead of [tex]|\Psi(x,t)|^2dxdt[/tex] being the probability of observing the particle in the region (x,x+dx) during the time interval (t,t+dt). And of course this is just another way of saying that time is a parameter and not a dynamic variable in the non-relativistic theory.

    However, this asymmetry seems to carry over too much into relativistic theory. At least in spirit if not in principle. Hence, we still deal with equal-time commutators in QFT just like in the Heisenberg picture of QM.
  14. Apr 25, 2008 #13


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    Consider the set of all Hilbert space-valued functions of time.

    Some things to note:
    [itex]e^{itH}[/itex] is not the time translation operator.
    The energy operator E is, in fact, [itex]i \hbar \partial / \partial t[/itex].
    We do not have the equality H = E.

    Now, consider the subset consisting of functions satisfying the dynamics of QM.

    [itex]e^{itH}[/itex] is the time translation operator.
    The energy operator E is, in fact, [itex]i \hbar \partial / \partial t[/itex].
    We do have the equality H = E.
    In fact, we have the equality [itex]H = i \hbar \partial / \partial t[/itex].
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