Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy-time uncertainty relation

  1. Sep 4, 2004 #1
    Is there an operator which corresponds to time whose commutator with the Hamiltonian equals ih? I mean, when you collapse the wave function by taking an energy measurement, then at that instant the uncertainty in energy is 0. But then as times goes on the state will be that energy eigenstate and vary with time exp(-iEt/h), and then you get uncertainty in energy times that of time >= h/2? You see, I don't really get that. Is this the reason that for a harmonic oscillator (whose energy levels are all discrete) you can excite a change in energy levels even if your photon doesn't have a frequency equal to the natural frequency of the oscillator, violating conservation of energy?

    P.S. h=planck's constant divided by 2pi
  2. jcsd
  3. Sep 4, 2004 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There is not. A good, concise explanation is found on the second page of this document:

    Introduction to Quantum Field Theory
    Last edited: Sep 4, 2004
  4. Sep 4, 2004 #3
    In non-relativistic quantum mechanics, time is an independent variable so there is no such thing as a measurement of time. There is, however, a meaningful way to introduce time measurements, and that is by tracking the rate of change of the expectations of some other observable Q:

    (1) (d/dt) <Q> = i/h <[H,Q]> + <dQ/dt>

    The second term on the RHS denotes an operator's explicit time dependence which seldom appears in the Schroedinger picture, so we'll drop that. From the generalized UP:

    (2) deltaH*deltaQ >= 1/2i [H,Q]

    Combining (1) and (2):

    (3) deltaH*deltaQ >= h/2 |d<Q>/dt|

    So we can define deltaT = deltaQ / |d<Q>/dt| which has dimension of time

    If we use standard deviations instead of delta's, this means that deltaT is the amount of time for the expectation of Q to change significantly (by one standard deviation) - it depends on the obervable Q. A possible interpretation is that the expectations of observables change at a rate proportional to the uncertainty in the energy; an eigenstate of the hamiltonian thus has the property that the expectations of other dynamical variables do not change with time.

    Source: D. Griffiths, "Introduction to QM"

    Apologies for not using Latex, I am in a rush :smile:
  5. Sep 4, 2004 #4
    Tom Mattson, thanks for the link, but I don't think I'm ready for that high-level stuff...yet. Need more training!


    I never thought I'd see Ehrenfest's theorem again.

    I also recognize the 2nd equation as the form the generalized uncertainty relation takes when the commutator of the two Hermitian operators equals
    a pure imaginary number, but what does "UP" mean?

    It all still seems a little fuzzy to me (though I understood everything). Shankar does a poor job of talking about it - he didn't even bring up Ehrenfest's theorem but just argued that for eigenstates |E>e^(-iEt/h), when existing for finite time, there's wave packet rather than an infinitely long wave (this really came out of nowhere is there a wavefunction in time space), so there's uncertainty in energy. Griffith's explanation was much better.

    Oh and I don't know LaTex.
  6. Sep 4, 2004 #5

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's up to you of course, but I think you should give it a shot. The explanation on page 2 of that document is a simple argument from QM, not QFT (in fact, it's part of the motivation for introducing QFT). But all you need to understand are commutators, exponential functions, and why the consequences of a t-operator are contradictory to QM.

    Specifically, a t-operator would violate the requirement that energy be bounded from below, and it would violate energy quantization.
  7. Sep 5, 2004 #6
    Sorry... UP is a common acronym (around here anyway) for Uncertainty Principle.
    If there's anything you still need explaining, let me know.

    As for LaTex, that's a scripting language used to make nice looking formulas. You'll probably come across it here sooner or later. Here's one of the equations rewritten in Latex:
    [tex] \frac{d<\hat{Q}>}{dt} = \frac{i}{\hbar} <[\hat{H},\hat{Q}]> + <\frac{d\hat{Q}}{dt}>[/tex]

    Tom: I now see what you mean. I had looked on the third page of the document (the second as far as article's body is concerned). Silly me.
    Last edited: Sep 5, 2004
  8. Sep 6, 2004 #7
    Actually I don't know 4-vectors so I can't even read the 1st page.

    Still [tex]\psi=|E> \frac{e^-iEt}{\hbar}[/tex] has zero uncertainty in energy if you do [tex]\Delta E=<\psi|(H-<H>)^2|\psi>^\frac{1}{2}[/tex]. But I guess there's uncertainty in energy. I guess Griffith's explanation for me can work as another nice proof that the eigenstates of the Hamiltonian keep the same probability distribution. Oh and I should've known UP meant uncertainty principle. Actually, if you had UR, I would've immediately come up with uncertainty relation. You see, at the time, I didn't regard the UP as a fundamental postulate to QM. Instead I had [tex]X->x, P->-i\hbar\frac{d}{dx}[/tex] in the X-basis as one of my postulates, and the uncertainty principle is derived from that. But now I know better and that you can actually have [tex]X->x, P->-i\hbar\frac{d}{dx}+f(x)[/tex]
    so that really [tex] [X,P]=i\hbar[/tex], the uncertainty principle, is more fundamental.

    Okay that took me forever to write, but I guess it's good practice for using LaTex. So for LaTex you need a forward slash before everything that you type in that will appear differently in LaTex, except for power ^ and subscript _?
    Last edited: Sep 6, 2004
  9. Sep 8, 2004 #8

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Pardon my stubbornness, but I still think you can get it. :smile:

    You don't need to know anything about 4-vectors to get the argument against the T-operator. Let me see if I can distill it.

    First, assume there exists an operator T whose eigenvalue is the time, and it satisfies the following commutation relation with the operator H:


    (I'm in natural units, so h-bar=1).

    From this commutation relation, you can derive the following:

    (2).....exp(-iaT) H exp(iaT)

    where a is a continuous parameter with the dimension of energy. Now because the Hamiltonian is the generator of time translations, we have:

    (3).....exp(-iaT) H exp(iaT)=H-a

    What does this mean? First of all, from the fact that a can be any real number, the energy of a system is not bounded from below, because it can go to negative infinity. Second, it means that energy in general is not quantized, because a is a continuous parameter. Both of these contradict observational evidence, so assumption (1) must be false.

    The 4-vector stuff just shows what must be done in a relativistic theory. If time and space are to be treated on the same footing (as SR demands) and time cannot be an operator (as QM demands), then in a relativistic quantum field theory, position cannot be an operator, either.
    Last edited: Sep 9, 2004
  10. Sep 20, 2004 #9
    Comments on “Is there an operator which corresponds to time ?"

    I Have some comments on the questions and answers to the question of redX “Is there an operator which corresponds to time whose commutator with the Hamiltonian equals ih?”

    At first, the demo given by Tom Mattson seems ok. This quick demo is used in a lot of scientific papers and books, see arxiv for details. He shows, at least the demo conclusion, that a time operator cannot be defined in classical quantum theory. This kind of demo uses mainly the requirement, “from the hat”, that the H operator has lower bounded eigenvalues (e.g. H=P^2/2m => E>=0). But this demo also assumes, implicitly, several facts like the following ones:
     Time is a linear operator.
     H is defined on a Hilbert space generated by the P,Q operators (i.e. [P,Q]=iћ and finite norm vectors).
     Etc …

    In non relativistic quantum mechanics, we must not forget that the evolution of a closed conservative system is given by the equation |psi(t)>=U(t,to)|psi(to)>.
    U(t,to)= exp(-iHt/hbar) is the unitary time evolution of the closed system where H, the Hamiltonian, is the generator of time translations as P is the generator of spatial translations Q.

    So we are free (mathematically) to define an abstract Hilbert space where the operators H and T are linear and where the relation [H,T]=ihbar is true. [H,T]=ihbar thus define this Hilbert space where we impose the finite norm value of vectors and call it “Time” hilbert space.
    In this model, we may add another Hilbert space defined by the the P,Q operators (call it “Space” Hilbert space).
    Then we may define the tensorial product of Hilbert spaces: MySpace=Time x Space and extend the domain of the above operators on this new Hilbert space with the commutators:
    [H,T]=[P,Q]= ihbar

    Thus we may define the hamiltonian operator Ho of the system as (e.g free particule) :
    Ho= P^2/2m + H . We thus have [Ho,T]=[H,T]=ihbar (what is required by the RedX question).

    Note that other Hamiltonian forms are possible as long as we keep the “time” commutation relations (this is mainly due to the gauge freedom on the choice of the Hilbert spaces – i.e. the choice of the P,Q, H, T variables)

    Now when we look for the eigenvalues “Eo” of this Ho operator, we can search them on the separate spaces (due to the form of the Hamiltonian):

    Ho|E_space>|E_time>=Eo |E_space>|E_time>

    So if we take |E_space>=|P_space> the eigenvector of P (eigenvalue p, positive or negative) and |E_time>= eigenvector of H (eigen value E_time, positive or negative) we have:

    Ho|E_space>|E_time>= (p^2/2m+E_time)|E_space>|E_time>

    i.e. Eo = p^2/2m+E_time

    Where p^2/2m is a positive real value and E_time is either a positive or negative value.

    We see that we can define freely define a linear “time” operator that has the property [Ho,T]=ihbar if we extend the Hilbert space (“time x space”). We also see that the Hamiltonian Ho is defined with eigenvalues that are either positive or negative depending on the relative state of the time Hilbert space. The eigenvalues Eo of Ho are the same of a classical H, but translated by an Energy offset (E_time). This is not important because all energy measures are relative (Eo_a - Eo_b) and as long as we keep the same state in the time Hilbert space.

    This formulation is not a new one. It may be recovered with the group theory and more precisely with the Galilean group invariance that applies in the non-relativistic formulation of quantum mechanics.

    I just have written this quick answer to show that a hypothesis such as “H operator has lower bounded eigenvalues” is very difficult to understand as a requirement (where does it come from?). And to my (small) opinion it does not apply to the question of RedX “Is there an operator which corresponds to time whose commutator with the Hamiltonian equals ih?”.

    Tom Mattson and Zefram c have tried to answer in interpreting the word “time” (implicit association). However, time interpretation (as well as the measurement process) is one of the most difficult topics of quantum mechanics. So it is always dangerous to add an external restriction to make a demo (such as “H operator has lower bounded eigenvalues”) involving time.


    P.S. Sorry for the formulas, next time, I’ll try to use Latex :smile: .
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook